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The following LP formulation represents a transportation problem where raw material (in tons) is transported from Supplier i, i={1,2,3) to Plant j, j={1,2,3). The problem is solved and the optimal objective function value is 5200. Using the sensitivity report shown below, what would be the optimal total cost if the cost of transporting 1 ton from Supplier 2 to Plant 3 becomes 5? Min Total cost = 1x11 + 3x12 + 5x13+3.5x21 +4x22+4.8x23 +3.5x31+3.6x32 +3.2x33 Subiect to

The following LP formulation represents a transportation problem where raw material (in tons) is transported from Supplier i, i={1,2,3) to Plant j, j={1,2,3). The problem is solved and the optimal objective function value is 5200. Using the sensitivity report shown below, what would be the optimal total cost if the cost of transporting 1 ton from Supplier 2 to Plant 3 becomes 5? Min Total cost = 1x11 + 3x12 + 5x13+3.5x21 +4x22+4.8x23 +3.5x31+3.6x32 +3.2x33 Subiect to

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter6: Linear Systems
Section6.8: Linear Programming
Problem 40E
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The following LP formulation represents a transportation problem where raw material (in tons) is transported from Supplier i, i={1,2,3) to Plant j, j={1,2,3). The problem is solved and the optimal objective function value is 5200. Using the sensitivity report shown below, what would be the optimal total cost if the cost of transporting 1 ton from Supplier 2 to Plant 3 becomes 5? Min Total cost = 1x11 + 3x12 + 5x13+3.5x21 +4x22+4.8x23 +3.5x31+3.6x32 +3.2x33 Subject to x31+x32+x33 <= 500 x11+x21+x31 = 500 x12+x22+x32= 700 x13+x23+x33 = 600 xij<=200, for i= {2,3} &j= {1, 2, 3) xij >= 0, for i= {(1,2,3) X &j= {1, 2, 3) Final Reduced Objective Allowable Allowable Name Value Cost Coefficient Increase Decrease 500 0 1 2.5 1E+30 700 0 3 0.6 1E+30 0.2 200 5 0 2.5 1E+30 1E+30 1E+30 0.2 1E+30 1 IE+30 2.5 1E+30 0.6 1E+30 1.8 Variables X11 X12 X13 X21 X22 X23 X31 X32 X33 0 200 0 0 200 0 0 0 0 0 0 0 3.5 4 4.8 3.5 3.6 3.2
Transcribed Image Text:The following LP formulation represents a transportation problem where raw material (in tons) is transported from Supplier i, i={1,2,3) to Plant j, j={1,2,3). The problem is solved and the optimal objective function value is 5200. Using the sensitivity report shown below, what would be the optimal total cost if the cost of transporting 1 ton from Supplier 2 to Plant 3 becomes 5? Min Total cost = 1x11 + 3x12 + 5x13+3.5x21 +4x22+4.8x23 +3.5x31+3.6x32 +3.2x33 Subject to x31+x32+x33 <= 500 x11+x21+x31 = 500 x12+x22+x32= 700 x13+x23+x33 = 600 xij<=200, for i= {2,3} &j= {1, 2, 3) xij >= 0, for i= {(1,2,3) X &j= {1, 2, 3) Final Reduced Objective Allowable Allowable Name Value Cost Coefficient Increase Decrease 500 0 1 2.5 1E+30 700 0 3 0.6 1E+30 0.2 200 5 0 2.5 1E+30 1E+30 1E+30 0.2 1E+30 1 IE+30 2.5 1E+30 0.6 1E+30 1.8 Variables X11 X12 X13 X21 X22 X23 X31 X32 X33 0 200 0 0 200 0 0 0 0 0 0 0 3.5 4 4.8 3.5 3.6 3.2
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