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For the Min LP min w = 24 y1-15 y2 + 27 3 s.t. C1: 8y1 + 6y2 - 4y3 >= 48 C2: y18y2 + 10y3 >= 20 C3: y1 + y2 + y3 <= 15 we use the Two Phase method, and at the end of Phase I we obtain this tableau with BV = {y1, y3, s3}: w 1 0. 0 0 yr Y2 93 0 0 0 1 1/3 0 0 -5/6 1 0 3/2 0 €1 0 in -5/42 1/84 3/28 €2 83 0 0 -1/21 0 -2/21 0 1/7 1 -3/28 -1/7 N a1 -1 At the START of Phase II, the simplex tableau (i.e. fixed rowo) corresponding to BV = {y1, y3, s3} will have a starting value of w equal to -1 5/42 1/21 20 -1/84 2/21 4 (i.e. rhs of rowo) X/N2

For the Min LP min w = 24 y1-15 y2 + 27 3 s.t. C1: 8y1 + 6y2 - 4y3 >= 48 C2: y18y2 + 10y3 >= 20 C3: y1 + y2 + y3 <= 15 we use the Two Phase method, and at the end of Phase I we obtain this tableau with BV = {y1, y3, s3}: w 1 0. 0 0 yr Y2 93 0 0 0 1 1/3 0 0 -5/6 1 0 3/2 0 €1 0 in -5/42 1/84 3/28 €2 83 0 0 -1/21 0 -2/21 0 1/7 1 -3/28 -1/7 N a1 -1 At the START of Phase II, the simplex tableau (i.e. fixed rowo) corresponding to BV = {y1, y3, s3} will have a starting value of w equal to -1 5/42 1/21 20 -1/84 2/21 4 (i.e. rhs of rowo) X/N2

Practical Management Science
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter2: Introduction To Spreadsheet Modeling
Section: Chapter Questions
Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...
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For the Min LP min w = 24 y1 - 15 y2 + 27 y3 s.t. C1: 8y1 + 6y2 - 4y3 >= 48 C2: y1 8y2 + 10y3 >= 20 C3: y1+ y2 + y3 <= 15 we use the Two Phase method, and at the end of Phase I we obtain this tableau with BV = {y1, y3, s3}: 'W' y/1 1 0 0 1 0 0 -5/6 0 0 3/2 Y2 0 €1 0 0 1/3 0 -5/42 1 1/84 03/28 At the START of Phase II, the simplex tableau (i.e. fixed row0) corresponding to BV = {y1, y3, s3} will have a starting value of w equal to this same tableau is optimal N €2 83 a1 0 0 -1 ( -1/21 0 5/42 1/21 20 -2/21 0 -1/84 2/21 2/21 4/ 1/7 1 -3/28 -1/7 7 02 −1 (i.e. rhs of rowO) Y/N? rł
Transcribed Image Text:For the Min LP min w = 24 y1 - 15 y2 + 27 y3 s.t. C1: 8y1 + 6y2 - 4y3 >= 48 C2: y1 8y2 + 10y3 >= 20 C3: y1+ y2 + y3 <= 15 we use the Two Phase method, and at the end of Phase I we obtain this tableau with BV = {y1, y3, s3}: 'W' y/1 1 0 0 1 0 0 -5/6 0 0 3/2 Y2 0 €1 0 0 1/3 0 -5/42 1 1/84 03/28 At the START of Phase II, the simplex tableau (i.e. fixed row0) corresponding to BV = {y1, y3, s3} will have a starting value of w equal to this same tableau is optimal N €2 83 a1 0 0 -1 ( -1/21 0 5/42 1/21 20 -2/21 0 -1/84 2/21 2/21 4/ 1/7 1 -3/28 -1/7 7 02 −1 (i.e. rhs of rowO) Y/N? rł
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