Please solve using Dynamic Programming. Thanks!
Consider an electronic system consisting of three components, each of which must function for
the system to function. The reliability of the system can be improved by installing several
parallel units in one or more of the components. The following table gives us the probability'
that the respective components will function if they consist of one, two or three parallel units.
The cost (in thousands) of installing 1, 2, or 3 parallel units in the respective components is given
by: Because of budget limitation, a maximum of Php 10,000 can be spent. Use DP to determine
the no. of parallel units to install in each of the 3 components to maximize the system's
reliability.
Solution
Let pi(x) be the probability that component i functions if there are x parallel units placed in it and
let cr(x) be the cost of putting x parallel units into component i.
Let r(w) be the maximum reliability of a system of components r, r + 1, . . . , 3 if one has enough
money to put in w parallel components altogether. We wish to compute 1(9).
In general 3(w) = p3(w) r = max x (pr(x)r+1(w cr(x)))
Therefore solution is :
x1 = 3, x2 = 1, x3 = 1
w 1 x1 2 x2 3 x3
0 0 0
1 0 0
2 0 .24 1
3 0 .24 1
4 0 .35 1
5 .210 1 .35 1
6 .294 1 .49 1
7 .374 1 .63 2

8 .374 1 .63 3
9 .3024 3 .439 2 .72 3