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Problems’ Solutions Project Evaluation Chapter 4 MARKET AND DEMAND ANALYSIS 1. We have to estimate the parameters a and b in the linear relationship Y t = a + bT Using the least squares method. According to the least squares method the parameters are: T Y n T Y b = T 2 n T 2 a = Y bT The parameters are calculated below: Calculation in the Least Squares Method T Y TY T 2 1 2,000 2,000 1 2 2,200 4,400 4 3 2,100 6,300 9 4 2,300 9,200 16 5 2,500 12,500 25 6 3,200 19,200 36 7 3,600 25,200 49 8 4,000 32,000 64 9 3,900 35,100 81 10 4,000 40,000 100 11 4,200 46,200 121 12 4,300 51,600 144 13 4,900 63,700 169 14 5,300 74,200 196 T = 105 Y = 48,500 TY = 421,600 T 2 = 1,015 T = 7.5 Y = 3,464 T Y n T Y 421,600 – 14 x 7.5 x 3,464 b = = T 2 n T 2 1,015 – 14 x 7.5 x 7.5 57,880 = = 254 227.5 a = Y bT = 3,464 – 254 (7.5) = 1,559 Thus linear regression is Y = 1,559 + 254 T 2. In general, in exponential smoothing the forecast for t + 1 is F t + 1 = F t + α e t Where F t + 1 = forecast for year ) α = smoothing parameter e t = error in the forecast for year t = S t = F t F 1 is given to be 2100 and α is given to be 0.3 1 By M. Iftikhar Mubbashir The forecasts for periods 2 to 14 are calculated below:

Project evaluation and feasibility analysis

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This is the soultuon manual of Parsana Chandara book " Project evaluation and feasibility analysis", this manual is created by Mr. Iftikhar Mubbashir, Instructor at Paf-Kiet city campus.

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Page 1: Project evaluation and feasibility analysis

Problems’ Solutions Project Evaluation

Chapter 4 MARKET AND DEMAND ANALYSIS 1. We have to estimate the parameters a and b in the linear relationship Yt = a + bT Using the least squares method. According to the least squares method the parameters are: ∑ T Y – n T Y b = ∑ T 2 – n T 2 a = Y – bT The parameters are calculated below: Calculation in the Least Squares Method

T Y TY T 2

1 2,000 2,000 1 2 2,200 4,400 4 3 2,100 6,300 9 4 2,300 9,200 16 5 2,500 12,500 25 6 3,200 19,200 36 7 3,600 25,200 49 8 4,000 32,000 64 9 3,900 35,100 81 10 4,000 40,000 100 11 4,200 46,200 121 12 4,300 51,600 144 13 4,900 63,700 169 14 5,300 74,200 196

∑ T = 105 ∑ Y = 48,500 ∑ TY = 421,600 ∑ T 2 = 1,015 T = 7.5 Y = 3,464

∑ T Y – n T Y 421,600 – 14 x 7.5 x 3,464 b = = ∑ T 2 – n T 2 1,015 – 14 x 7.5 x 7.5 57,880 = = 254 227.5 a = Y – bT = 3,464 – 254 (7.5) = 1,559 Thus linear regression is Y = 1,559 + 254 T 2. In general, in exponential smoothing the forecast for t + 1 is Ft + 1 = Ft + α et Where Ft + 1 = forecast for year ) α = smoothing parameter et = error in the forecast for year t = St = Ft F1 is given to be 2100 and α is given to be 0.3

1 By M. Iftikhar Mubbashir

The forecasts for periods 2 to 14 are calculated below:

Page 2: Project evaluation and feasibility analysis

Problems’ Solutions Project Evaluation

Period t Data (St) Forecast (Ft)

Error (et St =Ft)

Forecast for t + 1 (Ft + 1 = Ft + α et)

1 2,000 2100.0 -100 F2 = 2100 + 0.3 (-100) = 2070 2 2,200 2070 130 F3 = 2070 + 0.3(130) = 2109 3 2,100 2109.0 -9 F4 = 2109 + 0.3 (-9) = 2111.7 4 2,300 2111.7 188.3 F5 = 2111.7 + 0.3(188.3) = 2168.19 5 2,500 2168.19 331.81 F6 = 2168.19 + 0.3(331.81) = 2267.7 6 3,200 2267.7 932.3 F7 = 2267.7 + 0.3(9332.3) = 2547.4 7 3,600 2547.4 1052.6 F8 = 2547.4 + 0.3(1052.6) = 2863.2 8 4,000 2863.2 1136.8 F9 = 2863.2 + 0.3(1136.8) = 3204.24 9 3,900 3204.24 695.76 F10 = 33204.24 + 0.3(695.76) = 3413.0 10 4,000 3413 587.0 F11 = 3413.0 + 0.3(587) = 3589.1 11 4,200 3589.1 610.9 F12 = 3589.1 + 0.3(610.9) = 3773.4 12 4,300 3772.4 527.6 F13 = 3772.4 + 0.3(527.6) = 3930.7 13 4,900 3930.7 969.3 F14 = 3930.7 + 0.3(969.3) = 4221.5

3. According to the moving average method St + S t – 1 +…+ S t – n +1

Ft + 1 = n where Ft + 1 = forecast for the next period St = sales for the current period n = period over which averaging is done Given n = 3, the forecasts for the period 4 to 14 are given below:

Period t Data (St) Forecast (Ft)

Forecast for t + 1 Ft + 1 = (St+ S t – 1 + S t – 2)/ 3

1 2,000 2 2,200 3 2,100 F4 = (2000 + 2200 + 2100)/3 = 2100 4 2,300 2100 F5 =(2200 + 2100 + 2300)/3= 2200 5 2,500 2200 F6 = (2100 + 2300 + 2500)/3 = 2300 6 3,200 2300 F7 = (2300 + 2500 + 3200)/3= 2667 7 3,600 2667 F8 = (2500 + 3200 + 3600)/3 = 3100 8 4,000 3100 F9 = (3200 + 3600 + 4000)/3 = 3600 9 3,900 3600 F10 = (3600 + 4000 + 3900)/3 = 3833 10 4,000 3833 F11 = (4000 + 3900 + 4000)/3 =3967 11 4,200 3967 F12 =(3900 + 4000 + 4200)/3 = 4033 12 4,300 4033 F13 = (4000 + 4200 + 4300)/3 = 4167 13 4,900 4167 F14 = (4200 + 4300 + 4900) = 4467 14 5,300 4467

4.

Q1 = 60 Q2 = 70 I1 = 1000 I2 = 1200

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Problems’ Solutions Project Evaluation

Q1 – Q2 I1 + I2 Income Elasticity of Demand E1 = x

I2 - I1 Q2 – Q1

E1 = Income Elasticity of Demand Q1 = Quantity demanded in the base year Q2 = Quantity demanded in the following year I1 = Income level in base year I2 = Income level in the following year

70 – 60 1000 + 1200 E1 = x 1200 – 1000 70 + 60 22000 E1 = = 0.846 26000 5. P1 = Rs.40 P2 = Rs.50 Q1 = 1,00,000 Q2 = 95,000 Q2 – Q1 P1 + P2

Price Elasticity of Demand = Ep = x P2 –P1 Q2 + Q1

P1 , Q1 = Price per unit and quantity demanded in the base year P2, Q2 = Price per unit and quantity demanded in the following year Ep = Price Elasticity of Demand

95000 - 100000 40 + 50 Ep = x 50 - 40 95000 + 100000 - 45 Ep = = - 0.0231 1950

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Problems’ Solutions Project Evaluation

Chapter 6 FINANCIAL ESTIMATES AND PROJECTIONS 1. Projected Cash Flow Statement (Rs. in million)

Sources of Funds Profit before interest and tax 4.5 Depreciation provision for the year 1.5 Secured term loan 1.0 Total (A) 7.0 Disposition of Funds Capital expenditure 1.50 Increase in working capital∗ 0.35 Repayment of term loan 0.50 Interest 1.20 Tax 1.80 Dividends 1.00 Total (B) 6.35 Opening cash balance 1.00 Net surplus (deficit) (A – B) 0.65 Closing cash balance 1.65 Projected Balance Sheet (Rs. in million) Liabilities Assets Share capital 5.00 Fixed assets 11.00 Reserves & surplus 4.50 Investments .50 Secured loans 4.50 Current assets 12.85 Unsecured loans 3.00 * Cash 1.65 Current liabilities 6.30 * Receivables 4.20

& provisions 1.05 * Inventories 7.00 24.35 24.35 2. Projected Income Statement for the 1st Operating Year Rs. Sales 4,500 Cost of sales 3,000 Depreciation 319 Interest 1,044 Write off of Preliminary expenses 15 Net profit 122 Projected Cash Flow Statements

Construction period 1st Operating yearSources Share capital 1800 - Term loan 3000 600 Short-term bank borrowing 1800

∗ Working capital here is defined as : (Current assets other than cash) – (Current liabilities other than bank borrowings) In this case inventories increase by 0.5 million, receivables increase by 0.2 million and current liabilities and provisions increase by 0.35 million. So working capital increases by 0.35 million

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Problems’ Solutions Project Evaluation

Profit before interest and tax 1166 Depreciation 319 Write off preliminary expenses 15 4800 3900 Uses Capital expenditure 3900 - Current assets (other than cash) - 2400 Interest - 1044 Preliminary expenses 150 - Pre-operative expenses 600 - 4650 3444 Opening cash balance 0 150 Net surplus / deficit 150 456 Closing balance 150 606

Projected Balance Sheet Liabilities 31/3/n+1 31/3/n+2 Assets 31/3/n+1 31/3/n+2 Share capital 1800 1800 Fixed assets (net) 4500 4181 Reserves & surplus - 122 Secured loans : Current assets - Term loan 3000 3600 - Cash 150 606 - Short-term bank borrowing

1800 Other current assets 2400

Unsecured loans - - Miscellaneous expenditures & losses

Current liabilities and provisions

- Preliminary expenses

150 135

4800 7322 4800 7322 Notes : i. Allocation of Pre-operative Expenses: Rs.

