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Solution
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Solution provided by AtoZmath.com
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Replacement Model-2 calculator
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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 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
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Running Cost | 10,000 | 10,000 | 10,000 | 10,000 | 10,000 | 13,000 | 16,000 | 19,000 | 22,000 | 25,000 |
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2. A company is buying mini computers. It costs Rs 5 lakh, and its running and maintenance costs are Rs 60,000
for each of the first five years, increasing by Rs 20,000 per year in the sixth and subsequent years.
If the money is worth 10 percent per year, What is the optimal replacement period?
Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
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Running Cost | 60,000 | 60,000 | 60,000 | 60,000 | 60,000 | 80,000 | 1,00,000 | 1,20,000 | 1,40,000 | 1,60,000 |
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Example1. 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 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
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Running Cost | 10,000 | 10,000 | 10,000 | 10,000 | 10,000 | 13,000 | 16,000 | 19,000 | 22,000 | 25,000 |
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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 Service `n` (1) | Running Cost `R_n` (2) | Discounted factor `d^(n-1)` `(3) = 0.9091^(n-1)` | Discounted Cost `R_n * d^(n-1)` `(4)=(2)xx(3)` | Summation of Discounted Cost `Sigma R_i d^(i-1)` (5)=`Sigma`(4) | `C+Sigma R_i d^(i-1)` (6)=60,000+(5) | Summation of Discount `Sigma d^(i-1)` (7)=`Sigma`(3) | Weighted Average Cost `W(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.
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