An electronic device consists of three main components. The failure of one of the components results in the failure of the whole device because the three components are arranged in series. Therefore, it is decided that the reliability (prob. of not failure) of the device can be increased by installing parallel units on each component. Each component may be installed at most 3 parallel units. The total capital (in thousands) available for the device is 10. Consider the following data:
m_{i} | i = 1 | i =2 | i = 3 | |||
---|---|---|---|---|---|---|
r_{1}m_{1} | c_{1}m_{1} | r_{2}m_{2} | c_{2}m_{2} | r_{3}m_{3} | c_{3}m_{3} | |
1 | .5 | 2 | .7 | 3 | .6 | 1 |
2 | .7 | 4 | .8 | 5 | .8 | 2 |
3 | .9 | 5 | .9 | 6 | .9 | 3 |
Here m_{i} is the number of parallel units placed with i^{th} component, r_{i}m_{i} is the reliability of the i^{th} component and c_{i}m_{i} is the cost for the i^{th} component. Determine m_{i} which will maximize the total reliability of the system without exceeding the given capital.
Solution.
In this example, we consider each component as one stage. Let the
state x_{i} be defined as the capital allocated stages 1, 2,
..., i. The reliability r_{i}m_{i} is a function of
c_{i}m_{i}, i.e., r_{i}m_{i} (c_{i}m_{i}).
In general the recursive equation is ƒ_{i}(x_{i})
= Max. {r_{i}m_{i} (c_{i}m_{i}) ƒ_{i
-1}(x_{i} – c_{i}m_{i})}
where m_{i} = 1, 2, 3.
0 ≤ c_{i}m_{i} ≤ x_{i}, i = 1, 2, 3.
There is one table for each possible stage n, namely, n = 1, 2, and
3. We summarize this information in the format below:
State | Evaluations of feasible alternatives ƒ_{1}(x_{1}) = r_{1}m_{1} (c_{1}m_{1}) |
Maximum reliability | ||||||
m_{1} = 1 | m_{1} = 2 | m_{1} = 3 | ||||||
x_{1} | r_{1}m_{1} = .5 | c_{1}m_{1} = 2 | r_{1}m_{1} = .7 | c_{1}m_{1} = 4 | r_{1}m_{1} = .9 | c_{1}m_{1} = 5 | ƒ_{1}(x_{1}) | m_{1}* |
0 | - | - | - | - | - | |||
1 | - | - | - | - | - | |||
2 | .5 | - | - | .5 | 1 | |||
3 | .5 | - | - | .5 | 1 | |||
4 | .5 | .7 | - | .7 | 2 | |||
5 | .5 | .7 | .9 | .9 | 3 | |||
6 | .5 | .7 | .9 | .9 | 3 | |||
7 | .5 | .7 | .9 | .9 | 3 | |||
8 | .5 | .7 | .9 | .9 | 3 | |||
9 | .5 | .7 | .9 | .9 | 3 | |||
10 | .5 | .7 | .9 | .9 | 3 |
The analysis for n = 2 appears in the following table.
State | ƒ_{2}(x_{2}) = r_{2}m_{2} (c_{2}m_{2}). ƒ_{1}(x_{2} – c_{2}m_{2}) | Maximum reliability | ||||||
m_{2} = 1 | m_{2} = 2 | m_{2} = 3 | ||||||
x_{2} | r_{2}m_{2} = .7 | c_{2}m_{2} = 3 | r_{2}m_{2} = .8 | c_{2}m_{2} = 5 | r_{2}m_{2} = .9 | c_{2}m_{2} = 6 | ƒ_{2}(x_{2}) | m_{2}* |
0 | - | - | - | - | - | |||
1 | - | - | - | - | - | |||
2 | - | - | - | - | - | |||
3 | .7 X ( - ) = - | - | - | - | - | |||
4 | .7 X ( - ) = - | - | - | - | - | |||
5 | .7 X .5 = .35 | .8 X ( - ) = ( - ) | - | .35 | 1 | |||
6 | .7 X .5 = .35 | .8 X ( - ) = ( - ) | .9 X ( - ) = ( - ) | .35 | 1 | |||
7 | .7 X .7 = .49 | .8 X .5 = .40 | .9 X ( - ) = ( - ) | .49 | 1 | |||
8 | .7 X .9 = .63 | .8 X .5 = .40 | .9 X .5 = .45 | .63 | 1 | |||
9 | .7 X .9 = .63 | .8 X .7 = .56 | .9 X .5 = .45 | .63 | 1 | |||
10 | .7 X .9 = .63 | .8 X .9 = .72 | .9 X .7 = .63 | .72 | 2 |
Note that in case m_{2} = 1, _{ƒ1}(x_{2} - c_{2}m_{2}) has no value until x_{2} – c_{2}m_{2} ≤ 1 or x_{2} ≤ 1+ c_{2}m_{2} = 1 + 3 = 4. Similar is the case with other columns in this table.
State | ƒ_{3}(x_{3}) = r_{3}m_{3} (c_{3}m_{3}). ƒ_{2}(x_{3} - c_{3}m_{3}) | Maximum reliability | ||||||
m_{3} = 1 | m_{3} = 2 | m_{3} = 3 | ||||||
x_{3} | r_{3}m_{3} = .6 | c_{3}m_{3} = 1 | r_{3}m_{3} = .8 | c_{3}m_{3} = 2 | r_{3}m_{3} = .9 | c_{3}m_{3} = 3 | ƒ_{3}(x_{3}) | m_{3}* |
0 | - | - | - | - | - | |||
1 | .6 X ( - ) = ( - ) | - | - | - | - | |||
2 | .6 X ( - ) = ( - ) | .8 X ( - ) = ( - ) | - | - | - | |||
3 | .6 X ( - ) = ( - ) | .8 X ( - ) = ( - ) | .9 X ( - ) = ( - ) | - | - | |||
4 | .6 X ( - ) = ( - ) | .8 X ( - ) = ( - ) | .9 X ( - ) = ( - ) | - | - | |||
5 | .6 X ( - ) = ( - ) | .8 X ( - ) = ( - ) | .9 X ( - ) = ( - ) | - | - | |||
6 | .6 X .35 = .210 | .8 X ( - ) = ( - ) | .9 X ( - ) = ( - ) | .210 | 1 | |||
7 | .6 X .35 = .210 | .8 X .35 = .280 | .9 X ( - ) = ( - ) | .280 | 2 | |||
8 | .6 X .49 = .294 | .8 X .35 = .280 | .9 X .35 = .315 | .315 | 3 | |||
9 | .6 X .63 = .378 | .8 X .49 = .392 | .9 X .35 = .315 | .392 | 2 | |||
10 | .6 X .63 = .378 | .8 X .63 = .504 | .9 X .49 = .441 | .504 | 2 |
The maximum reliability is .504. for the m_{3}* = 2. Also for this, the optimal _{ƒ2}(x_{3} – c_{3}m_{2}) = _{ƒ2}(8) = .63. Hence the corresponding m_{2}* = 2. Similarly, the corresponding m_{1}* = 3.