{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

View Full Document Right Arrow Icon
C&O 350 Linear Optimization – Winter 2009 – Solutions to Assignment 4 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
Image of page 1

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

View Full Document Right 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 B and using N = { j | j 6∈ B } , and denoting the entering index by k and the leaving index by r : z - c k x k - j N, j 6 = k c j x j = v x r + a rk x k + j N, j 6 = k a rj x j = b r x i + a ik x k + j N, j 6 = k a ij x j = b i , i B, i 6 = r After the pivot on ( r, k ), the z -row of the new tableau is obtained by adding c k / a rk
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern