Lecture 5 Break Even Analysis

NED University of Engineering and Technology A P P L I E D E C ON ON O M I C S F OR E N G I N E E R S

TOPIC # 5

BREAK EVEN ANALYSES

 Learning Outcome: 

Concept of Breakeven Analysis (BEA)

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 Behavior of Costs Costs & Revenue Revenue

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 Numerical and and graphical presentations

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 Practical Applications Applications

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 BEA as a Management Management Tool

 This  Thi s chapte hapter cove covers rs the basics basics of brea break-eve k-even n analysis nalysis which which i s the simplest simplest analytica nalyticall tool in management. It details what break-even analysis is, what it is used for, what definitions are used in break-even analysis and how break-even analysis can be helpful in decision-making of professionals in construction industry . In construction industry, break-even analysis can be a handy tool to find answers to questions such as: How many years should I operate the facility to recover the initial investment and annual operating costs? H ow much much does our company company need need to sell to reach reach the desir desire ed profi pr ofi tability? What should be the better option between alternatives? 

 

An enterprise nterpri se,, whethe whetherr or not a prof it ma m aximiz ing organiz organiz ation, always always wants wants to know: Whatt price Wha pr ice or output outpu t le l eve vell must mu st be for total re r eve venue nue to just ju st equal equal to total cost? analysis. The an swer is:  is: a breakeven analysis. Stri ctly speakin speaking g, this thi s analysis is i s to de dete termi rmine ne the minimum level level of output that th at all all ows the firm to break even, even, but it could be used to compare and analyze various project options and alternatives.  neither profi t nor l oss; also : a financial re r esult refle refl ecting neith neithe er Break Even Means: Means: neither profit nor loss Break- even Point Point :  The origins of break-e break-eve ven n point can can be found in the economic conomic concepts of “the point of indifference.” In simple words, the break-even point can be defined as a point where total costs (expenses) and total sales (revenue) are equal. In simple words it can be described as a point where there is no net profit or loss. The firm just “breaks even.” Another way to look at it is that the break-even point is the point at which your product stops costing you money to produce and sell, and starts to genera generate te a profit profi t f or your co company mpany..

Break- Even Analysis: In its simplest form, it facilitates an insight into the fact whether the revenue from a project, service or product incorporates the ability to cover the relevant production cost of that particular project, product or service or not. On the surface, break-even analysis is a tool to calculate a t w h a t p r o d u ct i o n v o l u m e t h e variable and fixed costs of producing your product will be recovered . Break-even analysis can also be used to solve other management problems, including setting prices, and evaluating the best strategies to follow. Break-even analysis is done by using a parameter (or variable) which is an amount of revenue, cost, suppl y, demand, etc. for 1) one project or between 2) tw o alternatives.

BREAKEVEN ANALYSIS FOR A SINGLE PROJECT Basically, break-even analysis determines the “break-even point”, at which operations neither make money nor lose money (Paek 2000, Blank and Tarquin 2008 ). At the break-even point, there is no gain or l oss; hence costs or expenses are equal to revenues/ incomes . Costs and Revenue:  As far as costs and revenue are concerned we have already discussed them in detail in previous lectures. Here is only a memory refresher. Q or Q BE Quantity or Breakeven Quantity P or A R Price or Average Revenue T FC Total Fixed Cost (Costs that do not change in short-term) v  Variable Cost per Unit of production or sale    

   

T VC

Total Variable Cost(v   .QBE)

VC TC TR

Variable Cost Total Cost (FC + VC) Total Revenue (P.QBE)

F i x e d c o s t  represents the expenses that are not related with the volume of production

(or activity level) over a feasible range of operations. Examples include buildings, insurance expenses, depreciation, overheads, and cost of information systems ( Blank and Tarquin 2008). It is the sum of all costs to produce the first unit of a product. Another example could be the cost of excavation equipment regardless of the excavation work perf ormed on diff erent projects. V a r i a b l e c o st  represents the cost items that change with the volume of production or

construction. Input materials and time to produce a unit affect variable costs. Examples include direct labor costs, fuel costs, material types (e.g., a certain type of paint used for painting a facility), and marketing costs ( Blank and Tarquin 2008, Paek 2000). T o t a l c o st   is the sum of the fixed and total variable costs for any production or construction. T o t a l r ev en u e  is   the product of expected unit sales and the unit price of each unit.

