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Unformatted text preview: IOE 310  Problem Set 07 Solutions Suggested reading: Section 6.3 (p.275278: Changing Objctive Coeﬃcient of NonBasic Variable) (p.282284: Changing RHS of ConstraintEﬀect on Decision Variables and Z ), Section 6.5 Problem 1
Section 6.2 Problem 2 (p.275) 0 1 and B −1 = 1 −1 0 1 −1 and cBV = [1 0] and cBV B = [1 0] = [0 1] 1 −1 2 −1 Coeﬃcient of x1 in row0 = cBV B a1 − c1 = [0 1] +1=2 1 0 Coeﬃcient of s2 in row0 = cBV B −1 =1 1 4 RHS of row0 = cBV B −1 b = [0 1] =2 2 0 1 2 1 Column for x1 = B −1 a1 = = 1 −1 1 1 0 1 Column for s2 = B −1 = 1 −1 0 1 4 2 −1 RHS of Optimal Tableau = B b = = 1 −1 2 2 Thus optimal tableau is z + 2x1 + s2 = 2 x 1 + x2 + s 2 = 2 x1 + s1 + s2 = 2 BV = {x2 , s1 } and B = 1 1 1 0 Which can be written as: z 1 0 0 x1 2 1 1 x2 0 1 0 s1 0 0 1 s2 1 1 1 rhs 2 2 2 Problem 2
Section 6.3 Problem 6 (p.288) We are given the follwoing LP: IOE 310 – Problem Set 07 Page 2 of 6 max s.t z = 3x1 + 7x2 + 5x3 x1 + x2 + x3 2x1 + 3x2 + x3 x1 , x2 , x3 ≤ 50 (Sugar) ≤ 100 (Chocolate) ≥0 (a) x1 is nonbasic so changing the coeﬃcient of x1 in the objective function will only change the coeﬃcient of x1 in the optimal row0 . Let the new coeﬃcient of x1 in the objective function be 3 + ∆. The new coeﬃcient of x1 in the optimal row0 will be cBV B −1 a1 − (3 + ∆) = 3 − ∆. Thus if 3 − ∆ ≥ 0 or ∆ ≤ 3 the current basis remains optimal. Thus if proﬁt for a Type1 Candy Bar is ≤ 6 cents the current basis remains optimal. 3/2 −1/2 −1 (b)Changing Candy Bar Type2 proﬁt to 7 + ∆ changes cBV B to [5 7 + ∆] = [4 − −1/2 1/2 ∆/21 + ∆/2] Then coeﬃcient of x1 in row0 = [4 − ∆/2 1 + ∆/2] = 3 + ∆/2. Thus row0 of optimal tableau is now z + (3 + ∆/2)x1 + (4 − ∆/2)s1 + (1 + ∆/2)s2 = ? Thus current basis remains optimal if (1) − (3) are met: (1)3 + ∆/2 ≥ 0(or ≥ −6) (2)4 − ∆/2 ≥ 0(or∆ ≤ 8) (3)1 + ∆/2 ≥ 0(or∆ ≥ −2) Thus if −2 ≤ ∆ ≤ 8 the current basis remains optimal. Thus if proﬁt for Type2 Candy Bar is between 7 − 2 = 5 and 7 + 8 = 15 cents the current basis remains optimal. (c) If the amount of sugar available is changed to 50 + ∆ the current basis remains optimal if f 3/2 −1/2 −1/2 1/2 50 + ∆ 100 25 + 3∆/2 25 − ∆/2 = Thus current basis remains optimal if f (1) − (2) hold (1)25 + 3∆/2 ≥ 0(or∆ ≥ 50/3) (2)25 − ∆/2 ≥ 0(or∆ ≤ 50) Thus current basis remains optimal if f 100/3 = 50 − 50/3 ≤ (Amount of Available Sugar) ≤ 50 + 50 = 100. 60 (d) After this change the current basis is still optimal. New proﬁt = cBV B −1 b = [41] = $ 3.40 100 New values of decision variables are found from x3 x2 = B −1 b = 3/2 −1/2 −1/2 1/2 60 100 40 20 = Thus 40 Type3 Candy Bars, 20 Type2 Candy Bars, and 0 Type1 candy bars would now be manufactured. If only 30 ounces of sugar were available the current basis would no longer be optimal and we would have to resolve the problem to ﬁnd the new optimal solution. 1/2 (e) Coeﬃcient of Type1 Candy Bar in Row0 is now [4 1] − 3 = −0.5. Thus the current basis is 1/2 no longer optimal and the new optimal solution would manufacture Type1 Candy Bars. 3 (f)The coeﬃcient of Type4 Candy Bars in Row0 will now be [4 1] − 17 = −1. Thus x4 should be 4 entered into the basis and the current basis is no longer optimal. The new optimal solution will make Type4 Candy Bars. These problem set solutions were developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Problem Set 07 Page 3 of 6 Problem 3
Section 6.5 Problem 1 (p.301) We are given the following LP: max s.t z = 2x1 + x2 − x1 + x2 x1 + x2 x1 − 2x2 x1 , x2 ≤1 ≤3 ≤4 ≥0 The dual of this problem is: min s.t w = y1 + 3y2 + 4y 3 − y1 + y2 + y 3 y1 + y2 − 2y3 y1 , y 2 , y 3 ≥2 ≥1 ≥0 Problem 4
Section 6.5 Problem 4 (p.301) We are given the following LP: min s.t w = 4y1 + 2y2 − y3 y1 + 2y 2 y 1 − y2 + 2y 3 y 1 , y2 ≥ 0 ≤6 =8 y3 urs The dual of this problem is: max s.t z = 6x1 + 8x2 x1 + x2 2x 1 − x 2 2x2 x1 ≤ 0 ≤4 ≤2 = −1 x2 urs Problem 5
Suppose that you and your friend are going to start a business venture. You are going to make care packages which parents can order online to send to their students at school. You are going to produce four diﬀerent packages. The contents of each and the prices are seen in the table include below. You want to know how many of each you should produce in order to optimize your proﬁt. You both took IOE310 together and realized you could write an LP formulation that would tell you how many to produce each day. You also remembered that you can solve this type of problem using Microsoft Excel Solver. While playing in Excel, you discovered that you can generate Sensitivity Analysis reports