Module 6. Replacement theory

Lesson 13

REPLACEMENT OF ITEMS DETERIORATING WITH TIME

13.1 Introduction

In any establishment, sooner or later equipment needs to be replaced, particularly when new equipment gives more efficient or economical service than the old one. In some cases, the old equipment might fail and work no more or is worn out. In such situations it needs more expenditure on its maintenance than before. The problem in such situation is to determine the best policy to be adopted with respect to replacement of the equipment. The replacement theory provides answer to this question in terms of optimal replacement period. Replacement theory deals with the analysis of materials and machines which deteriorate with time and fix the optimal time of their replacement so that total cost is the minimum.

13.2 Replacement Decisions

The problem is to decide the best policy to adopt with regard to replacement. The need for replacement arises in a number of different following situations so that different types of decisions may have to be taken.

· It may be necessary to decide whether to wait for a certain item to fail which might cause some loss or to replace earlier at the expense of higher cost of the item.

· The item can be considered individually to decide whether to replace now or if not when to reconsider the item in question.

· It is necessary to decide whether to replace by the same item or by a different type of item.

13.3 Types of Replacement Problems

i) Replacement policy for items, efficiency of which declines gradually with time without change in money value.

ii) Replacement policy for items, efficiency of which declines gradually with time but with change in money value.

iii) Replacement policy of items breaking down suddenly

a) Individual replacement policy

b) Group replacement policy

iv) Staff replacement

In this lesson we confine ourselves to first two situations only

13.4 Replacement of Items that Deteriorate with Time

There are certain items which deteriorate gradually with usage and such items decline in efficiency over a period of time. Generally, the maintenance cost of certain items always increase gradually with time and a stage comes when the maintenance cost becomes so large that it is better and economical to replace the item with a new one. There may be number of alternatives and we may have a comparison between various alternatives by considering the costs due to waste, scrap, loss of output, damage to equipment and safety risks etc.

13.4.1 Replacement of items whose maintenance cost increases with time and the value of money remains same during the period

The following costs are considered in such decisions:

C: Capital cost of a certain item say a machine

s(t): The selling or scrap value of the item after t years

f(t): Operating (or maintenance) cost of the item at time t

n: Optimal replacement period of the item

The operating cost function f(t) is assumed to be strictly positive. It may be continuous or discrete.

13.4.2 When t is a continuous variable

 Now the annual cost of the machine at time t is given by C – S(t) + f(t) and since the total maintenance cost incurred on the machine during n years is . 

Total cost T incurred on machine during n years is given by

Thus the average annual cost incurred on the machine per year during n years is given by

To determine the value of optimal period (n), the principle of minima will be employed.

Clearly                                

Therefore, it can be seen that A(n) or T_A = f(n) is a minimum for T provided that f(t) is non–decreasing and f(0) = 0. Hence, if time is measured continuously, then the average annual cost will be minimized by replacing the item when the average cost becomes equal to the current maintenance cost.

13.4.3  When time is a discrete variable

Here the period of time is considered as fixed and n takes values 1,2,3, … then

 By using finite differences, A(n) will be a minimum for that value of n for which

or          

For this, we write

Similarly, it can be shown

This suggests the replacement policy that if time is measured in discrete units, then the average annual cost will be minimized by replacing item when the next period’s maintenance cost become greater than the current average cost. Hence, replace the equipment at the end of n years, if the maintenance cost in the (n+1)^th year is more than the average total cost in the (n)^th year and the (n)^th year’s maintenance cost is less than the previous year’s average total cost. The following examples will illustrate this methodology

Example 1  A milk plant is considering replacement of a machine whose cost price is Rs. 12,200 and the scrap value Rs. 200. The running (maintenance and operating) 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?