Cost/ expense and revenue/ income relations are commonly assumed as linear; however non-linear relations are more reali stic with more revenue for larger volumes (Blank and  Tarquin 2008). Here we will confine our discussion only to linear r elationship. Below is given the algebraic expression and explanation through which breakeven point is identified as QBE which is determined using linear math relations for revenue and cost. Mathematically, the formula for break-even point can be shown as: TR = TC or Profit = 0 Where T R represents the total revenues and T C represents total costs or expenses for an operation. TR = TC Expected unit sales (Q) x Unit pri ce (P) = Fixed cost (FC) + T otal v ariable cost (VC) Q x P = FC + Variable unit cost ( v  ) x Unit sales(Q) Q x P = FC + (v   x Q) (Q x P) – (v   x Q) = FC Q (P -

) = FC Q = FC / (P -

v 

v 

)

Let QBE denote the break-even output. The difference “P - v ” is often called the average contribution margin (ACM)  because it represents the portion of selling price that "contributes" to paying the fixed costs. Relationship between FC, TC, TR and Q EB Cost/ Revenue

 TR Profit

 TC  TVC

 TFC Quantity

QBE  Table below shows the summary of eff ect on QBE

Variable  Total Fixed Cost (e.g., cost of equipment) Average Variable Cost (e.g., cost of material) Product Price

Dir ection of Change Up Down

Break- even Output Up Down

Up Down

Up Down

Up Down

Down Up

 The break-even diagram can be employed to see the eff ects of various exogenous changes on the break-even point. Here are a few scenarios Init ial Change

Curv e Af fected

Increase in Output price

The TR curve counterclockwise

moves

Af fect on QBE Decrease

Increase in the price of one the  TVC and TC curves, variable input both counterclockwise

Increase

Higher TFC

Increase

TFC curve, parallel-shift up

Effects of TFC, AVC, and P on Q BE Cost/Revenue TR1

TR

TR TC  VC FC

TC1

TC

 VC

FC

Quantity

QBE

Ex a m p l e 1 :  Assume that as an investor, you are planning to enter the construction

industry as a panel formwork supplier. Given the size of the construction industry and the potential number of forthcoming projects, you forecasted that within two years, your fixed cost for producing formworks is Rs. 3,000,000. The variable unit cost for making one panel is Rs. 1500. The sale price for each panel you charge will be Rs. 2,500. How many panels you need to sell in total, in order to start making money?

Solution:

Variable unit cost = Rs. 1,500/ panel  Total fixed cost =Rs. 3,000,000 Pri ce per unit = Rs. 2,500  TC = TR VC + FC = TR 1,500 x Q BE + 3,000 000 = Q BE x 2,500 (Q BE  r efer s to the nu m ber of pan els) QBE = 3,000 000 / (2,500-1,500) = 3,000 panels Ex a m p l e 2 :  Calculate the break-even output f or FC = $20,000, P = $7, and AVC = $5

Solution :

QBE =

, 

,

=



− = 10,000

Ex a m p l e 3 :  A manufacturing company supplies its products to construction job sites.

 The average monthly fixed cost per site is $ 4500, while each unit costs $ 35 to produce, and selling price is $ 50. a) Determine the monthly volume of supplies to j ob sites in order to break-even. b)  The company has to modify the selling prices due to severe competition. In this case, the fixed cost and production costs will be the same, but the sales price per unit will be $ 50 for the first 200 units and $ 40 for all above this threshold level. Determine the monthly breakeven volume. Solution:

(a) Q = 4500 / (50-35) =300 units, where Q refers to the number of units per month (b) At 200 units, the profits is negative at -$ 1500, as determined by Profit = Revenue – cost Profit at 200 units pr oduction = 200x50 – (4500 + 35x200) = 10 000 – 4500- 7000 = -1500. 50 x 200 + 40 x Q = 4500 + 35x (200 + Q) QBE = (4500 + 7000 -10 000) / (40 – 35) = 1500 / 5 = 300 units per month Hence the required volume is 500 units per month, the point at which revenue and total cost break even at $ 22 000 Ex a m p l e 4 :  Suppose TFC = $10,000, P = $5, AVC = $2. What is the output necessary to

earn $5000 total profit? What is the “average contribution margin”? Answer : QBE =

,, 

,

=



− = 5,000

Ex a m p l e 5 :  Suppose that product price decreases from P1 to P2. Show on the diagram

how much output would change to m a i n t a i n t h e sa m e lev e l of p r o f i t t a r g et

