5. Consider the transportation problem having the following parameter table: Destination 1 Destination 2 Destination 3 Source 1 Source 2 Demand The initial BF solution given by Source 1 Source 2 Demand 3 2.9 3 X11 = 3, X12 = 2, X22=2, X23 = 2. Interactively apply the transportation simplex method to obtain an optimal solution. The initial transportation simplex tableau is Destination 1 Destination 2 2.7 Vj 3 2.9 3 3 2.7 2.8 4 2.8 4 2 2 Destination 3 0 0 0 2 0 2 2 supply 5 4 supply 5 4 U₁ For each iteration tableau, clearly mark: U₁, V, values, C₁j+ or negative values, the loop, the entering BV, the leaving BV. At the end of each iteration, also write out clearly what the new BF solution, and the new objection value.

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Chapter2: Introduction To Spreadsheet Modeling
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5. Consider the transportation problem having the following parameter table:
Destination 1
Destination 2
Destination 3
Source 1
Source 2
Demand
The initial BF solution given by
Source 1
Source 2
Demand
Vj
3
2.9
3
X11 = 3, X12 = 2, X22= 2, X23 = 2.
Interactively apply the transportation simplex method to obtain an optimal solution.
The initial transportation simplex tableau is
Destination 1
Destination 2
3
2.9
3
3
2.7
2.8
4
2.7
2.8
4
2
2
0
0
0
2
Destination 3 supply
5
0
2
supply
5
4
2
4
Ui
For each iteration tableau, clearly mark: U₁, V₁ values, č¡¡ + or negative values, the loop, the
entering BV, the leaving BV. At the end of each iteration, also write out clearly what the new BF
solution, and the new objection value.
Transcribed Image Text:5. Consider the transportation problem having the following parameter table: Destination 1 Destination 2 Destination 3 Source 1 Source 2 Demand The initial BF solution given by Source 1 Source 2 Demand Vj 3 2.9 3 X11 = 3, X12 = 2, X22= 2, X23 = 2. Interactively apply the transportation simplex method to obtain an optimal solution. The initial transportation simplex tableau is Destination 1 Destination 2 3 2.9 3 3 2.7 2.8 4 2.7 2.8 4 2 2 0 0 0 2 Destination 3 supply 5 0 2 supply 5 4 2 4 Ui For each iteration tableau, clearly mark: U₁, V₁ values, č¡¡ + or negative values, the loop, the entering BV, the leaving BV. At the end of each iteration, also write out clearly what the new BF solution, and the new objection value.
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