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MODULE 1 •

Definition of Terms

Engineering Economy – uses mathematical formulas to account for the time value of money and to balance current and future revenues and costs. Economics – is the science that deals with the production, allocation and use of goods and services. The two major subdivisions of economics are: a.

Macroeconomics is the study of the entire system

b. Microeconomics is the study of how the systems the economic system. •

of economics. affect one business or parts of

Necessities and Luxuries

Necessities – are products or services that are required to support human and activities that will be purchased in somewhat the same quantity even though the prices vary considerably. Luxuries – are products and services that are desired by humans and will be purchased if money is available after the required necessities have been obtained. •

Consumer and Producer Goods and Services

Goods – is defined as anything that anyone wants or needs. Services – would be the performance of any duties or work for another; helpful or professional activity. Marketing – refers to the distribution of goods and services. Marketing a Product – refers to the advertising, and other efforts to promote a products sale. •

Different Types of Goods

1.

Consumer Goods – are those such as food and clothing that satisfy human wants and needs.

2.

Producer Goods – are those such as raw goods.

3.

Capital Goods – are the machinery, used in the production of commodities in producer goods. •

materials and tools, used to make consumer

Supply and Demand

Supply – refers to how many of a certain good or services are available for people to purchase. Demand – means how many people wish to buy that good or service. Law of Supply and Demand

Under conditions of perfect competition, the price at which a given product will be supplied and purchased is the price that will result in the supply and demand being equal. •

Demand

Demand – it refers to the people’s willingness to buy a product or service. Demand Curve – is the plot or graph of the quantity demanded versus the price. Demand Schedule – is the schedule or table listing of the quantity demanded with the corresponding price. • 1. a

Types of Demand Elastic Demand – exists when there is a greater change in quantity demanded as a response to change in price.

2. Inelastic Demand – exists when there is a lesser response to a change in price.

change in quantity demanded as a

3. Unitary Demand – exists when there is an equal (increase or decrease).

change in price and quantity demanded

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Factors that Influence Demand

Factors that Influence Demand are: 1.

Income

2.

Population

3.

Taste and preference

4.

Price Expectation

5.

Price of Related Goods •

Supply

Supply – it is the willingness of a producer to manufacture goods. Supply Curve – is the plot or graph of the quantity supplied versus the price. Supply Schedule – is the schedule or table listing of the quantity supplied with the corresponding price. •

Factors that Influence Supply

Factors that Influence Supply are: 1.

Price of Goods

2.

Cost of Production

3.

Availability of Resources

4.

Number of Producer and Sellers

5.

Technological Advancement

6.

Taxes

7.

Subsidies •

Relationship of Supply and Demand

Shortage – the supply is less than the demand. Surplus – the supply exceeds the demand. Equilibrium Point – the supply is equal to the demand. •

Market Structures

Market – is the place where the vendors and buyers meet to transact. Perfect Competition – occurs in a situation where a commodity or service is supplied by a number of vendors and there is nothing to prevent additional vendors entering the market. Perfect Monopoly – exist when a unique product or services is available from a single vendor and that the vendor can prevent the entry of all others into the market. Oligopoly – exist when there are so few suppliers of a product or service that action by one will almost inevitably result in similar action by the others.

MODULE 2 Module 2 INTEREST & MONEY – TIME RELATIONSHIPS

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Simple Interest

Interest – is the return on capital or cost of using capital. It is the amount of money paid for the use of borrowed capital or the income produced by money, which has been loaned. Simple Interest – is calculated using the principal only, ignoring any interest that had been accrued in preceding period. •

Types of Simple Interest

1. Ordinary Simple Interest – simple interest in which it is assumed that each month contains 30 days and consequently each year has 360 days.

1 month = 30 days 1 year = 360 days (banker’s year) 2. used.

Exact Simple Interest – simple interest in which the

exact number of days per month is

1 ordinary year = 365 days 1 leap year = 366 days •

Sample Problems on Simple Interest

1. A loan of Php50,000 is made for a period of 13 months, from April 1 to April 30 of the following year, at a simple interest rate of 20%. What future amount is due at the end of the loan period? 2. What is the principal amount if the amount of interest at the end of 2 ½ years is Php450 for a simple interest rate of 6% per annum? 3. Determine the exact simple interest of Php25,000 for the period of December 27, 2002 to March 23, 2003, if the rate of interest is 10%. 4. What is the interest due on a Php1,500 loan for 4 years and three months if it bears 12% ordinary simple interest? 5. Determine the exact simple interest of Php4,000 for the period of Feb. 14, 1984 to November 30, 1984, if the rate of interest is 18%.

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Compound Interest

Compound Interest – the interest for an interest period is calculate on the principal plus total amount of interest accumulated in the previous period. Compound interest means “the interest on top of interest.”

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Rates of Interest

Rate of Interest – it is the cost of borrowing money. Nominal Rate of Interest – it specifies the rate of interest and a number of interest periods in one year. Effective Rate of Interest – it is the actual or exact rate of interest on the principal during one year. •

Values of “m”

m = 1 for compounded annually (every 12 months) m = 2 for compounded semi-annually (every 6 months) m = 4 for compounded quarterly (every 3 months) m = 6 for compounded bi-monthly (every 2 months) m = 8 for compounded semi-quarterly (every 1 1/2 months) m = 12 for compounded monthly (every month) m = 24 for compounded semi-monthly (every 1/2 month) •

F/P and P/F Factors: Notation and Equations

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Sample Problems on Compound Interest

1. What rate of interest compounded annually must be received if an investment of Php5,400 made now will result in a receipt of Php7,200 5 years hence? 2.

What amount will be accumulated by Php4,100 in 10 years at 6% compounded annually?

3.

What effective annual interest rate corresponds to the following situations? a.

nominal interest rate of 10% compounded semi-annually

b.

nominal interest rate of 6% compounded monthly

c.

nominal interest rate of 8% compounded quarterly

4. How much should Engr. Cruz deposit now, if after 10 years, this will amount to Php100,000. Interest rate is 12% compounded semiannually?

5. If Php1,000 becomes Php5,734 after 15 years, when invested at an unknown rate of interest compounded semi-annually, determine the unknown nominal rate and corresponding effective rate. •

Cash Flow Diagram

Cash Flow Diagram – depicts the timing and amount of expenses (negative, downward) and revenues (positive, upward) for engineering projects. •

Types of Cash Flow Diagrams

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Equation of Value

An equation of value is obtained by setting the sum of the values on a certain comparison or local date (or focal date) of one set of obligations equal to the sum of the values on the same date of another set of obligations. •

Sample Problems on Equation of Value

1. Jay wishes his son, Jason, to receive Php1,000,000 twenty years from now. What amount should he invest now, if it will earn interest of 12% compounded annually during the first five years and 10% compounded monthly for the remaining years. 2. Find the present worth of a future payment of Php300,000 to be made in 10 years with an interest rate of 10% compounded annually. What will be the amount if it will be paid 5 years later (on the 15th year)? •

Discrete Payments

The solution of discrete payments or number of transactions occurring at different periods is taking each transaction to the base year and equating each value. •

Sample Problems on Discrete Payments

1. Acosta Holdings borrowed Php9,000 from Smith Corporation on January 1, 1998 and Php12,000 on January 1, 2000. Acosta Holdings made a partial payment of Php7,000 on January 1, 2001. It was agreed that the balance of the loan would be amortized by two payments, one on January 1, 2002 and one on January 1, 2003, the second being 50% larger than the first. If the interest rate is 12%, what is the amount of each payment? 2. A contract has been signed to lease a restaurant at Php20,000 per year with annual increase of Php1,500 for 8 years. Payments are to be made at the end of each year, starting one year from now. The prevailing rate is 7%. What lump sum paid today would be equivalent to the 8 year lease program? 3. Mr. Cruz buys a second hand car worth Php150,000 if paid in cash. On installment basis, he pays Php50,000 downpayment, Php30,000 at the end of one year, Php40,000 at the end of two years and a final payment at the end of four years. Find the final payment if interest is 14%.

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Continuous Compounding Interest

The solution for interest compounded continuously can be derived thru differential equations and can be found as: •

Sample Problems on Continuous Compounding Interest

1. Philip invested $100 on a bank. The bank offers 5% interest compounded continuously in a savings account. Determine (a) how long will it require for him to earn $5 (b) the equivalent simple interest rate for 1 year of the bank. 2. Which is more advisable to invest Php5,000 for five (5) years, to bank A that offers 5% compounded continuously or to bank B that offers 10% simple interest? •

Banker’s Discount

Certain banks lend money in such a way that they deduct the interest on the money. They actually don’t lend you money you asked for. This type of computing money is called banker’s discount. The money received by the borrower after the discount has been deducted is called proceeds. •

Sample Problems on Banker’s Discount

1. Ms. Glydel Marquez borrowed money from a bank. She received from the bank Php1,342 and promised to repay Php1,500 at the end of 9 months. Determine the following: (a) simple interest rate (b) discount rate or often referred as Banker’s discount.

Module 3 ANNUITY

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Annuity

Annuity – it is a series of equal cash flows occurring each period over a range of periods. Types of Annuity: 1.

Ordinary Annuity

2.

Deferred Annuity

3.

Annuity Due

4.