Cost/Revenue

TR

TR1 TC

 A

Profit

FC Quantity

QBE

QBE

A n s w e r :   Draw a line parallel to TC through A. This line cuts the TR1 line at B, drop a

vertical line from B to determine the new output level. [Note: The solution may not be feasible if we are told that the market size cannot accommodate this quantity.] Ex a m p l e 6 :  Given TR and TC, a firm is currently operating at 50% of i ts capacity at

some profit target. How much of a price drop would cause the firm to operate at 75% capacity at the same level of profit? Answer:  Cost/Revenue TR1

TR TC1

B

TC

 A

FC Quantity QBE Answer:  The vertical dotted line on the right marks the firm’s capacity. Current output

is half of that. Draw the line through A and parallel to TC. From 75%capacity output point, draw the vertical line. B is the intersection. Draw the line TR2 through B. The slope of TR2 gives the required output price.

BREAKEVEN ANALYSIS BETWEEN TWO ALTERNATIVES OR PROJECT

Break-even analysis can also be used to select among alternatives (e.g., projects or construction processes). In order to perform break-even analysis between alternatives, there needs to be a parameter (e.g., cost or revenue variables) that is common in both alternatives. When two alternatives are compared, the break-even point represents the point of indif fer ence between the alternatives (i.e., the point at which two alternatives are equally desirable) (Badiru 1996).  The steps to find the point of indiff erence between alternatives: 

Find th e com m on var iable betw een th e altern atives

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Expr ess the total cost of each alter nat ive as a f un ction of the com m on va r iable

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Equ ate expr essions and solve for th e poin t of in dif fer ence

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Select the alter na tiv e w ith hi gh er va r iable cost (lar ger slope) if t he expected level is below t he point of in diff er ence, and select th e altern ative with lower v ar iable cost if th e level is above th e point of in dif fer ence.

For the example provided below, profit functions are graphed. This graph shows that Project B  is favorable over the other alternatives if the production is between 0 and 100 units, Project A is favorable if the production is between 100 and 200 units, and Project C is favorable if the production is larger than 200 units. Profit ($)

Project C Project A 

Project B Units (X) -200 - 300

100

150

200

Selection of alternative is based on anticipated value of common variable:  

Value BELOW breakeven; select the alternate with higher variable cost (Alt 1) Value ABOVE breakeven; select the alternate with lower variable cost (Alt 2)

Example 5:  There exist two alternative locations for an asphalt mixing plant to transport materials from. Characteristics of these two locations and associated costs are tabulated below. Which location is best for the asphalt mixing plant, the cheaper Location A or closer Location B? (Thi s example ha s been ad opted fr om M IT Engin eeri ng Econom ics lectur e n o t es , c o p y r i g h t © M I T , En v i r o n m e n t & C i v i l E n g i n e er i n g D e p a r t m e n t)  .

 Tr ansportation distance  Tr ansportation expense Monthly rental expense Set-up cost Workmanship costs  Total volume available  Time to use the location

Location A 6 km $ 1.15/ m3-km $ 1000/ month $ 15 000 0 50000 m3 4 months (85 days)

Location B 4.3 km $ 1.15/ m3-km $ 5000/ month $ 25 000 $ 96/ day 50000 m3 4 months (85 days)

Solution:

First obtain the total cost functions for all alternatives Location A Rental expense Set-up cost Workmanship costs  Transportation Total Cost

Fixed Costs 4 month x $1000/ month =$ 4000

Location B 4 month x $5000/ month =$ 20 000

$ 15 000 $ 25 000 0 85 days x $96/ day =$ 8160 Tr ansportation/ Variable costs 6 km x $1.15 x Q 4.3 km x $1.15 x Q $ 19 000 +6.9 Q $ 53 160 +4.945 Q

Equate the total cost functions to solve for volume to be transported for break-even point $ 19 000 + 6.9 Q BE = $ 53 160 + 4.945 Q BE QBE = 17473 m3. At 17473 m 3  of material usage, both sites are equally desirable. If less m ater ial i s , and if m ore volum e is t r a n sp o r t e d t h a n 1 7 4 7 3 m  3 , th en selecting location A is fa vor able  expected to be tr ansported th an 1 7473 m 3, then selecting location B is m or e favor able 

with less variable cost.

