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TUTORIALS.
Tutorial 2 : Money-Time Relationships 1.a. What will be the amount accumulated by the following present investment?
R5 000 in 8 years at 14% compounded annually. (R14263). F = ? 0
8 years
5 000 i = 14% compounded annually Formule:
n 8 F P1 i 50001 0.14 R14263
Tabels: F = P(F/P,i,n) = 5000(F/P,14,8) =5000(2.8526) = R14263
2.a. For an interest rate of 14% compounded annually, find how much can be loaned now if R2 000 will be repaid at the end of 3 years? (R1350). F = 2000 0
3 years
P = ? i = 14% compounded annually
Formula: P
2000 F R1350 n (1 i ) (1 0.14) 3
Tables: P = F(P/F,i,n) = 2000(P/F,14,3) = 2000(0.6750) = R 1350
3.a. What is the accumulated value of each of the following series of payments: R600 at the end of each year for 5 years at 14% compounded annually? (R3 996).
F
0 1 2 3 4 5years 600 600 600 600 600 i = 14% compounded annually Formula: F
(1 0.14) 5 1 (1 i ) n 1 600 A R 3966 i 0.14
Tables: F = A(F/A,i,n) = 600(F/A,14,5) = 600(6.6101) = R3966 4.
What equal series of payments must be put into a sinking fund to accumulate the following amount: R65 000 in 15 years at 14% compounded annually when payments are annual? (R1482.65).
F = 65000 0 1 15 years A
A
i = 14% compounded annually Formula:
i 0.14 A F 65000 (1 0.14) 15 1 R1482.58 n i ( 1 ) 1
Tables: A = F(A/F,i,n) = 65000(A/F,14,15) = 65000(0.02281) = R1482.65
5.
What is the present value of the following series of prospective receipts? R1 500 a year for 15 years at 14% compounded annually? (R9213.3).
1500 1500 0
1 15
P = ? i = 14% compounded annually Formula: n 15 1 P A (1 i ) n 1 1500 (1 0.14) 15 R 9213.3 i (1 i ) 0.14(1 0.14)
Tables: P = A(P/A,i,n) = 1500(P/A,14,15) = 1500(6.1422) = R9213.3 Tables: F = P(F/P,i,n) 23670 = 10000(F/P,i,10) (F/P,i,10) = 2.3670 i = 9% (F/P,9,10) = 2.3670 i = 8% (F/P,8,10) = 2.1589 After interpolation: i = 0.08998 = 9% 7.
What series of equal payments is necessary to repay the following present amount? R5 000 in 5 years at 14% compounded annually with annual payments? (R1 456.4).
5000
0 5 years A A
i = 14% compounded annually Formula:
i (1 i ) n 0.14(1 0.14) 5 A P 5000 R1456.37 n 5 (1 i ) 1 (1 0.14) 1
Tables: A = P(A/P,i,n) = 5000(A/P,14,5) = 5000(0.29128) = R1456.4
Multiple Factors in a Cash Flow 1. Convert the cash flow to an equivalent uniform annual amount , using an interest rate of 12% per year. (P = + 127279 AW = + 23888 ) Year 1 2 3 4 5 6 7 8 9 Cash 20000 20000 20000 25000 25000 25000 30000 30000 30000 flow
30000 25000
20000
0 1 2 3 4 5 6 7 8 9 AE = + 20000 + 5000(P/A,12,6)(P/F,12,3)(A/P,12,9)+ 5000(P/A,12,3)(P/F,12,6)(A/P,12,9) = + 20000 + 5000(4.1114)(0.7118)(0.18768) + 5000(2.4018)(0.5066)(0.18768) = + 23888
2. Convert the cash flow to an equivalent uniform annual amount , using an interest rate of 12% per year. (P = + 7559 AW = +1419 ) Year 2 3 4 5 6 7 8 9 Cash 1200 1200 1200 2000 2000 2000 3000 2000 flow
3000 2000 2000 1200 0 1 2 3 4 5 6 7 8 9
AE = [1200(P/A,12,8)(P/F,12,1) + 800(P/A,12,3)(P/F,12,4) + 1800(P/F,12,8) +800(P/F,12,9)](A/P,12,9) = [+ 1200(4.9676)(0.8929) + 800(2.4018)(0.6355) +1800(0.4039)+800(0.3606)](0.18768) = + 1418.72
Tutorial 1: Cost Estimation. 2.1. Manufacturing equipment that was purchased in 2005 for R500 000 must now be replaced at the end of 2010. What is the estimated cost of the replacement based on the following equipment cost index ?
YEAR 2005 2006 2007 2008 2009 2010
C 2005 500000 C 2010 ?
I 2005 223 I 2010 293 I 2010 I 2005
C 2010 C 2005
Index 223 238 247 257 279 293
293 500000 656950.6 223
2.2. Prepare a composite(weighted) index for housing construction costs in 2010 using the following data: Type of housing
Percent
Reference year 2005 Cost (R/sq m) 2010 (R/sq m) Index = 100 Single units 70% 4100 6200 Duplex units 5% 3800 5700 Multiple units 25% 3300 5300 How would a change in the percentage of the mix of single‐,duplex‐, and multiple units influence the value of the composite index ? What mix would you prefer to be the more advantageous for your company ?
6200 5700 5300 Composite Index 0.7 0.05 0.25 100 153.5 3800 3300 4100 2.3. If a plant that produces 500 000 Kg per year cost R2 500 000 to construct 8 years ago, what would a 1 500 000Kg per year plant cost now ? .The construction cost has increased at an average rate of 12% per year for the past eight years and that the cost capacity factor is 0.65.
Q C 2 C1 2 Q1
x
1500000 25000001 0.128 500000
0.65
12641900
2.4. The purchase price of a boiler with a capacity X was R181 000 eight years ago. Another similar boiler with capacity 1.42X is considered. If the boiler is purchased some extra features would increase the purchasing price by R28 000.The cost index was 162 for this equipment when the smaller boiler was purchased and is 221 today. The cost capacity factor is 0.8. What is the estimated price for the new boiler ?.
Q C 2 C1 2 Q1
x
I2 I1
221 1.42 X 28000 181000 162 X
0.8
354878.61
2.5. The structural engineering design section within the engineering department of a regional electrical utility corporation has developed several standard designs for a group of similar transmission line towers. The detailed design for each tower is based on one of the standard designs. A transmission line project involving 50 towers has been approved. The estimated number of engineering hours needed to accomplish the first detailed tower design is 126. Assuming a 95% learning curve, a. what is your estimate of the number of engineering hours needed to design the eight tower and to design the last tower in the project. b. what is your estimate of the cumulative average hours required for the first five designs ?
a. Z 8 1268
1268
log 0.95 log 2
Z 50 12650
log 0.95 log 2
0.073999
12650
108.02 hours
0.073999
94.33 hours
b.
T 5 126 10.073999 2 0.073999 3 0.07399 4 0.07399 5 0.07399 587.42
C 5
587.42 117.48 hours per unit 5
2.6. The cost of building a supermarket is related to the total area of the building. Data for the last 10 supermarkets built are shown in the table below: Building No. 1 2 3 4 5 6 7 8 9 10
Area(sq m) 14 500 15 000 17 000 18 500 20 400 21 000 25 000 26 750 28 000 30 000
Cost (R) 8 000 000 8 250 000 8 750 000 9 720 000 10 740 000 12 500 000 13 070 000 15 340 000 14 755 000 15 250 000
a. Develop a CER for the construction of supermarkets. Use the CER to estimate the cost of the next store that has a planned area of 23 000sq m. b. Compute the standard error and the correlation coefficient for the CER developed in part (a). 1 2 3 4 5 6 7 8
xi 14500 15000 17000 18500 20400 21000 25000 26750
yi 8000000 8250000 8750000 9720000 10740000 12500000 13070000 15340000
x i2 210250000 225000000 289000000 342250000 416160000 441000000 625000000 715562500
xi y i 1.16E+11 1.2375E+11 1.4875E+11 1.7982E+11 2.19096E+11 2.625E+11 3.2675E+11 4.10345E+11
9 10
28000 30000 216150
14755000 15250000 116375000 b = a =
784000000 900000000 4948222500
4.1314E+11 4.575E+11 2.65765E+12
514.9751802 506311.4803
b.
xi
i
yi
Cost i
(
y i ‐ Costi)^2 xi ‐ x
yi ‐ y
i
( xi x)( y i y )
( xi ‐
x )^2 (( y i ‐ y )^2
1
14500
8000000
7973400
707560000
‐7115
‐3637500
2.5881E+10
50623225
1.32E+13
2
15000
8250000
8230970
362140900
‐6615
‐3387500
2.2408E+10
43758225
1.15E+13
3
17000
8750000
9260900
2.61019E+11
‐4615
‐2887500
1.3326E+10
21298225
8.34E+12
4
18500
9720000
10033300
98156890000
‐3115
‐1917500
5973012500
9703225
3.68E+12
5
20400
10740000
11011810
73880676100
‐1215
‐897500
1090462500
1476225
8.06E+11
6
21000
12500000
11320700
1.39075E+12
‐615
862500
‐530437500
378225
7.44E+11
7
25000
13070000
13380600
96472360000
3385
1432500
4849012500
11458225
2.05E+12
8
26750
15340000
14281870
1.11964E+12
5135
3702500
1.9012E+10
26368225
1.37E+13
9
28000
14755000
14925580
29097536400
6385
3117500
1.9905E+10
40768225
9.72E+12
10
30000
15250000
15955500
4.9773E+11
8385
3612500
3.0291E+10
70308225
1.31E+13
21615
11637500
3.56781E+12
1.4221E+11
276140250
7.68E+13
R =
0.97649641
SE =
597311.7955
2.7. An electronics manufacturing company is planning to introduce a new product in the market. The best competitor sells a similar product at R420 per unit. Other data are: Direct labor cost: R15.00/hour Factory overheads : 120% of direct labor Production materials : R300.00 per unit Packaging costs : 20% of direct labor
It has been found that an 85% learning curve applies to the labor required. The time to complete the first unit has been estimated to be 5.26 hours. The company decides to use the time required to complete the 20th unit as a standard for cost estimation purposes. The profit margin is based on the total manufacturing costs. a. Based on the above information, determine the maximum profit margin that the company can have so as to remain competitive. b. If the company desires a profit margin of 15%, can the target cost be achieved ? If not, suggest two ways in which the target cost can be achieved. a.