Perpetuity

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Ordinary Annuity

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P/A and A/P Factors: Notation and Equations

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F/A and A/F Factors: Notation and Equations

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Sample Problems on Ordinary Annuity

1. What is the current value of a $50 payment to be made at the end of each of the next three years if the prevailing rate of interest is 7% compounded annually? 2. An obligation of Php20,000 is to be repaid in uniform annual amounts each of which included repayment of the debt and interest over a period of 5 years. If interest is 10% per year, what is the annual payment? 3. Maintenance cost for a small bridge expected to last for 60 years is estimated to be Php1,000 each for the first 5 years, followed by a Php10,000 expenditure in the 15th year and th Php10,000 in the 30 year. If interest is 10% per year, what is the equivalent uniform annual cost over the 60 year period? •

Sample Problems on Ordinary Annuity

4. What is the equivalent previous worth of Php500 years 72 years ago, if annual interest is 1%?

annuity to be paid constantly in 60

5. Find the annual payment to extinguish a debt of interest.

Php10,000 payable for 5 years at 12%

6. A savings loan is made between a man and banker. What should be the uniform monthly payment that the man should make if he is to borrow Php50,000 and he is to pay in 10 years? Interest is taken as 6% compounded quarterly. 7. What annuity is required over 10 years to equate with the future amount of Php15,000. Assume i = 5%. •

Deferred Annuity

Deferred Annuity – are annuities that are computed on different present year and/or future year. It is annuity where the first payment is made several periods after the beginning of the annuity. •

Methods of Solving Deferred Annuity Problems

1.

Draw the cash flow diagram.

2.

Select any convenient focal date.

Temporary focal date is used to convert deferred annuity to

Final focal date is used to obtained the required value.

3.

Project all values to temporary focal date.

4.

Obtain the final value. •

ordinary annuity

Sample Problems on Deferred Annuity

1. Find the value of x in the cash flow diagram, given that would make the equivalent present worth of the cash flow diagram to Php22,000 and interest rate is at 13% per year. •

Sample Problems on Deferred Annuity

2. Determine the uniform annual payments which would be equivalent to the cash flow diagram given. Interest rate of 12% per year. •

Annuity Due

Annuity Due – is a series of equal payments or receipts occurring over a specified number of periods with the payments or receipts occurring at the beginning of each period. •

Sample Problems on Annuity Due

1. What is the current value of a $50 payment to be made at the beginning of each year, for three years if the prevailing rate of interest is 7% compounded annually? 2. What is the accumulated value of a $25 payment to be of the next three years if the prevailing rate of interest is 9%

made at the beginning of each compounded annually?

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Perpetuity

Perpetuity – are uniform payments which are done infinitely. It is also called as perpetual annuity. •

Types of Perpetuity

1.

Ordinary Perpetuity – first payment is done one period after the focal date.

2.

Deferred Perpetuity – first payment is done several periods after the focal date. •

Sample Problems on Perpetuity

1. How much should Mr. Sy invest on a bank that offers Php1,000 each year in perpetuity.

10% interest so that he would earn

2. Don Jose deposited Php5,000,000 on a bank that earns 10% compounded annually. Five years later he died. His will states that his beneficiary is an orphanage which will be receiving the money in perpetuity a year after he died. How much is the yearly fund the orphanage will be receiving? 3. If money is worth 8% compounded quarterly, compute the present value of the perpetuity of Php1,000 payable quarterly.

Module 4 DEPRECIATION

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Definition of Terms

Depreciation – it is the decrease in the value of a fixed asset, or the value of physical property, with the passage of time. Value – is the present worth of all the future profits that are to be received through the ownership of a particular property. Market Value of a Property – is the amount, which a willing buyer will pay to a willing seller for the property where each has equal advantage and is under no compulsion to buy and sell. •

Definition of Terms

Utility or Use Value of Property – is what the property is worth to the owner as an operating unit. Fair Value – is the value which is usually determined by the disinterested third party in order to established a price that is fair to both seller and buyer. Book Value – is the worth of the property as shown in the accounting records of an enterprise. It is sometimes called as depreciated book value. Salvage or Resale Value – is the price that can be obtained from the property after it has been used. Salvage Year is the year when scrap value is equal to book value. Scrap Value or Junk Value – is the price that can be recovered if an asset is disposed as a junk. •

Purposes of Depreciation

1.

To provide for the recovery of capital

2.

To enable the cost of depreciation to be charged to the cost of producing products or services that results from the use of property. •

which has been invested in physical

property.

Causes of Depreciation

Physical Depreciation – it is due to wear and tear of the asset. Functional Depreciation – it is due to the obsolescence of the asset. Depletion – refers to the decrease in the value of a property due to the gradual extraction of its contents. Monetary Depreciation – depreciation due to changes in price level. •

Physical and Economic Life

Physical Life of a Property – is the length of time during which it is capable of performing the function for which it was designed and manufactured. Economic Life or Useful Life – is the length of time during which the property may be operated at a profit. •

Methods Used to Determine Depreciation

1.

Straight Line Method

2.

Declining Balance Method

3.

Double Declining Balance Method

4.

Sum-of-Years’ Digit Method

5.

Sinking Fund Method

6.

Hour Output Method

7.

Service Output Method •

Straight Line Method

The straight line method is the simplest way in computing for depreciation. In this method, the depreciation each year is constant and the interest rate is being neglected. •

Sample Problems on Straight Line Method

1. Prepare a depreciation table for an asset which was bought at Php15,000 and useful for a period of 5 years. Estimated salvage value is 10% of its original cost. 2. Equipment bought for Php60,000 is expected to last for 30 years. If the book value after 20 years is Php20,000, how much is the depreciation each year? Find the book value after 10 years. 3. A machine which costs Php10,000 was sold as scrap after being used for 10 years. If the scrap value was Php500, determine the total depreciation and book value at the end of th the 5 year. •

Sample Problems on Straight Line Method

4. Delivery jeeps purchased by KH Company cost Php180,000 each. Past records indicate that jeeps should have an economic life of 10 years. They can be sold from an average of Php50,000 each year after 10 years of use. The company receives 9% interest on investment funds. Using straight line method: Determine: a.

the depreciation charge during 3 years

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b.

the depreciation reserve accumulated at the end of 3

C.

book value at the end of 3 years.

years

Declining Balance Method or Reducing Balance Method

In this method, the net book value at the end of each period can be simply computed by multiplying the original market price by a fix percentage repeatedly until it reaches the salvage value. This method is also called Matheson’s Formula. •

Sample Problems on Declining Balance Method

1. A machine worth Php800,000 is bought from China. Freight charges amount to Php200,000. If the scrap value of the machine is Php50,000 that occurs at the end of 17 years. Compute (a) the depreciation and (b) book value at the end of 11 years. 2. Equipment bought for Php60,000 is expected to last years is Php20,000. How much is the depreciation for year •

for 30 years. If the scrap value after 20 10?

Double Declining Balance Method

This is the same as declining balance method except that k is replaced by 2/n. Sample Problem on Double Declining Balance Method A plant bought a calciner for Php220,000 and used it for 10 years, the life span of the equipment. What is the book value of the calciner after 5 years of use? Assume a scrap value of Php20,000 for straight-line method; Php22,000 for declining balance method and Php20,000 for double-declining balance method. •

Sum-of-Years’ Digit Method

This method uses the year’s digit (in reverse order) in computing for the depreciation. •

Sample Problem on Sum of Years’ Digit Method

An industrial plant bought a generator set for Php90,000. Other expenses including installation amounted to Php10,000. The generator set is to have a life of 17 years with a salvage value at the end of life at Php5,000. Determine the depreciation charge during the 13th year and the book value at the end of 13 years by:

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(a)

Declining balance method

(b)

Double declining balance method

(c)

Sum-of-Years’ digit method

Sinking Fund Method

Sinking fund method presents the idea of annuity in computing for the depreciation. The interest rate for the worth of money is being considered so as to have the depreciable value.

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Sample Problem on Sinking Fund Method

A broadcasting company purchased an equipment for Php53,000 and paid Php1,500 for freight and delivery charges to the job site. The equipment has a normal life of 10 years with a trade-in value of Php5,000 against the purchase of a new equipment at the end of the life. Determine the annual depreciation cost using:

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(a)

straight line method

(b)

sinking fund method assuming interest is

6% compounded annually

Hour Output method

In this method, the functionality period and the period the machine has been used is considered. Depreciation is computed based on the wear and tear of the machine. •

Service Output method

Similar to the hour output method, this method based its computation on how much the asset has been used. •

Sample Problems on Service and Hour Output Method

1. An electric bulb bought for Php100 is guaranteed to be useful for 50 hours. A certain company uses the said bulb for 10 hours a day. If there is no scrap value for the bulb. Compute the daily depreciation and create the depreciation table throughout its economic life. 2. A tire bought for Php1,000 is expected to be useful in traveling 100 km after which it can be sold as scrap for Php50. (a) If the pedometer displays a value of 85 km, what is the book value of the tire? (b) How much did the owner need to travel for the tire to amount to Php80?

