The efficiency of equipment/items that deteriorate with time will be getting low. As the items deteriorate, gradual failure sets in. the gradual failure is progressive in nature which affects the efficiency and resulted in:
(a) A decrease in its production capacity;
(b) The increasing maintenance and operating costs; and
(c) Decrease in the value of the re-sale price (or salvage) of the item.
Due to the above effects, it is economical to replace a deteriorating equipment/item with a new one. As the repair and maintenance costs are the determining factors in replacement policy, the following two policies are to be considered.
(a) Replacement of items that Deteriorate and whose maintenance and repair costs increase with time, ignoring changes in the value of money during the period.
This is a simple case of minimizing the average annual cost of an equipment when the maintenance cost is an increasing function time but the time value of money remains constant.
In order to determine the optional replacement age of a deteriorating equipment/item under the above conditions, the following mathematical formulae shall be used:
Let C = Capital or purchase cost of the equipment/item S = Scrap (or salvage) value of the equipment/item at
the end of t years.
TC(t) = Total cost incurred during t years.
ATC = Average annual total cost of the equipment/item
n = Replacement age of the equipment/item; i.e.
number of equipment years the equipment/item is to be in use.
f(t) = Operating and maintenance cost of the equipment/
item at time t.
Case 1: (When time “t” is a continuous variable)
Here, we need to find the value of n that minimizes the total cost incurred during the first n years.
TC(t) = Capital cost – scrap + maintenance cost n
= C – S + f(t) dt ………20.2.1 0
But for average annual total cost, we have n
ATC(n) = 1/n C – S + f(t) dt ……….. 20.2.2 o
To obtain optimal value for n for which ATC(n) is minimum, we differentiate ATC(n) with respect to n and equate the result to zero so that.
n
f(n) = 1/n C – S + f(t) dt , n 0 o
= ATC(n) ………. 20.2.3
Hence, the equipment/item is to be replaced when the average annual cost equals the current maintenance cost.
Case 2 (When time “t” is a discrete variable).
In a similar manner, the total cost incurred during n years is n n
TC(n) = C – S + f(t) = 12 C S f(t)dt ………20.2.4 t =0
while the average annual cost on item is given as n
TC(n) = 1/n C – S + f(t) ……….20.2.5 t =0
o
0
461 dn
n TC
d[ ( )] = 2 2 (2) n
t f n
S n
C ………. 20.2.6
For this case, we will use a tabular method for obtaining the value of n.
ILLUSTRATION 20-1
An owner of a grinding machine estimates from his past records that cost per year of operating his machine is given below:
Year 1 2 3 4 5 6 7 8
Operating Cost
(N) 250 550 850 1250 1850 2550 3250 4050 If the cost price is N12,300 and the scrap value is N250, when should the machine be replaced?
SUGGESTED SOLUTION 20-1 C = N12300 and S = N250
Year of Service
n
Running Cost (N)
f(n)
Cumulative Running
Cost (N) f(n)
Depreciation Cost Price =
C - S
Total Cost
(N) TC Average Cost (N) ATC (n)
Col 1 Col 2 Col 3 Col 4 Col 3 +
Col 4
= col5
Col 6 = Col5/n
1 250 250 12050 12300 12300
2 550 800 12050 12850 6425
3 850 1650 12050 13700 4566.67
4 1250 2900 12050 14950 3737.50
5 1850 4750 12050 16800 3360
6 2550 7300 12050 19350 3225 *
7 3250 10550 12050 22600 3228.57
8 4050 14600 12050 26650 3331.25
It is observed from the table that the average annual cost ATC(n) is minimum in the sixth year. Hence, the machine should be replaced at the end of sixth year of usage.
ILLUSTRATION 20-2
A company is considering replacement of a machine which cost N70000. The maintenance cost and resale values per year of the said machine are given below:
Year 1 2 3 4 5 6 7 8
Maintenance
Cost (N) 9000 12000 16000 21000 28000 37000 47000 59000 Resale Cost (N) 40000 20000 12000 6000 5000 4000 4000 4000
When should the machine be replaced?
SUGGESTED SOLUTION 20-2
Year of
Service n Resale value (N) (S)
Purchase Price resale value (N) C–S
Annual Maintenance Cost f(t)
Cumulation of Maintenance
cost (N) n
f(t) 0
Total Cost (N) n C-S + f(t) 0
Average Cost (N)
Col 1 Col 2 Col 3 = 70000 – Col 2
Col 4 Col 5 Col 6 = col3 +
col5 Col 7 =
Col6/n
1 40000 30000 9000 9000 39000 39000
2 20000 50000 12000 21000 71000 35500
3 12000 58000 16000 37000 95000 31666.67
4 6000 64000 21000 58000 122000 30500
5 5000 65000 28000 86000 151000 30200 *
6 4000 66000 37000 123000 189000 31500
7 4000 66000 47000 170000 236000 33714.29
8 4000 66000 59000 229000 295000 36875
From the table, it is observed that the average cost ATC(n) is minimum in the fifth year. Hence, the machine should be replaced by the end of 5th year.
(b) Replacement of items that deteriorate and whose maintenance cost increases with time; and value of money also changes with time.
This replacement policy can be seen as value of money criterion. Here, the replacement decision will be based on the equivalent annual cost whenever we have time value of money effect.
Whenever the value of money decreases at constant rate, the issue of depreciation factor or ratio comes in as in the computation of present value (or worth). For example, if the interest rate on N100 is r percent
463
per year, the present value (or worth) of N100 to be spent after n years will be:
r n
D 100
100 ……… 20.2.7
Where D is the discount rate or depreciation value. With the principle of the discount rate, determine the critical age at which an item should be replaced.
ILLUSTRATION 20-4
The yearly cost of two machines A and B, when money value is neglected, is given below:
Year 1 2 3
Machine A (N) 1400 800 1000 Machine B (N) 24000 300 1100
If the money value is 12% per year, find the cost patterns of the two machines and which of the machine is more economical?
SUGGESTED SOLUTION 20-4
The discount rate per year = (d) = 1 = 0.89 1 + 0.12
The discounted cost patterns for machines A and B are shown below:
1 1 (1 + 0.12)1 , (1 + 0.12)2
Year 1 2 3 Total Cost (N)
Machine A
(Discounted Cost in N) 1400 800 x 0.89
= 712 1000 x (0.89)2
= 792.1 2,904.1
Machine B
(Discounted cost in N) 24000 300 x 0.89
= 267 1100 x (0.89)2
= 871.31 25,138.3 Decision: Machine A is more economical because its total cost is lower.