Type Costs before allocation

Allocation Costs after allocation

Land 120 19 139 Building 630 97 727 Plant & machinery 2700 415 3115 Miscellaneous fixed assets 450 69 519 3900 600 4500

ii. Depreciation Schedule:

Land Building Plant & machinery M. Fixed assets Total (Rs.) Opening balance 139 727 3115 519 4500 Depreciation - 25 252 42 319 Closing balance 139 702 2863 477 4181

iii. Interest Schedule: Interest on term loan of Rs.3600 @20% = Rs.720 Interest on short term bank borrowings of Rs,1800 @ 18% = Rs.324 = Rs.1044

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Problems’ Solutions Project Evaluation

Chapter 7 THE TIME VALUE OF MONEY 1. Value five years hence of a deposit of Rs.1,000 at various interest rates is as follows: r = 8% FV5 = 1000 x FVIF (8%, 5 years) = 1000 x 1.469 = Rs.1469 r = 10% FV5 = 1000 x FVIF (10%, 5 years) = 1000 x 1.611 = Rs.1611 r = 12% FV5 = 1000 x FVIF (12%, 5 years) = 1000 x 1.762 = Rs.1762 r = 15% FV5 = 1000 x FVIF (15%, 5 years) = 1000 x 2.011 = Rs.2011 2. Rs.160,000 / Rs. 5,000 = 32 = 25

According to the Rule of 72 at 12 percent interest rate doubling takes place approximately in 72 / 12 = 6 years So Rs.5000 will grow to Rs.160,000 in approximately 5 x 6 years = 30 years 3. In 12 years Rs.1000 grows to Rs.8000 or 8 times. This is 23 times the initial deposit. Hence doubling takes place in 12 / 3 = 4 years. According to the Rule of 69, the doubling period is: 0.35 + 69 / Interest rate Equating this to 4 and solving for interest rate, we get Interest rate = 18.9%. 4. Saving Rs.2000 a year for 5 years and Rs.3000 a year for 10 years thereafter is equivalent to

saving Rs.2000 a year for 15 years and Rs.1000 a year for the years 6 through 15. Hence the savings will cumulate to: 2000 x FVIFA (10%, 15 years) + 1000 x FVIFA (10%, 10 years) = 2000 x 31.772 + 1000 x 15.937 = Rs.79481. 5. Let A be the annual savings. A x FVIFA (12%, 10 years) = 1,000,000 A x 17.549 = 1,000,000 So A = 1,000,000 / 17.549 = Rs.56,983. 6. 1,000 x FVIFA (r, 6 years) = 10,000

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FVIFA (r, 6 years) = 10,000 / 1000 = 10

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Problems’ Solutions Project Evaluation

From the tables we find that FVIFA (20%, 6 years) = 9.930 FVIFA (24%, 6 years) = 10.980 Using linear interpolation in the interval, we get: 20% + (10.000 – 9.930) r = x 4% = 20.3% (10.980 – 9.930) 7. 1,000 x FVIF (r, 10 years) = 5,000 FVIF (r,10 years) = 5,000 / 1000 = 5 From the tables we find that FVIF (16%, 10 years) = 4.411 FVIF (18%, 10 years) = 5.234 Using linear interpolation in the interval, we get: (5.000 – 4.411) x 2% r = 16% + = 17.4% (5.234 – 4.411) 8. The present value of Rs.10,000 receivable after 8 years for various discount rates (r ) are: r = 10% PV = 10,000 x PVIF(r = 10%, 8 years) = 10,000 x 0.467 = Rs.4,670 r = 12% PV = 10,000 x PVIF (r = 12%, 8 years) = 10,000 x 0.404 = Rs.4,040 r = 15% PV = 10,000 x PVIF (r = 15%, 8 years) = 10,000 x 0.327 = Rs.3,270 9. Assuming that it is an ordinary annuity, the present value is: 2,000 x PVIFA (10%, 5years) = 2,000 x 3.791 = Rs.7,582 10. The present value of an annual pension of Rs.10,000 for 15 years when r = 15% is: 10,000 x PVIFA (15%, 15 years) = 10,000 x 5.847 = Rs.58,470 The alternative is to receive a lump-sum of Rs.50,000. Obviously, Mr. Jingo will be better off with the annual pension amount of Rs.10,000. 11. The amount that can be withdrawn annually is: 100,000 100,000 A = ------------------ ------------ = ----------- = Rs.10,608 PVIFA (10%, 30 years) 9.427

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Problems’ Solutions Project Evaluation

12. The present value of the income stream is: 1,000 x PVIF (12%, 1 year) + 2,500 x PVIF (12%, 2 years) + 5,000 x PVIFA (12%, 8 years) x PVIF(12%, 2 years) = 1,000 x 0.893 + 2,500 x 0.797 + 5,000 x 4.968 x 0.797 = Rs.22,683. 13. The present value of the income stream is: 2,000 x PVIFA (10%, 5 years) + 3000/0.10 x PVIF (10%, 5 years) = 2,000 x 3.791 + 3000/0.10 x 0.621 = Rs.26,212 14. To earn an annual income of Rs.5,000 beginning from the end of 15 years from now, if the

deposit earns 10% per year a sum of Rs.5,000 / 0.10 = Rs.50,000 is required at the end of 14 years. The amount that must be deposited to get this sum is: Rs.50,000 / PVIF (10%, 14 years) = Rs.50,000 / 3.797 = Rs.13,165 15. Rs.20,000 =- Rs.4,000 x PVIFA (r, 10 years) PVIFA (r,10 years) = Rs.20,000 / Rs.4,000 = 5.00 From the tables we find that: PVIFA (15%, 10 years) = 5.019 PVIFA (18%, 10 years) = 4.494 Using linear interpolation we get: 5.019 – 5.00 r = 15% + ---------------- x 3% 5.019 – 4.494 = 15.1% 16. PV (Stream A) = Rs.100 x PVIF (12%, 1 year) + Rs.200 x PVIF (12%, 2 years) + Rs.300 x PVIF(12%, 3 years) + Rs.400 x PVIF (12%, 4 years) + Rs.500 x PVIF (12%, 5 years) + Rs.600 x PVIF (12%, 6 years) + Rs.700 x PVIF (12%, 7 years) + Rs.800 x PVIF (12%, 8 years) + Rs.900 x PVIF (12%, 9 years) + Rs.1,000 x PVIF (12%, 10 years) = Rs.100 x 0.893 + Rs.200 x 0.797 + Rs.300 x 0.712 + Rs.400 x 0.636 + Rs.500 x 0.567 + Rs.600 x 0.507 + Rs.700 x 0.452 + Rs.800 x 0.404 + Rs.900 x 0.361 + Rs.1,000 x 0.322 = Rs.2590.9 Similarly, PV (Stream B) = Rs.3,625.2

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Problems’ Solutions Project Evaluation

PV (Stream C) = Rs.2,851.1 17. FV5 = Rs.10,000 [1 + (0.16 / 4)]5x4

= Rs.10,000 (1.04)20

= Rs.10,000 x 2.191 = Rs.21,910 18. FV5 = Rs.5,000 [1+( 0.12/4)] 5x4

= Rs.5,000 (1.03)20

= Rs.5,000 x 1.806 = Rs.9,030 19. A B C Stated rate (%) 12 24 24 Frequency of compounding 6 times 4 times 12 times Effective rate (%) (1 + 0.12/6)6- 1 (1+0.24/4)4 –1 (1 + 0.24/12)12-1 = 12.6 = 26.2 = 26.8 Difference between the effective rate and stated rate (%) 0.6 2.2 2.8 20. Investment required at the end of 8th year to yield an income of Rs.12,000 per year from

the end of 9th year (beginning of 10th year) for ever: Rs.12,000 x PVIFA(12%, ∞ ) = Rs.12,000 / 0.12 = Rs.100,000 To have a sum of Rs.100,000 at the end of 8th year , the amount to be deposited now is: Rs.100,000 Rs.100,000 = = Rs.40,388 PVIF(12%, 8 years) 2.476 21. The interest rate implicit in the offer of Rs.20,000 after 10 years in lieu of Rs.5,000 now is: Rs.5,000 x FVIF (r,10 years) = Rs.20,000 Rs.20,000 FVIF (r,10 years) = = 4.000 Rs.5,000 From the tables we find that

FVIF (15%, 10 years) = 4.046 This means that the implied interest rate is nearly 15%. I would choose Rs.20,000 for 10 years from now because I find a return of 15% quite acceptable. 22. FV10 = Rs.10,000 [1 + (0.10 / 2)]10x2

= Rs.10,000 (1.05)20

= Rs.10,000 x 2.653 = Rs.26,530

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Problems’ Solutions Project Evaluation

If the inflation rate is 8% per year, the value of Rs.26,530 10 years from now, in terms of the current rupees is: Rs.26,530 x PVIF (8%,10 years) = Rs.26,530 x 0.463 = Rs.12,283 23. A constant deposit at the beginning of each year represents an annuity due. PVIFA of an annuity due is equal to : PVIFA of an ordinary annuity x (1 + r)

To provide a sum of Rs.50,000 at the end of 10 years the annual deposit should be Rs.50,000 A = FVIFA(12%, 10 years) x (1.12) Rs.50,000 = = Rs.2544 17.549 x 1.12 24. The discounted value of Rs.20,000 receivable at the beginning of each year from 2005 to

2009, evaluated as at the beginning of 2004 (or end of 2003) is: Rs.20,000 x PVIFA (12%, 5 years) = Rs.20,000 x 3.605 = Rs.72,100. The discounted value of Rs.72,100 evaluated at the end of 2000 is Rs.72,100 x PVIF (12%, 3 years) = Rs.72,100 x 0.712 = Rs.51,335 If A is the amount deposited at the end of each year from 1995 to 2000 then A x FVIFA (12%, 6 years) = Rs.51,335 A x 8.115 = Rs.51,335 A = Rs.51,335 / 8.115 = Rs.6326 25. The discounted value of the annuity of Rs.2000 receivable for 30 years, evaluated as at the

end of 9th year is: Rs.2,000 x PVIFA (10%, 30 years) = Rs.2,000 x 9.427 = Rs.18,854 The present value of Rs.18,854 is: Rs.18,854 x PVIF (10%, 9 years) = Rs.18,854 x 0.424 = Rs.7,994 26. 30 percent of the pension amount is 0.30 x Rs.600 = Rs.180 Assuming that the monthly interest rate corresponding to an annual interest rate of 12% is 1%, the discounted value of an annuity of Rs.180 receivable at the end of each month for 180 months (15 years) is: Rs.180 x PVIFA (1%, 180) (1.01)180 - 1 Rs.180 x ---------------- = Rs.14,998 .01 (1.01)180

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Page 11: Project evaluation and feasibility analysis