Module 5 BREAK-EVEN ANALYSIS

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Definition of Terms

Break-Even Analysis – it involves estimating the level of sales necessary to operate a business on a break-even basis. Break-Even Point (BEP) – is defined as the point where sales or revenues equal total expenses. Break-Even Margin – is a ratio that shows the gross-margin factor for a break-even condition. The formula is total expenses divided by net revenues multiplied by 100 to get a percentage. •

Break-Even Graph

Break-Even Chart – shows the graph of fixed cost, variable cost and expected income from sales for different production levels. •

Ways to Lower the Break-Even Point

Lower direct costs, which will raise the gross margin. Exercise cost controls on your fixed expenses, and lower the necessary total expenses. Raise prices. •

Key Break-Even Factors

Fixed Costs – these costs remain constant (or nearly so) within the projected range of sales levels. These can include facilities costs, certain general and administrative costs, and interest and depreciation expenses. Variable Costs – these costs vary in proportion to sales levels. They can include direct material and labor costs, the variable part of manufacturing overhead, and transportation and sales commission expenses. Contribution Margin – this is equal to sales revenues less variable costs. This amount is available to offset fixed expenses and (hopefully) produce an operating profit for the business. •

Appraisal of Break-Even Analysis

Advantages of Break-Even Analysis It points out the relationship between cost, production volume and returns. Limitations of Break-Even Analysis It is best suited to the analysis of one product at a time.

It may be difficult to classify a cost as all variable or all fixed. There may be a tendency to continue to use a break-even analysis after the cost and income functions have changed. •

Sample Problems on Break Even Analysis

1. An entrepreneur at the location in the United States is planning to enter the gourmet soy-based burger market. The forecast expected unit sales of 200,000 burgers in 18 months. The variable cost for making one burger is $0.85 and the fixed cost of making burgers for 18 months will be a total of $165,000 which covers for the rent, phone bill, and insurance coverage – these items tend not to vary in amount per month over the term of one year. The best estimate of what the average consumer will pay for the soy burger is $1.95. How many burgers will he have to sell to break even? •

Sample Problems on Break Even Analysis

2. Toyota Motor Phils. Corporation Sta. Rosa Plant has a production capacity of 700 cars per month and its fixed cost is Php100,000,000 monthly. The variable cost per car is Php300,000 and each car can be sold for Php650,000. Due to cost reduction program, fixed cost will be reduced to 10% and variable cost by 20%. Determine the new and old break break-even point. •

Sample Problems on Break Even Analysis

3. A farmer wants to buy a new combine rather than hire a custom harvester. The total fixed costs for the desired combine are $21,270 per year. The variable cost (not counting the operator’s labor) are $8.75 per hour. The farmer can harvest 5 acres per hour. The custom harvester charges $16.00 per acre. How many acres must be harvested per year to breakeven?

Module 6 COMPARISON OF ALTERNATIVES •

Definition of Terms

1. Comparison of Alternatives – deals with situations in which one has more than one choice and using engineering economic principles, one needs to decide between the alternatives so as to go with the one that is most economically justified. 2.

Methods used in the Selection of Alternatives

Present Worth Method

Annual Cost/Worth Method

Equivalent Uniform Annual Cost (EUAC) Method

Rate of Return (ROR) Method

Payback (Payout) Period Method

Capitalized Cost

Benefit/Cost Ratio Analysis

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Present Worth Method

This method involves finding the equivalent value of each alternative at the present time, identified as time 0. If only costs are involved, we can select the alternative with the smallest present worth of costs. If cost and revenues are involved, we select the alternative with the greatest present worth on net revenues. To perform present worth or annual worth analysis, an interest rate and a study period must be specified. The interest rate is usually the minimum acceptable rate of return (MARR) of the organization.

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Present Worth Method

Present worth analysis can only be used when the alternatives have the same lives. If the alternatives have different lives, some mechanism must be used to compare them over the common study period. •

Sample Problems on Present Worth Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors

will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

interest

Sample Problems on Present Worth Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Present Worth Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Present Worth Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Present Worth Method

3. A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit- return estimates shown below. Rate of interest is 15%. •

Sample Problems on Present Worth Method

3.

A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Annual Cost (Worth) Method

The annual worth method involves finding the equivalent end-of-period value of each alternative. The periods are usually in years. If only costs are involved, we can select the alternative with the smallest equivalent uniform annual cost (EUAC) or net annual cost (NAC). If costs and revenues are involved, we can select the alternative with the greatest equivalent uniform annual benefit, or net annual worth (NAW).

If two alternatives have the same annual worth, then the one with the greatest investment is preferred. The extra investment makes exactly the required MARR. •

Annual Cost (Worth) Method

Problems that can be solved by the present worth method can also be solved by the annual worth method. An annual worth analysis is sometimes preferred over a present worth analysis because people think better in terms of annual amounts than an equivalent amount to time zero. Both methods yield the same results. Annual worth analysis is the easiest method when the alternatives have different lives. No specific study period need be specified, but the implicit assumption that the alternative are compared over the least common multiple of the lives. •

Sample Problems on Annual Worth Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Annual Worth Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Annual Worth Method

2. Perform an annual worth analysis of equal service if the MARR is 10% per year. Revenues for all three the same. •

machines with the costs shown below, alternatives are expected to be

Sample Problems on Annual Worth Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Annual Worth Method

3. A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Sample Problems on Annual Worth Method

3.

A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Equivalent Uniform Annual Cost (EUAC) Method

In this method, all cash flow (irregular or uniform) must be converted to an equivalent uniform annual cost, that is, a year-end amount which is the same each year. The alternative with the least EUAC is preferred. When the EUAC method is used, the EUAC of the alternatives must be calculated for one life cycle only. This method is flexible and can be used for any type of alternative selection problems. The method is a modification of the annual cost method. •

Sample Problems on Equiv. Uniform Annual Cost Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Equiv. Uniform Annual Cost Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Equiv. Uniform Annual Cost Method

2. Perform an equivalent uniform annual cost analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same.

•

Sample Problems on Equiv. Uniform Annual Cost Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Equiv. Uniform Annual Cost Method

3. A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Sample Problems on Equiv. Uniform Annual Cost Method

3.

A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Rate of Return (ROR) Method

If ROR > MARR, select the alternative with the bigger investment ROR < MARR, select the alternative with the smaller investment •

Rate of Return (ROR) Method

The rate of return on the capital invested is •

Sample Problems on Rate of Return Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Rate of Return Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be

operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Rate of Return Method

2. Perform an annual worth analysis of equal service if the MARR is 10% per year. Revenues for all three the same. •

machines with the costs shown below, alternatives are expected to be

Sample Problems on Rate of Return Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Rate of Return Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Rate of Return Method

3. A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Sample Problems on Rate of Return Method

3.

A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Sample Problems on Rate of Return Method

4. A firm is considering purchasing equipment that will reduce costs by Php40,000. The equipment costs Php300,000 and has a salvage value of Php50,000 and a life of 7 years. The annual maintenance cost is Php6,000. While not used by the firm, the equipment can be rented to others to generate an income of Php10,000 per year. If money can be invested for an 8% return, is the firm justified in buying the equipment? •

Sample Problems on Rate of Return Method

4.

A firm is considering purchasing equipment that will reduce costs by Php40,000. The equipment costs Php300,000 and has a salvage value of Php50,000 and a life of 7 years. The annual maintenance cost is Php6,000. While not used by the firm, the equipment can be rented to others to generate an income of Php10,000 per year. If money can be invested for an 8% return, is the firm justified in buying the equipment? •

Payback (Payout) Period Method

In this method, the payback period of each alternative is computed. The alternative with shortest payback period is adopted. This method is seldom used. Payback Period – is the length of time required to recover the first cost of an investment from the net cash flow produced by that investment for an interest rate of zero. •

Sample Problem on Payback Period Method

In a marble block quarrying operation, hand rock drills, costing Php50,000 each, are used. It has a drilling rate of 10 cm per minute, produces 10 cubic meters of block per month and consumes 60 liters of diesel fuel for compressor drive, per rock drill per cubic meter produced utilizing 1 worker per drill. A modern equipment quarry bar mounted rock drill is being offered for Php180,000 per unit and has a drilling rate of 60 per minute that will produce 60 cubic meters of block per month, but consumes 120 liters of diesel fuel for the compressor drive, per 6 cubic meters of block utilized, utilizing 2 workers per quarry bar drill. Consider diesel fuel at Php6.00 per liter at the quarries, worker earning Php80.00 per day, 25 days per month, 5 years life of both drills with 20% salvage value, neglecting cost of money, other cost at Php500 per cubic meter and marble blocks sold at Php2,000 per cubic meter. Would you recommend the purchase of the new equipment? •

Sample Problem on Payback Period Method

In a marble block quarrying operation, hand rock drills, costing Php50,000 each, are used. It has a drilling rate of 10 cm per minute, produces 10 cubic meters of block per month and consumes 60 liters of diesel fuel for compressor drive, per rock drill per cubic meter produced utilizing 1 worker per drill. A modern equipment quarry bar mounted rock drill is being offered for Php180,000 per unit and has a drilling rate of 60 per minute that will produce 60 cubic meters of block per month, but consumes 120 liters of diesel fuel for the compressor drive, per 6 cubic meters of block utilized, utilizing 2 workers per quarry bar drill. Consider diesel fuel at Php6.00 per liter at the quarries, worker earning Php80.00 per day, 25 days per month, 5 years life of both drills with 20% salvage value, neglecting cost of money, other cost at Php500 per cubic meter and marble blocks sold at Php2,000 per cubic meter. Would you recommend the purchase of the new equipment? •