Z 20 5.2620
log 0.85 log 2
5.2620
0.234465
2.61 hours
Direct labour = 2.61(15) = R39.086 Overheads = 1.2(39.086) = R46.90 Material = R300 Packaging = 0.2(39.086) = R7.8172 Total cost = 393.80 Maximum profit = 420 – 393.80 = R26.20 b. Target cost = 420/1+0.15 = R365.21 Actual profit % =26.20/393.80 = 0.066 = 6.6% 2.8. A personal computer company is trying to bring a new model of a PC to the market. According to the marketing department the best selling price for a similar model from a world‐class competitor is R2 500. The company wants to sell at the same price as its best competitor. The cost breakdown of the new model is: Assembly time for first unit : 1.00hour Handling time: 10% of assembling time Direct labor: R15.00 per hour Planning labour : 10% of direct labour Quality control : 50% of direct labour Factory overheads: 200% of total labour General & admin. Expense : 300% of total labour Direct material cost : R200 per computer Outside manufacture: R2 000 per computer Packing cost : 10% of total labour Facility rental : 10% of total labour
Profit : 20% of total manufacturing cost Number of units : 20 000 The company management is of the opinion that the average assembly time for the first 8 computers can be used for the feasibility study. Since the company mainly produces sub‐ assemblies purchased from other manufacturers and repackages the product, the direct material cost is estimated at R200.00 per computer. Direct labor time consists of handling time and assembling time . The company estimates the learning curve for assembling the new model is 95%. Compute the total manufacturing cost for 20 000 of these computers and determine the unit selling price. How can the company reduce its costs to meet the target cost.
log 0.95 0.022276 0.073999 log 2 0.301029
T8 110.073999 2 0.073999 3 0.073999 4 0.073999 5 0.073999 6 0.073999 7 0.073999 8 0.073999
= 1[1+0.95+0.9219+0.9025+0.88772+0.87582+0.86589+0.857377] = 7.2612 hours
T
7.2612 0.90765 hours 8
x 0.2 2500 x
x 416.66
T arg et cos t 2500 416.66 2083.34
Cost/unit Factor Estimate
Direct Estimate
Total
1
Total hours
0.90765hours/unit
18 153 hours
2
Factory labour
1.1hours/unit
R15/hour
R330 000
3
Planning labour
R1.50/hour
R27 202.5
4
Quality control
R7.50/hour
R136 147.5
5
Total labour
R493 350
6
Factory rental
10% of total labour
R49 335
7
Overheads
R30/hour
R544 590
8
General & Adm
R45/hour
R816 885
9
Production material
R200/computer
R4 000 000
10 Outside Manufacture
R2000/computer
R40 000 000
11 Sub total
R45 904 160
12 Packing cost
R1.50/hour
R27229.50
15 Total manufacturing cost
R45 931 389.50
16 Quantity/lot size
20 000
17 Manufacturing cost /unit
R2296.56
18 Profit
20% of total manufacturing cost
R9186277.90
19 Unit selling price
R2755.88
2.9. You have been requested to prepare a quick estimate of the construction cost for a coal‐fired electricity generating plant and facilities. A work breakdown structure(level3) is shown below. You have the following information available to make your decision: A coal‐fired generating plant twice the size of the one you are estimating was built in 1977. The 1977 boiler(1.2) and boiler support system (1.3) cost R110 million. The cost index for boilers was 110 in 1977; it is 492 in 2000. The cost capacity factor for similar boilers and support systems is 0.9. The 600 acre site is property you already own, but improvements (1.1.1) and roads (1.1.2) ,will cost R 2 000 per acre and railroads (1.1.3) will cost R3 000 000. Project integration (1.9) is projected to cost 3% of all other construction costs. The security systems (1.5.4) are expected to cost R1 500 per acre, based on recent(2000) construction of similar plants. All other support facilities and equipment (1.5) elements are to be built by Index Engineering. Index Engineering has built the support facilities and equipment elements for two similar generating plants. Their experience is expected to reduce labor requirements substantially; a 90% learning curve can be assumed. Index Engineering built the support facilities and equipment on their first job in 95 000 hours. For this project Index `s labor will be billed to you at R60 per hour. Index estimates materials for the construction of the support facilities and equipment elements(except (1.5.4) will cost you R15 000 000.