when doing so. The Excel ﬁle and the sensitivity output are shown below (and can be downloaded from CTools if you wish to play along). These problem set solutions were developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Problem Set 07 Page 4 of 6 Sensitivity Report These problem set solutions were developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Problem Set 07 Page 5 of 6 For each of the situations below, assume that the ONLY thing changed from the original problem is the item discussed. Use the Sensitivity Report to help you answer the questions. You will not need to resolve the problem in order to answer them, but you may do so if you wish to see for yourself (Excel File is posted on CTools). (a) What is the optimal number of each type of care package to produce? Produce 6 Exam Times, 11 Movie Nights, 0 Spoiled Child and 0 Race Day packages. (b) What is the optimal objective value? The optimal objective is $ 37. (c) What happens if the Race Day package only sells for $ 9? Nothing will happen. The objective coeﬃcient will become 0.5 so we will be gettin less proﬁt for something we already don’t produce so the optimum solution does not change. (d) Would the optimal solution change if the Race Day package price increases? Yes, the optimal solution will change if the Race Day package price increases because there is an allowable increase of 0. (e) By how much must the price of the Spoiled Student care package increase by in order for it to have a positive value in the optimal solution? the Spoiled Student package must increase its price by at least $ 2.50 in order for it to be included in the optimal solution (see allowable increase). (f) What happens if we increase the price of the Exam Time package to $ 20? Doing so will increase the value of the objective function. (g) Would you be willing to purchase additional VIA (beyond the now 24) at $ 0.75 each? No, we are now currently not using all 24 so we would not want to buy more. Also, the shadow price listed is 0 so we are willing to pay at most $ 0 for more VIA. (h) If we increase the quantity of Protein Bars available from 12 to 17: (i) Will the number of each package produced change? Yes (ii) Will the packages which we produce change (i.e. if we now produce xE and xM will that change)? No, the allowable increase is 6, so increasing by only 5 will change the quantities of each produced, but not what types we produce. (iii) Will the optimal objective value change? Yes, the optimal objective value will change because the number of each type of package produced changes. These problem set solutions were developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Problem Set 07 Page 6 of 6 Problem 6
Let us consider a problem that is not in standard form to start. First, let us consider the case where the linear constraints are equalities (and the variables are nonnegative): max s.t n cx nj =1 j j aij xj j =1 xj = bi ≥0 ∀i = 1, 2..m ∀j = 1, 2..n The above problem has an equality constraint. We have learned, however, that this type of problem can be rewritten using inequality constraints: one ≥ and one ≤. We see this below: max s.t n cx nj =1 j j ax j =1 ij j n j =1 aij xj xj ≤ bi ≥ bi ≥0 ∀i = 1, 2..m ∀i = 1, 2..m ∀j = 1, 2..n We can next write this problem as a normal max problem (see text p.295 for a description of normal max problems) using only ≤ constraints: max s.t n cx nj =1 j j ax n j =1 ij j j =1 −aij xj xj ≤ bi ≤ − bi ≥0 ∀i = 1, 2..m ∀i = 1, 2..m ∀j = 1, 2..n Now that we have a normal max problem, we can write out its dual. We know that there are two sets of m inequality constraints so we will need two sets of m dual variables. For uniformity, let’s denote the dual + variables associated with the ﬁrst set of m constraints by yi where i = 1, 2..m and the rest of the dual − variables by yi where i = 1, 2..m. Using this notation, write the dual problem: (a) min s.t m m + − m i=1+bi yi − m =1 bi yi i − i=1 yi aij − i=1 yi aij + − y i , yi ≥ cj ≥0 j = 1, 2..n i = 1, 2..m + − Looking at your solution and given that yi = yi − yi , where i = 1, 2..m reduce the dual problem to a simpler form: (b) min s.t m i m =1 bi yi i=1 yi aij yi ≥ cj ≥0 j = 1, 2..n i = 1, 2..m These problem set solutions were developed for IOE 310, fall semester 2010. This document may not be copied or distributed. ...
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This note was uploaded on 12/20/2010 for the course IOE 310 taught by Professor Saigal during the Fall '08 term at University of Michigan.
 Fall '08
 Saigal

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