Capitalized Cost

Capitalized Cost – is the present worth of an alternative that will last “forever.” Public sector projects such as bridges, dams, irrigation systems and railroad fall into this category. •

Sample Problems on Capitalized Cost

Two methods of conveying eater are being studied. Method A requires a tunnel, first cost Php180,000, life perpetual, annual operation and upkeep is Php3,000. Method B requires a ditch plus flume; first cost of ditch is Php40,000, life perpetual, annual depreciation and upkeep is Php1,500, first cost of flume is Php30,000, life 10 years, salvage value is Php5,000, annual operation and upkeep is Php4,000. If money is worth 6%, determine which method is to be recommended? •

Sample Problems on Capitalized Cost

Two methods of conveying eater are being studied. Method A requires a tunnel, first cost Php180,000, life perpetual, annual operation and upkeep is Php3,000. Method B requires a ditch plus flume; first cost of ditch is Php40,000, life perpetual, annual depreciation and upkeep is Php1,500, first cost of flume is Php30,000, life 10 years, salvage value is Php5,000, annual operation and upkeep is Php4,000. If money is worth 6%, determine which method is to be recommended? •

Benefit/Cost Ratio Analysis

Benefit/Cost Ratio Analysis – it is the most commonly used method by government agencies for analyzing the desirability of public projects. If B/C ≥ 1.0, accept the project as economically acceptable for the estimates and discount rate applicable. •

Benefit/Cost Ratio Analysis

Cost – estimated expenditures to the government entity for construction, operation, and maintenance of the project less any expected salvage value. Benefits – advantages to be experienced by the owners, the public. Disbenefits – expected undesirable or negative consequences to the owners if the alternative is implemented. Disbenefits may be indirect economic disadvantages of the alternative. •

Benefit/Cost Ratio Analysis

The benefit cost ratio on the capital invested is •

Significant Differences in the Characteristics of Public and Private Sector Alternatives

•

Sample Problems on Benefit/Cost Ratio Analysis

1. The National Government intends to build a dam and hydroelectric project in the Cagayan Valley at a total cost of Php455,500,000. The project will be financed by soft foreign loan with an interest of 5% per year. The annual cost for operation, maintenance, distribution,

facilities and others be Php56,500,000.

would total Php15,100,000. Annual revenues

If the structures are expected to last for 50 project economically acceptable? •

and benefits are estimated to

years with no salvage value, is the

Sample Problems on Benefit/Cost Ratio Analysis

1. The National Government intends to build a dam and hydroelectric project in the Cagayan Valley at a total cost of Php455,500,000. The project will be financed by soft foreign loan with an interest of 5% per year. The annual cost for operation, maintenance, distribution, facilities and others would total Php15,100,000. Annual revenues and benefits are estimated to be Php56,500,000. If the structures are expected to last for 50 years with no salvage project economically acceptable? •

value, is the

Sample Problems on Benefit/Cost Ratio Analysis

2. Two routes are under construction for a new highway. Route A would be located about 5 miles from the central business district and would require longer travel distances by local commuter traffic. Route B would pass directly through the downtown area and although its construction cost would be higher, it would reduce the travel time and distance for local commuters. The costs for the two roads are as follows: •

Sample Problems on Benefit/Cost Ratio Analysis

2. Two routes are under construction for a new highway. Route A would be located about 5 miles from the central business district and would require longer travel distances by local commuter traffic. Route B would pass directly through the downtown area and although its construction cost would be higher, it would reduce the travel time and distance for local commuters. The costs for the two roads are as follows: •

Sample Problems on Benefit/Cost Ratio Analysis

3. Four alternatives for providing electric power supply to identified with the following annual benefits and costs: •

a small town have been

Sample Problems on Benefit/Cost Ratio Analysis

3. Four alternatives for providing electric power supply to a small town have been identified with the following annual benefits and costs:

Module 7 CAPITAL FINANCING

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Equity and Borrowed Capital

Equity Capital or Ownership Forms – are those supplied and used by the owners of an enterprise in the expectation that profit will be earned. Borrowed Funds or Capital – are those supplied by others on which a fixed rate of interest must be paid at a specified time. •

Types of Business Ownership

Individual Ownership or Sole Proprietorship – is one which is owned and run by one individual and where there is no legal distinction between the owner and the business. Partnership – is an association of two or more persons for the purpose of engaging in business for a profit. Corporation – is a fictitious being, recognized by law, that can engage in almost any type of business transaction in which a real person could occupy himself. It operates under a charter that is granted by the government. •

Advantages and Disadvantages of Sole Proprietorship

Advantages: It is easy to organize. The owner has full control of the enterprise. The owner is entitled to whatever benefits and profits that accrue from the business. It is easy to dissolve. Disadvantages: The amount of equity capital which can be accumulated is limited. The organization ceases upon the death of the owner. It is difficult to obtain borrowed capital owing to the uncertainty of the life of the organization. The liability of the owner for his debts are unlimited. •

Advantages and Disadvantages of Partnership

Advantages:

More capital may be obtained by the partners pooling their resources together. It is bound by few legal requirements as to its accounts, procedures, tax forms and other items. Dissolution of a partnership may take place at any time by mere agreement of the partners. It provides an easy method whereby two or more persons may enter into business each carrying those burdens that he can best handle. Disadvantages: The amount of capital that can be accumulated is definitely limited. The life of the partnership is determined by the life of the individual partners. When any partner dies, the partnership automatically ends. There may be serious disagreement among individual partners. Each partner is liable for debts of the other partnership. •

Advantages and Disadvantages of Corporation

Advantages: It enjoys perpetual life without regard to any change in the person of its owners, the stockholders. The stockholders of the corporation are not liable for the debts of the corporation. It is relatively easier to obtain large amounts of money for expansion due to its perpetual life. The ownership in the corporation is readily transferred. Authority is easily delegated by the hiring of the managers. Disadvantages: The activities of the corporation are limited to those stated in its charter. It is relatively complicated in formation and administration. There is greater degree of government control as compared to other types of business organization. •

Capitalization of a Corporation

The capital of a corporation is acquired through the sale of stock. There are two principal types of stock. Common Stock – represents ordinary ownership without special guarantees of return. Preferred Stock – are guaranteed a definite dividend on their stocks. In the case the corporation is dissolved, the assets must be used to satisfy the claims of the preferred stockholders before those of the common stockholders. Preferred stockholders usually have the right to vote in meetings, but not always.

•

Rights of Common Stockholders

Vote at stockholder’s meeting. Elect directors and delegates to them power to conduct the affairs of the business. Sell or dissolve the corporation. Make and amend the by-laws of the corporation. Subject to government approval, amend or change the charter or capital structure. Participate in profits. Inspect the book of the corporation. •

Financing of Bonds

Bond – is a certificate of indebtedness of a corporation usually for a period of not less than 10 years and guaranteed by mortgage on certain assets of the corporation or its subsidiaries. Bonds are issued when there is a need for more capital such as for expansion of the plant or the services rendered by the corporation. •

Classification of Bonds

Registered Bond – the name of the owner of this bond is recorded on the record books of the corporation and interest payments are sent to the owner periodically without any action. •

Classification of Bonds

According to the Security Behind the Bonds: •

Methods of Bond Retirement

The corporation may issue another set of bonds equal to the amount of bonds due for redemption. The corporation may set up a sinking fund into which periodic deposits of equal amount are made. The accumulated amount in the sinking fund is equal to the amount needed to retire the bond at the time they are due. The corporation may issue callable bonds. These bonds permit repayment of the principal before maturity. •

Value of a Bond

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Sample Problems on Bonds

1. A 10 year corporate bond has a face value of Php5,000 and a bond rate of 8% payable quarterly. A prospective buyer desires to earn a nominal rate of 12% quarterly on investment. What purchase price would the buyer be willing to pay? (php 4,573.49)

2. A bond with a par value of Php1,000 and with a bond rate of 16% payable annually is sold now for Php1,050. If the yield is to be 14%, how much should be the redemption price at the end of seven years? (Php 910.50) •

Sample Problems on Bonds

3. A bond issue of Php200,000 in 10 year bonds, Php1,000 units, paying 16% nominal interest in semiannual payments, must be retired by the use of a sinking fund that earns 12% compounded semiannually. What is the total semiannual expense? 4. A man wants to make a 14% nominal interest compounded semiannually on a bond investment. How much should the man be willing to pay now for a 12% compounded semiannually, Php10,000 bond that will mature in 10 years and pays interest semiannually? (Php 8,909.89) 5. A Php1,000 bond which will mature in 10 years and with a bond rate of 8% payable annually is to be redeemed at par at the end of this period. If it is sold at Php1,030, determine the yield at this price. (7.56%)

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Definition of Terms

Engineering Economy – uses mathematical formulas to account for the time value of money and to balance current and future revenues and costs. Economics – is the science that deals with the production, allocation and use of goods and services. The two major subdivisions of economics are: a.