Problems’ Solutions Project Evaluation

If Mr. Ramesh borrows Rs.P today on which the monthly interest rate is 1% P x (1.01)60 = Rs.14,998 P x 1.817 = Rs.14,998 Rs.14,998 P = ------------ = Rs.8254 1.817 27. Rs.300 x PVIFA(r, 24 months) = Rs.6,000 PVIFA (4%,24) = Rs.6000 / Rs.300 = 20 From the tables we find that: PVIFA(1%,24) = 21.244 PVIFA (2%, 24) = 18.914 Using a linear interpolation 21.244 – 20.000 r = 1% + ---------------------- x 1% 21.244 – 18,914

= 1.53% Thus, the bank charges an interest rate of 1.53% per month. The corresponding effective rate of interest per annum is [ (1.0153)12 – 1 ] x 100 = 20% 28. The discounted value of the debentures to be redeemed between 8 to 10 years evaluated at

the end of the 5th year is: Rs.10 million x PVIF (8%, 3 years) + Rs.10 million x PVIF (8%, 4 years) + Rs.10 million x PVIF (8%, 5 years) = Rs.10 million (0.794 + 0.735 + 0.681) = Rs.2.21 million If A is the annual deposit to be made in the sinking fund for the years 1 to 5, then

A x FVIFA (8%, 5 years) = Rs.2.21 million A x 5.867 = Rs.2.21 million

A = 5.867 = Rs.2.21 million A = Rs.2.21 million / 5.867 = Rs.0.377 million

29. Let `n’ be the number of years for which a sum of Rs.20,000 can be withdrawn annually. Rs.20,000 x PVIFA (10%, n) = Rs.100,000 PVIFA (15%, n) = Rs.100,000 / Rs.20,000 = 5.000 From the tables we find that PVIFA (10%, 7 years) = 4.868 PVIFA (10%, 8 years) = 5.335 Thus n is between 7 and 8. Using a linear interpolation we get

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Problems’ Solutions Project Evaluation

5.000 – 4.868 n = 7 + ----------------- x 1 = 7.3 years 5.335 – 4.868 30. Equated annual installment = 500000 / PVIFA(14%,4) = 500000 / 2.914 = Rs.171,585

Loan Amortization Schedule

Beginning Annual Principal Remaining Year amount installment Interest repaid balance 1 500000 171585 70000 101585 398415 2 398415 171585 55778 115807 282608 3 282608 171585 39565 132020 150588 4 150588 171585 21082 150503 85* (*) rounding off error 31. Define n as the maturity period of the loan. The value of n can be obtained from the

equation. 200,000 x PVIFA(13%, n) = 1,500,000 PVIFA (13%, n) = 7.500 From the tables or otherwise it can be verified that PVIFA(13,30) = 7.500 Hence the maturity period of the loan is 30 years. 32. Expected value of iron ore mined during year 1 = Rs.300 million Expected present value of the iron ore that can be mined over the next 15 years assuming a price escalation of 6% per annum in the price per ton of iron 1 – (1 + g)n / (1 + i)n

= Rs.300 million x ------------------------ i - g = Rs.300 million x 1 – (1.06)15 / (1.16)15

0.16 – 0.06 = Rs.300 million x (0.74135 / 0.10) = Rs.2224 million

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Page 13: Project evaluation and feasibility analysis

Problems’ Solutions Project Evaluation

Chapter 8 INVESTMENT CRITERIA

1.(a) NPV of the project at a discount rate of 14%. 100,000 200,000 = - 1,000,000 + ---------- + ------------ (1.14) (1.14)2

300,000 600,000 300,000 + ----------- + ---------- + ---------- (1.14)3 (1.14)4 (1.14)5

= - 44837

(b) NPV of the project at time varying discount rates = - 1,000,000 100,000 + (1.12) 200,000 + (1.12) (1.13) 300,000 + (1.12) (1.13) (1.14) 600,000 + (1.12) (1.13) (1.14) (1.15) 300,000 + (1.12) (1.13) (1.14)(1.15)(1.16) = - 1,000,000 + 89286 + 158028 + 207931 + 361620 + 155871 = - 27264 2. Investment A a) Payback period = 5 years b) NPV = 40000 x PVIFA (12%,10) – 200 000 = 26000

c) IRR (r ) can be obtained by solving the equation: 40000 x PVIFA (r, 10) = 200000 i.e., PVIFA (r, 10) = 5.000 From the PVIFA tables we find that PVIFA (15%,10) = 5.019

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Problems’ Solutions Project Evaluation

PVIFA (16%,10) = 4.883 Linear interpolation in this range yields r = 15 + 1 x (0.019 / 0.136) = 15.14% d) BCR = Benefit Cost Ratio = PVB / I = 226,000 / 200,000 = 1.13

Investment B a) Payback period = 9 years b) NP V = 40,000 x PVIFA (12%,5) + 30,000 x PVIFA (12%,2) x PVIF (12%,5) + 20,000 x PVIFA (12%,3) x PVIF (12%,7)

- 300,000 = (40,000 x 3.605) + (30,000 x 1.690 x 0.567) + (20,000 x 2.402 x 0.452) – 300,000 = - 105339 c) IRR (r ) can be obtained by solving the equation 40,000 x PVIFA (r, 5) + 30,000 x PVIFA (r, 2) x PVIF (r,5) + 20,000 x PVIFA (r, 3) x PVIF (r, 7) = 300,000 Through the process of trial and error we find that

r = 1.37% d) BCR = PVB / I = 194,661 / 300,000 = 0.65

Investment C a) Payback period lies between 2 years and 3 years. Linear interpolation in

This range provides an approximate payback period of 2.88 years. b) NPV = 80.000 x PVIF (12%,1) + 60,000 x PVIF (12%,2) + 80,000 x PVIF (12%,3) + 60,000 x PVIF (12%,4)

+ 80,000 x PVIF (12%,5) + 60,000 x PVIF (12%,6) + 40,000 x PVIFA (12%,4) x PVIF (12%,6)

- 210,000 = 111,371

c) IRR (r) is obtained by solving the equation 80,000 x PVIF (r,1) + 60,000 x PVIF (r,2) + 80,000 x PVIF (r,3)

+ 60,000 x PVIF (r,4) + 80,000 x PVIF (r,5) + 60,000 x PVIF (r,6) + 40000 x PVIFA (r,4) x PVIF (r,6) = 210000 Through the process of trial and error we get

r = 29.29%

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Problems’ Solutions Project Evaluation

d) BCR = PVB / I = 321,371 / 210,000 = 1.53

Investment D a) Payback period lies between 8 years and 9 years. A linear interpolation in this range

provides an approximate payback period of 8.5 years. 8 + (1 x 100,000 / 200,000)

b) NPV = 200,000 x PVIF (12%,1) + 20,000 x PVIF (12%,2) + 200,000 x PVIF (12%,9) + 50,000 x PVIF (12%,10)

- 320,000 = - 37,160 c) IRR (r ) can be obtained by solving the equation 200,000 x PVIF (r,1) + 200,000 x PVIF (r,2) + 200,000 x PVIF (r,9) + 50,000 x PVIF (r,10) = 320000 Through the process of trial and error we get r = 8.45% d) BCR = PVB / I = 282,840 / 320,000 = 0.88

Comparative Table Investment A B C D

a) Payback period (in years) 5 9 2.88 8.5 b) NPV @ 12% 26000 -105339 111371 -37160 c) IRR (%) 15.14 1.37 29.29 8.45 d) BCR 1.13 0.65 1.53 0.88

Among the four alternative investments, the investment to be chosen is ‘C’ because it has the a. Lowest payback period b. Highest NPV c. Highest IRR d. Highest BCR 3. IRR (r) can be calculated by solving the following equations for the value of r. 60000 x PVIFA (r,7) = 300,000 i.e., PVIFA (r,7) = 5.000 Through a process of trial and error it can be verified that r = 9.20% p.a. 4. The IRR (r) for the given cash flow stream can be obtained by solving the following

equation for the value of r. -3000 + 9000 / (1+r) – 3000 / (1+r) = 0 Simplifying the above equation we get r = 1.61, -0.61; (or) 161%, (-)61% Note: Given two changes in the signs of cash flow, we get two values for the IRR of the cash flow stream. In such cases, the IRR rule breaks down. 5. Define NCF as the minimum constant annual net cash flow that justifies the purchase of the

given equipment. The value of NCF can be obtained from the equation

15 By M. Iftikhar Mubbashir

NCF x PVIFA (10%,8) = 500000

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Problems’ Solutions Project Evaluation

NCF = 500000 / 5.335 = 93271 6. Define I as the initial investment that is justified in relation to a net annual cash inflow of

25000 for 10 years at a discount rate of 12% per annum. The value of I can be obtained from the following equation

25000 x PVIFA (12%,10) = I i.e., I = 141256 7. PV of benefits (PVB) = 25000 x PVIF (15%,1) + 40000 x PVIF (15%,2) + 50000 x PVIF (15%,3) + 40000 x PVIF (15%,4) + 30000 x PVIF (15%,5) = 122646 (A) Investment = 100,000 (B) Benefit cost ratio = 1.23 [= (A) / (B)] 8. The NPV’s of the three projects are as follows: Project P Q R Discount rate 0% 400 500 600 5% 223 251 312 10% 69 40 70 15% - 66 - 142 - 135 25% - 291 - 435 - 461 30% - 386 - 555 - 591 9. NPV profiles for Projects P and Q for selected discount rates are as follows: (a) Project P Q Discount rate (%) 0 2950 500 5 1876 208 10 1075 - 28 15 471 - 222 20 11 - 382 b) (i) The IRR (r ) of project P can be obtained by solving the following equation for `r’. -1000 -1200 x PVIF (r,1) – 600 x PVIF (r,2) – 250 x PVIF (r,3) + 2000 x PVIF (r,4) + 4000 x PVIF (r,5) = 0 Through a process of trial and error we find that r = 20.13%

(ii) The IRR (r') of project Q can be obtained by solving the following equation for r' -1600 + 200 x PVIF (r',1) + 400 x PVIF (r',2) + 600 x PVIF (r',3) + 800 x PVIF (r',4) + 100 x PVIF (r',5) = 0

16 By M. Iftikhar Mubbashir

Through a process of trial and error we find that r' = 9.34%.

