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Replacement of items which deteriorate with time without considering the change in money value

Replacement of items which deteriorate with time without considering the change in money value

Most of the machinery and equipment having moving parts deteriorate in their performance with passage of time. The cost of maintenance and repair keeps increasing with passage of time and a stage may reach when it is more economical (in overall analysis) to replace the item with a new one. For example, a passenger car is bound to wear out with time and its repair and maintenance cost may go to such level that the owner has to replace it with a new one.

Let C = Capital cost of the item

S (t)=Scrap value of the item after t years of use,

O (t) = Operating and maintenance cost of the equipment at time t.

n = number of years the item can be used.

TC (n) = Total cost of using the equipment for n years.

Time ‘t’ in this case is a discrete variable.

In this case as long as the average TC (n) is minimum, the equipment can remain in use for that number of years. If average total cost keeps decreasing up to ith year and starts increasing from (i+1)th year then ith year may be considered as most economic year for replacement of the equipment.

The: concept-of depreciation cost also must be understood here. As the years pass by, the cost of the equipment or items keeps decreasing. How much the cost keeps decreasing can be calculated by two methods commonly used i.e. straight line depreciation method and the diminishing value method.

Example 8.1. A JCB excavator operator purchases the machine for Rs 15,00,000. The Operating cost and the resale value of the machine is given below.

Year

1

2

3

4

5

6

7

8

Operating Cost in Rs.

30000

32000

36000

40000

45000

52000

60000

70000

Resale value (in lakhs of Rs.)

12

10

8

5

4.5

4

3

2

When should the machine be replaced?

Sol.

C=15,00,000

O (t) = Operating cost

S (t) = Resale value

t=Time

n = Number of years after when the machine is to replaced.

Let us draw a table showing the various variables required to make decision. This is shown in the table

Year

O (t) (in thousand of rupees)

Comulative O (t)

Resale value S (t) in thousands of rupees

Depreciation C-S (t) in thousands of Rupees

Total cost TC (n) thousands of Rupees

Average TC (n) Thousands of Rupees

1

30

30

1200

300

330

330

2

32

62

1000

500

562

281

3

36

98

800

700

798

266

4

40

138

500

1000

1138

284.5

5

45

183

450

1050

1233

246.6

6

52

235

400

1100

1335

222.5

7

60

295

300

1200

1495

213.6

8

70

365

200

1300

1665

208

 

In 3rd year the minimum average cost is 2,66,000 as shown in the table above, So replacement should take place at the end of 3rd year.

Example 8.2 A taxi owner estimates from his past records that the cost per year for operating a taxi whose purchase price when new is Rs 60,000 are as given below.

Age 1 2 3 4 5
Operating Cost (Rs.) 10000 12000 15000 18000 20000

After 5 years, the operating cost if Rs 6000k where k = 6, 7, 8, 9, 10 (k denoting age in years). If the resale value decreases 10% of purchase price per year, what is the best replacement policy? Cost of money is zero.

Sol. Since the depreciation of the taxi is 10% of Rs 60,000, it means the resale value of the taxi decreases by Rs 6000 every year. Average annual cost can be calculated as shown in the table below.

Year of Service

Operating Cost O (t)

Cumulative O (t)

Resale value S (t)

Depreciation C-S (t)

Total cost TC (n)

Average TC (n)

1

10000

10000

54000

6000

16000

16000

2

12000

22000

48000

12000

34000

17000

3

15000

37000

42000

18000

55000

18333

4

18000

55000

36000

24000

79000

19750

5

20000

75000

30000

30000

105000

21000

6

360000

111000

24000

36000

147000

24500

7

42000

153000

18000

42000

195000

27857

8

48000

201000

12000

48000

249000

31125

9

54000

255000

6000

54000

309000

34333

10

60000

315000

0

60000

375000

37500

 

Total cost TC (n) = operating cost O (t) + Depreciation C – S (t)

It can be seen that average Total cost is minimum after one year. Hence the taxi should be replaced after one year.

Example 8.3 A truck owner finds from his past records that the manufacturing cost of a truck (whose purchase is Rs 3,00,000) during the first 8 years of its life and the resale price at the end of each year is as follows.