Macroeconomics is the study of the entire system

b. Microeconomics is the study of how the systems the economic system. •

of economics. affect one business or parts of

Necessities and Luxuries

Necessities – are products or services that are required to support human and activities that will be purchased in somewhat the same quantity even though the prices vary considerably. Luxuries – are products and services that are desired by humans and will be purchased if money is available after the required necessities have been obtained. •

Consumer and Producer Goods and Services

Goods – is defined as anything that anyone wants or needs. Services – would be the performance of any duties or work for another; helpful or professional activity. Marketing – refers to the distribution of goods and services. Marketing a Product – refers to the advertising, and other efforts to promote a products sale. •

Different Types of Goods

1.

Consumer Goods – are those such as food and clothing that satisfy human wants and needs.

2.

Producer Goods – are those such as raw goods.

3.

Capital Goods – are the machinery, used in the production of commodities in producer goods. •

materials and tools, used to make consumer

Supply and Demand

Supply – refers to how many of a certain good or services are available for people to purchase. Demand – means how many people wish to buy that good or service. Law of Supply and Demand

Under conditions of perfect competition, the price at which a given product will be supplied and purchased is the price that will result in the supply and demand being equal. •

Demand

Demand – it refers to the people’s willingness to buy a product or service. Demand Curve – is the plot or graph of the quantity demanded versus the price. Demand Schedule – is the schedule or table listing of the quantity demanded with the corresponding price. • 1. a

Types of Demand Elastic Demand – exists when there is a greater change in quantity demanded as a response to change in price.

2. Inelastic Demand – exists when there is a lesser response to a change in price.

change in quantity demanded as a

3. Unitary Demand – exists when there is an equal (increase or decrease).

change in price and quantity demanded

•

Factors that Influence Demand

Factors that Influence Demand are: 1.

Income

2.

Population

3.

Taste and preference

4.

Price Expectation

5.

Price of Related Goods •

Supply

Supply – it is the willingness of a producer to manufacture goods. Supply Curve – is the plot or graph of the quantity supplied versus the price. Supply Schedule – is the schedule or table listing of the quantity supplied with the corresponding price. •

Factors that Influence Supply

Factors that Influence Supply are: 1.

Price of Goods

2.

Cost of Production

3.

Availability of Resources

4.

Number of Producer and Sellers

5.

Technological Advancement

6.

Taxes

7.

Subsidies •

Relationship of Supply and Demand

Shortage – the supply is less than the demand. Surplus – the supply exceeds the demand. Equilibrium Point – the supply is equal to the demand. •

Market Structures

Market – is the place where the vendors and buyers meet to transact. Perfect Competition – occurs in a situation where a commodity or service is supplied by a number of vendors and there is nothing to prevent additional vendors entering the market. Perfect Monopoly – exist when a unique product or services is available from a single vendor and that the vendor can prevent the entry of all others into the market. Oligopoly – exist when there are so few suppliers of a product or service that action by one will almost inevitably result in similar action by the others.

MODULE 2 Module 2 INTEREST & MONEY – TIME RELATIONSHIPS

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Simple Interest

Interest – is the return on capital or cost of using capital. It is the amount of money paid for the use of borrowed capital or the income produced by money, which has been loaned. Simple Interest – is calculated using the principal only, ignoring any interest that had been accrued in preceding period. •

Types of Simple Interest

1. Ordinary Simple Interest – simple interest in which it is assumed that each month contains 30 days and consequently each year has 360 days.

1 month = 30 days 1 year = 360 days (banker’s year) 2. used.

Exact Simple Interest – simple interest in which the

exact number of days per month is

1 ordinary year = 365 days 1 leap year = 366 days •

Sample Problems on Simple Interest

1. A loan of Php50,000 is made for a period of 13 months, from April 1 to April 30 of the following year, at a simple interest rate of 20%. What future amount is due at the end of the loan period? 2. What is the principal amount if the amount of interest at the end of 2 ½ years is Php450 for a simple interest rate of 6% per annum? 3. Determine the exact simple interest of Php25,000 for the period of December 27, 2002 to March 23, 2003, if the rate of interest is 10%. 4. What is the interest due on a Php1,500 loan for 4 years and three months if it bears 12% ordinary simple interest? 5. Determine the exact simple interest of Php4,000 for the period of Feb. 14, 1984 to November 30, 1984, if the rate of interest is 18%.

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Compound Interest

Compound Interest – the interest for an interest period is calculate on the principal plus total amount of interest accumulated in the previous period. Compound interest means “the interest on top of interest.”

•

Rates of Interest

Rate of Interest – it is the cost of borrowing money. Nominal Rate of Interest – it specifies the rate of interest and a number of interest periods in one year. Effective Rate of Interest – it is the actual or exact rate of interest on the principal during one year. •

Values of “m”

m = 1 for compounded annually (every 12 months) m = 2 for compounded semi-annually (every 6 months) m = 4 for compounded quarterly (every 3 months) m = 6 for compounded bi-monthly (every 2 months) m = 8 for compounded semi-quarterly (every 1 1/2 months) m = 12 for compounded monthly (every month) m = 24 for compounded semi-monthly (every 1/2 month) •

F/P and P/F Factors: Notation and Equations

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Sample Problems on Compound Interest

1. What rate of interest compounded annually must be received if an investment of Php5,400 made now will result in a receipt of Php7,200 5 years hence? 2.

What amount will be accumulated by Php4,100 in 10 years at 6% compounded annually?

3.

What effective annual interest rate corresponds to the following situations? a.

nominal interest rate of 10% compounded semi-annually

b.

nominal interest rate of 6% compounded monthly

c.

nominal interest rate of 8% compounded quarterly

4. How much should Engr. Cruz deposit now, if after 10 years, this will amount to Php100,000. Interest rate is 12% compounded semiannually?

5. If Php1,000 becomes Php5,734 after 15 years, when invested at an unknown rate of interest compounded semi-annually, determine the unknown nominal rate and corresponding effective rate. •

Cash Flow Diagram

Cash Flow Diagram – depicts the timing and amount of expenses (negative, downward) and revenues (positive, upward) for engineering projects. •

Types of Cash Flow Diagrams

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Equation of Value

An equation of value is obtained by setting the sum of the values on a certain comparison or local date (or focal date) of one set of obligations equal to the sum of the values on the same date of another set of obligations. •

Sample Problems on Equation of Value

1. Jay wishes his son, Jason, to receive Php1,000,000 twenty years from now. What amount should he invest now, if it will earn interest of 12% compounded annually during the first five years and 10% compounded monthly for the remaining years. 2. Find the present worth of a future payment of Php300,000 to be made in 10 years with an interest rate of 10% compounded annually. What will be the amount if it will be paid 5 years later (on the 15th year)? •

Discrete Payments

The solution of discrete payments or number of transactions occurring at different periods is taking each transaction to the base year and equating each value. •

Sample Problems on Discrete Payments

1. Acosta Holdings borrowed Php9,000 from Smith Corporation on January 1, 1998 and Php12,000 on January 1, 2000. Acosta Holdings made a partial payment of Php7,000 on January 1, 2001. It was agreed that the balance of the loan would be amortized by two payments, one on January 1, 2002 and one on January 1, 2003, the second being 50% larger than the first. If the interest rate is 12%, what is the amount of each payment? 2. A contract has been signed to lease a restaurant at Php20,000 per year with annual increase of Php1,500 for 8 years. Payments are to be made at the end of each year, starting one year from now. The prevailing rate is 7%. What lump sum paid today would be equivalent to the 8 year lease program? 3. Mr. Cruz buys a second hand car worth Php150,000 if paid in cash. On installment basis, he pays Php50,000 downpayment, Php30,000 at the end of one year, Php40,000 at the end of two years and a final payment at the end of four years. Find the final payment if interest is 14%.

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Continuous Compounding Interest

The solution for interest compounded continuously can be derived thru differential equations and can be found as: •

Sample Problems on Continuous Compounding Interest

1. Philip invested $100 on a bank. The bank offers 5% interest compounded continuously in a savings account. Determine (a) how long will it require for him to earn $5 (b) the equivalent simple interest rate for 1 year of the bank. 2. Which is more advisable to invest Php5,000 for five (5) years, to bank A that offers 5% compounded continuously or to bank B that offers 10% simple interest? •

Banker’s Discount

Certain banks lend money in such a way that they deduct the interest on the money. They actually don’t lend you money you asked for. This type of computing money is called banker’s discount. The money received by the borrower after the discount has been deducted is called proceeds. •

Sample Problems on Banker’s Discount

1. Ms. Glydel Marquez borrowed money from a bank. She received from the bank Php1,342 and promised to repay Php1,500 at the end of 9 months. Determine the following: (a) simple interest rate (b) discount rate or often referred as Banker’s discount.

Module 3 ANNUITY

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Annuity

Annuity – it is a series of equal cash flows occurring each period over a range of periods. Types of Annuity: 1.

Ordinary Annuity

2.

Deferred Annuity

3.

Annuity Due

4.