The coal storage facility (1.4) for the coal‐fired generating plant built in 1977 cost R5 million. Although your plant is smaller, you require the same size coal storage facility as the 1977 plant. You assume you can apply the cost index for similar boilers to the coal storage facility. What is your estimated 2000 cost for building the coal fired generating facility ?. Summarize your calculations in a cost estimating worksheet, and state the assumptions you make. WBS:
Line No. 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031
Title Coal‐fired power plant Site Land improvements Roads, parking and paved area Railroads Boiler Furnace Pressure vessel Heat exchange system Generators Boiler support system Coal transport system Coal pulverizing system Instrumentation & control Ash disposal system Transformers & distribution Coal storage facility Stockpile reclaim system Rail car dump Coal handling equipment Support facilities & equipment Hazardous waste systems Support equipment Utilities& communication system Security systems Project integration Project management Environmental management Project safety Quality assurance Test, start‐up & transition management
WBS Element code 1. 1.1. 1.1.1 1.1.2. 1.1.3. 1.2. 1.2.1. 1.2.2. 1.2.3. 1.2.4 1.3. 1.3.1 1.3.2. 1.3.3. 1.3.4. 1.3.5. 1.4. 1.4.1. 1.4.2. 1.4.3 1.5. 1.5.1. 1.5.2. 1.5.3. 1.5.4 1.9. 1.9.1. 1.9.2. 1.9.3. 1.9.4. 1.9.5.
Tutorial3: Risk Analysis. 1. A machine costing R10 000 will produce net cash savings of R4 000 per year. The useful life is 5 years. A major overhaul is planned after 3 years of operation. These repairs will cost R5 000. If the company`s MARR is 10%, would this project be economically viable ? Analyze the sensitivity of the economic viability of the project if the once of overhaul cost changes by ±20%. 4 000 0 1 3 5 5 000 10 000 MARR = 10% PV(R5000) = ‐10 000 +4 000(P/A,10,5) ‐ 5000(P/F,10,3)
= ‐10 000 +4 000(3.7908) ‐ 5000(0.7513)
= 1406.7 PV(R3000) = ‐10 000 +4 000(P/A,10,5) ‐ 3000(P/F,10,3)
= ‐10 000 +4 000(3.7908) ‐ 3000(0.7513)
= 2909.3 PV(R7000) = ‐10 000 +4 000(P/A,10,5) ‐ 7000(P/F,10,3)
= ‐10 000 +4 000(3.7908) ‐ 7000(0.7513)
= ‐ 95.9
PV(R6000) = ‐10 000 +4 000(P/A,10,5) ‐ 6000(P/F,10,3)
= ‐10 000 +4 000(3.7908) ‐ 6000(0.7513)
= 655.4 After interpolation between 6000 and 7000 x = 6872.35 (Maximum value for upgrade) 2. It is desired to determine the optimal height for a proposed building that is expected to last 40 years. The net market value after being demolished is zero Rand. Analyze the sensitivity of the decision due to changes in estimates of the MARR value between 10%, 15% and 20%. Use the present value calculation . Ignore tax payable.
Number of floors
2
3
4
5
Capital investment 200 000 250 000 320 000 400 000 Annual revenue
40 000
60 000
85 000
100 000
Annual cost
15 000
25 000
25 000
45 000
PV=‐Po+A(P/A,i,40) Number of floors
2
3
4
5
MARR
10%
PV 44477.50 92268.5
15%
PV ‐ 33955
‐ 17316.50 78886
20%
PV ‐ 75085
‐ 75119
266746 137850.50 ‐ 34354.50
‐ 20204 ‐ 125187
PV(10%)(2floors)=‐200000+25000(P/A,10,40) =‐200000+25000(9.7791)= 44477.50 PV(10%)(3floors)=‐250000+35000(P/A,10,40) =‐250000+35000(9.7791)= 92268.5 PV(10%)(4floors)=‐320000+60000(P/A,10,40) =‐320000+60000(9.7791)= 266746 PV(10%)(5floors)=‐400000+55000(P/A,10,40) =‐400000+55000(9.7791)= 137850.50 PV(15%)(2floors)=‐200000+25000(P/A,15,40) =‐200000+25000(6.6418)= ‐ 33955 PV(15%)(3floors)=‐250000+35000(P/A,15,40) =‐250000+35000(6.6481)= ‐ 17316.50 PV(15%)(4floors)=‐320000+60000(P/A,15,40) =‐320000+60000(6.6481)= 78886 PV(15%)(5floors)=‐400000+55000(P/A,15,40) =‐400000+55000(6.6481)= ‐ 34354.50 PV(20%)(2floors)=‐200000+25000(P/A,20,40) =‐200000+25000(4.9966)= ‐ 75085 PV(20%)(3floors)=‐250000+35000(P/A,20,40) =‐250000+35000(4.9966)= ‐ 75119 PV(20%)(4floors)=‐320000+60000(P/A,20,40) =‐320000+60000(4.9966)= ‐ 20204 PV(20%)(5floors)=‐400000+55000(P/A,20,40) =‐400000+55000(4.9966)= ‐ 125187
3. A bridge is to be constructed as part of a new road. Analysis has shown that traffic density on the new road will justify a two lane bridge at the present time. Because of uncertainty regarding future use of the road, the time at which an extra two lanes will be required is currently being studied. The estimated probabilities of having to widen the bridge to four lanes at various times in the future are: Widen bridge in Probability 3 years