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c) From (a) we find that at a cost of capital of 10% NPV (P) = 1075 NPV (Q) = - 28 Given that NPV (P), NPV (Q) and NPV (P) > 0, I would choose project P. From (a) we find that at a cost of capital of 20% NPV (P) = 11 NPV (Q) = - 382 Again NPV (P) > NPV (Q); and NPV (P) > 0. I would choose project P. d) Project P PV of investment-related costs = 1000 x PVIF (12%,0) + 1200 x PVIF (12%,1) + 600 x PVIF (12%,2) + 250 x PVIF (12%,3) = 2728 TV of cash inflows = 2000 x (1.12) + 4000 = 6240 The MIRR of the project P is given by the equation: 2728 = 6240 x PVIF (MIRR,5) (1 + MIRR)5 = 2.2874 MIRR = 18% (c) Project Q

PV of investment-related costs = 1600 TV of cash inflows @ 15% p.a. = 2772 The MIRR of project Q is given by the equation: 16000 (1 + MIRR)5 = 2772 MIRR = 11.62% 10. (a) Project A NPV at a cost of capital of 12% = - 100 + 25 x PVIFA (12%,6) = Rs.2.79 million IRR (r ) can be obtained by solving the following equation for r. 25 x PVIFA (r,6) = 100 i.e., r = 12,98% Project B NPV at a cost of capital of 12% = - 50 + 13 x PVIFA (12%,6) = Rs.3.45 million IRR (r') can be obtained by solving the equation 13 x PVIFA (r',6) = 50 i.e., r' = 14.40% [determined through a process of trial and error] (b) Difference in capital outlays between projects A and B is Rs.50 million

Difference in net annual cash flow between projects A and B is Rs.12 million. NPV of the differential project at 12%

= -50 + 12 x PVIFA (12%,6)

17 By M. Iftikhar Mubbashir

= Rs.3.15 million

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IRR (r'') of the differential project can be obtained from the equation 12 x PVIFA (r'', 6) = 50 i.e., r'' = 11.53% 11. (a) Project M The pay back period of the project lies between 2 and 3 years. Interpolating in this range we get an approximate pay back period of 2.63 years. Project N The pay back period lies between 1 and 2 years. Interpolating in this range we

get an approximate pay back period of 1.55 years. (b) Project M Cost of capital = 12% p.a PV of cash flows up to the end of year 2 = 24.97 PV of cash flows up to the end of year 3 = 47.75 PV of cash flows up to the end of year 4 = 71.26

Discounted pay back period (DPB) lies between 3 and 4 years. Interpolating in this range we get an approximate DPB of 3.1 years.

Project N Cost of capital = 12% per annum PV of cash flows up to the end of year 1 = 33.93 PV of cash flows up to the end of year 2 = 51.47 DPB lies between 1 and 2 years. Interpolating in this range we get an approximate DPB of 1.92 years. (c) Project M Cost of capital = 12% per annum NPV = - 50 + 11 x PVIFA (12%,1) + 19 x PVIF (12%,2) + 32 x PVIF (12%,3) + 37 x PVIF (12%,4) = Rs.21.26 million Project N Cost of capital = 12% per annum NPV = Rs.20.63 million Since the two projects are independent and the NPV of each project is (+) ve, both the projects can be accepted. This assumes that there is no capital constraint. (d) Project M Cost of capital = 10% per annum NPV = Rs.25.02 million Project N Cost of capital = 10% per annum NPV = Rs.23.08 million

18 By M. Iftikhar Mubbashir

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Since the two projects are mutually exclusive, we need to choose the project with the higher NPV i.e., choose project M. Note : The MIRR can also be used as a criterion of merit for choosing between the two projects because their initial outlays are equal. (e) Project M Cost of capital = 15% per annum NPV = 16.13 million Project N Cost of capital: 15% per annum NPV = Rs.17.23 million Again the two projects are mutually exclusive. So we choose the project with the higher NPV, i.e., choose project N. (f) Project M Terminal value of the cash inflows: 114.47 MIRR of the project is given by the equation 50 (1 + MIRR)4 = 114.47 i.e., MIRR = 23.01% Project N Terminal value of the cash inflows: 115.41 MIRR of the project is given by the equation 50 ( 1+ MIRR)4 = 115.41 i.e., MIRR = 23.26% 12. The internal rate of return is the value of r in the equation

2,000 1,000 10,000 2,000 8000 = - + + (1+r) (1+r)2 (1+r)3 (1+r)4

At r = 18%, the right hand side is equal to 8099 At r = 20%, the right hand side is equal to 7726 Thus the solving value of r is: 8,099 – 8,000 18% + x 2% = 18.5% 8,099 – 7,726 Un-recovered Investment Balance

Year Un-recovered investment balance at

the beginning Ft-1

Interest for the year Ft-1 (1+r)

Cash flow at the end of the year CFt

Un-recovered investment balance at the end of the year Ft-1

(1+r) + CFt1 -8000 -1480 2000 -7480 2 -7480 -1383.8 -1000 -9863.8 3 -9863.8 -1824.80 10000 -1688.60 4 -1688.60 -312.39 2000 0

19 By M. Iftikhar Mubbashir

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13. Rs. in lakhs Year 1 2 3 4 5 6 7 8 Sum Average

Investment 24.0 21.0 18.0 15.0 12.0 9.0 6.0 3.0 108 13.500 Depreciation 3.0 3.0 3.0 3.0 3.0 3.0 3.0 3.0 24.0 3.000 Income before interest and tax

6.0 6.5 7.0 7.0 7.0 6.5 6.0 5.0 51.0 6.375

Interest 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 20.0 2.500 Income before tax 3.5 4.0 4.5 4.5 4.5 4.0 3.5 2.5 31.0 3.875 Tax - 1.0 2.5 2.5 2.5 2.2 1.9 1.4 14.0 1.750 Income after tax 3.5 3.0 2.0 2.0 2.0 1.8 1.6 1.1 17.0 2.125

Measures of Accounting Rate of Return A. Average income after tax 2.125 = = 8.9% Initial investment 24

B. Average income after tax 2.125 = = 15.7% Average investment 13.5 C. Average income after tax but before interest 2.125 + 2.5 = = 19.3% Initial investment 24

D. Average income after tax but before interest 2.125 + 2.5 = = 34.3% Average investment 13.5

E. Average income before interest and taxes 6.375 = = 26.6% Initial investment 24

F. Average income before interest and taxes 6.375 = = 47.2% Average investment 13.5 G. Total income after tax but before Depreciation – Initial investment 17.0 + 24.0 – 24.0 = (Initial investment / 2) x Years (24 / 2) x 8 = 17.0 / 96.0 = 17.7%

20 By M. Iftikhar Mubbashir

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Chapter 9 PROJECT CASH FLOWS 1. (a) Project Cash Flows (Rs. in million) Year 0 1 2 3 4 5 6 7 1. Plant & machinery (150) 2. Working capital (50) 3. Revenues 250 250 250 250 250 250 250 4. Costs (excluding de- preciation & interest) 100 100 100 100 100 100 100 5. Depreciation 37.5 28.13 21.09 15.82 11.87 8.90 6.67 6. Profit before tax 112.5 121.87 128.91 134.18 138.13 141.1 143.33 7. Tax 33.75 36.56 38.67 40.25 41.44 42.33 43.0 8. Profit after tax 78.75 85.31 90.24 93.93 96.69 98.77 100.33 9. Net salvage value of plant & machinery 48 10. Recovery of working 50 capital 11. Initial outlay (=1+2) (200) 12. Operating CF (= 8 + 5) 116.25 113.44 111.33 109.75 108.56 107.67 107.00 13. Terminal CF ( = 9 +10) 98 14. NCF (200) 116.25 113.44 111.33 109.75 108.56 107.67 205 (c) IRR (r) of the project can be obtained by solving the following equation for r -200 + 116.25 x PVIF (r,1) + 113.44 x PVIF (r,2) + 111.33 x PVIF (r,3) + 109.75 x PVIF (r,4) + 108.56 x PVIF (r,5) +107.67 x PVIF (r,6) + 205 x PVIF (r,7) = 0

Through a process of trial and error, we get r = 55.17%. The IRR of the project is 55.17%.

21 By M. Iftikhar Mubbashir

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2. Post-tax Incremental Cash Flows (Rs. in million) Year 0 1 2 3 4 5 6 7 1. Capital equipment (120) 2. Level of working capital 20 30 40 50 40 30 20 (ending) 3. Revenues 80 120 160 200 160 120 80 4. Raw material cost 24 36 48 60 48 36 24 5. Variable mfg cost. 8 12 16 20 16 12 8 6. Fixed operating & maint. 10 10 10 10 10 10 10 cost 7. Variable selling expenses 8 12 16 20 16 12 8 8. Incremental overheads 4 6 8 10 8 6 4 9. Loss of contribution 10 10 10 10 10 10 10 10.Bad debt loss 4 11. Depreciation 30 22.5 16.88 12.66 9.49 7.12 5.34 12. Profit before tax -14 11.5 35.12 57.34 42.51 26.88 6.66 13. Tax - 4.2 3.45 10.54 17.20 12.75 8.06 2.00 14. Profit after tax - 9.8 8.05 24.58 40.14 29.76 18.82 4.66 15. Net salvage value of capital equipments 25 16. Recovery of working 16 capital 17. Initial investment (120) 18. Operating cash flow 20.2 30.55 41.46 52.80 39.25 25.94 14.00 (14 + 10+ 11) 19. ∆ Working capital 20 10 10 10 (10) (10) (10) 20. Terminal cash flow 41 21. Net cash flow (140) 10.20 20.55 31.46 62.80 49.25 35.94 55.00 (17+18-19+20) (b) NPV of the net cash flow stream @ 15% per discount rate = -140 + 10.20 x PVIF(15%,1) + 20.55 x PVIF (15%,2) + 31.46 x PVIF (15%,3) + 62.80 x PVIF (15%,4) + 49.25 x PVIF

(15%,5) + 35.94 x PVIF (15%,6) + 55 x PVIF (15%,7)

= Rs.1.70 million 3. (a) A. Initial outlay (Time 0) i. Cost of new machine Rs. 3,000,000 ii. Salvage value of old machine 900,000 iii Incremental working capital requirement 500,000 iv. Total net investment (=i – ii + iii) 2,600,000

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B. Operating cash flow (years 1 through 5) Year 1 2 3 4 5 i. Post-tax savings in manufacturing costs 455,000 455,000 455,000 455,000 455,000 ii. Incremental depreciation 550,000 412,500 309,375 232,031 174,023 iii. Tax shield on incremental dep. 165,000 123,750 92,813 69,609 52,207 iv. Operating cash flow ( i + iii) 620,000 578,750 547,813 524,609 507,207

C. Terminal cash flow (year 5) i. Salvage value of new machine Rs. 1,500,000 ii. Salvage value of old machine 200,000 iii. Recovery of incremental working capital 500,000 iv. Terminal cash flow ( i – ii + iii) 1,800,000