Perpetuity

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Ordinary Annuity

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P/A and A/P Factors: Notation and Equations

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F/A and A/F Factors: Notation and Equations

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Sample Problems on Ordinary Annuity

1. What is the current value of a $50 payment to be made at the end of each of the next three years if the prevailing rate of interest is 7% compounded annually? 2. An obligation of Php20,000 is to be repaid in uniform annual amounts each of which included repayment of the debt and interest over a period of 5 years. If interest is 10% per year, what is the annual payment? 3. Maintenance cost for a small bridge expected to last for 60 years is estimated to be Php1,000 each for the first 5 years, followed by a Php10,000 expenditure in the 15th year and th Php10,000 in the 30 year. If interest is 10% per year, what is the equivalent uniform annual cost over the 60 year period? •

Sample Problems on Ordinary Annuity

4. What is the equivalent previous worth of Php500 years 72 years ago, if annual interest is 1%?

annuity to be paid constantly in 60

5. Find the annual payment to extinguish a debt of interest.

Php10,000 payable for 5 years at 12%

6. A savings loan is made between a man and banker. What should be the uniform monthly payment that the man should make if he is to borrow Php50,000 and he is to pay in 10 years? Interest is taken as 6% compounded quarterly. 7. What annuity is required over 10 years to equate with the future amount of Php15,000. Assume i = 5%. •

Deferred Annuity

Deferred Annuity – are annuities that are computed on different present year and/or future year. It is annuity where the first payment is made several periods after the beginning of the annuity. •

Methods of Solving Deferred Annuity Problems

1.

Draw the cash flow diagram.

2.

Select any convenient focal date.

Temporary focal date is used to convert deferred annuity to

Final focal date is used to obtained the required value.

3.

Project all values to temporary focal date.

4.

Obtain the final value. •

ordinary annuity

Sample Problems on Deferred Annuity

1. Find the value of x in the cash flow diagram, given that would make the equivalent present worth of the cash flow diagram to Php22,000 and interest rate is at 13% per year. •

Sample Problems on Deferred Annuity

2. Determine the uniform annual payments which would be equivalent to the cash flow diagram given. Interest rate of 12% per year. •

Annuity Due

Annuity Due – is a series of equal payments or receipts occurring over a specified number of periods with the payments or receipts occurring at the beginning of each period. •

Sample Problems on Annuity Due

1. What is the current value of a $50 payment to be made at the beginning of each year, for three years if the prevailing rate of interest is 7% compounded annually? 2. What is the accumulated value of a $25 payment to be of the next three years if the prevailing rate of interest is 9%

made at the beginning of each compounded annually?

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Perpetuity

Perpetuity – are uniform payments which are done infinitely. It is also called as perpetual annuity. •

Types of Perpetuity

1.

Ordinary Perpetuity – first payment is done one period after the focal date.

2.

Deferred Perpetuity – first payment is done several periods after the focal date. •

Sample Problems on Perpetuity

1. How much should Mr. Sy invest on a bank that offers Php1,000 each year in perpetuity.

10% interest so that he would earn

2. Don Jose deposited Php5,000,000 on a bank that earns 10% compounded annually. Five years later he died. His will states that his beneficiary is an orphanage which will be receiving the money in perpetuity a year after he died. How much is the yearly fund the orphanage will be receiving? 3. If money is worth 8% compounded quarterly, compute the present value of the perpetuity of Php1,000 payable quarterly.

Module 4 DEPRECIATION

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Definition of Terms

Depreciation – it is the decrease in the value of a fixed asset, or the value of physical property, with the passage of time. Value – is the present worth of all the future profits that are to be received through the ownership of a particular property. Market Value of a Property – is the amount, which a willing buyer will pay to a willing seller for the property where each has equal advantage and is under no compulsion to buy and sell. •

Definition of Terms

Utility or Use Value of Property – is what the property is worth to the owner as an operating unit. Fair Value – is the value which is usually determined by the disinterested third party in order to established a price that is fair to both seller and buyer. Book Value – is the worth of the property as shown in the accounting records of an enterprise. It is sometimes called as depreciated book value. Salvage or Resale Value – is the price that can be obtained from the property after it has been used. Salvage Year is the year when scrap value is equal to book value. Scrap Value or Junk Value – is the price that can be recovered if an asset is disposed as a junk. •

Purposes of Depreciation

1.

To provide for the recovery of capital

2.

To enable the cost of depreciation to be charged to the cost of producing products or services that results from the use of property. •

which has been invested in physical

property.

Causes of Depreciation

Physical Depreciation – it is due to wear and tear of the asset. Functional Depreciation – it is due to the obsolescence of the asset. Depletion – refers to the decrease in the value of a property due to the gradual extraction of its contents. Monetary Depreciation – depreciation due to changes in price level. •

Physical and Economic Life

Physical Life of a Property – is the length of time during which it is capable of performing the function for which it was designed and manufactured. Economic Life or Useful Life – is the length of time during which the property may be operated at a profit. •

Methods Used to Determine Depreciation

1.

Straight Line Method

2.

Declining Balance Method

3.

Double Declining Balance Method

4.

Sum-of-Years’ Digit Method

5.

Sinking Fund Method

6.

Hour Output Method

7.

Service Output Method •

Straight Line Method

The straight line method is the simplest way in computing for depreciation. In this method, the depreciation each year is constant and the interest rate is being neglected. •

Sample Problems on Straight Line Method

1. Prepare a depreciation table for an asset which was bought at Php15,000 and useful for a period of 5 years. Estimated salvage value is 10% of its original cost. 2. Equipment bought for Php60,000 is expected to last for 30 years. If the book value after 20 years is Php20,000, how much is the depreciation each year? Find the book value after 10 years. 3. A machine which costs Php10,000 was sold as scrap after being used for 10 years. If the scrap value was Php500, determine the total depreciation and book value at the end of th the 5 year. •

Sample Problems on Straight Line Method

4. Delivery jeeps purchased by KH Company cost Php180,000 each. Past records indicate that jeeps should have an economic life of 10 years. They can be sold from an average of Php50,000 each year after 10 years of use. The company receives 9% interest on investment funds. Using straight line method: Determine: a.

the depreciation charge during 3 years

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b.

the depreciation reserve accumulated at the end of 3

C.

book value at the end of 3 years.

years

Declining Balance Method or Reducing Balance Method

In this method, the net book value at the end of each period can be simply computed by multiplying the original market price by a fix percentage repeatedly until it reaches the salvage value. This method is also called Matheson’s Formula. •

Sample Problems on Declining Balance Method

1. A machine worth Php800,000 is bought from China. Freight charges amount to Php200,000. If the scrap value of the machine is Php50,000 that occurs at the end of 17 years. Compute (a) the depreciation and (b) book value at the end of 11 years. 2. Equipment bought for Php60,000 is expected to last years is Php20,000. How much is the depreciation for year •

for 30 years. If the scrap value after 20 10?

Double Declining Balance Method

This is the same as declining balance method except that k is replaced by 2/n. Sample Problem on Double Declining Balance Method A plant bought a calciner for Php220,000 and used it for 10 years, the life span of the equipment. What is the book value of the calciner after 5 years of use? Assume a scrap value of Php20,000 for straight-line method; Php22,000 for declining balance method and Php20,000 for double-declining balance method. •

Sum-of-Years’ Digit Method

This method uses the year’s digit (in reverse order) in computing for the depreciation. •

Sample Problem on Sum of Years’ Digit Method

An industrial plant bought a generator set for Php90,000. Other expenses including installation amounted to Php10,000. The generator set is to have a life of 17 years with a salvage value at the end of life at Php5,000. Determine the depreciation charge during the 13th year and the book value at the end of 13 years by:

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(a)

Declining balance method

(b)

Double declining balance method

(c)

Sum-of-Years’ digit method

Sinking Fund Method

Sinking fund method presents the idea of annuity in computing for the depreciation. The interest rate for the worth of money is being considered so as to have the depreciable value.

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Sample Problem on Sinking Fund Method

A broadcasting company purchased an equipment for Php53,000 and paid Php1,500 for freight and delivery charges to the job site. The equipment has a normal life of 10 years with a trade-in value of Php5,000 against the purchase of a new equipment at the end of the life. Determine the annual depreciation cost using:

•

(a)

straight line method

(b)

sinking fund method assuming interest is

6% compounded annually

Hour Output method

In this method, the functionality period and the period the machine has been used is considered. Depreciation is computed based on the wear and tear of the machine. •

Service Output method

Similar to the hour output method, this method based its computation on how much the asset has been used. •

Sample Problems on Service and Hour Output Method

1. An electric bulb bought for Php100 is guaranteed to be useful for 50 hours. A certain company uses the said bulb for 10 hours a day. If there is no scrap value for the bulb. Compute the daily depreciation and create the depreciation table throughout its economic life. 2. A tire bought for Php1,000 is expected to be useful in traveling 100 km after which it can be sold as scrap for Php50. (a) If the pedometer displays a value of 85 km, what is the book value of the tire? (b) How much did the owner need to travel for the tire to amount to Php80?

Module 5 BREAK-EVEN ANALYSIS

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Definition of Terms

Break-Even Analysis – it involves estimating the level of sales necessary to operate a business on a break-even basis. Break-Even Point (BEP) – is defined as the point where sales or revenues equal total expenses. Break-Even Margin – is a ratio that shows the gross-margin factor for a break-even condition. The formula is total expenses divided by net revenues multiplied by 100 to get a percentage. •

Break-Even Graph

Break-Even Chart – shows the graph of fixed cost, variable cost and expected income from sales for different production levels. •

Ways to Lower the Break-Even Point

Lower direct costs, which will raise the gross margin. Exercise cost controls on your fixed expenses, and lower the necessary total expenses. Raise prices. •

Key Break-Even Factors

Fixed Costs – these costs remain constant (or nearly so) within the projected range of sales levels. These can include facilities costs, certain general and administrative costs, and interest and depreciation expenses. Variable Costs – these costs vary in proportion to sales levels. They can include direct material and labor costs, the variable part of manufacturing overhead, and transportation and sales commission expenses. Contribution Margin – this is equal to sales revenues less variable costs. This amount is available to offset fixed expenses and (hopefully) produce an operating profit for the business. •

Appraisal of Break-Even Analysis

Advantages of Break-Even Analysis It points out the relationship between cost, production volume and returns. Limitations of Break-Even Analysis It is best suited to the analysis of one product at a time.