0.1
4 years
0.2
5 years
0.3
6 years
0.4
The present estimated cost of the two lane bridge is R2 000 000. If constructed now , the four lane bridge will cost R3 500 000. The future cost of widening the bridge will be an extra R2 000 000 plus R250 000 for every year that widening is delayed. a. If the relevant MARR = 12% what should the decision be? 4 Lane bridge if constructed now = R3 500 000 2 Lane bridge if constructed now = R2 000 000 2 lane bridge
PV(2lane +3Y)
0.7118 [2 x10 6 2.75 x10 6 ]0.1 P / F ,12,3
=395 745
PV(2lane +4Y)
0.6355 [2 x10 6 3x10 6 ]0.2 P / F ,12,4`
=781 300
PV(2lane +5Y)
0.5674 [2 x10 6 3.25 x10 6 ]0.3 P / F ,12,5
=1 153 215
PV(2lane +6Y)
0.5066 [2 x10 6 3.5 x10 6 ]0.4 P / F ,12,6
=1 509 240
EPV =
3 839 500
b. Determine how sensitive the choice of a four lane bridge built now versus a four lane bridge constructed in 2 stages to the relevant MARR value. 2 lane bridge
PV(2lane +3Y)
=3 80812.5 0.6575 [2 x10 6 2.75 x10 6 ]0.1 P / F ,15,3
PV(2lane +4Y)
0.5718 [2 x10 6 3x10 6 ]0.2 P / F ,15,4`
=743080
PV(2lane +5Y)
0.4972 [2 x10 6 3.25 x10 6 ]0.3 P / F ,15,5
=1 084 770
PV(2lane +6Y)
0.4323 [2 x10 6 3.5 x10 6 ]0.4 P / F ,15,6
=1 405 220
EPV =
3 613 882.5
c. Will a MARR=15% change the decision ? No d. At what MARR would constructing the two lane bridge now be preferred? 2 lane bridge
MARR = 16% MARR = 17%
PV(2lane +3Y) 376192.5
371 710
PV(2lane +4Y) 731380
720 220
PV(2lane +5Y) 1 064 197.5
1 044 697.5
PV(2lane +6Y) 1 374 560
1 345 720
EPV =
3 546 330 3 482 347.5
After interpolation between 16% and 17% MARR = 16.72% 4. A ski resort is considering buying a new ski lift for R900 000. Expenses for operating and maintenance are estimated atR1 500 per day when operating. It is estimated that there is a 60% probability of 80 days of skiing weather per year, 30% probability of 100 days per year and probability of 10% of 120 days per year. The owner of the resort estimates that during the first 80 days of operation an average of 500 people will use the lift per day at a fee of R10 each. If 20
additional days are available, the lift will be used by 400 people per day for the extra 20 days. If another 20 days can be added 300 people will be using the lift during the additional 20 days. The owner wants to recover his investment over a period of 5 years if his MARR = 25%. Is this project economically viable ?.
80 days/year
80+20days/year
80+20+20days/year
Po
‐900 000
‐900 000
‐900 000
Income /year
80(500)(10)
80(500)(10)+20(400)(10) 80(500)(10)+20(400)(10)+20(300)(10)
O&M cost/year
80(1500)
(80+20)1500
(80+20+20)1500
Probability
60%
30%
10%
MARR
25%
25%
25%
PV(80days/5years)=‐900000+[80(500)(10)‐120000](P/A,25,5)
=‐900000+[80(500)(10)‐120000](2.6893)
= ‐146996 PV(80+20days/5years)=‐900000+[80(500)(10)+20(400)(10)‐150000](P/A,25,5)
=‐900000+[80(500)(10)+20(400)(10)‐150000](2.6893)
= ‐12531 PV(80+20days/5years)=‐900000+[80(500)(10)+20(400)(10)+20(300)(10)‐180000](P/A,25,5) =‐900000+[80(500)(10)+20(400)(10)+20(300)(10)‐180000](2.6893) = 68148 EPV = (‐146996)(0.6) +(‐12531)(0.3) +(68148)(0.1) = ‐ 85142.1
Tutorial 6: Capital investment decisions.
1.
Engineering projects A, B1, B2 and C are being considered with cash flows estimated over 10 years as shown in the table below. Projects B1 and B2 are mutually exclusive. Project
C depends upon B2, and Project A depends upon B1. The budget limit is R100 000. The MARR is 12%.
A B1 B2 C First cost 25 000 20 000 50 000 70 000 Installation cost 3 000 1 000 10 000 10 000 System test 2 000 1 000 10 000 2 000 Annual sales 100000 80 000 120000 130 000 Annual cost of 92 000 74 000 106 000 112 000 production Salvage value 3 000 2 000 5 000 7 000 1.1 Identify ALL possible alternatives. 1.2 Develop the cash flows for ALL FEASIBLE alternatives. 1.3 Which investment alternative should be selected? Use the Present Value Method. 1.4 Verify your decision by using the annual equivalent cost method. (Answer: Limited budget A+B1 Unlimited budget B2 + C)
1.
Total Investment Approach.
