11-3-17 In Class Problem Based on Winston Problem 6.6 A group of kids in your ne
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Question
11-3-17 In Class Problem Based on Winston Problem 6.6 A group of kids in your neighborhood have decided to have a bake sale where they will sell different kinds of fudge but they have asked you for help deciding how much of each type of fudge to make. All of the fudge will made with sweetened condensed milk (SCM) and chocolate. The composition of each below. type of fudge and the profit made from selling one piece of each type of fudge are given in the table Fudge Type Ounces of SCM Ounces of Chocolate Profit (cents) 7 They have only 50 ounces of SCM and 100 ounces of chocolate available to make the fudge Let xi be the number of pieces of fudge of type i that the kids will make 1) Formulate an LP that you can use to determine how much of each type of fudge the kids should make. Clearly define your variables and constraints. 2) Write the dual of this problem formulation. 3) Use the simplex method to solve the LP. a. How much of each type of fudge should they make? b. What is their total profit from the bake sale if they make the amount of fudge suggested in (a)? c. What are the optimal solution and objective function value for the dual problem? For what values of profit of fudge type 1 does the current basis (solution) remain optimal? If the profit for a piece of fudge type 1 were 7 cents, what would be the new optimal solution to the bake sale problem? or what values of profit of fudge type 2 does the current basis (solution) remain optimal? if the 5) F rofit for a piece of fudge type 2 were 13 cents, what would be the new optimal solution to the bake sale problem? 6) For what a amount of available SCM would the curre nt basis remain optimal? If 60 ounces of SCM ho many pieces of each type of fudge should the kids make and what would be the profit from the bake sale? e of type 1 fudge only used 0.5 ounces of SCM and 0.5 ounces of chocolate Should the kids make and sell type 1 fudge?Explanation / Answer
(1)
max Z = 3x1 + 7x2 + 5x3
Subject to,
1x1 + 1x2 + 1x3 <= 50 (SCM)
2x1 + 3x2 + 1x3 <= 100 (Chocolate)
x1, x2, x3 >= 0
(2)
Dual
min W = 50y1 + 100y2
Subject to,
1y1 + 2y2 >= 3
1y1 + 3y2 >= 7
1y1 + 1y2 >= 5
y1, y2 >= 0
(3)
Standard Form of the primal
max Z = 3x1 + 7x2 + 5x3 + 0s1 + 0s2
Subject to,
1x1 + 1x2 + 1x3 + 1s1 + 0s2 = 50
2x1 + 3x2 + 1x3 + 0s1 + 1s2 = 100
x1, x2, x3, s1, s2 >= 0
(a) x1=0, x2=25, x3=25
(b) Z = 300
(c) Optimal solution of dual = Shadow price of primal i.e. y1=4, and y2=1; Objective will be same i.e. W=Z=300
(4)
The current solution will remain optimal up to the profit of fudge 1 being 6 (since the reduced cost is 3 from the final optimal tableau)
For price of fudge 1 to be 7, the new optimal solution will be x1=50, x2=x3=0, and Z=350
Initial Simplex Tableau CBi Cj 3 7 5 0 0 Solution Ratio Basic variables x1 x2 x3 s1 s2 0 s1 1 1 1 1 0 50 50.0 0 s2 2 3 1 0 1 100 33.3 Zj 0 0 0 0 0 0 Cj - Zj 3 7 5 0 0 First iteration CBi Cj 3 7 5 0 0 Solution Ratio Basic variables x1 x2 x3 s1 s2 0 s1 0.333 0 0.667 1 -0.333 16.67 25 7 x2 0.667 1 0.333 0 0.333 33.33 100 Zj 4.667 7 2.333 0 2.333 233.33333 Cj - Zj -1.667 0 2.667 0 -2.333 Second and final iteration CBi Cj 3 7 5 0 0 Solution Basic variables x1 x2 x3 s1 s2 5 x3 0.5 0 1 1.5 -0.5 25 7 x2 0.5 1 0 -0.5 0.5 25 Zj 6 7 5 4 1 300 Cj - Zj -3 0 0 -4 -1 <--- All <= 0Related Questions
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