IOE310_AdditionalProblems_B_Solutions

IOE310_AdditionalProblems_B_Solutions - IOE 310 -...

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Unformatted text preview: IOE 310 - Additional Problems B For additional practice only. Not collected. Problem 1 Consider the following LP: max s.t z = 5x1 + 3x2 + x3 2 x1 + x2 + x3 x 1 + 2x 2 + x 3 x1 , x2 , x3 ≤6 ≤7 ≥0 (a) Graphically solve this problem. To graph this in two-dimensions we need to find the dual: min s.t. w = 6y1 2y1 y1 y1 y1 , + 7y2 + y2 + 2y2 + y2 y2 ≥5 ≥3 ≥1 ≥0 IOE 310 – Additional Problems B Page 2 of 9 (b) Use complementary slackness to solve the max problem. Graphically we find the dual’s optimal solution to be w = 49/3, y1 = 7/3, y2 = 1/3. The fact that e3 > 0, y1 > 0, and y2 > 0 shows that the optimal primal solution must have both constraints binding and have x3 = 0. Thus the optimal values of x1 and x2 may be found by solving 2x1 + x2 = 6 and x1 + 2x2 = 7. This yields the optimal primal solution z = 49/3, x1 = 5/3, x2 = 8/3, x3 = 0. Note that the optimal objective function values are equal. Problem 2 Consider the following LP and its optimal tableau: max s.t z = 5x1 + x2 + 2x3 x1 + x2 + x3 6x1 + +x3 x2 + x3 x1 , x2 , x3 x3 0 0 0 1 s1 0 1 0 0 s2 5/6 -1/6 1/6 0 ≤6 ≤8 ≤2 ≥0 s3 7/6 -5/6 -1/6 1 rhs 9 3 1 2 z 1 0 0 0 x1 0 0 1 0 x2 1/6 1/6 -1/6 1 (a) Find the dual to this LP and the optimal solution to the dual. min s.t. w = 6y1 y1 y1 y1 y1 , + 8y2 + 6y2 + y2 y2 + 2y3 + y3 + y3 y3 ≥ 0 ≥5 ≥1 ≥2 Optimal dual solution is w = 9, y1 = 0, y2 = 5/6, y3 = 7/6. (b) Find the range of values of c2 for which the current basis remains optimal. Since x2 is non-basic, c2 can increase by at most the reduced cost for the current basis to remain optimal. Thus current basis remains optimal as long as the reduced cost of c2 is positive. This can be expressed ¯ as c2 = cBV V −1 a2 − c2 . We can also write this expression as: c2 = [y1 y2 y3 ][a12 a22 a32 ]T − c2 . This is then ¯ ¯ c2 = [0 5/6 7/6][1 0 1]T − c2 . In order for this expression to be ≥ 0 we see that c2 must be ≤ 7/6. ¯ This additional problem set was developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Additional Problems B Page 3 of 9 Problem 3 Consider the following LP and its optimal tableau: max s.t z = 3x1 + x2 − x3 2 x1 + x2 + x3 4x1 + x2 − x3 x1 , x2 , x3 x2 0 1 0 x3 1 3 -1 s1 1/2 2 -1/2 ≤8 ≤ 10 ≥0 rhs 9 6 1 z 1 0 0 x1 0 0 1 s2 1/2 -1 1/2 (a) Find the dual to this LP and the optimal solution to the dual. min s.t. w = 8y1 2y1 y1 y1 + 10y2 + 4y2 + y2 - y2 ≥3 ≥1 ≥1 The optimal dual solution is w = 9, y1 = y2 = 1/2. (b) Find the range of values of b2 for which the current basis remains optimal. If b2 = 12, what is the new optimal solution? ￿￿ ￿ 8 0 B ≥ 10 + ∆ 0 or ￿￿ ￿￿ ￿￿ ￿ −1 8 6−∆ 0 = ≥ 1/2 10 + ∆ 1 + ∆/2 0 or −2 ≤ ∆ ≤ 6 or 8 ≤ b2 ≤ 16. −1 ￿ ￿ 2 −1/2 If b2 = 10, then ∆ = 2 and the new optimal solution may be obtained from ￿ x2 x1 ￿ = ￿ 6−2 1 + (2/2) ￿ = ￿ 4 2 ￿ You can see that we can read B −1 directly from the optimal tableau. The matrix values are given by the ￿ ￿ 2 −1 values in the tableau for the slack variables s1 and s2 and give us . −1/2 1/2 See Section 6.7 in the text book for more on this topic (Dual Theorem) (p.307-313). This additional problem set was developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Additional Problems B Page 4 of 9 Problem 4 Use the Theorem of Complementary Slackness to find the optimal solution to the following LP and its dual: max s.t z = 3x1 + 4x2 + x3 + 5x4 x 1 + 2x 2 + x 3 + 2x 4 2x 1 + 3 x 2 + x 3 + 3 x 4 x1 , x2 , x3 , x4 ≤5 ≤8 ≥0 The dual problem is: min s.t. w = 5y1 y1 2y1 y1 2y1 y1 , + 8y2 + 2y2 + 3y2 + y2 + 3y2 y2 , ≥3 ≥4 ≥1 ≥5 ≥0 (1) (2) (3) (4) Trying (1) and (2) binding yields an infeasible solution. Trying (1) and (3) binding yields an infeasible solution. Trying (1) and (4) binding yields y1 = y2 = 1, w = 13. This solution has e2 = e3 = 0. Thus we try a primal solution with both constraints binding and x2 = x3 = 0. This yields the feasible primal solution x1 = 1, x2 = x3 = 0, x4 = 2, z = 13. Since the z and w values match we know that we have found the optimal solution to both the primal and dual. Note that we only need to consider 2 constraints at a time to find an intersection (corner) in 2 Dimensions which will be a feasible solution. We do not need to go through all the combinations once we find one that works because then we have found the optimal already. Problem 5 Find the dual of the following problem: max s.t z = 4x1 + x2 x 1 + 2x 2 x1 − x2 2x1 + x2 x1 , x2 =6 ≥3 ≤ 10 ≥0 + 10y3 + 2y3 + y3 y3 ≥ 0 The dual is: min s.t. w = 6y1 y1 2y1 y1 urs, + 3y2 + y2 + y2 y2 ≤ 0, ≥4 ≥1 This additional problem set was developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Additional Problems B Page 5 of 9 Problem 6 Vivian’s Gem Company produces two types of gems: Types 1 and 2. Each Type 1 gem contains 2 rubies and 4 diamonds. A Type 1 gem sells for $ 10 and costs $ 5 to produce. Each Type 2 gem contains 1 ruby and 1 diamond. A Type 2 gem sells for $ 6 and costs $ 4 to produce. A total of 30 rubies and 50 diamonds are available. All gems that are produced can be sold, but marketing considerations dictate that at least 11 Type 1 gems be produced. Let x1 = number of Type 1 gems produced and x2 = number of Type 2 gems produced. Assume that Vivian wants to maximize profit. We can formulate this problem as: max s.t 5x 1 + 2 x 2 2x1 + x2 4x1 + x2 x1 ≥0 x1 , x2 ≤ 30 ≤ 50 ≥ 11 Plugging this into Excel and getting the Sensitivity Report we see: (a) What would Vivian’s profit be if 46 diamonds were available? Allowable Decrease is 6, so New Profit = 67 − 4(2) = $ 59 (b) What would Vivian’s profit be if at least 12 Type 1 gems had to be produced? Allowable Increase = 1.5 so we may conclude that New Profit = 67 + 1(3) = $ 64. This additional problem set was developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Additional Problems B Page 6 of 9 Problem 7 Find the dual of the following LP: max s.t z = 4x1 + x2 + 2x3 8 x 1 + 3x 2 + x 3 6x1 + x2 + x3 x1 , x2 , x3 ≤2 ≤8 ≥0 The dual is: min s.t. w = 2y1 8y1 3y1 y1 y1 , + 8y2 + 6y2 + y2 + y2 y2 ≥4 ≥1 ≥2 ≥0 Problem 8 Consider the following LP: max s.t c 1 x1 + c 2 x2 3x1 + 4x2 2x1 + 3x2 x1 , x2 ≤6 ≤4 ≥0 The optimal tableau for this LP is: z 1 0 0 x1 0 1 0 x2 0 0 1 s1 1 3 -2 s2 2 -4 3 rhs 14 2 0 Without doing any pivots, determine c1 and c2 . ￿ 3 −4 = [1 2] or −2 3 3c1 − 2c2 = 1 4c 1 + 3 c 2 = 2 Solving these equations we find that c1 = 7 and c2 = 10. cBV B −1 = [12]. Thus [c1 c2 ] ￿ This additional problem set was developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Additional Problems B Page 7 of 9 Problem 9 Find the dual of the following LP: max s.t x 1 + 2x 2 z = −4x1 − x2 4 x 1 + 3x 2 ≤3 3x1 + x2 x1 , x2 ≥6 =3 ≥0 The dual is: min s.t. w = 6y1 4y1 3y1 y1 ≤ 0, + 3y2 + y2 + 2y2 y2 ≥ 0, + 3y3 + 3y3 + y3 y3 urs ≥ −4 ≥ −1 Problem 10 Let’s pretend that we are in a 1 dimensional world. What this means is that the only distance we travel is along a line. You and two of your friends live along this line in our one dimensional world. You all decide that one day you want to meet up for lunch. You are trying to decide where you meet for lunch. You want to choose one location along this line that minimizes the total distance you all travel. You each live at locations l1 , l2 , and l3 along this line. Point x is somewhere on this line (it can be between you and one of your friends or between them, or wherever). Let’s say that a ≤ x ≤ b; this is the only constraint. The objective function for this problem is then: min |x − l1 | + |x − l2 | + |x − l3 |. Reformulate this problem as an LP. (Hint: There are more than one alternatives for re-writing the objective function here.) min |x − l1 | + |x − l2 | + |x − l3 | Can be rewritten using artificial variables and adding constraints associated with them. (Note there are more ways to reformulate this slightly differently.) min u+ + u− + u+ + u− + u+ + u− 1 1 2 2 3 3 s.t. u+ ≥ x − l1 1 u+ ≥ 0 1 u− ≥ − x + l1 1 u− ≥ 0 1 u+ ≥ x − l2 2 u+ ≥ 0 2 u− ≥ − x + l2 2 u− ≥ 0 2 u+ ≥ x − l3 3 u+ ≥ 0 3 u− ≥ − x + l3 3 u− ≥ 0 3 a≤x≤b This additional problem set was developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Additional Problems B Page 8 of 9 Problem 11 Happy Birthday! You and your roommates are planning to have a dinner party for your birthday this weekend. You are inviting 18 guests (you and your roommate included). You are going to make 5 different drinks to entertain your guests for the evening. These are: Simosas, Lutinent Cokes, Minty Hot Coco, Two-tone Lemonade, and a drink your roommate coined called the Kitchen Sink. The ingredients for these are listed (including the quantites needed and the amount of each of them that you have in your house) in the screen shot of the excel file used to solve the problem below. You know that all of your guests will at least need one drink and 1/3 of them will for sure want another. Any extra drinks you make WILL be finished, and you cannot run out. Your goal for the night, even though it is your birthday, is to maximize the enjoyment of your guests. Each drink you make serves a different number of guests and brings each of those guests a certain amount of enjoyment. To help yourself out, you quickly put together an LP in Excel which maximizes the enjoyment of you and your guests and solved it with Solver you got the following solution. This additional problem set was developed for IOE 310, fall semester 2010. This document may not be copied or distributed. IOE 310 – Additional Problems B Page 9 of 9 You did a sensitivity analysis too and found the following: (a) What is the optimal solution (how many of each drink to make) and the optimal objective value? Make 2 Simosas, 2 Lutinent Cokes, 1.5 Two-Toned Lemonades, and 5 servigs of the Kitchen Sink. These will bring your guests 99 enjoyment ’points.’ (b) Your roommate called from Kroger and said Sprite is on sale for $ 1. Do you tell him to buy any? Yes, absolutely! (c) What if it wasn’t on sale and was regular price for $ 2.50? No, that price is above the shadow price so you would be paying more than the enjoyment you deliver. (d) What is the economic interpretation of the shadow price of the mint cream syrup? You are willing to pay up to $ 6.50 for one more part of the mint cream syrup. (e) If of the 8 units of the lemonade you have available, you slip and spill 2. Does this change which drinks you will make? (not how many, but which) No, the allowable decrease for Lemonade is 3 so spilling 2 units won’t change which drinks you make, however, you will make different amounts of the ones you do make already. (f) If Mindy Hot Coco brought each guest 3 enjoyment points rather than 1, would you have made it instead? No. (See allowable increase.) This additional problem set was 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.

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