Review Problems Solutions MATH 3300A (01) Optimization Fall Graphically, the optimal solution is found to be x 1 = 36/7, x 2 = 66/7, z = 69/7.

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1 Review Problems Solutions MATH 3300A (01) Optimization Fall 2016 Chapter 3 2. Let x 1 = number of chocolate cakes baked x 2 = number of vanilla cakes baked Then we should solve max z = x x 2 s.t. 20x x x 1 + x 2 30 x 1, x 2 0 Graphically, the optimal solution is found to be x 1 = 36/7, x 2 = 66/7, z = 69/ Let D = Daytime calls and E = Evening calls. Then the correct formulation is min z = 2D + 5E s.t..30d +.30E 150 (WIVES CONSTRAINT).10D +.30E 120 (HUSBANDS CONSTRAINT).10D +.15E 100 (SINGLE MALE CONSTRAINT).10D +.20E 110 (SINGLE FEMALE CONSTRAINT) E/ (E+D) 1/2 OR E-D 0 D 0, E Let W1 = Pounds of wheat used to produce Feed 1. W2 = Pounds of wheat used to produce Feed 2. A1 = Pounds of alfalfa used to produce Feed 1. A2 = Pounds of alfalfa used to produce Feed 2. Then the correct formulation is max z = 1.5(W1+A1)+1.3(W2+A2)-.5(W1+W2)-.4(A1+A2) s.t. W1+W2 1,000 (CAN PURCHASE AT MOST 1,000 POUNDS OF WHEAT) A1+A2 800 (CAN PURCHASE AT MOST 800 POUNDS OF ALFALFA)

2 W1/(A1+W1).8 OR.2W1-.8A1 0 (FEED 1 AT LEAST 80% WHEAT) A2/(A2+W2).6 OR.4A2-.6W2 0 (FEED 2 AT LEAST 60% ALF) W1, W2, A1, A Let W1',W2',A1', and A2'represent the number of pounds of wheat and alfalfa used to produce the feed sold at the discount price and let W1,W2,A1, and A2 represent the wheat and alfalfa used to produce the feeds sold at the non-discounted price. Then a correct formulation is max z = 1.5(W1+A1)+1.3(W2+A2)+1.25(W1'+A1')+(W2'+A2') -.5(W1+W2+W1'+W2')-.4(A1+A2+A1'+A2') s.t. A1+W1 300 (CAN SELL AT MOST 300 POUNDS OF FEED 1 AT HIGH PRICE) A2+W2 300(CAN SELL AT MOST 300 POUNDS OF FEED 2 AT HIGH PRICE) W1+W2+W1'+W2' 1,000 A1+A2+A1'+A2' 800 (W1+W1')/(A1+A1'+W1+W1').8 OR -.8A1-.8A1'+.2W1+.2W1` 0 (A2+A2')/(A2+A2'+W2+W2').6 OR -.6W2-.6W2'+.4A2+.4A2' 0 ALL VARIABLES 0 Note that the act of maximizing ensures that W1'= A1' = 0 unless A1+W1 = 300. Similarly the act of maximizing ensures that W2' = A2' = 0 unless A2+W2 = 300. Chapter 4 1. z x 1 x 2 x 3 s 1 s 2 RHS /5 9/5 1-1/5 3

3 0 1 3/5 6/5 0 1/5 3 This is an optimal tableau with the optimal solution being z = 15, x 1 =3, x 2 =0. Pivoting x 2 into the basis yields the alternative optimal solution z = 15, x 1 =0, x 2 = z x 1 x 2 s 1 s 2 RHS Basic Variable z= s 1 = s 2 =3 z x 1 x 2 s 1 s 2 RHS Basic Variable z= x 1 = s 2 =2 Unbounded LP (look at x 2 column). 6. max z= x 1 + x 2 - Ma 1 s.t. 2x 1 + x 2 - e 1 + a 1 = 3 3x 1 + x 2 + s 2 = 7/2 x 1 + x 2 + s 3 = 1 After eliminating a 1 from Row 0 (by adding -(Row 1) to Row 0) we obtain z x 1 x 2 e 1 a 1 s 2 s 3 RHS Basic Variable 1-2M-1-1-M M M z=-3m a 1 = /2 s 2 =7/ s 3 =1