Alternative A: 3 000 8 000 0 1 10 30 000 NPV 30000 8000 5.6502 3000 0.3220 16167 .6 P / A,12,10
P / F ,12,10
Alternative B1: 2 000 6 000
0 1 10 22 000 NPV
5 . 6502 0 . 3220 22000 6000 2000 12545 . 2 P / A ,12 ,10 P / F ,12 ,10
Alternative B2: 5 000 14 000 0 1 10 70 000 NPV
5 . 6502 0 . 3220 70000 14000 5000 10712 . 8 P / A ,12 ,10 P / F ,12 ,10
Alternative C: 70 000 8 000 0 1 10 82 000 NPV
0 . 3220 5 . 6502 82000 18000 21957 . 6 7000 P / F ,12 ,10 P / A ,12 ,10
A B1 B2 C Investment NPV
FV
AE
AO 0 0
0
0 0
0
A1
1
30000
16167.6 50214
2861
A2
1
22000
12545.2 38962
2220
A3
1 82000
21957.6 68197
3886
A4
1 1
52000
28712.8 89174
5081 Limited
A5
1
70000
10712
1896
A6
1
1 152000
32669.6 101456 5780 Unlimited
A7
1
1
A8
1
A9
1
1
1
A11 1 1
1
A12
1
1
1
A13 1 1
1
1
A14 1 1
1
A15 1
1
1
A10
1
1
0
33270
0
2.
A small company has surplus funds that it wishes to invest in new revenue producing projects. Three independent sets of mutually exclusive projects have been developed. At most one project from each one of the three different sets can be selected.
If the MARR = 12% nominal compounded annually:
Set
Project
1 2 3
2.1 2.2
A1 A2 B1 B2 C1 C2
R5 000 R7 000 R12 000 R18 000 R14 000 R18 000
NPW
INVESTMENT
A1 +1258.3
5000
A2 ‐ 211
7000
Not considered
B1 ‐2764
12000
Not considered
B2 +472
14000
C1 +6064.4
14000
C2 +4572.45 18000
Useful life of project
R1 500 R1 600 R2 000 R4 000 R4 000 R4 500
Salvage value
5 years 5 years 6 years 6 years 7 years 7 years
R1 500 R1 800 R2 000 R4 000 R4 000 R4 500
Use the annual equivalent cost concept to determine which projects should be selected if the available funds are unlimited. (Answer : A1+B2+C1 ) Use the annual equivalent cost concept to determine which projects should be selected if the available funds are limited to R20 000? (Answer: Limited budget A1+C1 )
Net Annual benefits
First Cost
A1 B2 C1 C2 NPW
INVESTMENT
A1
1
+1258.3
5000
A2
1
+472
14000
A3
1
+4572.45 18000
A4
1
1
+1730.3
19000
A5
1
+6064.4
14000
A6
1
1
A7
1
1
+7322.7
19000 limited
A8
1
1
+5044.45 36000
A9
1
1
+5830.75 23000
A10
1
1
+6536.4
A11 1
1
1
+6302.75 41000
A12
1
1
1
A13 1
1
1
1
A14 1
1
1
+7794.7
37000 unlimit.
A15 1
1
1
32000
Tutorial 4: Decision between alternatives. 1.a. Which alternative of A and B should be recommended if the MARR = 20% and the requirement is short term? A: 0 1 2 3 4 5 6 1 000 1 500 10 000 2 000 2 500 3 000 3 500
B 0 1 2 3 4 2 500 2 500 2 500 2 500 9 000
PV(A) = ‐ 10000 – 1000(P/A,20,4) – 500(P/G,20,4) = ‐ 10000 – 1000(2.5887) – 500(3.2986) = ‐ 14238 PV(B) = ‐ 9000 – 2500(P/A,20,4) = ‐ 9000 – 2500(2.5887) = ‐ 15471.75 Alternative A most economical
1.b. Which alternative of A and B should be recommended if the MARR = 20% and the requirement is short term? The salvage value decreases by R1 000 per year. Salvage value 5 500 A: 0 1 2 3 4 5 6 2 000 2 000 2 000 2 000 2 000 2 000 10 000 Salvage value 4 500 B: 0 1 2 3 4 1 500 1 500 1 500 1 500 9 000 PV(A) = ‐10000 – 2000(P/A,20,4)+ 7500(P/F,20,4)
= ‐10000 – 2000(2.5887) + 7500(0.4823) = ‐11560.15
PV(B) = ‐9000 – 1500(P/A,20,4) + 4500(P/F,20,4) = ‐9000 – 1500(2.5887) + 4500(0.4823) = ‐ 10712.7 Alternative B most economical one 1.c. Which alternative of A and B should be recommended if the MARR = 20% and the requirement is short term? The salvage value decreases by R1 000 per year. Salvage value 5 500 A: 2 500 2 500 2 500 2 500 2 500 2 500 0 1 2 3 4 5 6 10 200 Salvage value 5 000 2 500 2 500 2 500 2 500 B : 0 1 2 3 4 8 000
PV(A) = ‐10200 + 2500(P/A,20,6) + 5500(P/F,20,6) = ‐10200 + 2500(3.3255) + 5500(0.3349) = ‐10000 +8313.75 + 1841.95 = ‐ 44.3 PV(A) = ‐10200 + 2500(P/A,20,4) + 7500(P/F,20,4) = ‐10200 + 2500(2.5887) + 7500(0.4823) = ‐10200 +6471.75 + 3617.25 = ‐ 111
