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STEPS INVOLVED IN CALCULATION OF REPLACEMENT POUCY WHEN TIME VALUE CHANGES

STEPS INVOLVED IN CALCULATION OF REPLACEMENT POUCY WHEN TIME VALUE CHANGES

Find out the present value factor at the given rate and multiply it with the
operating/maintenance cost of the equipment/items for different years.

Step I. Find out the present value factor at the given rate and multiply it with the operating/maintenance cost of the equipment/items for different years.

Step II. Work out the total cost by adding the cumulative present value to the original cost for all the year

Step III. Cumulate the discount factors.

Step IV. Divide the total cost by corresponding value of the cumulated discount factor for every.

Step V. Find out the value of last column that exceeds the total cost. Equipment/item will be replaced in the latest year.

These steps will be explained with the help of an example.

Example 8.8 The yearly cost of two machines X and Y, when money value is neglected is shown below. Find which machines is more economical if money value is 10% per year.”

Year

1

2

3

Machine X (Rs)

2400

1600

1800

Machine Y (Rs)

3200

800

1800

 

Sol. It may be seen that the total cost for each machine X and Y is Rs 5800 (2400 + 1600 + 1800) or (3200 + 800 + 1800). When the money value is not discounted the machines are equally good, total cost wise, when money value is not changed with time, with money value 10% per year, the discount rate, it changes as follows.

12=0.9091 “>

Discounted costs are obtained by multiplying the original costs with O· 9091 after one year. Total costs, of machines X and Y are calculated as shown below.

Year

1

2

3

Total cost (Rs)

Machine X

2400

1600×0.9091= 1440

1800×0.9081 =1620

5460

Machine Y

3200

800×0.9091 =720

1620

5540

 

The total cost of machine X is less than that of machine Y, machine X is more economical.

Example 8.9 The cost of a new machine is Rs 5000 the maintenance cost during nth year is given by Mn = Rs 500 (n – 1) where n = 1, 2, 3,…. If the discount rate per year is 0.05, after how many years will it be economical to replace the machine by a new one?

Sol. The discounted rate is given as 0·05 i.e. 5% then the present value. 

12=0.9523 after one year. “>

After 2 years it will be (.9523)2 and so on.

Year

Maintenance Cost

Discounted Cost

Discounted Maintenance Cost

Cumulative Total Discounted Cost

Average Total Cost

1

2

3

4

5

6

0

500

1000

1500

2000

2500

1.0

0.9523

0.9070

0.8638

0.8227

0.7835

0

467

907

1296

1645

1959

5000

5476

6383

7679

9324

11283

5000

2738

2127

1919

1865

1880

 

From above it is clear that it will be economical to replace the machine after 5th year.

Example 8.10 The original cost of the machine is Rs 10,000. Maintenance costs vary as given below.

Year 1 2 3 4 5 6 7
Maintenance Cost 500 800 1200 1500 2000 2500 3000

If the money is discounted at 10% per year, what is the optimum replacement policy?

Year

Maintenance Cost OC

1

V at 10%

2

PV of OC

3

Cumulative PV of OC

4

C-S  (t)

5

TC(n) 6=4+5

 

Cumulative value of V

Weighted Average

1

500

1

500

500

10000

10500

1

10500

2

800

.9523

761.6

1261.6

10000

11261.6

1.25

3

1200

.9070

1088.4

2345.6

10000

123456

2.85

4

1500

.8638

1294.5

3640

10000

13640

3.71

5

2000

.8227

164.5

5258

10000

15285

4.53

6

2500

.7835

195.8

7243

10000

17243

5.35

3223

7

30000

0.7325

2197.5

9440.5

10000

19440

6.05

3213

 

Since the average cost is less in the 7th year, optimum replacement policy is at the end at 7th year.

Example 8.11 An engineering company is offered a material handling equipment A. A is purchased for Rs. 69,000 originally and maintenance costs are estimated to be Rs 10,000 for each of the first five years and increasing every year by Rs 3000 from the sixth and subsequent years. The company expects a return if 10% all of its investments. What is the optimum replacement period? Assume that maintenance cost is increased at tile end of the year.

Sol. Here C = 6,0000

r= 10% 

Let us work out the weighted cost with the help of following table.

Year

Operating Cost OC

1

Discounted factor

 

2

PV of OC

3

Total cost Cumulative

4

Cumulative PV factor

5

Weighted Average Cost 6=4+5

1

2

3

4

5

6

7

8

9

10

10000

10000

10000

10000

10000

13000

16000

19000

22000

25000

0.909

0.826

0.751

0.683

0.621

0.564

0.513

0.466

0.424

0.385

90090

8260

7510

6830

6210

7332

8208

8854

9328

9625

69090

77350

84860

91690

97900

105232

113440

122294

131622

141247

1.91

2.73

3.48

4.16

4.79

5.35

5.86

6.33

6.75

7.14

36192

24282

24343

21993

20438

19655

19335

19311

19479

19777

 

It can be seen above that the weighted average cost is minimum at the end of 8 years. Hence optimum replacement period is 8 years.