Solution The computations can be summarized in the following tabular form:

Table 13.1 Calculations for average cost of machine

	(In Rupees)
Year (1)	Running Cost (2)	Cumulative Running Cost (3)	Depreciation Cost (4)	Total Cost TC (5) = (3) + (4)	Average Cost (6) = (5)/(1)
1	200	200	12000	12200	12200
2	500	700	12000	12700	6350
3	800	1500	12000	13500	4500
4	1200	2700	12000	14700	3675
5	1800	4500	12000	16500	3300
6	2500	7000	12000	19000	3167
7	3200	10200	12000	22200	3171
8	4000	14200	12000	26200	3275

From the table it is noted that the average total cost per year, A(n) is minimum in the 6^th year (Rs. 3167). Also the average cost in 7^th year (Rs.3171) is more than the cost in 6^th year. Hence the machine should be replaced after every 6 years.

Example 2

A Machine owner finds from his past records that the maintenance costs per year of a machine whose purchase price is Rs. 8000 are as given below:

Year:	1	2	3	4	5	6	7	8
Maintenance Cost:	1000	1300	1700	2200	2900	3800	4800	6000
Resale Price:	4000	2000	1200	600	500	400	400	400

Determine at which time it is profitable to replace the machine.

Solution C = Rs. 8000. Table 13.2 shows  the average cost per year during the life of machine. Here, The computations can be summarized in the following tabular form:

Table 13.2 Calculations for average cost of machine

Year	f(t)	Cumulative maintenance cost	Scrap value	Total cost
1	1000	1000	4000	5000	5000
2	1300	2300	2000	8300	4150
3	1700	4000	1200	10800	3600
4	2200	6200	600	13600	3400
5	2900	9100	500	16600
6	3800	12900	400	20500	3417
7	4800	17700	400	25300	3614
8	6000	23700	400	31300	3913

The above table shows that the value of T_A during fifth year is minimum. Hence the machine should be replaced after every fifth year.

Example 3

The cost of a machine is Rs. 6100 and its scrap value is only Rs.100. The maintenance costs are found to be

Year:	1	2	3	4	5	6	7	8
Maintenance Cost (in Rs.):	100	250	400	600	900	1250	1600	2000

When should the Machine be replaced?

Solution

 C = 6100, s(t) = 100 The computations can be summarized in the following tabular form:

Table 13.3 Calculations for average cost of machine

Replace at the end of year		Cumulative maintenance cost	Total cost
1	100	100	6100	6100
2	250	350	6350	3175
3	400	750	6750	2250
4	600	1350	7350	1737.50
5	900	2250	8250	1650
6	1250	3500	9500
7	1600	5100	11100	1585.7
8	2000	7100	13100	1637.50

It is now observed that the machine should be replaced at the end of sixth year otherwise the average cost per year will start to increase.

13.4.4 Replacement of items whose maintenance cost increases with time and the money value changes at a constant rate

To understand this let us define the following terms:

Money Value

Since money has a value over time, therefore the explanation of the statement: ‘Money is worth 10% per year’ can be given in two ways:

(a) In one way, spending Rs.100 today would be equivalent to spend Rs.110 in year’s time. In other words if we plan to spend Rs.110 after a year from now, we could spend Rs.100 today and an investment which would be worth Rs.110 next year.

(b) Alternatively if we borrow Rs.100 at the interest of 10% per year and spend Rs.100 today, we have to pay Rs.100 after one year (next year).

Thus, we conclude that Rs.100 is equal to Rs.110 a year from now. Consequently Rs. 1 from a year now is equal to (1+0.1)^-1 rupee today.

Present Worth Factor

As we have seen, a rupee a year from now will be equivalent to (1+0.1)^-1 rupee today at the discount rate of 10% per year. So, one rupee in n years from now will be equal to (1+0.1)^-n. Therefore, the quantity (1+0.1)^-n is called the Present Worth Factor (PWF) or Present Value (PV) of one rupee spent in n years from now. In general, if r is the rate of interest, then (1+r)^-n is called PWF or PV of one rupee spent in n years from now onwards. The expression (1+r)^-n is known as compound amount factor of one rupee spent in n years.