PV(B) = ‐8000 + 2500(P/A,20,4) + 5000(P/F,20,4) = ‐8000 + 2500(2.5887) + 5000(0.4823) = 883.25 Alternative B most economical one 1.d. Which alternative of A and B should be recommended if the MARR = 20% and the requirement is short term? The salvage value decreases by R1 000 per year. Salvage value 5 500 A: 3 000 3 000 3 000 3 000 3 000 3 000 0 1 2 3 4 5 6 10 000 Salvage value 5 000 2 500 2 500 2 500 2 500 B: 0 1 2 3 4 8 000 PV(A) = ‐10000 +3000(P/A,20,6) + 5500(P/F,20,6)
= ‐10000 +3000(3.3255) + 5500(0.3349) = 1818.45 PV(B) = ‐8000 + 2500(P/A,20,4) + 5000(P/F,20,4) = ‐8000 + 2500(2.5887) + 5000(0.4823) = 883 Alternative A most economical one
AE(B) = ‐ 10712.7(A/P,20,4) = ‐ 10712.7(0.38629) = ‐ 4138.20 Alternative B most economical one
Tutorial5: Depreciation and Income Taxes Loans: 1.An individual is borrowing R100 000 at 10% compounded annually. The loan is to be repaid in equal annual payments over 10 years. However , just after the fifth payment is made , the bank increases the interest rate to 15% per year compounded annually. Calculate : a. The annual payments if the interest rate is 10%. (16275) b. What amount is being paid as interest with the fifth payment.(7088.25) c. What amount is being paid as capital with the fifth payment (9187.23) d. What is the balance of the principal amount after the fifth payment. (61689.49) e. What is the balance of the principal amount after the 6th payment.(52539.91) f. What amount is being paid as interest with the 6th payment.(9253.68) g. What amount is being paid as capital with the 6th payment (9150)
100000 i = 10% i = 15% 0 1 5 6 10y
16275 18403.2 a. Annual payments i = 10%: A = 100000(A/P,10,10) = 100000(0.16275) = 16275 b. Interest component of 5th payment: I(t=5) = A(P/A,i,n‐t+1)i = 16275(P/A,10,6)0.10 = 16275(4.3553)0.10 = 7088.25 Alternative calculation: U(4) = 100000(F/P,10,4) – 16275(F/A,10,4) = 100000(1.4641) – 16275(4.6410) = 70877.7 I(t=5) = U(4)i = 70877.72(0.10) = 7087.7
c. Capital component of 5th payment: B(t=5) = A(P/F,i,n‐t+1) = 16275(P/F,10,6) = 16275(0.5645) = 9187.23 Alternative calculation: B(t=5) = A – I(t=5) = 16275 – 7088.25 = 9186.75 d. Balance of principal amount at t=5: U(5) = 100000(F/P,10,5) – 16275(F/A,10,5) = 100000(1.6105) – 16275(6.1051) = 61689.49 Alternative calculation: U(5) = 16275(P/A,10,5) = 16275(3.7908) = 61695.27 e. Annual payment i = 15%: A(y6‐y10) = 61689.49(A/P,15,5) = 61689.49(0.29832) = 18403.2 U(t=6) = 61689.49(F/P,15,1) – 18403.2(F/A,15,1) = 61689.49(1.15) – 18403.2(1.0000) = 52539.91 Alternative calculation: U(6) = 18403.2(P/A,15,4) = 18403.2(2.8550) = 52541.13 f. Interest component of 6th payment: I(t=6) = A(P/A,i,n‐t+1)i = 18403.2(P/A,15,5)0.15 = 18403.2(3.3522)0.15 = 9253.68 Alternative calculation: I(t=6) = U(5)0.15 = 61689.49(0.15) = 9253.42 g. Capital component of 6th payment: B(t=6) = A(P/F,i,n‐t+1) = 18403.2(P/F,15,5) = 18403.2(0.4972) = 9150
After tax cash flow calculations.
1. A lathe can be purchased new for R18 000. It will have an 8‐year useful life and a zero salvage value after the 8 years. Reductions in operating costs (savings) from the machine will be R8 000 for the first 4 years and R3 000 for the last 4 years. Depreciation will be by the 200% Declining Balance Method and using a tax life of 5 years. A used lathe can be bought for R8 000 and will have a salvage of R3 000 after 5 years. It will save a constant R3 000 per year over the 8‐year period , a market value of R3000 after 8 years , and will be depreciated by the straight‐line method over a period of 5 years. The effective income tax rate is 40%. The before tax MARR = 15%. Should the company buy the new lathe? Note: .(Answer: NPV new machine = 7543 NPV used machine = 5024)
New Machine: Y
BTCF
Depr.
