asst4sol - C&O 350 Linear Optimization Winter 2009...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Question # Max. marks Part marks 1 5 1 1 1 2 2 5 2 3 3 7 3 4 4 43 18 13 12 5 No Credit 6 No Credit 7 No Credit Total 60 1. 5 marks = 1+1+1+2 The following is a tableau of a maximization problem, where the entries α , β , γ , δ and ε in the tableau are unknown parameters. z + δx 4 + γx 5 = 0 x 2 + αx 4 = 3 x 3 - 2 x 4 + εx 5 = 2 x 1 + 2 x 5 = β For each of the following statements, find conditions on the values of α , β , γ , δ and ε such that the statement is true. (a) The basic solution is not feasible. (b) The basic solution is feasible, but the tableau is not optimal. (c) The basic solution is feasible and the next pivot, under the smallest subscript rule (see Section 6.6), indicates that the problem is unbounded. (d) The basic solution is feasible, x 5 is a candidate for entering the basis, and when x 5 enters the basis, then x 3 leaves the basis. Solution : (a) β < 0. The basic solution x * defined by x * i = b i for i B and x * i = 0 for i / B is feasible if and only if x * 0. (b) β 0 and either δ < 0 or γ < 0. A tableau is optimal if and only if it is feasible and c j 0 for all j . (c) β 0, δ < 0 and α 0. To satisfy the simplex “test” for unboundedness, the entering column k must have a ik 0 for all i B . Thus we must have k = 4. (d) β 0, γ < 0, ± > 0 and β± 4. Since x 5 enters, we must have c 5 = - γ > 0. Since x 3 leaves, we must have a 35 = ± > 0, and by the ratio test b 3 / a 35 = min { b i / a i 5 : i B, a i 5 > 0 } . 2. 5 marks = 2+3 (Based on Exercises 6.9.5 and 6.9.6 in the course notes) 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(a) Give an example to show that the entering variable on one iteration of the simplex method can become the leaving variable on the next iteration. (b) Is it possible for the leaving variable on one iteration of the simplex method to become an entering variable on the next iteration? Justify your answer. Hint : Suppose that x k enters and x r leaves on an iteration of the simplex method. Derive a formula for the new value of c r (the reduced cost of x r in the new tableau) in terms of the coefficients of the old tableau. Solution : (a) The following tableau has basis B = { 2 } . z - x 1 - 2 x 3 = 0 x 1 + x 2 + x 3 = 1 We choose x 1 as the entering variable. Then x 2 is the leaving variable, and we get a new tableau with basis B = { 1 } : z + x 2 - x 3 = 1 x 1 + x 2 + x 3 = 1 Now, the entering variable is x 3 , and the leaving variable is x 1 . This shows that the entering variable on one iteration ( x 1 in this example) can be the leaving variable on the next iteration. (b) No. The new tableau necessarily has a negative reduced cost for the leaving variable of the previous iteration, so that variable cannot be chosen as an entering variable. To see this in detail, write the formula for the old tableau, denoting the old basis by
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 11

asst4sol - C&amp;O 350 Linear Optimization Winter 2009...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online