Explanation Consider a linear programming problem which is as below: Minimize: M = x + y which is subjected to the constraints: { 7 x + 5 y ≥ 40 x + 4 y ≥ 9 x ≥ 0 y ≥ 0 The minimization problem of simplex can be solved as below: Since, this is minima problem, so convert the minima problem into maxima problem by multiplying ( − 1 ) on both sides of objective function which is: ( − 1 ) × M = − x − y . So, the objective function which has to maximize is M = − x − y . And, revised constraints are: { − 7 x − 5 y ≤ − 40 − x − 4 y ≤ − 9 x ≥ 0 y ≥ 0 Write the above constraints in the form of linear equations by introducing slack variables, u , and v which are: − 7 x − 5 y + u = − 40 − x − 4 y + v = − 9 x + y + M = 0 Step 1: Write the initial tableau and find the feasible solution. In the matrix form it can be written as: x y u v M [ − 7 − 5 1 0 0 − 40 − 1 − 4 0 1 0 − 9 1 1 0 0 1 0 ] So, the above simplex tableau is initial tableau, therefore one feasible solution is: Step 2: Make the all elements of last column positive. The choice to make positive is either − 40 or − 9 . But choose − 40 because it is most negative. Step 3: Pivot element will be that element which can make minimum positive. So, ( − 40 − 7 = 5.714 ) and ( − 40 − 5 = 8 ) , the minimum positive is 5.714. In this case the pivot element is − 7 . u v M [ − 7 − 5 1 0 0 − 40 − 1 − 4 0 1 0 − 9 1 1 0 0 1 0 ] Step 4: Now to make the pivot element unity, divide all the element of pivot row by − 7 and make all other elements of pivot column to zero by row operation which is as below: x y u v M [ 1 5 7 − 1 7 0 0 40 7 − 1 − 4 0 1 0 − 9 1 1 0 0 1 0 ] R 2 + 1 × R 1 R 3 − 1 × R 1 → [ 1 5 7 − 1 7 0 0 40 7 0 − 23 7 − 1 7 1 0 − 23 7 0 2 7 1 7 0 1 −

Finite Mathematics & Its Applications (12th Edition)

12th Edition

ISBN: 9780134437767

Author: Larry J. Goldstein, David I. Schneider, Martha J. Siegel, Steven Hair

Publisher: PEARSON

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Finite Mathematics & Its Applications (12th Edition)

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