Year

Maintenance Cost (Rs.)

Resale Price (Rs.)

1

36000

200000

2

48000

150000

3

60000

100000

4

72000

80000

5

84000

70000

6

96000

60000

7

108000

50000

8

120000

40000

 

Sol. The average cost per year is calculated in the following table.

Year of Service

Operating Cost O (t)

Cumulative O (t)

Resale value S (t)

Depreciation C-S (t)

Total cost TC (n)

Average TC (n)

1

36000

36000

200000

100000

136000

136000

2

48000

84000

150000

150000

234000

117000

3

60000

144000

100000

200000

344000

114667

4

72000

216000

80000

220000

436000

109000

5

84000

300000

70000

230000

530000

106000

6

96000

396000

60000

240000

636000

106000

7

108000

504000

50000

250000

754000

107714

8

120000

624000

40000

260000

884000

110500

 

Since average total cost is minimum in the 5th & 6th year, the truck should be replaced at the end of six years. He gets no advantage by replacing it at the end of 5 years.

Example 8.4. A new tempo costs Rs 1,00,000 and may be sold at the end of year at the following prices.

Year 1 2 3 4 5 6
Selling Price (Rs.) 60000 45000 32000 22000 10000 2000

The corresponding annual operating costs are

Year 1 2 3 4 5 6
Cost / Year (Rs.) 10000 12000 15000 20000 30000 45000

It is not only possible to sell the tempo after use but also to buy a second hand tempo. It may be cheaper to do so than to buy a new tempo.

Age of tempo 0 1 2 3 4 5
Purchase Price (Rs.) 100000 60000 45000 33000 20000 10000

What is the age to buy and to sell so as to minimize average annual cost?

Sol. Cost of new tempo = Rs 1,00,000

Let us find out the average cost per year of the new tempo,

Year of Service

(1)

Operating Cost O (t)

(2)

Cumulative O (t)

(3)

Resale value S (t)

(4)

Depreciation C-S (t)

(5)

Total cost TC (n)

6=3+5

Average TC (n)

7=3/1

1

10000

10000

60000

40000

50000

50000

2

12000

22000

45000

55000

77000

38500

3

15000

37000

32000

68000

105000

35000

4

20000

57000

22000

78000

135000

33750

5

30000

87000

10000

90000

177000

35400

6

45000

132000

2000

98000

230000

35000

 

Average cost is minimum at the end of 4 years, hence the new tempo should be replaced after 4 years.

Let us now find out the average total cost of second hand tempo.

Year of Service

(1)

Operating Cost O (t)

(2)

Cumulative O (t)

(3)

Resale value S (t)

(4)

Depreciation C-S (t)

(5)

Total cost TC (n)

6=3+5

Average TC (n)

7=3/1

0

100000

100000

100000

1

10000

10000

60000

40000

50000

50000

2

12000

22000

45000

55000

77000

38500

3

15000

37000

32000

68000

105000

35000

4

20000

57000

20000

78000

137000

33750

5

30000

87000

10000

90000

177000

35400

 

The tempo may be replaced by second hand tempo at the end of third year and the owner can save Rs (35000-34666) i.e. Rs 334 instead of buying a new one.

Example 8.5. A machine type A costs Rs 50,000 and the operating costs are estimated at Rs 1200 for the first year and increased by Rs 8000 in second and subsequent years. Another machine type B costs Rs 60,000 and operating costs are Rs 1500 for the first year and increasing by Rs 5000 per year. Should machine A be replaced by machine B assuming that both machines have no resale value and cost of money does not change with time?

Machine A

Year of Service

 

Operating Cost O (t)

(1)

Cumulative O (t)

(2)

Depreciation C-S (t)

(3)

Total cost TC (n)

4=2+3

Average TC (n)

 

1

1200

1200

50000

51200

51200

2

9200

10400

50000

60400

30200

3

17200

27600

50000

77600

25866

4

25200

52800

50000

102800

25700

5

33200

86000

50000

136000

27200

6

41200

127200

50000

177000

29500

 

Machine B

Year of Service

 

Operating Cost O (t)

(1)

Cumulative O (t)

(2)

Depreciation C-S (t)

(3)

Total cost TC (n)

4=2+3

Average TC (n)

 

1

1500

1500

60000

61500

61500

2

6500

8100

60000

68100

34050

3

11500

19600

60000

796000

26533

4

16500

36100

60000

96100

24025

5

21500

57600

60000

117600

23520

6

26500

84100

60000

144100

24016

 

It can be exactly seen that the lowest average cost of machine B (Rs. 23520) is lesser than that of machine A (Rs. 25700), machine A should be replaced by machine B at the end of 4th year.

