FUTURE WORTH METHOD

PROBLEMS

PROBLEM 1• A suburban taxi company is considering buying taxies with diesel

engines instead of petrol engines. The cars average usage is 50,000 km a year with a useful life of three years for the taxi with a petrol engine and four years for the diesel engine. Other information are as follows. Using a 18% interest rate determine which alternative should be selected based on future worth method of comparison.

Information Diesel taxi Petrol taxiVehicle cost Rs.5,00,000 Rs.4,00,000

Fuel cost /litre Rs. 9.00 Rs. 24.00

Mileage in km/litre 30 30

Annual insurance premium Rs.500 Rs.500

Salvage value at the end of vehicle life

Rs. 70,000 Rs. 1,00,000

SOLUTION• Alternative 1 :

Vehicle cost : Rs. 5,00,000Life : 4 yearsNumber of litres/year : 50,000/30 = 1666.67 litFuel cost/year : 1666.67 x 9 = Rs. 15,000.03Fuel cost + Premium/year = 15000.03 + 500 = Rs.15,500.03Salvage value = Rs . 70,000

SOLUTION

70,000

5,00,000

15,500.03 15,500.03 15,500.03

0 1 2 3 4

15,500.03

SOLUTION• FW = 5,00,000(F/P,18%,4) + 15,000.03(F/A,18%,4)-70,000

5,00,000 X 1.939 + 15000.03X 5.215 – 70000 = Rs. 9,80,332.33/-

Alternative 2 :Vehicle cost : Rs. 4,00,000

Life : 3 yearsNumber of litres/year : 50,000/20 = 2500 litFuel cost/year : 2500 x 24 = Rs. 60,000Fuel cost + Premium/year = 60,000 + 500 = Rs.60,500Salvage value = Rs . 1,00,000

SOLUTION

1,00,000

4,00,000

60500 60,500

0 1 2 3

60500

SOLUTION• FW = 4,00,000(F/P,18%,3) + 60500(F/A,18%,3)-1,00,000

4,00,000 X 1.643 + 60500X 3.572 – 1,00,000 = Rs. 7,73,306/-• The future worth of alternative 2 is < alternative 1 so alternative 2 is

the best choice.

PROBLEM 2• A Company must decide to buy machine a or machine b. At 15% rate

of interest which machine should be selected based on future method of comparison

Information Machine A Machine B

Initial cost 4,00,000 8,00,000

Useful life 5 5

Salvage value 2,00,000 5,50,000

Annual maintenance cost 40,000 0

SOLUTION

2,00,000

4,00,000

4000040000 40000 40000

0 1 2 3 4 5

MACHINE A

40000

SOLUTION• Fw = 4,00,000 (F/P,15%,5) + 40000(F/A,15%,5)-200000• = 4,00,000 * 2.011 + 40000 * 6.742 – 200000 = Rs. 8,74,080/-

MACHINE B :

5,50,000

8,00,000

0 1 2 3 4 5

SOLUTION• Fw = 8,00,000 (F/P,15%,5) -5,50,000• = 8,00,000 * 2.011– 550000 = Rs. 10,58,800/-• FW machine A is < FW of machine B. so machine A is selected.

PROBLEM 3

1,00,000

5,00,000

50,000

1,50,0002,00,000

0 1 2 3 4

2,50,0003,00,000

• For the given diagram determine the future worth assuming interest rate of 9% compounded annually and comment

5 6

SOLUTION• P = Rs. 5,00,000/-• A1 = Rs. 50,000/-• G = Rs. 50,000/- , n = 6 years, i = 9%

• FW = -P (F/P,9%,6) + [A1 + G (A/G,i,n)] (F/A,9%,6) = -50,000 × 1.677 + [50,000,+ 50,000 (2.2498)] × 7.523 = Rs. 3,83,912.27/-

PROBLEMS ON ANNUAL EQUIVALENT METHOD

• A company has three proposals for expanding its business operations. The details are as follows Assuming an interest rate of 15% compounded annually determine the best alternative using the annual equivalent method.

Alternative Initial cost Annual revenue Life

1 25,00,000 8,00,000 10

2 20,00,000 6,00,000 10

3 30,00,000 10,00,000 10

SOLUTION• Alternative 1 :

25,00,000

8,00,000

0 1 2 3 4 10

8,00,000 8,00,000 8,00,000 8,00,000

SOLUTION• AE = PW(i) x (A/P,i,n)• AE = [- P + A (P/A,i,n)+S (P/F,i,n)] x (A/P,i,n)• AE = -P(A/P,i,n) + A +S(A/F,i,N) • AE = - 25,00,000 (0.1993) + 8,00,000 = Rs. 3,01,750/-

• Alternative 2 :

20,00,000

6,00,000

0 1 2 3 4 10

6,00,000 6,00,000 6,00,000 6,00,000

SOLUTION• AE = -P(A/P,i,n) + A + S(A/F,i,N)• AE = - 20,00,000 (0.1993) + 6,00,000 = Rs. 2,01,400/-• Alternative 3 :

30,00,000

10,00,000

0 1 2 3 4 10

10,00,000 10,00,000 10,00,000 10,00,000

AE = -P(A/P,i,n) + A + S(A/F,i,N)AE = - 30,00,000 (0.1993) + 10,00,000 = Rs. 15,97,900/-Alternative 3 is > 2 & 1. So alternative 3 is the best choice

PROBLEM 2• Two possible routes for laying a power line are under study. Data on

the routes are as follows, if interest rate of 15% is used should the power line be routed around the lake or under the lake.

