# GE330_lect8 - Lecture 8 Algebraic Sensitivity Analysis...

This preview shows pages 1–6. Sign up to view the full content.

Lecture 8 Algebraic Sensitivity Analysis February 17, 2009

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

View Full Document
Algebraical Sensitivity Analysis TOYCO Model: TOYCO assembles three types of toys–trains, trucks and cars–using three operations. The daily limits on the available times for the three operations are 430,460, and 420 minutes, respectively, and the revenues per unit of toy train, truck, and car are \$3, \$2, and \$5, respectively. The assembly times per train at the three operations are 1, 3, 1 minutes, re- spectively. The corresponding times per train and per car are (2,0,4) and (1,2,0) minutes (a zero time indicates that the op- eration is not used.) LP Model: Let x 1 , x 2 , and x 3 be the daily number of units assembled of trains, trucks, and cars, respectively. max z = 3 x 1 + 2 x 2 + 5 x 3 s.t. 2 x 1 + x 2 + x 3 430 (Operation 1) x 1 + 2 x 3 460 (Operation 2) x 1 + 4 x 2 420 (Operation 3) x 1 , x 2 , x 3 0 9
TOYCO Model: Optimal Tableau Introduce slack variables x 4 , x 5 , and x 6 , the initial tableau is: Basic x 1 x 2 x 3 x 4 x 5 x 6 Solution z -3 -2 -5 0 0 0 0 x 4 1 2 1 1 0 0 430 x 5 3 0 2 0 1 0 460 x 6 1 4 0 0 0 1 420 The optimal tableau is: Basic x 1 x 2 x 3 x 4 x 5 x 6 Solution z 4 0 0 1 2 0 1350 x 2 - 1 4 1 0 1 2 - 1 4 0 100 x 3 3 2 0 1 0 1 2 0 230 x 6 2 0 0 -2 1 1 20 10

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

View Full Document
Changes in the Right-Hand Side After adding slack variables, rewrite the constraints as follows: x 1 + 2 x 2 + x 3 = 430 - x 4 (Operation 1) 3 x 1 + 2 x 3 = 460 - x 5 (Operation 2) x 1 + 4 x 2 = 420 - x 6 (Operation 3) We can say that a one-unit decrease in the slack variables is equivalent to a one-unit increase in the resource (operations time). On the other hand, from the optimal tableau we know that: z + 4 x 1 + x 4 + 2 x 5 + 0 x 6 = 1350 , which is equivalent to z = 1350 - 4 x 1 + 1 × (increase in operation 1 time) + 2 × (increase in operation 2 time) + 0 × (increase in operation 3 time) Note: 1. In this case, the shadow price for a resource is actually the corresponding z -row coefficient in the optimal tableau! 2. The shadow price of operation 3 is zero, this is reasonable because this resource is already abundant (the slack is positive). 11
Determine the Feasibility Ranges The idea: Find the range of the changes in the right-hand side so that the basis of the optimal solution remains the same, i.e., x 2 , x 3 , and x 6 are still the basic variables in the optimal tableau after the changes.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern