quiz6solutions

# quiz6solutions - x 1,x 2 and two constraints is solved by...

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640:354:01, 04/25/07 Quiz #6, with solutions 1. Consider the transportation problem with the following input: C = 5 2 3 6 2 7 7 4 1 3 6 9 , s = 100 80 140 , d = 60 60 80 120 . Let x be a basic feasible solution given by x 12 = 60 , x 13 = 40 , x 24 = 80 , x 31 = 60 , x 33 = 40 , x 34 = 40 (the remaining variables have value 0). Given this x , determine which variable would be entering and which one leaving in a step of the transportation algorithm (justify). Do not compute the next solution. Solution: Set up and solve the system v i + w j = c ij for the basic variables x ij . Setting v 1 = 0, the solution is v 1 = 0, v 2 = - 2, v 3 = 3, w 1 = - 2, w 2 = 2, w 3 = 3, w 4 = 6. The values of v i + w j - c ij for the nonbasic variables are: x 11 : -7 x 14 : 0 x 21 : -6 x 22 : -7 x 23 : -6 x 32 : 2 Therefore, the entering variable is x 32 . It forms the loop x 32 - x 12 - x 13 - x 33 with the basic variables. On this loop, x 12 and x 33 appear with a minus sign, and x 33 has the smaller value. Therefore, x 33 is the leaving variable. 2. A certain LPP in standard form, with decision variables

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Unformatted text preview: x 1 ,x 2 and two constraints, is solved by introducing slack variables u 1 ,u 2 for the respective constraints. The ﬁnal tableau is x 1 x 2 u 1 u 2 x 1 1 2-1 16 x 2 1-1 3 12 1 2 28 The original right sides were 60 and 40 for the ﬁrst and the second constraint, respec-tively. If the original problem is modiﬁed by changing the 60 to 75, will optimality still occur with the same variables being basic? Justify your answer. Solution . Δ b 1 = 75-60 = 15, and the new values of x 1 and x 2 in a basic solution are • x 1 x 2 ‚ = • 16 12 ‚ + 15 · • 2-1 ‚ = • 46-3 ‚ . Therefore, the basic solution to the modiﬁed problem with x 1 and x 2 being basic is not feasible. Optimality cannot occur with x 1 and x 2 being basic. 2...
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quiz6solutions - x 1,x 2 and two constraints is solved by...

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