D. Net cash flows associated with the replacement project (in Rs) Year 0 1 2 3 4 5 NCF (2,600,000) 620000 578750 547813 524609 307207 (b) NPV of the replacement project = - 2600000 + 620000 x PVIF (14%,1) + 578750 x PVIF (14%,2) + 547813 x PVIF (14%,3) + 524609 x PVIF (14%,4) + 2307207 x PVIF (14%,5) = Rs.267849 4. Tax shield (savings) on depreciation (in Rs) Depreciation Tax shield PV of tax shield Year charge (DC) =0.4 x DC @ 15% p.a. 1 25000 10000 8696 2 18750 7500 5671 3 14063 5625 3699 4 10547 4219 2412 5 7910 3164 1573 -------- 22051 Present value of the tax savings on account of depreciation = Rs.22051

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5. A. Initial outlay (at time 0) i. Cost of new machine Rs. 400,000 ii. Salvage value of the old machine 90,000 iii. Net investment 310,000

B. Operating cash flow (years 1 through 5)

Year 1 2 3 4 5 i. Depreciation of old machine 18000 14400 11520 9216 7373 ii. Depreciation of new machine 100000 75000 56250 42188 31641 iii. Incremental depre- ciation ( ii – i) 82000 60600 44730 32972 24268 iv. Tax savings on inc- remental depreciation ( 0.35 x (iii)) 28700 21210 15656 11540 8494 v. Operating cash flow 28700 21210 15656 11540 8494

C. Terminal cash flow (year 5) i. Salvage value of new machine Rs. 25000 ii. Salvage value of old machine 10000 iii. Incremental salvage value of new machine = Terminal cash flow 15000

D. Net cash flows associated with the replacement proposal. Year 0 1 2 3 4 5 NCF (310000) 28700 21210 15656 11540 23494 6. Net Cash Flows Relating to Equity (Rs. in million)

Particulars Year 0 1 2 3 4 5 6 1. Equity funds (100) 2. Revenues 500 500 500 500 500 500 3. Operating costs 320 320 320 320 320 320 4. Depreciation 83.33 55.56 37.04 24.69 16.46 10.97 5. Interest on working capital

advance 18.00 18.00 18.00 18.00 18.00 18.00

6. Interest on term loan 30.00 28.50 22.50 16.50 10.50 4.50 7. Profit before tax 48.67 77.94 102.46 120.81 135.04 146.53 8. Tax 24.335 38.97 51.23 60.405 67.52 73.265 9. Profit after tax 24.335 38.97 51.23 60.405 67.52 73.265 10. Preference dividend 11. Net salvage value of fixed assets 200 12. Net salvage value of current - 40 40 40 40 40

24 By M. Iftikhar Mubbashir

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Problems’ Solutions Project Evaluation

assets 13. Repayment of term-loans 14. Redemption of preference capital 15. Repayment of short-term bank

borrowings 100

16. Retirement of trade creditors 50 17. Initial investment (1) (100) 18. Operating cash flows (9-10+4) 107.665 94.53 88.27 85.095 83.98 84.235 19. Liquidation and retirement cash

flows (11+12-13-14-15-16) 107.665 54.53 48.27 45.095 43.98 90

20. Net cash flows (17+18+19) (100) 107.665 54.53 48.27 45.095 43.98 174.235

Net Cash Flows Relating to Long-term Funds (Rs. in million) Particulars Year 0 1 2 3 4 5 6 1. Fixed assets (250) 2. Working capital margin (50) 3. Revenues 500 500 500 500 500 500 4. Operating costs 320 320 320 320 320 320 5. Depreciation 83.33 55.56 37.04 24.69 16.46 10.97 6. Interest on working capital

advance 18.00 18.00 18.00 18.00 18.00 18.00

7. Interest on term loan 30.00 28.50 22.50 16.50 10.50 4.50 8. Profit before tax 48.67 77.94 102.46 120.81 135.04 146.53 9. Tax @ 50% 24.335 38.97 51.23 60.405 67.52 73.265 10. Profit after tax 24.335 38.97 51.23 60.405 67.52 73.265 11. Net salvage value of fixed assets 80 12. Net recovery of working capital

margin 50

13. Initial investment (1+2) (300) 14. Operating cash inflow (9+5+7

(1-T) ) 122.665 108.78 99.52 93.345 89.23 86.845

15. Terminal cash flow (11+12) 130.00 16. Net cash flow (13+14+15) (300) 122.665 108.78 99.52 93.345 89.23 216.485

Cash Flows Relating to Total Funds (Rs. in million)

Year 0 1 2 3 4 5 6 1. Total funds (450) 2. Revenues 500 500 500 500 500 500 3. Operating costs 320 320 320 320 320 320 4. Depreciation 83.33 55.56 37.04 24.69 16.46 10.97 5. Interest on term loan 30.00 28.50 22.50 16.50 10.50 4.50 6. Interest on working capital

advance 18.00 18.00 18.00 18.00 18.00 18.00

7. Profit before tax 48.67 77.94 102.46 120.81 135.04 146.53 8. Tax 24.34 38.97 51.23 60.41 67.52 73.265 9. Profit after tax 24.34 38.97 51.23 60.41 67.52 73.265 10. Net salvalue of fixed assets 80 11. Net salvage value of current assets 200 12. Initial investment (1) (450) 13. Operating cash inflow 9+4+6 (1-t)

+ 5(1-t) 131.67 117.78 108.52 102.35 98.23 95.485

14. Terminal cash flow (10+11) 280 15. Net cash flow (12+13+14) (450) 131.67 117.78 108.52 102.35 98.23 375.485

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Chapter 11 RISK ANALYSIS OF SINGLE INVESTMENTS

1. (a) NPV of the project = -250 + 50 x PVIFA (13%,10) = Rs.21.31 million (b) NPVs under alternative scenarios: (Rs. in million) Pessimistic Expected Optimistic Investment 300 250 200 Sales 150 200 275 Variable costs 97.5 120 154 Fixed costs 30 20 15 Depreciation 30 25 20 Pretax profit - 7.5 35 86 Tax @ 28.57% - 2.14 10 24.57 Profit after tax - 5.36 25 61.43 Net cash flow 24.64 50 81.43 Cost of capital 14% 13% 12% NPV - 171.47 21.31 260.10 Assumptions: (1) The useful life is assumed to be 10 years under all three scenarios. It is also assumed that the salvage value of the investment after ten years is zero.

(2) The investment is assumed to be depreciated at 10% per annum; and it is also assumed that this method and rate of depreciation are acceptable to the IT (income tax) authorities.

(3) The tax rate has been calculated from the given table i.e. 10 / 35 x

100 = 28.57%.

(4) It is assumed that only loss on this project can be offset against the taxable profit on other projects of the company; and thus the company can claim a tax shield on the loss in the same year.

(c) Accounting break even point (under ‘expected’ scenario) Fixed costs + depreciation = Rs. 45 million Contribution margin ratio = 60 / 200 = 0.3 Break even level of sales = 45 / 0.3 = Rs.150 million Financial break even point (under ‘expected’ scenario) i. Annual net cash flow = 0.7143 [ 0.3 x sales – 45 ] + 25 = 0.2143 sales – 7.14 ii. PV (net cash flows) = [0.2143 sales – 7.14 ] x PVIFA (13%,10) = 1.1628 sales – 38.74

26 By M. Iftikhar Mubbashir

iii. Initial investment = 200

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iv. Financial break even level of sales = 238.74 / 1.1628 = Rs.205.31 million 2. (a) Sensitivity of NPV with respect to quantity manufactured and sold: (in Rs) Pessimistic Expected Optimistic Initial investment 30000 30000 30000 Sale revenue 24000 42000 54000 Variable costs 16000 28000 36000 Fixed costs 3000 3000 3000 Depreciation 2000 2000 2000 Profit before tax 3000 9000 13000 Tax 1500 4500 6500 Profit after tax 1500 4500 6500 Net cash flow 3500 6500 8500 NPV at a cost of capital of 10% p.a and useful life of 5 years -16732 - 5360 2222 (b) Sensitivity of NPV with respect to variations in unit price. Pessimistic Expected Optimistic Initial investment 30000 30000 30000 Sale revenue 28000 42000 70000 Variable costs 28000 28000 28000 Fixed costs 3000 3000 3000 Depreciation 2000 2000 2000 Profit before tax -5000 9000 37000 Tax -2500 4500 18500 Profit after tax -2500 4500 18500 Net cash flow - 500 6500 20500 NPV - 31895 (-) 5360 47711 (c) Sensitivity of NPV with respect to variations in unit variable cost. Pessimistic Expected Optimistic Initial investment 30000 30000 30000 Sale revenue 42000 42000 42000 Variable costs 56000 28000 21000 Fixed costs 3000 3000 3000 Depreciation 2000 2000 2000 Profit before tax -11000 9000 16000 Tax -5500 4500 8000 Profit after tax -5500 4500 8000 Net cash flow -3500 6500 10000 NPV -43268 - 5360 7908 (d) Accounting break-even point i. Fixed costs + depreciation = Rs.5000

27 By M. Iftikhar Mubbashir

ii. Contribution margin ratio = 10 / 30 = 0.3333

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iii. Break-even level of sales = 5000 / 0.3333 = Rs.15000 Financial break-even point

i. Annual cash flow = 0.5 x (0.3333 Sales – 5000) = 2000 ii. PV of annual cash flow = (i) x PVIFA (10%,5) = 0.6318 sales – 1896 iii. Initial investment = 30000 iv. Break-even level of sales = 31896 / 0.6318 = Rs.50484 2. Define At as the random variable denoting net cash flow in year t. A1 = 4 x 0.4 + 5 x 0.5 + 6 x 0.1 = 4.7 A2 = 5 x 0.4 + 6 x 0.4 + 7 x 0.2 = 5.8 A3 = 3 x 0.3 + 4 x 0.5 + 5 x 0.2 = 3.9 NPV = 4.7 / 1.1 +5.8 / (1.1)2 + 3.9 / (1.1)3 – 10 = Rs.2.00 million σ1

2 = 0.41 σ2

2 = 0.56 σ3

2 = 0.49 σ1

2 σ22 σ3

2

σ2 NPV = + + (1.1)2 (1.1)4 (1.1)6

= 1.00 σ (NPV) = Rs.1.00 million 3. Expected NPV 4 At

= ∑ - 25,000 t=1 (1.08)t

= 12,000/(1.08) + 10,000 / (1.08)2 + 9,000 / (1.08)3

+ 8,000 / (1.08)4 – 25,000 = [ 12,000 x .926 + 10,000 x .857 + 9,000 x .794 + 8,000 x .735] - 25,000 = 7,708 Standard deviation of NPV 4 σt

