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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:
F P i Rn 1 5000 1 014 142638.
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:
PF
iRn
( ) ( . )1
2000
1 01413503
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 Ai
iR
n
( ) ( . )
.
1 1600
1 014 1
0143966
5
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:
A Fi
iRn
( )
.
( . ).
1 165000
014
1 014 11482 5815
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:
P Ai
i iR
n
n
( )
( )
( . )
. ( . ).
1 1
11500
1 014 1
014 1 01492133
15
15
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 00
5000 0 in 5 years at 14% compounded annually with annual payments? (R1 456.4).
0 5 years
A A
i = 14% compounded annually
Formula:
A P
i i
iR
n
n
( )
( )
. ( . )
( . ).
1
1 15000
014 1 014
1 014 11456 37
5
5
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
flow
20000 20000 20000 25000 25000 25000 30000 30000 30000
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
flow
1200 1200 1200 2000 2000 2000 3000 2000
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 Index
2005 223
2006 238
2007 247
2008 257
2009 279
2010 293
223500000 20052005 IC
293? 20102010 IC
6.656950223
293500000
2005
201020052010
I
ICC
2.2. Prepare a composite(weighted) index for housing construction costs in 2010 using the
following data:
Type of housing Percent Reference year 2005 (R/sq m) Index = 100
Cost (R/sq m) 2010
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 ?
5.1531003300
530025.0
3800
570005.0
4100
62007.0
IndexComposite
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.
12641900500000
150000012.012500000
65.08
1
212
x
Q
QCC
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 ?.
61.35487842.1
162
22118100028000
8.0
1
2
1
212
X
X
I
I
Q
QCC
x
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.
hoursZ 02.10881268126 073999.02log
95.0log
8
hoursZ 33.945012650126 073999.02log
95.0log
50
b.
42.587543211265 07399.007399.007399.0073999.0073999.0 T
unitperhoursC 48.1175
42.5875
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. Area(sq m) Cost (R)
1 14 500 8 000 000
2 15 000 8 250 000
3 17 000 8 750 000
4 18 500 9 720 000
5 20 400 10 740 000
6 21 000 12 500 000
7 25 000 13 070 000
8 26 750 15 340 000
9 28 000 14 755 000
10 30 000 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 14500 8000000 210250000 1.16E+11
2 15000 8250000 225000000 1.2375E+11
3 17000 8750000 289000000 1.4875E+11
4 18500 9720000 342250000 1.7982E+11
5 20400 10740000 416160000 2.19096E+11
6 21000 12500000 441000000 2.625E+11
7 25000 13070000 625000000 3.2675E+11
8 26750 15340000 715562500 4.10345E+11
ix iy 2ix ix iy
9 28000 14755000 784000000 4.1314E+11
10 30000 15250000 900000000 4.575E+11
216150 116375000 4948222500 2.65765E+12
b = 514.9751802
a = 506311.4803
b.
i ix iy Cost i ( ‐ Costi)^2 iy ix ‐
x iy ‐
y i ))((
yyxx ii ( ‐ )^2 ix
x (( ‐ )^2 iy
y
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. hoursZ 61.22026.52026.5 234465.02log
85.0log
20
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.
073999.0301029.0
022276.0
2log
95.0log
073999.0073999.0073999.0073999.0073999.0073999.0073999.0073999.08 876543211T
= 1[1+0.95+0.9219+0.9025+0.88772+0.87582+0.86589+0.857377]
= 7.2612 hours
hoursT 90765.08
2612.7
34.208366.4162500cosarg66.4162.02500
tetTxx
x
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.
