Explanation Given information: The objective function which has to minimize is: M = 20 x + 30 y which is subjected to the constraints: { x + 4 y ≥ 8 x + y ≥ 5 2 x + y ≥ 7 x ≥ 0 y ≥ 0 Formula used: The minimization problem of simplex can be solved as below: Step 1: The minima problem can be converted into maxima problem by multiplying ( − 1 ) on both sides Step 2: Write the initial tableau and find the feasible solutions. Step 3: Make the all elements of last column positive by row operations. Step 4: Find the pivot column, by finding maximum negative number in the last row of the tableau. Step 5: Now divide the all elements which are in the last column by their corresponding elements of pivot column to decide pivot elements. Step 6: Now to make the pivot element unity, divide all the element of pivot row by 1 and make all other elements of pivot column to zero by row operation. Step 7: If there is one or more than one negative number in the last row, then the above tableau is not optimal, it needs iteration. Calculation: 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 = − 20 x − 30 y . So, the objective function which has to maximize is M = − 20 x − 30 y . And, revised constraints are: { − x − 4 y ≤ − 8 − x − y ≤ − 5 − 2 x − y ≤ − 7 x ≥ 0 y ≥ 0 Write the provided constraints in the form of linear equations by introducing slack variables, u , v and w which are: − x − 4 y + u = − 8 − x − y + v = − 5 − 2 x − y + w = − 7 20 x + 30 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 w M [ − 1 − 4 1 0 0 0 − 8 − 1 − 1 0 1 0 0 − 5 − 2 − 1 0 0 1 0 − 7 20 30 0 0 0 1 0 ] Step 2: Make the all elements of last column positive till all the elements of last column become positive. The choices to make positive are − 8 , − 5 or − 7 . Step 3: Chose the 3 rd row as a pivot row, and pivot element can be any element which can make the element − 7 positive, so the choice can be − 2 . x y u v w M u v w M [ − 1 − 4 1 0 0 0 − 8 − 1 − 1 0 1 0 0 − 5 − 2 − 1 0 0 1 0 − 7 20 30 0 0 0 1 0 ] Step 4: Now to make the pivot element unity, divide all the element of pivot row by − 2 and make all other elements of pivot column to zero by row operation which is as below: x y u v w M [ − 1 − 4 1 0 0 0 − 8 − 1 − 1 0 1 0 0 − 5 1 1 2 0 0 − 1 2 0 7 2 20 30 0 0 0 1 0 ] R 1 + 1 × R 3 R 2 + 1 × R 3 R 4 − 20 × R 3 → [ 0 − 7 2 1 0 − 1 2 0 − 9 2 0 − 1 2 0 1 − 1 2 0 − 3 2 1 1 2 0 0 − 1 2 0 7 2 0 20 0 0 10 1 − 70 ] So, the next tableau for pivot − 7 is: u v x M [ 0 − 7 2 1 0 − 1 2 0 − 9 2 0 − 1 2 0 1 − 1 2 0 − 3 2 1 1 2 0 0 − 1 2 0 7 2 0 20 0 0 10 1 − 70 ] Now the choices are − 9 2 o r − 3 2 to make them positive

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ISBN: 9780134464053

Author: HAIR

Publisher: YUZU

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Chapter 4, Problem 7RE

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To calculate: The minimum value of objective function: M=20x+30y which is subjected to the constraints:

{x+4y≥8x+y≥52x+y≥7x≥0y≥0

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Chapter 4, Problem 6RE

Chapter 4, Problem 8RE

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The minimum value of objective function: M = 20 x + 30 y which is subjected to the constraints: { x + 4 y ≥ 8 x + y ≥ 5 2 x + y ≥ 7 x ≥ 0 y ≥ 0

Solution Summary: The author explains how the minimization problem of simplex can be solved by multiplying (-1) on both sides.

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To calculate: The minimum value of objective function: M=20x+30y which is subjected to the constraints:

{x+4y≥8x+y≥52x+y≥7x≥0y≥0

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Chapter 4 Solutions

The minimum value of objective function: M = 20 x + 30 y which is subjected to the constraints: { x + 4 y ≥ 8 x + y ≥ 5 2 x + y ≥ 7 x ≥ 0 y ≥ 0

Solution Summary: The author explains how the minimization problem of simplex can be solved by multiplying (-1) on both sides.

Videos

To calculate: The minimum value of objective function: M=20x+30y which is subjected to the constraints: {x+4y≥8x+y≥52x+y≥7x≥0y≥0

Want to see the full answer?

Chapter 4 Solutions

To calculate: The minimum value of objective function: M=20x+30y which is subjected to the constraints:

{x+4y≥8x+y≥52x+y≥7x≥0y≥0