Home > Operation Research calculators > Replacement Model-2 example

4. Model-2 example ( Enter your problem )
 Example-1Example-2 Other related methods

 3. Model-1.3 (Previous method) 2. Example-2 (Next example)

### 1. Example-1

1. An engineering company is offered a material handling equipment A. It is priced at Rs 60,000 includeing cost of installation. The costs for operation and maintenance are estimated to be Rs 10,000 for each of the first five years, increasing every year by Rs 3,000 in the sixth and subsequent years. The company expects a return of 10 percent on all its investment. What is the optimal replacement period?
 Year Running Cost 1 2 3 4 5 6 7 8 9 10 10,000 10,000 10,000 10,000 10,000 13,000 16,000 19,000 22,000 25,000

Solution:
Since money is worth 10 per cent per year, the discounted factor over a period of one year is given by:
d = 1/(1 + 10 / 100) = 0.9091

It is also given that C = Rs 60,000

The optimum replacement age must satisfy the condition R_n < W(n) < R_(n+1)

 Year of Servicen(1) Running CostR_n(2) Discounted factord^(n-1)(3) = 0.9091^(n-1) Discounted CostR_n * d^(n-1)(4)=(2)xx(3) Summation of Discounted CostSigma R_i d^(i-1)(5)=Sigma(4) C+Sigma R_i d^(i-1)(6)=60,000+(5) Summation of DiscountSigma d^(i-1)(7)=Sigma(3) Weighted Average CostW(n)(8)=(6)/(7) 1 10,000 1 1 = 0.9091^(1-1) 10,000 10000 = 10000 xx 1(4)=(2)xx(3) 10,000 10000 = 0 + 10000(5) = Previous(5) + (4) 70,000 70000 = 60000 + 10000(6) = 60000 + (5) 1 1 = 0 + 1(7) = Previous(7) + (3) 70,000 70000 = 70000 / 1(8)=(6)/(7) 2 10,000 0.9091 0.9091 = 0.9091^(2-1) 9,090.91 9090.91 = 10000 xx 0.9091(4)=(2)xx(3) 19,090.91 19090.91 = 10000 + 9090.91(5) = Previous(5) + (4) 79,090.91 79090.91 = 60000 + 19090.91(6) = 60000 + (5) 1.9091 1.9091 = 1 + 0.9091(7) = Previous(7) + (3) 41,428.57 41428.57 = 79090.91 / 1.9091(8)=(6)/(7) 3 10,000 0.8264 0.8264 = 0.9091^(3-1) 8,264.46 8264.46 = 10000 xx 0.8264(4)=(2)xx(3) 27,355.37 27355.37 = 19090.91 + 8264.46(5) = Previous(5) + (4) 87,355.37 87355.37 = 60000 + 27355.37(6) = 60000 + (5) 2.7355 2.7355 = 1.9091 + 0.8264(7) = Previous(7) + (3) 31,933.53 31933.53 = 87355.37 / 2.7355(8)=(6)/(7) 4 10,000 0.7513 0.7513 = 0.9091^(4-1) 7,513.15 7513.15 = 10000 xx 0.7513(4)=(2)xx(3) 34,868.52 34868.52 = 27355.37 + 7513.15(5) = Previous(5) + (4) 94,868.52 94868.52 = 60000 + 34868.52(6) = 60000 + (5) 3.4869 3.4869 = 2.7355 + 0.7513(7) = Previous(7) + (3) 27,207.5 27207.5 = 94868.52 / 3.4869(8)=(6)/(7) 5 10,000 0.683 0.683 = 0.9091^(5-1) 6,830.13 6830.13 = 10000 xx 0.683(4)=(2)xx(3) 41,698.65 41698.65 = 34868.52 + 6830.13(5) = Previous(5) + (4) 101,698.65 101698.65 = 60000 + 41698.65(6) = 60000 + (5) 4.1699 4.1699 = 3.4869 + 0.683(7) = Previous(7) + (3) 24,388.95 24388.95 = 101698.65 / 4.1699(8)=(6)/(7) 6 13,000 0.6209 0.6209 = 0.9091^(6-1) 8,071.98 8071.98 = 13000 xx 0.6209(4)=(2)xx(3) 49,770.63 49770.63 = 41698.65 + 8071.98(5) = Previous(5) + (4) 109,770.63 109770.63 = 60000 + 49770.63(6) = 60000 + (5) 4.7908 4.7908 = 4.1699 + 0.6209(7) = Previous(7) + (3) 22,912.86 22912.86 = 109770.63 / 4.7908(8)=(6)/(7) 7 16,000 0.5645 0.5645 = 0.9091^(7-1) 9,031.58 9031.58 = 16000 xx 0.5645(4)=(2)xx(3) 58,802.21 58802.21 = 49770.63 + 9031.58(5) = Previous(5) + (4) 118,802.21 118802.21 = 60000 + 58802.21(6) = 60000 + (5) 5.3553 5.3553 = 4.7908 + 0.5645(7) = Previous(7) + (3) 22,184.21 22184.21 = 118802.21 / 5.3553(8)=(6)/(7) 8 19,000 0.5132 0.5132 = 0.9091^(8-1) 9,750 9750 = 19000 xx 0.5132(4)=(2)xx(3) 68,552.22 68552.22 = 58802.21 + 9750(5) = Previous(5) + (4) 128,552.22 128552.22 = 60000 + 68552.22(6) = 60000 + (5) 5.8684 5.8684 = 5.3553 + 0.5132(7) = Previous(7) + (3) 21,905.77 21905.77 = 128552.22 / 5.8684(8)=(6)/(7) 9 22,000 0.4665 0.4665 = 0.9091^(9-1) 10,263.16 10263.16 = 22000 xx 0.4665(4)=(2)xx(3) 78,815.38 78815.38 = 68552.22 + 10263.16(5) = Previous(5) + (4) 138,815.38 138815.38 = 60000 + 78815.38(6) = 60000 + (5) 6.3349 6.3349 = 5.8684 + 0.4665(7) = Previous(7) + (3) 21,912.71 21912.71 = 138815.38 / 6.3349(8)=(6)/(7) 10 25,000 0.4241 0.4241 = 0.9091^(10-1) 10,602.44 10602.44 = 25000 xx 0.4241(4)=(2)xx(3) 89,417.82 89417.82 = 78815.38 + 10602.44(5) = Previous(5) + (4) 149,417.82 149417.82 = 60000 + 89417.82(6) = 60000 + (5) 6.759 6.759 = 6.3349 + 0.4241(7) = Previous(7) + (3) 22,106.42 22106.42 = 149417.82 / 6.759(8)=(6)/(7)

The calculations in table show that the average cost is lowest during the 8^(th) year (Rs 21,905.77).

Hence, the machine should be replaced after every 8^(th) years, otherwise the average cost per year for running the machine would start increasing.

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here

 3. Model-1.3 (Previous method) 2. Example-2 (Next example)

Dear user,