∑ t=1 (1.08)t

= 5,000 / (1.08) + 6,000 / (1.08)2 + 5,000 / (1,08)3 + 6,000 / (1.08)4

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= 5,000 x .926 + 6,000 x .857 + 5000 x .794 + 6,000 x .735 = 18,152 4. Expected NPV 4 At

= ∑ - 25,000 …. (1) t=1 (1.06)t

A1 = 2,000 x 0.2 + 3,000 x 0.5 + 4,000 x 0.3 = 3,100 A2 = 3,000 x 0.4 + 4,000 x 0.3 + 5,000 x 0.3 = 3,900 A3 = 4,000 x 0.3 + 5,000 x 0.5 + 6,000 x 0.2 = 4,900 A4 = 2,000 x 0.2 + 3,000 x 0.4 + 4,000 x 0.4 = 3,200 Substituting these values in (1) we get

Expected NPV = NPV = 3,100 / (1.06)+ 3,900 / (1.06)2 + 4,900 / (1.06)3 + 3,200 / (1,06)4

- 10,000 = Rs.3,044 The variance of NPV is given by the expression

4 σ2t

σ2 (NPV) = ∑ …….. (2) t=1 (1.06)2t

σ1

2 = [(2,000 – 3,100)2 x 0.2 + (3,000 – 3,100)2 x 0.5 + (4,000 – 3,100)2 x 0.3] = 490,000 σ2

2 = [(3,000 – 3,900)2 x 0.4 + (4,000 – 3,900)2 x 0.3 + (5,000 – 3900)2 x 0.3] = 690,000 σ3

2 = [(4,000 – 4,900)2 x 0.3 + (5,000 – 4,900)2 x 0.5 + (6,000 – 4,900)2 x 0.2] = 490,000 σ4

2 = [(2,000 – 3,200)2 x 0.2 + (3,000 – 3,200)2 x 0.4 + (4,000 – 3200)2 x 0.4] = 560,000 Substituting these values in (2) we get 490,000 / (1.06)2 + 690,000 / (1.06)4

+ 490,000 / (1.06)6 + 560,000 / (1.08)8

[ 490,000 x 0.890 + 690,000 x 0.792 + 490,000 x 0.705 + 560,000 x 0.627 ]

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= 1,679,150

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σ NPV = 1,679,150 = Rs.1,296 NPV – NPV 0 - NPV Prob (NPV < 0) = Prob. < σ NPV σ NPV 0 – 3044 = Prob Z < 1296 = Prob (Z < -2.35) The required probability is given by the shaded area in the following normal curve. P (Z < - 2.35) = 0.5 – P (-2.35 < Z < 0) = 0.5 – P (0 < Z < 2.35) = 0.5 – 0.4906 = 0.0094 So the probability of NPV being negative is 0.0094 Prob (P1 > 1.2) Prob (PV / I > 1.2) Prob (NPV / I > 0.2) Prob. (NPV > 0.2 x 10,000) Prob (NPV > 2,000) Prob (NPV >2,000)= Prob (Z > 2,000- 3,044 / 1,296) Prob (Z > - 0.81)

The required probability is given by the shaded area of the following normal curve:

P(Z > - 0.81) = 0.5 + P(-0.81 < Z < 0) = 0.5 + P(0 < Z < 0.81) = 0.5 + 0.2910 = 0.7910 So the probability of P1 > 1.2 as 0.7910 5. Given values of variables other than Q, P and V, the net present value model of Bidhan

Corporation can be expressed as: 5

∑ [Q(P – V) – 3,000 – 2,000] (0.5)+ 2,000 0 t=1 NPV = ---------------------------------------------------------- + ------- - 30,000 (1.1)t (1.1)5

5

∑ 0.5 Q (P – V) – 500 t=1 ------------------------------------ - 30,000 (1.1)t

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Problems’ Solutions Project Evaluation

= [ 0.5Q (P – V) – 500] x PVIFA (10,5) – 30,000 = [0.5Q (P – V) – 500] x 3.791 – 30,000 = 1.8955Q (P – V) – 31,895.5 Exhibit 1 presents the correspondence between the values of exogenous variables and the two digit random number. Exhibit 2 shows the results of the simulation.

Exhibit 1

Correspondence between values of exogenous variables and two digit random numbers QUANTITY PRICE VARIABLE COST Value

Prob

Cumulative Prob.

Two digit random numbers

Value

Prob

Cumulative Prob.

Two digit random numbers

Value

Prob

Cumulative Prob.

Two digit random numbers

800 0.10 0.10 00 to 09 20 0.40 0.40 00 to 39 15 0.30 0.30 00 to 29 1,000 0.10 0.20 10 to 19 30 0.40 0.80 40 to 79 20 0.50 0.80 30 to 79 1,200 0.20 0.40 20 to 39 40 0.10 0.90 80 to 89 40 0.20 1.00 80 to 99 1,400 0.30 0.70 40 to 69 50 0.10 1.00 90 to 99 1,600 0.20 0.90 70 to 89 1,800 0.10 1.00 90 to 99

Exhibit 2

Simulation Results

31 By M. Iftikhar Mubbashir

QUANTITY (Q) PRICE (P) VARIABLE COST (V) NPV Run Random

Number Corres-ponding Value

Random Number

Corres-ponding value

Random Number

Corres-ponding value

1.8955 Q(P-V)-31,895.5

1 03 800 38 20 17 15 -24,314 2 32 1,200 69 30 24 15 2,224 3 61 1,400 30 20 03 15 -18,627 4 48 1,400 60 30 83 40 -58,433 5 32 1,200 19 20 11 15 -20,523 6 31 1,200 88 40 30 20 13,597 7 22 1,200 78 30 41 20 -9,150 8 46 1,400 11 20 52 20 -31,896 9 57 1,400 20 20 15 15 -18,627 QUANTITY (Q) PRICE (P) VARIABLE COST (V) NPV Run Random

Number Corres-ponding Value

Random Number

Corres-ponding value

Random Number

Corres-ponding value

1.8955 Q(P-V)-31,895.5

10 92 1,800 77 30 38 20 2,224 11 25 1,200 65 30 36 20 -9,150 12 64 1,400 04 20 83 40 -84,970 13 14 1,000 51 30 72 20 -12,941 14 05 800 39 20 81 40 -62,224 15 07 800 90 50 40 20 13,597 16 34 1,200 63 30 67 20 -9,150 17 79 1,600 91 50 99 40 -1,568 18 55 1,400 54 30 64 20 -5,359 19 57 1,400 12 20 19 15 -18,627 20 53 1,400 78 30 22 15 7,910 21 36 1,200 79 30 96 40 -54,642 22 32 1,200 22 20 75 20 -31,896 23 49 1,400 93 50 88 40 -5,359 24 21 1,200 84 40 35 20 13,597 25 08 .800 70 30 27 15 -9,150 26 85 1,600 63 30 69 20 -1,568 27 61 1,400 68 30 16 15 7,910

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Problems’ Solutions Project Evaluation

28 25 1,200 81 40 39 20 13,597 29 51 1,400 76 30 38 20 -5,359 30 32 1,200 47 30 46 20 -9,150 31 52 1,400 61 30 58 20 -5,359 32 76 1,600 18 20 41 20 -31,896 33 43 1,400 04 20 49 20 -31,896 34 70 1,600 11 20 59 20 -31,896 35 67 1,400 35 20 26 15 -18,627 36 26 1,200 63 30 22 15 2,224 QUANTITY (Q) PRICE (P) VARIABLE COST (V) NPV Run Random

Number Corres-ponding Value

Random Number

Corres-ponding value

Random Number

Corres-ponding value

1.8955 Q(P-V)-31,895.5

37 89 1,600 86 40 59 20 28,761 38 94 1,800 00 20 25 15 -14,836 39 09 .800 15 20 29 15 -24,314 40 44 1,400 84 40 21 15 34,447 41 98 1,800 23 20 79 20 -31,896 42 10 1,000 53 30 77 20 -12,941 43 38 1,200 44 30 31 20 -9,150 44 83 1,600 30 20 10 15 -16,732 45 54 1,400 71 30 52 20 -5,359 46 16 1,000 70 30 19 15 -3,463 47 20 1,200 65 30 87 40 -54,642 48 61 1,400 61 30 70 20 -5,359 49 82 1,600 48 30 97 40 -62,224 50 90 1,800 50 30 43 20 2,224

Expected NPV = NPV 50 = 1/ 50 ∑ NPVi i=1 = 1/50 (-7,20,961) = 14,419 50 Variance of NPV = 1/50 ∑ (NPVi – NPV)2

i=1 = 1/50 [27,474.047 x 106] = 549.481 x 106

Standard deviation of NPV = 549.481 x 106

= 23,441 6. To carry out a sensitivity analysis, we have to define the range and the most likely values of

the variables in the NPV Model. These values are defined below Variable Range Most likely value I Rs.30,000 – Rs.30,000 Rs.30,000 k 10% - 10% 10% F Rs.3,000 – Rs.3,000 Rs.3,000 D Rs.2,000 – Rs.2,000 Rs.2,000 T 0.5 – 0.5 0.5 N 5 – 5 5 S 0 – 0 0

32 By M. Iftikhar Mubbashir

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Problems’ Solutions Project Evaluation

Q Can assume any one of the values - 1,400* 800, 1,000, 1,200, 1,400, 1,600 and 1,800 P Can assume any of the values 20, 30, 30** 40 and 50 V Can assume any one of the values 20* 15,20 and 40 ---------------------------------------------------------------------------------------- * The most likely values in the case of Q, P and V are the values that

have the highest probability associated with them ** In the case of price, 20 and 30 have the same probability of

occurrence viz., 0.4. We have chosen 30 as the most likely value because the expected value of the distribution is closer to 30

Sensitivity Analysis with Reference to Q The relationship between Q and NPV given the most likely values of other variables is given by 5 [Q (30-20) – 3,000 – 2,000] x 0.5 + 2,000 0 NPV = ∑ + - 30,000 t=1 (1.1)t (1.1)5

5 5Q - 500 = ∑ - 30,000 t=1 (1.1)t

The net present values for various values of Q are given in the following table: Q 800 1,000 1,200 1,400 1,600 1,800 NPV -16,732 -12,941 -9,150 -5,359 -1,568 2,224