Description Around the lake Under the lake

Length 15km 5km

Fixed cost (rs.) 1,50,000/km 7,50,000/km

Useful life (years) 15 15

Maintenance cost (Rs.) 6000/km/yr 12,000/km/yr

Salvage value (Rs.) 90,000/km 1,50,000/km

Yearly power loss (Rs.) 15,000/km 15,000/km

SOLUTION• Alternative 1:First cost = 150000 × 15 = Rs. 22,50,000Maintenance cost/yr = 6000 × 15 = Rs. 90,000Power Loss / Yr = 15,000 × 15 = Rs. 2,25,000Maintenance cost + Power Loss cost = 90000+225000 = 315000Salvage value = 90000 × 15 = Rs. 13,50,000

22,50,000315000

0 1 2 3 4 15

315000 315000 315000 315000

1350000

SOLUTION • AE = P(A/P,i,n) + A – S (A/F,i,n) = 22,50,000 (A/P,15%,15) + 315000 – 1350000 (A/F,15%,15) = 22,50,000 × 0.1710 + 315000 – 1350000 × 0.0210 = Rs. 6,71,400/- Alternative 2:First cost = 750000 × 5 = Rs. 37,50,000Maintenance cost/yr = 12000 × 5 = Rs. 60,000Power Loss / Yr = 15,000 × 5 = Rs. 75,000Maintenance cost + Power Loss cost = 60000+75000 = 135000Salvage value = 150000 × 5 = Rs. 7,50,000

SOLUTION

• AE = P(A/P,i,n) + A – S (A/F,i,n) = 37,50,000 (A/P,15%,15) + 135000 – 750000 (A/F,15%,15) = 37,50,000 × 0.1710 + 135000 – 750000 × 0.0210 = Rs. 7,60,500/-

Since this is cost dominating problem alternative 1 < alternative 2 Hence alternative 1 is chosen therefore select the route around the lake for laying the power line.

37,50,000 135000

0 1 2 3 4 15

135000 135000 135000

750000

135000

PROBLEM 3• A transport company has been looking for a new tyre for its truck and

has located the following alternatives. If the company feels that the warranty period is a good estimate of the tyre life and that a nominal interest rate of 12% is compounded annually which tyre should it buy

Brand Tyre Warranty (months) Price per tyre (Rs)

A 12 1,200

B 24 1,800

C 36 2,100

D 48 2,700

SOLUTION• Tyre A

1200

0 12

AE = P(A/P,i,n) + A – S (A/F,i,n)AE = 1200 (A/P,12%,1)AE = 1200 *1.1200 = Rs. 1344• Tyre B

1800

0 24

SOLUTION• AE = P(A/P,i,n) + A – S (A/F,i,n)• AE = 1800 (A/P,12%,2)• AE = 1800 *0.5917 = Rs. 1065.06/-Tyre C :

2100

0 36

AE = P(A/P,i,n) + A – S (A/F,i,n)AE = 2100 (A/P,12%,3)AE = 2100 *0.4163 = Rs. 874.23/-

SOLUTION

• Tyre D :

2700

0 48

AE = P(A/P,i,n) + A – S (A/F,i,n)AE = 2700 (A/P,12%,4)AE = 2700 *0.3292 = Rs. 888.84/-Since the AE value of tyre C is < A,B,D tyre C is selected as the best alternative

RATE OF RETURN METHOD

1,00,000

30,000

0 1 2 3 4 5

SOLUTION• PW(i) = -P + A(P/A,i,n)• PW(10%) = -100000+ 30,000(P/A,10%,5)

-100000 + 30,000(3.7908)• PW(10%) = Rs. 13,724• PW(15%) = -100000 + 30,000(3.3522)

Rs. 566PW(18%) = -100000 + 30,000(3.1272) = Rs. -6184/-

i = 15% + × 3% = 15.252%

PROBLEM 2• A Company is planning to expand its present business activity. It has

two alternatives for the expansion programme and the corresponding cash flows are tabulated below. Each alternative has a life of 5 years and a negligible salvage value. The minimum attractive rate of return for the company is 12%. Suggest the best alternative to the company

Alternative Initial Investment Yearly revenue

Alternative 1 5,00,000 1,70,000Alternative 2 8,00,000 2,70,000

SOLUTION

5,00,000

1,70,000

0 1 2 3 4 5

SOLUTION• PW(17%) = -500000 + 1,70,000 (3.1993) = Rs. 43,881• PW(20%) = -500000 + 1,70,000 (2.9906) = Rs. 8402• PW(22%) = -500000 + 1,70,000 (2.8636) = Rs. -13188• i = 20% + × 2% = 20.78%• Alternaitve 2• PW(20%) = -800000 + 2,70,000 (2.9906) = Rs. 7462• PW(22%) = -800000 + 2,70,000 (2.8636) = Rs. - 26828• i = 20% + × 2% = 20.43%•Alternative 1 > Alternative 2