(a) Explanation Given: (a) The distributor D 1 needs at least 3000 barrels of oil. (b) Distributor D 2 needs at least 5000 barrels of oil. (c) The two outlets s 1 and s 2 can each furnish up to 5000 barrels of oil. (d) The shipment cost per barrel is given by Outlets/Distributor D 1 D 2 s 1 $30 $20 s 2 $25 $22 (e) Southwestern Oil is willing to spend no more than $40,000 on shipment taxes. The shipment tax for each barrel is given by Outlets/Distributor D 1 D 2 s 1 $2 $6 s 2 $5 $4 Explanation: Let x 1 be the totaloil barrels supplied by s 1 to the distributor D 1 . Let x 2 be the total oil barrels supplied by s 1 to the distributor D 2 . Let x 3 be the total oil barrels supplied by s 2 to the distributor D 1 . Let x 4 be the totaloil barrels supplied by s 2 to the distributor D 2 . Use the given statements to form the following system of linear equalities: x 1 + x 3 ≥ 3000 x 2 + x 4 ≥ 5000 x 1 + x 2 ≤ 5000 x 3 + x 4 ≤ 5000 2 x 1 + 6 x 2 + 5 x 3 + 4 x 4 ≤ 40 , 000 The objective function to be minimized is 30 x 1 + 20 x 2 + 25 x 3 + 22 x 4 = z . Assume an objective function w such that w = − z . Then, the maximum value of w will be equal to the minimum value of z . Introduce surplus variables c 1 and c 2 . Introduce slack variables c 3 , c 4 , and c 5 . The system of linear inequalities becomes x 1 + x 3 − c 1 = 3000 x 2 + x 4 − c 2 = 5000 x 1 + x 2 + c 3 = 5000 x 3 + x 4 + c 4 = 5000 2 x 1 + 6 x 2 + 5 x 3 + 4 x 4 + c 5 = 40 , 000 w + 30 x 1 + 20 x 2 + 25 x 3 + 22 x 4 = 0 Also x 1 , x 2 , x 3 , x 4 ≥ 0 This produces a tableau given by [ x 1 x 2 x 3 x 4 c 1 c 2 c 3 c 4 c 5 z 1 0 1 0 − 1 0 0 0 0 0 3000 0 1 0 1 0 − 1 0 0 0 0 5000 1 1 0 0 0 0 1 0 0 0 5000 0 0 1 1 0 0 0 1 0 0 5000 2 6 5 4 0 0 0 0 1 0 40 , 000 30 20 25 22 0 0 0 0 0 1 0 ] _ In this case, the basic variable c 1 is negative, hence choose an element farthest to the left to make it a pivot element. Use the ratio test to find the pivot. 3000 1 = 3000 , 5000 1 = 5000 , 40 , 000 2 = 20 , 000 Therefore, 1 is the pivot as the least positive ratio is obtained for this element. Use the first element in the first row as the pivot. Use row operations R 3 − R 1 → R 3 , R 5 − 2 R 1 → R 5 , R 6 − 30 R 1 → R 6 to change all the remaining elements in the pivot column to 0. [ x 1 x 2 x 3 x 4 c 1 c 2 c 3 c 4 c 5 z 1 0 1 0 − 1 0 0 0 0 0 3000 0 1 0 1 0 − 1 0 0 0 0 5000 0 1 − 1 0 1 0 1 0 0 0 2000 0 0 1 1 0 0 0 1 0 0 5000 0 6 3 4 2 0 0 0 1 0 34 , 000 0 20 − 5 22 30 0 0 0 0 1 − 90 , 000 ] _ The basic variable c 2 is negative, hence choose the positive entry farthest on the left on the second row. Use ratio test to find the pivot. 5000 1 = 5000 , 2000 1 = 2000 , 34 , 000 6 = 5666.6 Therefore, 1 is the pivot as the least positive ratio is obtained for this element. Use the second element in the third row as the pivot. Use row operations R 2 − R 3 → R 2 , R 5 − 6 R 3 → R 3 , R 6 − 20 R 3 → R 6 to change all the remaining elements in the pivot column to 0. [ x 1 x 2 x 3 x 4 c 1 c 2 c 3 c 4 c 5 z 1 0 1 0 − 1 0 0 0 0 0 3000 0 0 1 1 − 1 − 1 − 1 0 0 0 3000 0 1 − 1 0 1 0 1 0 0 0 2000 0 0 1 1 0 0 0 1 0 0 5000 0 0 9 4 − 4 0 − 6 0 1 0 22 , 000 0 0 15 22 10 0 − 20 0 0 1 − 130 , 000 ] _ The basic variable c 2 is negative, hence choose the positive entry farthest on the left on the second row. Use ratio test to find the pivot. 3000 1 = 3000 , 3000 1 = 3000 , 2000 − 1 = − 2000 , 5000 1 = 5000 , 22 , 000 9 = 2444.4 Therefore, 9 is the pivot as the least positive ratio is obtained for this element. Use the third element in the fifth row as the pivot. Use row operation 1 9 R 5 → R 5 to change the pivot element to 1 and use row operations R 1 − R 5 → R 1 , R 2 − R 5 → R 2 , R 3 + R 5 → R 3 , R 4 − R 5 → R 4 , R 6 − 15 R 5 → R 6 to change all the remaining elements in the pivot column to 0. [ x 1 x 2 x 3 x 4 c 1 c 2 c 3 c 4 c 5 z 1 0 1 0 − 1 0 0 0 0 0 3000 0 0 1 1 − 1 − 1 − 1 0 0 0 3000 0 1 − 1 0 1 0 1 0 0 0 2000 0 0 1 1 0 0 0 1 0 0 5000 0 0 1 4 9 − 4 9 0 − 2 3 0 1 9 0 22 , 000 9 0 0 15 22 10 0 − 20 0 0 1 − 130 , 000 ] [ x 1 x 2 x 3 x 4 c 1 c 2 c 3 c 4 c 5 z 1 0 0 − 4 9 − 5 9 0 2 3 0 − 1 9 0 5000 9 0 0 0 5 9 − 5 9 − 1 − 1 3 0 − 1 9 0 5000 9 0 1 0 4 9 5 9 0 1 3 0 1 9 0 40 , 000 9 0 0 0 5 9 4 9 0 2 3 1 − 1 9 0 23 , 000 9 0 0 1 4 9 − 4 9 0 − 2 3 0 1 9 0 22 , 000 9 0 0 0 46 3 50 3 0 − 10 0 − 5 3 1 − 500 , 000 3 ] _ Still, the basic variable c 2 is negative, hence choose the positive entry farthest on the left on the second row (b) To determine The value of any nonzero slack or surplus variables and interpret it.

11th Edition

ISBN: 9780321979438

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 4.4, Problem 23E

(a)

To determine

The way how oil should be supplied so that the shipment cost is minimum.

(b)

To determine

The value of any nonzero slack or surplus variables and interpret it.

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Chapter 4.4, Problem 22E

Chapter 4.4, Problem 24E

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Chapter 4 Solutions

Finite Mathematics (11th Edition)

Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 2E Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 4E Ch. 4.1 - For Exercises 5-8. (a) determine the number of...Ch. 4.1 - Prob. 6E Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - Introduce slack variables as necessary, then...Ch. 4.1 - Introduce slack variables as necessary, then write...

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The way how oil should be supplied so that the shipment cost is minimum.

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The way how oil should be supplied so that the shipment cost is minimum.

The value of any nonzero slack or surplus variables and interpret it.

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