Sensitivity analysis with reference to P The relationship between P and NPV, given the most likely values of other variables is

defined as follows: 5 [1,400 (P-20) – 3,000 – 2,000] x 0.5 + 2,000 0 NPV = ∑ + - 30,000 t=1 (1.1)t (1.1)5

5 700 P – 14,500 = ∑ - 30,000 t=1 (1.1)t

The net present values for various values of P are given below : P (Rs) 20 30 40 50 NPV(Rs) -31,896 -5,359 21,179 47,716 8. NPV - 5 0 5 10 15 20 (Rs.in lakhs) PI 0.9 1.00 1.10 1.20 1.30 1.40 Prob. 0.02 0.03 0.10 0.40 0.30 0.15

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Problems’ Solutions Project Evaluation

6 Expected PI = PI = ∑ (PI)j P j j=1 = 1.24 6 Standard deviation = ∑ (PIj - PI) 2 P j o f P1 j=1 = √ .01156 = .1075 The standard deviation of P1 is .1075 for the given investment with an expected PI of 1.24. The maximum standard deviation of PI acceptable to the company for an investment with an expected PI of 1.25 is 0.30. Since the risk associated with the investment is much less than the maximum

risk acceptable to the company for the given level of expected PI, the company should accept the investment. 9. Investment A Outlay : Rs.10,000 Net cash flow : Rs.3,000 for 6 years Required rate of return : 12% NPV(A) = 3,000 x PVIFA (12%, 6 years) – 10,000 = 3,000 x 4.11 – 10,000 = Rs.2,333

Investment B Outlay : Rs.30,000

Net cash flow : Rs.11,000 for 5 years Required rate of return : 14% NPV(B) = 11,000 x PVIFA (14%, 5 years) – 30,000 = Rs.7763 10. The NPVs of the two projects calculated at their risk adjusted discount rates are as follows: 6 3,000 Project A: NPV = ∑ - 10,000 = Rs.2,333 t=1 (1.12)t

5 11,000

Project B: NPV = ∑ - 30,000 = Rs.7,763 t=1 (1.14)t

PI and IRR for the two projects are as follows:

Project A B PI 1.23 1.26 IRR 20% 24.3% B is superior to A in terms of NPV, PI, and IRR. Hence the company must choose B.

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Problems’ Solutions Project Evaluation

Chapter 13 SPECIAL DECISION SITUATIONS

1. PV Cost UAE = PVIFAr,n Cost of plastic emulsion painting = Rs.3,00,000 Life = 7 years Cost of distemper painting = Rs. 1,80,000 Life = 3 years Discount rate = 10% UAE of plastic emulsion painting = Rs.3,00,000 / 4.868 = Rs.61,627 UAE of distemper painting = Rs.1,80,000 / 2.487 = Rs.72,376 Since plastic emulsion painting has a lower UAE, it is preferable. 2. Present value of the operating costs : 3,00,000 3,60,000 4,00,000 4,50,000 5,00,000 = + + + + 1.13 (1.13)2 (1.13)3 (1.13)4 (1.13)5

= Rs.1,372,013 Present value of salvage value = 3,00,000 / (1.13)5 = Rs.162,828 Present value of costs of internal transportation = 1,500,000 –1,372,013

system – 162,828 = Rs.27,09,185 UAE of the internal transportation system = 27,09,185 / 3.517 = Rs.7,70,311 3. Cost of standard overhaul = Rs.500,000 Cost of less costly overhaul = Rs.200,000 Cost of capital = 14% UAE of standard overhaul = 500,000 / 3.889 = Rs.128,568 UAE of less costly overhaul = 200,000 / 1.647 = Rs.121,433 Since the less costly overhaul has a lower UAE, it is the preferred alternative 4. The details for the two alternatives are shown below :

Gunning plow Counter plow

1. Initial outlay Rs.2,500,000 Rs.1,500,000 2. Economic life 12 years 9 years 3. Annual operating and maintenance costs Rs.250,000 Rs.320,000 4. Present value of the stream of operating

and maintenance costs at 12% discount rateRs.1,548,500 Rs.1,704,960

5. Salvage value Rs.800,000 Rs.500,000 6. Present value of salvage value Rs.205,600 Rs.180,500 7. Present value of total costs (1+4-6) Rs.3,842,900 Rs.3,024,460 8. UAE of 7 Rs.3,842,900

PVIFA (12%,12) = 3,842,900 6.194 = Rs.620,423

Rs.3,024,460 PVIFA (12%,9) = 3,024,460 5.328 = Rs.567,654

The Counter plow is a cheaper alternative

35 By M. Iftikhar Mubbashir

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Problems’ Solutions Project Evaluation

5. The current value of different timing options is given below :

Time Net Future Value Current Value

Rs. in million Rs. in million 0 10 10 1 15 13.395 2 19 15.143 3 23 16.376 4 26 16.536 The optimal timing of the project is year 4. 6. Calculation of UAE (OM) for Various Replacement Periods (Rupees)

Time (t)

Operating and

maintenance costs

Post-tax operating & maintenance

costs

PVIF (12%,t)

Present value of

(3)

Cumulative present value

PVIFA (12%,t)

UAE (OM)

(1) (2) (3) (4) (5) (6) (7) (8) 1 20,000 12,000 0.893 10,716 10,716 0.893 12,0002 25,000 15,000 0.797 11,955 22,671 1.690 13,4153 35,000 21,000 0.712 14,952 37,623 2.402 15,6634 50,000 30,000 0.636 19,080 56,703 3.037 18,6715 70,000 42,000 0.567 23,814 80,517 3.605 22,335

Calculation of UAE (IO) for Various Replacement Periods

Time (t) Investment Outlay Rs. PVIFA (12%, t) UAE of investment outlay Rs. 1 80,000 0.893 89,586 2 80,000 1.690 47,337 3 80,000 2.402 33,306 4 80,000 3.037 26,342 5 80,000 3.605 22,191

Calculation of UAE (DTS) for Various Replacement Periods

Time (t)

Depreciation charge R.s.

Depreciation tax shield

PVIF (12%, t)

PV of depreciation

tax shield Rs..

Cumulative present

value Rs..

PVIFA (12%, t)

UAE of depreciation tax shield Rs.

(1) (2) (3) (4) (5) (6) (7) (8) 1 20,000 8,000 0.893 7,144 7,144 0.893 8,000 2 15,000 6,000 0.797 4,782 11,926 1.690 7,057 3 11,250 4,500 0.712 3,204 15,130 2.402 6,299 4 8,438 3,375 0.636 2,147 17,277 3.037 5,689 5 6,328 2,531 0.567 1,435 18,712 3.605 5,191

Calculation of UAE (SV) for Various Replacement Periods

Time Salvage value Rs.

PVIF (12%, t)

Present value of salvage value Rs.

PVIFA (12%, t)

UAE of salvage value Rs. (4) / (5)

(1) (2) (3) (4) (5) (6) 1 60,000 0.893 53,580 0.893 60,000 2 45,000 0.797 35,865 1.690 21,222 3 32,000 0.712 22,784 2.402 9,485

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Problems’ Solutions Project Evaluation

4 22,000 0.636 13,992 3.037 4,607 5 15,000 0.567 8,505 3.605 2,359

Summary of Information Required to Determine the Economic Life

Replacement period

UAE (OM) Rs.

UAE (IO) Rs.

UAE (DTS) Rs.

UAE (SV) Rs.

UAE (CC) Rs.

UAE (TC) Rs.

(1) (2) (3) (4) (5) (6) (7) 1 12,000 89,586 8,000 60,000 21,586 33,586 2 13,415 47,337 7,057 21,222 19,058 32,473 3 15,663 33,306 6,299 9,485 17,522 33,185 4 18,671 26,342 5,689 4,607 16,046 34,717 5 22,335 22,191 5,190 2,359 14,642 36,977

OM - Operating and Maintenance Costs IO - Investment Outlay DTS - Depreciation Tax Shield SV - Salvage Value CC - Capital Cost TC - Total Cost UAE (CC) = UAE (IO) – [UAE (DTS) + UAE (SV)] UAE (TC) = UAE (OM) + UAE (CC)

7. Calculation of UAE (OM) for Various Replacement periods

Time O&M costs Rs.

Post-tax O&M costs

Rs.

PVIF (12%,t)

PV of post-tax O&M costs Rs.

Cumulative present

value Rs.

PVIFA (12%, t)

UAE of O&M

costs Rs. (1) (2) (3) (4) (5) (6) (7) (8) 1 800,000 560,000 0.893 500,080 500,080 0.893 560,000 2 1,000,000 700,000 0.797 557,900 1,057,980 1.690 626,024 3 1,300,000 910,000 0.712 647,920 1,705,9000 2.402 710,200 4 1,900,000 1,330,000 0.636 845,880 2,551,780 3.037 840,230 5 2,800,000 1,960,000 0.567 1,111,320 3,663,100 3.605 1,016,117

Calculation of UAE (IO) for Various Replacement Periods

Time Investment outlay Rs. PVIFA (12%, t) UAE of investment outlay Rs. 1 4,000,000 0.893 4,479,283 2 4,000,000 1.690 2,366,864 3 4,000,000 2.402 1,665,279 4 4,000,000 3.037 1,317,089 5 4,000,000 3.605 1,109,570

Calculation of UAE (DTS) for Various Replacement Periods

Time (t)

Depreciation charge Rs.

Depreciaton tax shield

Rs.

PVIF (12%, t)

PV of depreciation tax shield Rs.

Cumulative present

value Rs.

PVIFA (12%, t)

UAE of depreciation tax shield Rs.

1 1,000,000 300,000 0.893 267,940 267,900 0.893 300,000 2 750,000 225,000 0.797 179,325 447,225 1.690 264,630 3 562,500 168,750 0.712 120,150 567,375 2.402 236,209 4 421,875 126,563 0.636 80,494 647,869 3.037 213,325 5 316,406 94,922 0.567 53,821 701,690 3.605 194,643

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Problems’ Solutions Project Evaluation

Calculation of UAE (SV) for Various Replacement Periods Time Salvage

value Rs. PVIF

(12%, t) Present value of salvage value Rs.

PVIFA (12%, t)

UAE of salvage value Rs. (4)/ (5)

(1) (2) (3) (4) (5) (6) 1 2,800,000 0.893 267,900 0.893 2,800,000 2 2,000,000 0.797 1,594,000 1.690 943,195 3 1,400,000 0.712 996,80 2.402 414,988 4 1,000,000 0.636 636,000 3.037 209,417 5 800,000 0.567 453,600 3.605 125,825

Summary of Information Required Determining the Economic Life

Replacement period

UAE (OM) Rs.

