To determine The solution of the given problem using simplex method. Explanation Given: Maximize z = x 1 + x 2 + 4 x 3 + 5 x 4 subject to x 1 + 2 x 2 + 3 x 3 + x 4 ≤ 115 2 x 1 + x 2 + 8 x 3 + 5 x 4 ≤ 200 x 1 + x 3 ≤ 50 x 1 ≥ 0 x 2 ≥ 0 x 3 ≥ 0 x 4 ≥ 0 Explanation: As there are 3 constraints, there will be 3 slack variables namely s 1 , s 2 and s 3 . Rewriting constraint 1 gives, x 1 + 2 x 2 + 3 x 3 + x 4 + s 1 = 115 Rewriting constraint 2 gives, 2 x 1 + x 2 + 8 x 3 + 5 x 4 + s 2 = 200 Rewriting constraint 3 gives, x 1 + x 3 + s 3 = 50 Subject to, x 1 ≥ 0 x 2 ≥ 0 x 3 ≥ 0 x 4 ≥ 0 And, s 1 ≥ 0 s 2 ≥ 0 s 3 ≥ 0 The required initial simplex tableau with indicators is given as x 1 x 2 x 3 x 4 s 1 s 2 s 3 z [ 1 2 3 1 1 0 0 0 115 2 1 8 5 0 1 0 0 200 1 0 1 0 0 0 1 0 50 − 1 − 1 − 4 − 5 0 0 0 1 0 ] Look at the indicators and choose the most negative one, which in this case is - 5 that belongs to column 4 . Thus, column 4 is pivot column. Now, for each positive entry in the pivot column, divide the number in the far-right column of the same row by the positive number in the pivot column. x 1 x 2 x 3 x 4 s 1 s 2 s 3 z [ 1 2 3 1 1 0 0 0 115 1 2 1 8 5 0 1 0 0 200 5 1 0 1 0 0 0 1 0 50 0 − 1 − 1 − 4 − 5 0 0 0 1 0 ] The row with the smallest quotient (in this case, the second row) is the pivot row. The entry in the pivot row and pivot column is the pivot which is 5 . To perform pivoting on the simplex tableau, start by multiplying each entry of row 2 by 1 5 in order to change the pivot to 1 . x 1 x 2 x 3 x 4 s 1 s 2 s 3 z [ 1 2 3 1 1 0 0 0 115 2 5 1 5 8 5 1 0 1 5 0 0 40 1 0 1 0 0 0 1 0 50 − 1 − 1 − 4 − 5 0 0 0 1 0 ] ⋅ 1 5 R 2 Now use row operations to make the entry in row 1 , column 4 , a 0

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Author: MULLINS

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Chapter 7.4, Problem 36E

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The solution of the given problem using simplex method.

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Chapter 7.4, Problem 35E

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