Line No. Title WBS Element code
001 Coal‐fired power plant 1.
002 Site 1.1.
003 Land improvements 1.1.1
004 Roads, parking and paved area 1.1.2.
005 Railroads 1.1.3.
006 Boiler 1.2.
007 Furnace 1.2.1.
008 Pressure vessel 1.2.2.
009 Heat exchange system 1.2.3.
010 Generators 1.2.4
011 Boiler support system 1.3.
012 Coal transport system 1.3.1
013 Coal pulverizing system 1.3.2.
014 Instrumentation & control 1.3.3.
015 Ash disposal system 1.3.4.
016 Transformers & distribution 1.3.5.
017 Coal storage facility 1.4.
018 Stockpile reclaim system 1.4.1.
019 Rail car dump 1.4.2.
020 Coal handling equipment 1.4.3
021 Support facilities & equipment 1.5.
022 Hazardous waste systems 1.5.1.
023 Support equipment 1.5.2.
024 Utilities& communication system 1.5.3.
025 Security systems 1.5.4
026 Project integration 1.9.
027 Project management 1.9.1.
028 Environmental management 1.9.2.
029 Project safety 1.9.3.
030 Quality assurance 1.9.4.
031 Test, start‐up & transition management 1.9.5.
WBS:
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 266746 137850.50
15% PV ‐ 33955 ‐ 17316.50 78886 ‐ 34354.50
20% PV ‐ 75085 ‐ 75119 ‐ 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) 1.0]
3,12,/
7118.01075.2102[ 66
FPxx
=395 745
PV(2lane +4Y) 2.0]
`4,12,/
6355.0103102[ 66
FPxx
=781 300
PV(2lane +5Y) 3.0]
5,12,/
5674.01025.3102[ 66
FPxx
=1 153 215
PV(2lane +6Y) 4.0]
6,12,/
5066.0105.3102[ 66
FPxx
=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) 1.0]
3,15,/
6575.01075.2102[ 66
FPxx
=3 80812.5
PV(2lane +4Y) 2.0]
`4,15,/
5718.0103102[ 66
FPxx
=743080
PV(2lane +5Y) 3.0]
5,15,/
4972.01025.3102[ 66
FPxx
=1 084 770
PV(2lane +6Y) 4.0]
6,15,/
4323.0105.3102[ 66
FPxx
=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
production
92 000 74 000 106 000 112 000
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
6.1616710,12,/
3220.03000
10,12,/
6502.5800030000
FPAPNPV
Alternative B1: 2 000
6 000
0 1 10
22 000
2.1254510,12,/
3220.02000
10,12,/
6502.5600022000
FPAPNPV
Alternative B2: 5 000
14 000
0 1 10
70 000
8.1071210,12,/
3220.05000
10,12,/
6502.51400070000
FPAPNPV
Alternative C: 70 000
8 000
0 1 10
82 000
6.21957
10,12,/
3220.07000
10,12,/
6502.51800082000
FPAPNPV
A B1 B2 C Investment NPV FV AE
AO 0 0 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 33270 1896
A6 1 1 152000 32669.6 101456 5780 Unlimited
A7 1 1
A8 1 1
A9 1 1
A10 1 1
A11 1 1 1
A12 1 1 1
A13 1 1 1 1
A14 1 1 1
A15 1 1 1
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 First Cost Net Annual
benefits
Useful life of
project Salvage value
1 A1 R5 000 R1 500 5 years R1 500
A2 R7 000 R1 600 5 years R1 800
2 B1 R12 000 R2 000 6 years R2 000
B2 R18 000 R4 000 6 years R4 000
3 C1 R14 000 R4 000 7 years R4 000
C2 R18 000 R4 500 7 years R4 500
2.1 Use the annual equivalent cost concept to determine which projects should be selected if
the available funds are unlimited. (Answer : A1+B2+C1 )
2.2 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 )
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
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 32000
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
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.
=0.4
BV beginning
of the year
Taxable
income
Tax
‐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
of the year
Taxable
income
Tax
‐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
α = 0.2
BV It Taxable
income
Tax Loan
repay‐
ment
ATCF
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
line depr.
BV 200%
DB
α=0.25
It Taxable
income
Tax Loan
repay‐
ment
ATCF
0 ‐57747.5 100000 ‐57747.5
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
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