Example 8.6. A firm is considering replacement of a machine, whose cost price is Rs 12.200 and the scrap value Rs 200. The running (maintenance and operation) costs in Rs, are found from experience to be as follows:

Year 1 2 3 4 5 6 7 8
Running Cost 200 500 800 1200 1800 2500 3200 4000

When should the machine be replaced?

Sol. We are given the following data.

C = Rs 12,200

S (t) = Rs200

O (t) is also given.

Let n be the number of years the item can be used.

Average total cost per year during the life of the machine is shown in the table below. 

The average total cost per year is minimum in 6th year i.e Rs 3167 And the average cost in 7th year is Rs 3171 which is more than the cost in 6th year. Hence, the machine should be replaced after 6 years.

Example 8.7 (a) Machine A costs Rs 9000. Annual operating costs are Rs 200 for the first year and then increases Rs 2000 every year. Determine the best age at which to replace the machine. If the optimum replacement policy is followed, what will be the average yearly cost of owning and operating the machine?

(b) Machine B costs Rs 10.000. Annual operating cost are Rs 400 for the first year and then increases by Rs 800 every year. You now have a machine of type A which is one year old. Should you replace it with B, if so when?

Sol. (a) Let us assume that there is no scrap value of the machine Average total cost can be computed as

Year (n)

Operating Cost O (t)

Cumulative 12Σ O(t)”>

Depreciation C-S(t)

Total Cost

Average Cost

1

2

3

4

5

200

2200

4200

6200

8200

200

2400

6600

12800

21000

9000 for all years

9200

11400

15600

21800

30000

9200

5700

5200

5450

6000

 

It can be seen that the best age of replacement is 3rd year.

(b) For machine B, the average cost be calculated as follows.

Year (n)

Operating Cost O (t)

Cumulative 12Σ O(t)”>

Depreciation C-S(t)

Total Cost

Average Cost

1

2

3

4

5

6

40

1200

2000

2800

3600

4400

400

1600

3600

6400

10000

14400

10000 for all years

10400

11600

13600

16400

20000

24400

10400

5800

4533

4100

4000

4066

 

Since the minimum average cost for machine B is lower than for machine A, machine B should be replaced by machine A Minimum average cost is (Rs 4,000), it should be replaced when it exceeds Rs 4000. In case of one year old machine Rs 2200/- will be spent next year and Rs 4200 the following year. We should keep machine A for one year.

Replacement policy of an equipment/item whose operating cost increases with time and money value also changes with time.

In previous examples, we assumed that the money value does not change and remains constant but it is well known that as the equipment deteriorates and operating costs keep increasing, the money value keeps decreasing with time. Hence we must calculate the Net Present Value (NPV) of the money to be spent a few years hence. Otherwise the resale value, the operating costs, which are to take place in future, will not be realistic and management will not be able to take optimal decisions.

Let C = initial cost of item/equipment

OC = operating cost

r = rate of interest

A rupee invested at present will be equivalent to (1 + r) a year after (1 + r)2 two years hence and (1 + r)n in 11 years time. It means that making a payment of one rupee after n years is equivalent to paying (1 + r)n now. The quantity (1 + r)-n is called the present worth or present value of one rupee spent n years from now.

Present value of a rupee V = (1 + r)-1 = 1/1 + r is called discount rate and is always less than 1.

Then year wise present value of expenditure in future years can be calculated as

Present value (n) = (c + oc1) + oc2 v + oc3 i + + ocn vn-1

+ (c + oc1) vn + oc2 vn+1+ oc3 vn+2 + ….   + ocn v2n-1

+ (c + oc1) v2n + oc2 v2n + 1 …. + ocn v3n-1