solve using dynamic programming a total of 6 units of water is to be allocated o
ID: 3199015 • Letter: S
Question
solve using dynamic programming
a total of 6 units of water is to be allocated optimally to three users. the allocation is made in discrete steps of one unit ranging from 1 to 6. with the three users denoted as user1, user2, and user 3 respectively. the returns obtained from the users for a given allocation are given in the following table. find allocations to the three users so that the total return is maximized.
Answer : X1 = 2 , X2 = 1 , X3 = 3 and Maximum return = 28
amount of water allocated return from return from return from user 1 user 1 user 1 B1(x) B2(x) B3(x) 0 0 0 0 1 5 5 7 2 8 6 12 3 9 3 15 4 8 -4 16 5 5 -15 15 6 0 -30 12Explanation / Answer
amount of water allocated return from return from return from user 1 user 2 user 3 B1(x) B2(x) B3(x) 0 0 0 0 1 5 5 7 2 8 6 12 3 9 3 15 4 8 -4 16 5 5 -15 15 6 0 -30 12 Each user is to to be allocated between 1 and 6 So maximum a user can be alloctaed is 4 with other user being allocated eah 1. So rearranging the matrix we get amount of water allocated return from return from return from user 1 user 2 user 3 B1(x) B2(x) B3(x) 0 0 0 0 1 5 5 7 2 8 6 12 3 9 3 15 4 8 -4 16 Possible allocations are as follows B1 B2 B3 Total Return Allocation 1 1 4 Return 5 5 16 26 Allocation 1 2 3 Return 5 6 15 26 Allocation 2 1 3 Return 8 5 15 28 Allocation 1 3 2 Return 5 3 12 20 Allocation 3 1 2 Return 9 5 12 Allocation 2 2 2 Return 8 6 12 26 Allocation 3 2 1 Return 9 6 7 22 Allocation 2 3 1 Return 8 3 7 17 Allocation 2 2 2 Return 8 6 12 26 Allocation 1 4 1 Return 5 -4 7 8 Allocation 4 1 1 Return 8 5 7 20 Best allocation is B1 B2 B3 2 1 3 and maximum return is 28
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