Example 8.12. A manufacturer is offered two machines X and Y. Machine X is priced at Rs 10.000 with running cost of Rs 1000 for first four years and increasing by 400 in fifth year and subsequent years Machine Y which has the same capacity and performance as X costs Rs 8000 but has maintenance cost of Rs 1200 per year for first five years increasing by Rs 400 in the sixth and subsequent years. If cost of money is 10% per year, which is a more economical machine? Assume running cost is incurred at the beginning of the year. 

Machine X, C=10,000

Year

OC

PV factors

PV of OC

C+ Cumulative PV of OC

Cumulative PV factor

Weighted Average Cost

1

2

3

4

5

6

7

8

9

10

1000

1000

1000

1000

1400

1800

2200

2600

3000

3400

1.00

0.909

0.836

0.751

0.683

0.621

0.564

0.513

0.466

0.424

1000

909

826

751

956

1116

1240

1326

1398

1429

11000

11909

12735

13486

14442

15558

16798

18124

19532

21951

1.00

1.909

2.735

3.486

4.169

4.790

5.355

5.868

6.334

6.759

11000

5984.5

4656.30

3868.6

3464

3248

3137

30886

3084

3247

 

Machine Y, C=8000

Year

OC

PV factors

PV of OC

C+ Cumulative PV of OC

Cumulative PV factor

Weighted Average Cost

1

2

3

4

5

6

7

8

9

10

1200

1200

1200

1200

1200

1600

2000

2400

2800

3200

1.00

0.909

0.826

0.751

0.683

0.621

0.564

0.513

0.466

0.424

1200

1090.8

991.2

901.2

819.6

993.6

1128

1231.8

1304.8

1356.8

9200

10290.8

11282

12183.2

13002.8

13996.4

15124.4

16355.6

17660.4

19017.2

1.00

1.909

2.735

3.486

4.169

4.790

5.355

5.868

6.334

6.759

9200

5390.67

4125

3498.8

3119

2922

2824.35

2787.25

2788.2

2813.6

 

It can be seen that weighted average cost of machine X is minimum i.e. Rs 3084 in 9th year: Where as the weighted average cost of machine Y is minimum in 8th year i.e. 2787·25 so it is advisable to purchase

Example 8.13 An executive has the option to buy car A or car B. Car A costs 6,50,000 and its running and maintenance cost is Rs 60,000 for each of the first five years increasing by Rs 20000 per year from 6th year. Car B is considered as good as car A but costs only 5,85,000. However, its running and maintenance cost is Rs 1,00,000 or each of the first 5 years and increases by 20,000 per year thereafter. If
the money is worth 9% per year which car should be purchased? What is the optimal replacement period for each car assuming that there is no salvage value for either of the cars?

Sol. We have to make choice between car A and car B hence we should, work out the average weighted cost for both cars 

Car A, cost C = 6.50,000

 

Year

OC

PV factors

PV of OC

C+ Cumulative PV of OC

Cumulative PV factor

Weighted Average Cost

1

2

3

4

5

6

7

8

9

10

60000

60000

60000

60000

60000

80000

100000

120000

140000

160000

1.000

0.917

0.841

0.772

0.708

0.650

0.596

0.546

0.501

0.459

60000

55020

50460

46320

42480

52000

59600

65520

70140

73440

710000

765020

815480

861800

904280

956280

1015880

1081400

1151540

1224980

1

1.917

2.758

3.530

4.238

4.888

4.484

6.03

6.531

6.99

71000

399071.46

295678

244136

213374

1956388

185244.3

179336.6

176319

175475

 

 

 

Car B, Cost C=5.85000

 

Year

OC

PV factors

PV of OC

C+ Cumulative PV of OC

Cumulative PV factor

Weighted Average Cost

1

2

3

4

5

6

7

8

9

10

10000

10000

10000

10000

10000

120000

140000

160000

180000

200000

1.000

0.917

0.841

0.772

0.708

0.650

0.596

0.546

0.501

0.459

100000

91700

84100

77200

70800

184615.3

234899.3

293040.3

359281.4

435729.6

685000

776700

860800

938000

1008800

1193415.3

1428314.6

1721354.9

2080636.3

2516366

1

1.917

2.758

3.530

4.238

4.888

5.484

6.03

6.531

6.99

685000

405164.3

312110.2

265722.3

238036.8

244152

260451.2

285465.1

318578.5

359995.1

 

 

 

It is obvious by comparing the weighted costs of two cars that Car A is far better. Car A should be replaced after 10 years and car B should be replaced after 9 years.