solutions tutorials

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


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


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 = ?


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


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?


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


1.d. Which alternative of A and B should be recommended if the MARR = 20% and the requirement

is short term?


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


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


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


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


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


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

solutions tutorials

Documents

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