BV beginning
Taxable
Tax
=0.4
of the year
income
‐40%
ATCF
0
‐18000
‐18000
1
+8000
7200
18000
+800
‐320
+7680
2
+8000
4320
10800
+3680
‐1472
+6528
3
+8000
2592
6480
+5408
‐2163
+5837
4
+8000
1555
3888
+6445
‐2578
+5422
5
+3000
933
2333
+2067
‐826.8
+2173
6
+3000
1400
+3000
‐1200
+1800
7
+3000
+3000
‐1200
+1800
8
+3000
+3000
‐1200
+1800
MV
0
1400
‐1400
+560
+ 560
After Tax MARR = (0.15)(1 – 0.40) = 0.09 = 9% NPV = – 18000 + 7680(P/F,9,1) + 6528(P/F,9,2) + 5837(P/F,9,3) + 5422(P/F,9,4) + 2173(P/F,9,5) + 1800(P/A,9,3)(P/F,9,5) + 560(P/F,9,8) = – 18000 + 7680(0.9174) + 6528(0.8417) + 5837(0.7722) + 5422(0.7084) + 2173(0.6499) + 1800(2.5313)(0.6499) + 560(0.5019) = 7543 Used Machine:
Y
BTCF
Depr.
BV beginning
Taxable
Tax
of the year
income
‐40%
ATCF
0
‐8000
‐8000
1
+3000
‐1000
8000
+2000
‐800
+2200
2
+3000
‐1000
7000
+2000
‐800
+2200
3
+3000
‐1000
6000
+2000
‐800
+2200
4
+3000
‐1000
5000
+2000
‐800
+2200
5
+3000
‐1000
4000
+2000
‐800
+2200
6
+3000
3000
+3000
‐1200
+1800
7
+3000
+3000
‐1200
+1800
8
+3000
+3000
‐1200
+1800
MV
+3000
3000
0
+3000
After Tax MARR = (0.15)(1 – 0.40) = 0.09 = 9% NPV = – 8000 + 2200(P/A,9,5) + 1800(P/A,9,3)(P/F,9,5) + 3000(P/F,9,8) = 5024 5. First cost: R75 000
Savings per year: R10 000 Project life : 6 years Tax life: 10 years Tax rate: 40% Residual value: R25 000 Loan : R25 000 Interest on loan: 12% Payback period : 2 years Market value end of project life: R30 000 Depreciation method: 200% declining balance method . Before tax MARR = 33.3333% By calculating the after tax NPV, determine if this project is economically feasible .
Y
BTCF
DB
BV
It
α = 0.2
Taxable
Tax
income
Loan
ATCF
repay‐ ment
0
‐50000
‐50000
1
10000
15000
75000
3000
‐8000
+3200
14792.50
‐1592.5
2
10000
12000
60000
1585
‐3585
+1434
14792.50
‐3358.5
3
10000
9600
48000
+400
‐160
+9840
4
10000
7680
38400
+2320
‐928
+9072
5
10000
5720
30720
+4280
‐1712
+8288
6
10000
+15000
‐6000
+34000
Mv
30000
25000
NPV = ‐ 28872.63 Not economically feasible A = 25000(A/P,12,2) = 25000(0.59170) = 14792.50 It(y1) = 14792.50(P/A,12,2)0.12 =14792.50(1.6901)0.12 = 3000 It(y2) = 14792.5(P/A,12,1)0.12 =14792.5(0.8929)0.12 = 1585
3. Index Engineering specialize in the manufacturing of low volume components for the health sector. Index Engineering received an order for components that will realize an annual profit of R20 000 per year for the next ten years. To execute this order Index Engineering will have to procure equipment worth R100 000. A loan worth R42252.50 to be repaid in two equal annual payments, the interest payable is 12% per year, has been negotiated with the local bank. The loan of R42252.50 forms part of the total amount of R100 000 to be invested. The following information is also available: Tax life: 8 years Depreciation method: The straight line or the 200% declining balance method can be applied which is the most advantageous for the company in each year. Residual value at the end of the tax life: R15 000 Effective tax rate: 40% Before tax MARR = 33.3333% Original project life : 10 years Market value: R10 000 from year 5 until year 10 The client of Index Engineering is now negotiating to terminate the contract at the end of year 5. Will the contract be economically viable for Index Engineering if they allow their client to change the contract accordingly ?
Y
BTCF
Straight
BV
line depr.
200%
It
DB
Taxable
Tax
income
Loan
ATCF
repay‐
α=0.25
ment
0
‐57747.5
100000
1
20000
10625
75000
25000
5070.3
‐10070.3
+4028.12
25000 ‐971.88
2
20000
8571.42
56250
18750
2678.7
‐1428.7
+571.48
25000 ‐4428.5
3
20000
6875
42188
14062
5938
‐2375.2
17624.8
4
20000
5437.6
31641
10547
9453
‐3781
16219
5
20000
4160.25
23731
7910
12090
‐4836
15164
Mv
10000
23731
‐13731
+5492.4
15492.4
A = 42252.50(A/P,12,2) = 42252.50(0.5917) = 25000
‐57747.5
I(t1) = 25000(P/A,12,2)0.12 = 5070.3 I(t2) = 25000(P/A,12,1)0.12 = 2678.7 After tax MARR = 33.3333(1‐0.4) = 20% NPV = ‐ 57747.5 – 971.88(P/F,20,1) ‐ 4428.52(P/F,20,2) ‐ 17624.8(P/F,20,3) + 16219(P/F,20,4) + 15164(P/F,20,5)+15492.4(P/F,20,5) = ‐ 31291.46 Not economically feasible
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