Discount Rate

Let r be the rate of interest. Therefore present worth factor of unit amount to be spent after one year is .

Then v is known as the discount rate. The optimum replacement policy for replacement of item where maintenance costs increase with time and money value changes with constant rate can be determined by following method:

Suppose that the item (which may be machine or equipment) is available for use over a series of time periods of equal intervals (say one year). Let

C = Purchase price of the item to be replaceds

R_t = Running (or maintenance) cost in the t^th year

R = Rate of interest  

is the present worth of a rupee to be spent in a year hence.

Let the item be replaced at the end of every n^th year. The year wise present worth of expenditure on the item in the successive cycles of n years can be calculated as follows:

Year	1	2----	n	n+1	n+2	----	2n	2n+1
Present worth	C+R1	R2v	Rnv^n-1	(C+R1)v^n	R2v^n+1		Rnv^2n-1	(C+R1)v^2n

Assuming that machines has no resale value at the time of replacement, the present worth of the machine in n years will be given by

Summing up the right-hand side, column-wise

,

using sum of an infinite G.P.

f(n) and f(n+1) given above at n = 0,1,2…, are called the weighted average cost of previous n years with weights 1,v, v², ----v^n-1respectively. P(n) is the amount of money required now to pay all the future costs of acquiring and operating the equipment when it is renewed every n years. However, if P (n) is less than P (n+1) then replacing the equipment each n year is preferable to replacing each n years is preferable to replacing each (n+1) years. Further, if the best policy is replacing every n years, then the two inequalities P (n+1) – P (n) > 0 and P (n-1) - P (n) < 0 must hold, without giving the proof we shall state the following two inequalities which holds good at n, the optimal replacement interval.

As a result of these two inequalities, rules for minimizing costs may be stated as follows:

1. Do not replace if the operating cost of next period is less than the weighted average of previous costs.

2. Replace if the operating cost of the next period is greater than the weighted average of the previous costs.

Working Procedure

The step-by-step procedure for solving the problem is stated as under:

1. Write in a column the running/maintenance costs of machine or equipment for different years, R_n.

2. In the next column write the discount factor indicating the present value of a rupee received after (i-1) years,

3. The two column values are multiplied to get present value of the maintenance costs, i.e.,.

4. These discounted maintenance costs are then cumulated to the i^th year to get.

5. The cost of machine or equipment is added to the values obtained in Step 4 above to  

Obtain C+ .

6. The discount factors are then cumulated to get .

7. The total costs obtained in (Step 5) are divided by the corresponding value of the accumulated discount factor for each of the years.

8. Now compare the column of maintenance costs which is constantly increasing with the last column. Replace the machine in the latest year that the last column exceeds the column of maintenance costs.

Example 4

A milk plant is offered an equipment A which is priced at Rs.60,000 and the costs of operation and maintenance are estimated to be Rs.10,000 for each of the first 5 years, increasing every year by Rs. 3000 per year in the sixth and subsequent years. If money carries the rate of interest 10% per annum what would the optimal replacement period?

Solution

Table 13.4 Determination of optimal replacement period

At the end of year (n)	Operating & maintenance cost Rn	Discounted factor	Discounted operation & maintenance cost	Cumulative Discounted operation & maintenance cost	Discounted total cost	Cumulative discounted factor	Weighted average annual cost
(1)	(2)	(3)	(4)=(2)x(3)	(5)	(6)=(5)+60000	(7)	(8)=(6)+(7)
1	10000	1.0000	10000.00	10000.00	70000.00	1.00	70000.00
2	10000	0.9091	9091.00	19091.00	79091.00	1.91	41428.42
3	10000	0.8264	8264.00	27355.00	87355.00	2.74	31933.83
4	10000	0.7513	7513.00	34868.00	94868.00	3.49	27207.75
5	10000	0.6830	6830.00	41698.00	101698.00	4.17	24389.18
6	13000	0.6209	8071.70	49769.70	109769.70	4.79	22913.08
7	16000	0.5645	9032.00	58801.70	118801.70	5.36	22184.36
8	19000	0.5132	9750.80	68552.50	128552.50	5.87	21905.89
9	22000	0.4665	10263.00	78815.50	138815.50	6.33	21912.82
10	25000	0.4241	10602.50	89418.00	149418.00	6.76	22106.52

From Table 13.4 we find the weighted cost is minimum at the end of 8^th year, hence the equipment should be replaced at the end of 8^th year.