UAE (IO) Rs.

UAE (DTS)

Rs.

UAE (SV)

Rs.

UAE (CC)

Rs.

UAE (TC)

Rs. 1 560,000 4,479,283 300,000 2,800,000 (-)1,379,283 -819,283 2 626,024 2,366,864 264,630 943,195 1,159,039 1,785,063 3 710,200 1,665,279 236,209 414,988 1,014,082 1,724,282 4 840,230 1,317,089 213,325 209,417 894,347 1,734,577 5 1,016,117 1,109,570 194,643 125,825 789,102 1,805,219

The economic life of the well-drilling machine is 3 years 8. Adjusted cost of capital as per Modigliani – Miller formula: r* = r (1 – TL) r* = 0.16 (1 – 0.5 x 0.6) = 0.16 x 0.7 = 0.112 Adjusted cost of capital as per Miles – Ezzell formula: 1 + r

r* = r – LrDT 1 + rD

1 + 0.16 = 0.16 – 0.6 x 0.15 x 0.5 x 1 + 0.15 = 0.115 9. a. Base case NPV = -12,000,000 + 3,000,000 x PVIFA (20%, b) = -12,000,000 + 3,000,000 x 3,326 = - Rs.2,022,000 b. Adjusted NPV = Base case NPV – Issue cost + Present value of tax shield. Term loan = Rs.8 million Equity finance = Rs.4 million Issue cost of equity = 12% Rs.4,000,000 Equity to be issued = = Rs.4,545,455 0.88 Cost of equity issue = Rs.545,455

38 By M. Iftikhar Mubbashir

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Problems’ Solutions Project Evaluation

Computation of Tax Shield Associated with Debt Finance Year (t) Debt outstanding

at the beginning Rs.

Interest

Rs.

Tax shield

Rs.

Present value of tax shield

Rs. 1 8,000,000 1,440,000 432,000 366,102 2 8,000,000 1,440,000 432,000 310,256 3 7,000,000 1,260,000 378,000 230,062 4 6,000,000 1,080,000 324,000 167,116 5 5,000,000 900,000 270,000 118,019 6 4,000,000 720,000 216,000 80,013 1,271,568

Adjusted NPV = - Rs.2,022,000 – Rs.545,455 + Rs.1,271,568 = - Rs.1,295,887 Adjusted NPV if issue cost alone is considered = Rs.2,567,455 Present Value of tax shield of debt finance = Rs.1,271,568 10. a. Base Case NPV = - 8,000,000 + 2,000,000 x PVIFA (18%, 6) = - 8,000,000 + 2,000,000 x 3,498 = - Rs.1,004,000 b. Adjusted NPV = Base case NPV – Issue cost + Present value of tax shield. Term loan = Rs.5 million Equity finance = Rs.3 million Issue cost of equity = 10% Rs.3,000,000

Hence, Equity to be issued = = Rs.3,333,333 0.90 Cost of equity issue = Rs.333,333 Computation of Tax Shield Associated with Debt Finance

Year Debt outstanding at the beginning

Interest

Tax shield

Present value of tax shield

1 Rs.5,000,000 Rs.750,000 Rs.300,000 Rs.260,869 2 5,000,000 750,000 300,000 226,843 3 4,000,000 600,000 240,000 157,804 4 3,00,000 450,000 180,000 102,916 5 2,000,000 300,000 120,000 59,66 6 1,000,000 150,000 60,000 25,940 843,033

Adjusted NPV = - 1004000 – 333333 + 834033 = - Rs.503,300 Adjusted NPV if issue cost of external equity alone is adjusted for = - Rs.1,004000 – Rs.333333 = Rs.1337333 c. Present value of tax shield of debt finance = Rs.834,033 11. Adjusted cost of capital as per Modigliani – Miller formula: r* = r (1 – TL) r* = 0.19 x (1 – 0.5 x 0.5) = 0.1425 = 14.25%

39 By M. Iftikhar Mubbashir

Page 40: Project evaluation and feasibility analysis

Problems’ Solutions Project Evaluation

Adjusted cost of capital as per Miles and Ezzell formula: 1 + r

r* = r – LrDT 1 + rD

1 + 0.19 = 0.19 – 0.5 x 0.16 x 0.5 x 1 + 0.16 = 0.149 = 14.9% 12. S0 = Rs.46 , rh = 11 per cent , rf = 6 per cent Hence the forecasted spot rates are : Year Forecasted spot exchange rate 1 Rs.46 (1.11 / 1.06)1 = Rs.48.17

2 Rs.46 (1.11 / 1.06)2 = Rs.50.44 3 Rs.46 (1.11 / 1.06)3 = Rs.52.82 4 Rs.46 (1.11 / 1.06)4 = Rs.55.31 5 Rs.46 (1.11 / 1.06)5 = Rs.57.92 The expected rupee cash flows for the project

Year Cash flow in dollars Expected exchange Cash flow in rupees (million) rate (million) 0 -200 46 -9200 1 50 48.17 2408.5 2 70 50.44 3530.8 3 90 52.82 4753.8 4 105 55.31 5807.6 5 80 57.92 4633.6

Given a rupee discount rate of 20 per cent, the NPV in rupees is :

2408.5 3530.8 4753.8 NPV = -9200 + + + (1.18)2 (1.18)3 (1.18)4

5807.6 4633.6 + + (1.18)5 (1.18)6

= Rs.3406.2 million

The dollar NPV is : 3406.2 / 46 = 74.05 million dollars

40 By M. Iftikhar Mubbashir

Page 41: Project evaluation and feasibility analysis

Problems’ Solutions Project Evaluation

Chapter 14 SOCIAL COST BENEFIT ANALYSIS

1. Social Costs and Benefits Nature Economic

value (Rs in million)

Explanation

Costs 1. Construction cost One-

shot 400

2. Maintenance cost Annual 3 Benefits

3. Savings in the operation cost of existing ships

Annual 40

4. Increase in consumer satisfaction

Annual 3.6 The number of passenger hours saved will be : (75,000 x 2 + 50,000 + 50,000 x 2) = 600000. Multiplying this by Rs.6 gives Rs.3.6 million

The IRR of the stream of social costs and benefits is the value of r in the

equation 50 40 + 3.6 – 3.0 50 40.6 400 = Σ = Σ

t=1 (1+r)t t=1 (1+r)t

The solving value r is about 10.1% 2. Social Costs and Benefits Costs Decrease in customer satisfaction as reflected Rs.266,667

in the opportunity cost of the extra time taken by bus journey 800 x (2/3) x 250 x Rs.2

Benefits 1. Resale value of the diesel train (one time) Rs.240,000 2. Avoidance of annual cash loss Rs.400,000

Fare collection = 1000 x 250 x Rs.4 = Rs.1,000,000 Cash operating expenses = Rs.1,400,000

3. The social costs and benefits of the project are estimated below: (Rs. in million) Costs & Benefits Time Economic

value Explanation

1. Construction cost 0 24 2. Land development cost 0 150 3. Maintenance cost 1-40 1 4. Labour cost 0 40 This includes the cost of

transport and rehabilitation 5. Labour cost 1-40 12 The shadow price of labour

equals what others are willing to pay.

6. Decrease in the value of the timber 2-40 4

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Problems’ Solutions Project Evaluation

output Benefits

7. Savings in the cost of shipping the agriculture produce

1-40 0.5

8. Income from cash crops 1-5 10 9. Income from the main crop 6-40 50 10. Increase in the value of timber output 1 20

Assuming that the life of the road is 40 years, the NPV of the stream of social costs and benefits at a discount rate of 10 percent is:

40 1 + 12 40 4 NPV = - 24 - 150 - 40 - Σ - Σ t=1 (1.1)t t=2 (1.1)t

40 0.5 5 10 40 50 20 + Σ + Σ + Σ + t=1 (1.1)t t=1 (1.1)t t=6 (1.1)t (1.1)1

= - Rs.9.93 million 4. Table 1 Social Costs Associated with the Initial Outlay Rs. in million

Item Financial cost

Basis of conversion

Tradeable value ab initio

T L R

Land 0.30 SCF = 1/1.5 0.20 Buildings 12.0 T=0.50, L=0.25

R=0.25 6.0 3.0 3.0

Imported equipment 15.0 CIF value 9.0 Indigeneous equipment 80.0 CIF value 60.0 Transport 2.0 T=0.65, L=0.25

R=0.10 1.3 0.5 0.2

Engineering and know-how fees

6.0 SCF=1.5 9.0

Pre-operative expenses 6.0 SCF=1.0 6.0 Bank charges 3.7 SCF=0.02 0.074 Working capital requirement

25.0 SCF=0.8 20.0

150.0 104.274 7.3 3.5 3.2 Table 2 Conversion of Financial Costs into Social Costs (Rs. in million)

Item Financial cost

Basis of conversion

Tradeable value ab initio

T L R

Indigeneous raw material and stores

85 SCF=0.8 68

Labour 7 SCF=0.5 3.5 Salaries 5 SCF=0.8 4.0 Repairs and maintenance 1.2 SCF=1/1.5 0.8 Water, fuel, etc 6 T=0.5, L=0.25

R=0.25 3 1.5 1.5

Electricity (Rate portion) 5 T=0.71, L=0.13 R=0.16

3.55 0.65 0.8

Other overheads 10 SCF=1/1.5 6.667 119.2 82.967 6.55 2.15 2.3

42 By M. Iftikhar Mubbashir

Page 43: Project evaluation and feasibility analysis

Problems’ Solutions Project Evaluation

As per table 1, the social cost of initial outlay is worked out as follows : Rs. in million Tradeable value ab initio 104.274 Social cost of the tradeable component 4.867 (7.3 / 1.5) Social cost of labour component 1.75 (3.5 x 0.5) Social cost of residual component 1.60 (3.2 x 0.5) Total 112.491

As per Table 2, the annual social cost of operation is worked out as follows :

Tradeable value ab initio 82.967 Social cost of the tradeable component 4.367 ( 6.55 x 1/1.5 ) Social cost of labor component 1.075 (2.15 x 0.5) Social cost of residual component 1.150 (2.3 x 0.5) Total 89.559

The annual CIF value of the output is Rs.110 million. Hence the annual social

net benefit will be : 110 – 89.559 = Rs.20.441 million Working capital recovery will be Rs.20 million at the end of the 20th year. Putting the above figures together the social flows associated with the project

would be as follows : Year / ’s Social flow (Rs. in million) 0 -112.491 1-19 20.441

43 By M. Iftikhar Mubbashir