Example: Perform a make/ buy analysis where the common variable is Q, the number of units produced each year. AW relations are: (N o t e :  In

examples like the one given below, deter m ine one of the par am eters P, A, F, i, or n , wi th others

constan t, that m akes tw o elements equal )

AWmake. =- 18,000(A/ P, 15%,6) +2,000(A/ F, 15%,6) – 0.4Q =- 18,000 x 0.2642 + 2,000 x 0.1142 – 0.4Q AWbuy  =- 1.5Q

AWbuy

10

8

AWmake

6

Solution: Equate AW relations, solve for Q 1.5Q = - 4528 - 0.4Q , Q = 4116 per year

4

2 0

1

2

3

4

5

If anticipated production > 4116, select AWmake alternative (lower variable cost)  TH E EXCEL SHEET HAS AN IN BUILT FUNCTION NA MED “GOAL SEEK” WHICH CAN BE USED FOR BREAK EVEN ANALYSIS: Data What- If A nalysis Goal Seek  If you want to learn it then arrange extra class other than regular classes, I will explain how to use it.

INTRODUCTION TO COMPOUND INTEREST FACTOR Notations Used in Compound Interest i = Interest rate per interest period*. n = Number of interest periods. P = A present sum of money. F = A future sum of money.  A  = An end-of-period cash receipt or disbursement in a uniform series continuing for n periods. G = Uniform period-by-period increase or decrease in cash receipts or disbursements. g = Uniform rate of cash flow increase or decrease from period to period; the geometric gradient. r = Nominal interest rate per interest period*. m = Number of compounding sub periods per periods*. Concept of Compound Interest Interest (i) applies to total amount (P + sum of all I) during each period. Consider the following Cash flow:  A $1000 deposit for 5 years at 10% / year would result in:  Amount accrued  Year Begin Year Int. End year 1 P = 1000 100 F1 = 1100 F1 = P + Pi 2 1100 110 F2 = 1210 F2 = P + Pi+ (P + Pi) i 3 1210 121 F3 = 1331 F3 = P + Pi+ (P + Pi) i + (P + Pi+ (P + Pi) i) i 4 1331 133 1464 = P + Pi+ Pi+Pi2 + Pi+Pi2 + Pi2 + Pi3 5 1464 146 1610 = P + 3Pi + 3Pi 2 + Pi3 = P(1+3i +3i2 + i3) = P(1 + i) 3 N N Hence F = P(1 + i) where (1 + i) is called Compound Amount Factor.

Solving Problems It is now easy to solve problems regarding the Equivalence of Present (P) and Future (F) values over time (N) with interest (i). These steps will help: 1) Identify cash flows. 2) Identify P, F, i & N. 3) Determine the missing value. 4) Solve for missing value using equation. E x a m p l e :   What

annual interest rate must you get if you need $7000 in 4 years and have $5000

to invest now? P = $5000,

F = $7000,

N = 4 years,

i =?

7000 = 5000(1 + i)4 7000 / 5000 = (1 + i) 4 (1.4)1/4 = 1 + i 1.0878 = 1 + i 0.0878 = i = 8.78% / year Functional Notation Standardized notation has been established to avoid writing the equation each time, and to give a logical method by which to find the correct factor to use.

Read the factor as saying "Find F given P at i% for N periods". E x a m p l e :  What

is F for $1000 deposit for 5 years @ 10% / year? F = 1000( F/P,10,5) (F/P,10,5) = 1.611 --- See table at the end of any book on Engineering Economy F = 1000(1.611) = $1611

i E x a m p l e : Find  

for P = $5000, F = $7000, N = 4 years Using F = P(F/P, i, N) 7000 = 5000(F/P, i, 4) 1.4 = (F/P, i, 4) Search the Compound Interest Factor tables (Pages 180 - 208) to find the Interest rate that matches the factor. i = 9% (F/P, 9, 4) = 1.412, i = 8% (F/P, 8, 4) =1.360 Since an even i value can't be found to match (F/P,i,4) = 1.4, you must interpolate to find the solution.

Interest Factor Formulas Compound Amount: To find F, given P (F/P, i, n) Present Worth: To find P, given F (P/F, i, n) Series Compound Amount: To find F, given A (F/A, i, n) Sinking Fund: To find A, given F (A/F, i, n) Capital Recovery: To find A, given P (A/P, i, n) Series Present Worth: To find P, given A (P/A, i, n)  Arithmetic Gradient Uniform Series: To find A, given G (A/G, i, n)  Arithmetic Gradient Present Worth: To find P, given G (P/G, i, n) Geometric Gradient: To find P, given A1, g (P/G, g, i, n)