Introduction:
Replacement model is the part of operation research mostly used in the industries when a purchased items like machinery, buildings efficiency is reduced or wear out due to much usage.
REPLACEMENT OF ITEMS WHOSE EFFICIENCY REDUCES :
Costs to be considered:
Example:
Consider a machine, in this case, the maintenance cost always increases with time and usage and a time comes when the maintenance cost becomes large enough, which indicates that it is better and economical to replace the machine with a new one. When we want to replace the machine, we may come across various alternative choices, where we have to compare the various cost elements such as running costs and maintenance costs to select optimal choice.
The various techniques we may come across to analyze the situation are:
(a) Replacement of items whose maintenance cost increases with time and value of money remains same during the period,
(b) Replacement of items whose maintenance cost increases with time and the value of money also changes with time, and
(c) To compare alternative choices, use of concept of present value.
Replacement of Items whose Maintenance Cost Increases with Time and the Value of Money Remains Same During the Period
Let C = Purchase cost or Capital cost of the item,
S = Scrap value or resale value of the item, it is assumed that this cost will remain constant over
time.
So, replace the item when the average annual cost reaches at the minimum that will always occur at a time when the average cost becomes equal to the current maintenance cost.
note:
If time is measured continuously then the average annual cost will be minimized by replacing the machine or item when the average cost to date becomes equal to the current maintenance cost.
Case 1.
Here time ‘t’ is considered as a discrete variable. In this case, the time period is taken as one year and ‘t’ can take the values of 1, 2, 3 …etc., then,
Total cost incurred on the item during ‘y’ years is T (y) = C M (y) - S = C - S Σ=ytu1(t) Average annual cost incurred during ‘y’ years is
G (y) = {T (y) / y} = {C M (y) - S} / y
G (y) will be minimum for that value of ‘y’, for which G (y 1) > G (y) and G (y – 1) > G (y) or say that
G (y 1) – G (y) > 0 and G (y–1) – G (y) > 0 This will exist at:
G (y 1) – G (y) > 0 if u (y 1) > G (y) and G (y–1) – G (y) > 0 if u (y) < G (y–1) ...(7.4)