It may be difficult to classify a cost as all variable or all fixed. There may be a tendency to continue to use a break-even analysis after the cost and income functions have changed. •

Sample Problems on Break Even Analysis

1. An entrepreneur at the location in the United States is planning to enter the gourmet soy-based burger market. The forecast expected unit sales of 200,000 burgers in 18 months. The variable cost for making one burger is $0.85 and the fixed cost of making burgers for 18 months will be a total of $165,000 which covers for the rent, phone bill, and insurance coverage – these items tend not to vary in amount per month over the term of one year. The best estimate of what the average consumer will pay for the soy burger is $1.95. How many burgers will he have to sell to break even? •

Sample Problems on Break Even Analysis

2. Toyota Motor Phils. Corporation Sta. Rosa Plant has a production capacity of 700 cars per month and its fixed cost is Php100,000,000 monthly. The variable cost per car is Php300,000 and each car can be sold for Php650,000. Due to cost reduction program, fixed cost will be reduced to 10% and variable cost by 20%. Determine the new and old break break-even point. •

Sample Problems on Break Even Analysis

3. A farmer wants to buy a new combine rather than hire a custom harvester. The total fixed costs for the desired combine are $21,270 per year. The variable cost (not counting the operator’s labor) are $8.75 per hour. The farmer can harvest 5 acres per hour. The custom harvester charges $16.00 per acre. How many acres must be harvested per year to breakeven?

Module 6 COMPARISON OF ALTERNATIVES •

Definition of Terms

1. Comparison of Alternatives – deals with situations in which one has more than one choice and using engineering economic principles, one needs to decide between the alternatives so as to go with the one that is most economically justified. 2.

Methods used in the Selection of Alternatives

Present Worth Method

Annual Cost/Worth Method

Equivalent Uniform Annual Cost (EUAC) Method

Rate of Return (ROR) Method

Payback (Payout) Period Method

Capitalized Cost

Benefit/Cost Ratio Analysis

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Present Worth Method

This method involves finding the equivalent value of each alternative at the present time, identified as time 0. If only costs are involved, we can select the alternative with the smallest present worth of costs. If cost and revenues are involved, we select the alternative with the greatest present worth on net revenues. To perform present worth or annual worth analysis, an interest rate and a study period must be specified. The interest rate is usually the minimum acceptable rate of return (MARR) of the organization.

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Present Worth Method

Present worth analysis can only be used when the alternatives have the same lives. If the alternatives have different lives, some mechanism must be used to compare them over the common study period. •

Sample Problems on Present Worth Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors

will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

interest

Sample Problems on Present Worth Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Present Worth Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Present Worth Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Present Worth Method

3. A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit- return estimates shown below. Rate of interest is 15%. •

Sample Problems on Present Worth Method

3.

A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Annual Cost (Worth) Method

The annual worth method involves finding the equivalent end-of-period value of each alternative. The periods are usually in years. If only costs are involved, we can select the alternative with the smallest equivalent uniform annual cost (EUAC) or net annual cost (NAC). If costs and revenues are involved, we can select the alternative with the greatest equivalent uniform annual benefit, or net annual worth (NAW).

If two alternatives have the same annual worth, then the one with the greatest investment is preferred. The extra investment makes exactly the required MARR. •

Annual Cost (Worth) Method

Problems that can be solved by the present worth method can also be solved by the annual worth method. An annual worth analysis is sometimes preferred over a present worth analysis because people think better in terms of annual amounts than an equivalent amount to time zero. Both methods yield the same results. Annual worth analysis is the easiest method when the alternatives have different lives. No specific study period need be specified, but the implicit assumption that the alternative are compared over the least common multiple of the lives. •

Sample Problems on Annual Worth Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Annual Worth Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Annual Worth Method

2. Perform an annual worth analysis of equal service if the MARR is 10% per year. Revenues for all three the same. •

machines with the costs shown below, alternatives are expected to be

Sample Problems on Annual Worth Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Annual Worth Method

3. A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Sample Problems on Annual Worth Method

3.

A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Equivalent Uniform Annual Cost (EUAC) Method

In this method, all cash flow (irregular or uniform) must be converted to an equivalent uniform annual cost, that is, a year-end amount which is the same each year. The alternative with the least EUAC is preferred. When the EUAC method is used, the EUAC of the alternatives must be calculated for one life cycle only. This method is flexible and can be used for any type of alternative selection problems. The method is a modification of the annual cost method. •

Sample Problems on Equiv. Uniform Annual Cost Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Equiv. Uniform Annual Cost Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Equiv. Uniform Annual Cost Method

2. Perform an equivalent uniform annual cost analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same.

•

Sample Problems on Equiv. Uniform Annual Cost Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Equiv. Uniform Annual Cost Method

3. A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Sample Problems on Equiv. Uniform Annual Cost Method

3.

A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Rate of Return (ROR) Method

If ROR > MARR, select the alternative with the bigger investment ROR < MARR, select the alternative with the smaller investment •

Rate of Return (ROR) Method

The rate of return on the capital invested is •

Sample Problems on Rate of Return Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Rate of Return Method

1. Motors from two different manufacturers are being considered for application. Both motors are 50 HP, 460 volts, 3-phase, 60 Hz, but motor A operates at 80% efficiency whereas motor B operates at 88% efficiency. The expected used for motors are 20 years. Motor A costs Php600,000 and motor B costs Php750,000. Electrical energy cost Php3.00 per kW-hr and the motors will be

operated at 8 hours per day, 250 days per year. Assume taxes are 5% and rate of interest is 10%. Which motor is purchased? Also, life of both motors is 15 years. •

Sample Problems on Rate of Return Method

2. Perform an annual worth analysis of equal service if the MARR is 10% per year. Revenues for all three the same. •

machines with the costs shown below, alternatives are expected to be

Sample Problems on Rate of Return Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Rate of Return Method

2. Perform a present worth analysis of equal service machines with the costs shown below, if the MARR is 10% per year. Revenues for all three alternatives are expected to be the same. •

Sample Problems on Rate of Return Method

3. A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Sample Problems on Rate of Return Method

3.

A project engineer with EnvironCare is assigned to start up a new office in a city where a 6 year contract has been finalized to take and analyze ozone-level readings. Two lease options are available, each with a first cost annual lease cost, and deposit-return estimates shown below. Rate of interest is 15%. •

Sample Problems on Rate of Return Method

4. A firm is considering purchasing equipment that will reduce costs by Php40,000. The equipment costs Php300,000 and has a salvage value of Php50,000 and a life of 7 years. The annual maintenance cost is Php6,000. While not used by the firm, the equipment can be rented to others to generate an income of Php10,000 per year. If money can be invested for an 8% return, is the firm justified in buying the equipment? •

Sample Problems on Rate of Return Method

4.

A firm is considering purchasing equipment that will reduce costs by Php40,000. The equipment costs Php300,000 and has a salvage value of Php50,000 and a life of 7 years. The annual maintenance cost is Php6,000. While not used by the firm, the equipment can be rented to others to generate an income of Php10,000 per year. If money can be invested for an 8% return, is the firm justified in buying the equipment? •

Payback (Payout) Period Method

In this method, the payback period of each alternative is computed. The alternative with shortest payback period is adopted. This method is seldom used. Payback Period – is the length of time required to recover the first cost of an investment from the net cash flow produced by that investment for an interest rate of zero. •

Sample Problem on Payback Period Method

In a marble block quarrying operation, hand rock drills, costing Php50,000 each, are used. It has a drilling rate of 10 cm per minute, produces 10 cubic meters of block per month and consumes 60 liters of diesel fuel for compressor drive, per rock drill per cubic meter produced utilizing 1 worker per drill. A modern equipment quarry bar mounted rock drill is being offered for Php180,000 per unit and has a drilling rate of 60 per minute that will produce 60 cubic meters of block per month, but consumes 120 liters of diesel fuel for the compressor drive, per 6 cubic meters of block utilized, utilizing 2 workers per quarry bar drill. Consider diesel fuel at Php6.00 per liter at the quarries, worker earning Php80.00 per day, 25 days per month, 5 years life of both drills with 20% salvage value, neglecting cost of money, other cost at Php500 per cubic meter and marble blocks sold at Php2,000 per cubic meter. Would you recommend the purchase of the new equipment? •

