**This model is represented by:**

- Increasing maintenance cost.
- Decreasing salvage value.

**Assumption**

- Increased age reduces efficiency

Generally, the criteria for measuring efficiency is the discounted value of all future costs associated with each policy.

**Let**

C = the 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 cost of the item at time t

n = optimal replacement period of the time

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 F(t) dt, the total cost T, incurred on the machine during n years is given by:

T = C – S(t) + F(t) dt

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

TA = | 1 —– n |
C – S(t) + F(t) dt |

To determine the optimal period for replacing the machine, the above function is differentiated with respect to n and equated to zero.

dTA —— dn |
= | -1 —– n ^{2} |
C – S(t) | -1 —– n ^{2} |
F(t) dt | + | F(n) —— n |

Equating | dTA —— dn |
= 0, we get |

F(n) = | 1 —– n |
C – S(t) + F(t) dt |

That is, F(n) = TA

Example

The initial cost of a machine is Rs. 7100 and scrap value is Rs. 100. The maintenance costs found from experience are as follows:

Year |
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

Maintenance |
200 | 350 | 500 | 700 | 1000 | 1300 | 1700 | 2100 |

When should the machine be **replaced**?

### Solution.

Year |
Runningcost |
Cumulative running cost |
Scrapvalue |
Difference between initial cost and scrap value |
Average investment cost / year |
Average running cost / year |
Average annualtotal cost |
||

A |
B |
C |
D |
E |
F = E/A |
G = C/A |
H = F + G |
||

1 | 200 | 200 | 100 | 7000 | 7000 | 200 | 7200 | ||

2 | 350 | 200 + 350 = 550 | 100 | 7000 | 3500 | 225 | 3775 | ||

3 | 500 | 550 + 500 = 1050 | 100 | 7000 | 2333.33 | 350 | 2683.33 | ||

4 | 700 | 1050 + 700 = 1750 | 100 | 7000 | 1750 | 437.5 | 2187.50 | ||

5 | 1000 | 1750 + 1000 = 2750 | 100 | 7000 | 1400 | 550 | 1950 | ||

6 | 1300 | 2750 + 1300 = 4050 | 100 | 7000 | 1166.67 | 675 | 1841.67 | ||

7 | 1700 | 4050 + 1700 = 5750 | 100 | 7000 | 1000 | 821.42 | 1821.42 | ||

8 | 2100 | 5750 + 2100 = 7850 | 100 | 7000 | 875 | 981.25 | 1856.25 | ||

This table shows that the average annual total cost during the seventh year is minimum. Hence, the machine should be **replaced** after the **7**** ^{th}** year.

## 3 thoughts on “Replacement of Assets that Deteriorate with Time”