Example 5

A Manufacturer is offered two machines A and B. Machine A is priced at Rs. 5000 and running cost is estimated at Rs. 800 for each of the first five years, increasing by Rs. 200 per year in the sixth and subsequent years. Machine B, with the same capacity as A, costs Rs. 2500, but has running cost of Rs. 1200 per year for six years, thereafter increasing by Rs. 200 per year. If money is worth 10% per year, which machine should be purchased? (Assume that the machines will eventually be sold for scrap at a negligible price).

Solution

Since money is worth 10% per year, therefore discount rate is

Table 13.5 Computation of weighted average cost for machine A

At the end of year (n)	Operating & maintenance cost Rn	Discounted factor	Discounted operation & maintenance cost	Cumulative Discounted operation & maintenance cost	Discounted total cost	Cumulative discounted factor	Weighted average annual cost
(1)	(2)	(3)	(4)=(2)x(3)	(5)	(6)=(5)+        5000	(7)	(8)=(6)+(7)
1	800	1.0000	800	800	5800	1	5800
2	800	0.9091	727	1527	6527	1.9091	3419.035
3	800	0.8264	661	2188	7188	2.7355	2627.819
4	800	0.7513	601	2789	7789	3.4868	2233.98
5	800	0.6830	546	3336	8336	4.1698	1999.098
6	1000	0.6209	621	3957	8957	4.7907	1869.61
7	1200	0.5645	677	4634	9634	5.3552	1799.025
8	1400	0.5132	718	5353	10353	5.8684	1764.13
9	1600	0.4665	746	6099	11099	6.3349	1752.043
10	1800	0.4241	763	6862	11862	6.759	1755.053

From table 13.5 we conclude that for machine A 1600<1752.043<1800. Since the running cost of 9^th year is 1600and that of 10^th year is 1800 and 1800>1752.043, it would be economical to replace machine A at the end of nine years.

Table 13.6 Computation of weighted average cost for machine B

At the end of year (n)	Operating & maintenance cost Rn	Discounted factor	Discounted operation & maintenance cost	Cumulative Discounted operation & maintenance cost	Discounted total cost	Cumulative discounted factor	Weighted average annual cost
(1)	(2)	(3)	(4)=(2)x(3)	(5)	(6)=(5)+        2500	(7)	(8)=(6)+(7)
1	1200	1.0000	1200.00	1200.00	3700.00	1.00	3700.00
2	1200	0.9091	1090.92	2290.92	4790.92	1.91	2509.52
3	1200	0.8264	991.68	3282.60	5782.60	2.74	2113.91
4	1200	0.7513	901.56	4184.16	6684.16	3.49	1916.99
5	1200	0.6830	819.60	5003.76	7503.76	4.17	1799.55
6	1200	0.6209	745.08	5748.84	8248.84	4.79	1721.84
7	1400	0.5645	790.30	6539.14	9039.14	5.36	1687.92
8	1600	0.5132	821.12	7360.26	9860.26	5.87	1680.23
9	1800	0.4665	839.70	8199.96	10699.96	6.33	1689.05
10	2000	0.4241	848.20	9048.16	11548.16	6.76	1708.56

In table13.6 we find that 1800<1689<2300 so it is better to replace the machine B after 8^th year. The equivalent yearly average discounted value of future costs is Rs. 1748.60 for machine A and it is 1680.23for machine B. Hence, it is more economical to buy machine B rather than machine A.