Sample Problem on Payback Period Method

In a marble block quarrying operation, hand rock drills, costing Php50,000 each, are used. It has a drilling rate of 10 cm per minute, produces 10 cubic meters of block per month and consumes 60 liters of diesel fuel for compressor drive, per rock drill per cubic meter produced utilizing 1 worker per drill. A modern equipment quarry bar mounted rock drill is being offered for Php180,000 per unit and has a drilling rate of 60 per minute that will produce 60 cubic meters of block per month, but consumes 120 liters of diesel fuel for the compressor drive, per 6 cubic meters of block utilized, utilizing 2 workers per quarry bar drill. Consider diesel fuel at Php6.00 per liter at the quarries, worker earning Php80.00 per day, 25 days per month, 5 years life of both drills with 20% salvage value, neglecting cost of money, other cost at Php500 per cubic meter and marble blocks sold at Php2,000 per cubic meter. Would you recommend the purchase of the new equipment? •

Capitalized Cost

Capitalized Cost – is the present worth of an alternative that will last “forever.” Public sector projects such as bridges, dams, irrigation systems and railroad fall into this category. •

Sample Problems on Capitalized Cost

Two methods of conveying eater are being studied. Method A requires a tunnel, first cost Php180,000, life perpetual, annual operation and upkeep is Php3,000. Method B requires a ditch plus flume; first cost of ditch is Php40,000, life perpetual, annual depreciation and upkeep is Php1,500, first cost of flume is Php30,000, life 10 years, salvage value is Php5,000, annual operation and upkeep is Php4,000. If money is worth 6%, determine which method is to be recommended? •

Sample Problems on Capitalized Cost

Two methods of conveying eater are being studied. Method A requires a tunnel, first cost Php180,000, life perpetual, annual operation and upkeep is Php3,000. Method B requires a ditch plus flume; first cost of ditch is Php40,000, life perpetual, annual depreciation and upkeep is Php1,500, first cost of flume is Php30,000, life 10 years, salvage value is Php5,000, annual operation and upkeep is Php4,000. If money is worth 6%, determine which method is to be recommended? •

Benefit/Cost Ratio Analysis

Benefit/Cost Ratio Analysis – it is the most commonly used method by government agencies for analyzing the desirability of public projects. If B/C ≥ 1.0, accept the project as economically acceptable for the estimates and discount rate applicable. •

Benefit/Cost Ratio Analysis

Cost – estimated expenditures to the government entity for construction, operation, and maintenance of the project less any expected salvage value. Benefits – advantages to be experienced by the owners, the public. Disbenefits – expected undesirable or negative consequences to the owners if the alternative is implemented. Disbenefits may be indirect economic disadvantages of the alternative. •

Benefit/Cost Ratio Analysis

The benefit cost ratio on the capital invested is •

Significant Differences in the Characteristics of Public and Private Sector Alternatives

•

Sample Problems on Benefit/Cost Ratio Analysis

1. The National Government intends to build a dam and hydroelectric project in the Cagayan Valley at a total cost of Php455,500,000. The project will be financed by soft foreign loan with an interest of 5% per year. The annual cost for operation, maintenance, distribution,

facilities and others be Php56,500,000.

would total Php15,100,000. Annual revenues

If the structures are expected to last for 50 project economically acceptable? •

and benefits are estimated to

years with no salvage value, is the

Sample Problems on Benefit/Cost Ratio Analysis

1. The National Government intends to build a dam and hydroelectric project in the Cagayan Valley at a total cost of Php455,500,000. The project will be financed by soft foreign loan with an interest of 5% per year. The annual cost for operation, maintenance, distribution, facilities and others would total Php15,100,000. Annual revenues and benefits are estimated to be Php56,500,000. If the structures are expected to last for 50 years with no salvage project economically acceptable? •

value, is the

Sample Problems on Benefit/Cost Ratio Analysis

2. Two routes are under construction for a new highway. Route A would be located about 5 miles from the central business district and would require longer travel distances by local commuter traffic. Route B would pass directly through the downtown area and although its construction cost would be higher, it would reduce the travel time and distance for local commuters. The costs for the two roads are as follows: •

Sample Problems on Benefit/Cost Ratio Analysis

2. Two routes are under construction for a new highway. Route A would be located about 5 miles from the central business district and would require longer travel distances by local commuter traffic. Route B would pass directly through the downtown area and although its construction cost would be higher, it would reduce the travel time and distance for local commuters. The costs for the two roads are as follows: •

Sample Problems on Benefit/Cost Ratio Analysis

3. Four alternatives for providing electric power supply to identified with the following annual benefits and costs: •

a small town have been

Sample Problems on Benefit/Cost Ratio Analysis

3. Four alternatives for providing electric power supply to a small town have been identified with the following annual benefits and costs:

Module 7 CAPITAL FINANCING

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Equity and Borrowed Capital

Equity Capital or Ownership Forms – are those supplied and used by the owners of an enterprise in the expectation that profit will be earned. Borrowed Funds or Capital – are those supplied by others on which a fixed rate of interest must be paid at a specified time. •

Types of Business Ownership

Individual Ownership or Sole Proprietorship – is one which is owned and run by one individual and where there is no legal distinction between the owner and the business. Partnership – is an association of two or more persons for the purpose of engaging in business for a profit. Corporation – is a fictitious being, recognized by law, that can engage in almost any type of business transaction in which a real person could occupy himself. It operates under a charter that is granted by the government. •

Advantages and Disadvantages of Sole Proprietorship

Advantages: It is easy to organize. The owner has full control of the enterprise. The owner is entitled to whatever benefits and profits that accrue from the business. It is easy to dissolve. Disadvantages: The amount of equity capital which can be accumulated is limited. The organization ceases upon the death of the owner. It is difficult to obtain borrowed capital owing to the uncertainty of the life of the organization. The liability of the owner for his debts are unlimited. •

Advantages and Disadvantages of Partnership

Advantages:

More capital may be obtained by the partners pooling their resources together. It is bound by few legal requirements as to its accounts, procedures, tax forms and other items. Dissolution of a partnership may take place at any time by mere agreement of the partners. It provides an easy method whereby two or more persons may enter into business each carrying those burdens that he can best handle. Disadvantages: The amount of capital that can be accumulated is definitely limited. The life of the partnership is determined by the life of the individual partners. When any partner dies, the partnership automatically ends. There may be serious disagreement among individual partners. Each partner is liable for debts of the other partnership. •

Advantages and Disadvantages of Corporation

Advantages: It enjoys perpetual life without regard to any change in the person of its owners, the stockholders. The stockholders of the corporation are not liable for the debts of the corporation. It is relatively easier to obtain large amounts of money for expansion due to its perpetual life. The ownership in the corporation is readily transferred. Authority is easily delegated by the hiring of the managers. Disadvantages: The activities of the corporation are limited to those stated in its charter. It is relatively complicated in formation and administration. There is greater degree of government control as compared to other types of business organization. •

Capitalization of a Corporation

The capital of a corporation is acquired through the sale of stock. There are two principal types of stock. Common Stock – represents ordinary ownership without special guarantees of return. Preferred Stock – are guaranteed a definite dividend on their stocks. In the case the corporation is dissolved, the assets must be used to satisfy the claims of the preferred stockholders before those of the common stockholders. Preferred stockholders usually have the right to vote in meetings, but not always.

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Rights of Common Stockholders

Vote at stockholder’s meeting. Elect directors and delegates to them power to conduct the affairs of the business. Sell or dissolve the corporation. Make and amend the by-laws of the corporation. Subject to government approval, amend or change the charter or capital structure. Participate in profits. Inspect the book of the corporation. •

Financing of Bonds

Bond – is a certificate of indebtedness of a corporation usually for a period of not less than 10 years and guaranteed by mortgage on certain assets of the corporation or its subsidiaries. Bonds are issued when there is a need for more capital such as for expansion of the plant or the services rendered by the corporation. •

Classification of Bonds

Registered Bond – the name of the owner of this bond is recorded on the record books of the corporation and interest payments are sent to the owner periodically without any action. •

Classification of Bonds

According to the Security Behind the Bonds: •

Methods of Bond Retirement

The corporation may issue another set of bonds equal to the amount of bonds due for redemption. The corporation may set up a sinking fund into which periodic deposits of equal amount are made. The accumulated amount in the sinking fund is equal to the amount needed to retire the bond at the time they are due. The corporation may issue callable bonds. These bonds permit repayment of the principal before maturity. •

Value of a Bond

•

Sample Problems on Bonds

1. A 10 year corporate bond has a face value of Php5,000 and a bond rate of 8% payable quarterly. A prospective buyer desires to earn a nominal rate of 12% quarterly on investment. What purchase price would the buyer be willing to pay? (php 4,573.49)

2. A bond with a par value of Php1,000 and with a bond rate of 16% payable annually is sold now for Php1,050. If the yield is to be 14%, how much should be the redemption price at the end of seven years? (Php 910.50) •

Sample Problems on Bonds

3. A bond issue of Php200,000 in 10 year bonds, Php1,000 units, paying 16% nominal interest in semiannual payments, must be retired by the use of a sinking fund that earns 12% compounded semiannually. What is the total semiannual expense? 4. A man wants to make a 14% nominal interest compounded semiannually on a bond investment. How much should the man be willing to pay now for a 12% compounded semiannually, Php10,000 bond that will mature in 10 years and pays interest semiannually? (Php 8,909.89) 5. A Php1,000 bond which will mature in 10 years and with a bond rate of 8% payable annually is to be redeemed at par at the end of this period. If it is sold at Php1,030, determine the yield at this price. (7.56%)

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