4 1 0 M M 0 0 2M+1 -M+1 z=-m a 1 = /2 s 2 =1/ x 1 =1 This is an optimal tableau. Since a 1 is positive, the original LP is infeasible. Chapter 6 1a. min w = 6y 1 + 3y y 3 s.t. y 1 + y 2 + 2y 3 4 2y 1 - y 2 + y 3 1 y 1 urs, y 2 0, y 3 0 Optimal dual solution is w = 58/3, y 1 = -2/3, y 2 = 0, y 3 = 7/3 2 / 3 0 1/ 3 6 (2 Δ) / 3 1b. B -1 = 1/ / 3 B = (14 + 2Δ) / Δ 1+ Δ Thus current basis remains optimal iff Δ 2, Δ -7 and Δ -1 or -1 Δ 2 or 9 b Substituting Δ = 1 we obtain z = 58/3 + Row 3 Shadow Price = 65/3 x 2 = (2-1)/3 = 1/3, x 1 = (14 + 2)/3 = 16/3, e 2 = = 2 3a. min w = 6y 1 + 8y 2 + 2y 3 s.t. y 1 + 6y 2 5 y 1 + y 3 1

5 y 1 + y 2 + y 3 2 y 1 0, y 2 0, y 3 0 Optimal dual solution is w = 9, y 1 = 0, y 2 = 5/6, y 3 = 7/6. 3b. c BV B -1 = [0 (5 +Δ)/6 (7-Δ)/6] _ Pricing out x 2 we obtain 1 c 2 = [0 (5 +Δ)/6 (7-Δ)/6] 0-1 = (1-Δ)/6. 1 Coefficient of s 2 in Optimal Row 0 = (5 + Δ)/6 Coefficient of s 3 in Optimal Row 0 = (7-Δ)/6 Thus current basis remains optimal iff Δ 1, Δ -5, Δ 7 or -5 Δ 1 or 0 c c. Since x 2 is non-basic, c 2 can increase by at most the reduced cost for the current basis to remain optimal. Thus current basis remains optimal as long as c /6 = 7/6. 5a. min w = 6y 1 + 4y 2 + 3y 3 s.t. 3y 1 + 2y 2 + y 3 4 y 1 + y 2 + y 3 1 y 1 0 y 2 0, y 3 urs The optimal dual solution is w = 12, y 1 = y 2 = 0, y 3 = 4. 5b. c (reduced cost for x 2 ) = = 4 5c. c BV B -1 = [4 + Δ 0 0] = [ Δ] The current basis remains optimal iff c 2 0. But _ 1 c 2 = [ Δ] 1-1 = 3 + Δ Now 3 + Δ 0 iff Δ -3 1 _

6 Thus current basis remains optimal for c = 1 6a. min w = 8y y 2 s.t. 2y 1 + 4y 2 3 y 1 + y 2 1 y 1 - y 2-1 The optimal dual solution is w = 9, y 1 = y 2 = 1/2. 6b. The current basis remains optimal iff B Δ or 0 1/ 2 1/ Δ or -2 Δ 6 or 8 b If b 2 = 10, then Δ = 2 optimal solution may be obtained from 6 Δ = 1 + Δ / 2 and the new 0 0 x = = x1 1 + (2 / 2) 2 The new z-value = 9 + 2(row 2 shadow price) = When the isoprofit lines are steeper than the 2x 1 + x 2 = 8 constraint, the optimal solution shifts to (1, 6). Thus if -c 1 /4<-2 or c 1 >8 the current basis is no longer optimal.

7 10a. min w = 8y y 2 s.t. 2y 1 + 3y 2 3 5y 1 + 7y 2 2 y 1 0, y 2 0 The optimal dual solution is w = 10, y 1 = 0, y 2 = / b. B -1 =, B / Δ (4 2Δ) / 3 = (10 + Δ) / 3 Thus current basis is optimal as long as 4-2Δ 0 (Δ 2) and 10 + Δ 0 (Δ -10) or 0 b If b 2 = 5 the new optimal solution is given by s1 x1 = 1 2 / 3 8 = 14 / 3 0 1/ / 3 z = 10 + (1) (-5) = 5, x 1 = 5/3, s 1 = 14/3, s 2 = x 2 = 0

8 Chapter 9 Section 9.3: 4.

9 6. Section 9.4: 1. Section 9.5: 1. We might choose up to eight "Item 1's" so we define items 1', 2',... 8' all identical to Item 1. We might choose up to six Item 2's, so we define items 9', 10',... 14' all identical to Item 2. Finally, we might choose up to five Item 3's, so we define items 15', 16',... 19' all identical to Item 3. Letting x i = 1 if item i' is chosen and x i = 0 otherwise yields a 0-1 knapsack problem. 3. Letting x i = 1 if item i is chosen and x i = 0 otherwise yields the following knapsack problem: max z = 5x 1 + 8x 2 + 3x 3 + 7x 4 st 3x 1 + 5x 2 + 2x 3 + 4x 4 6 x i = 0 or 1 We obtain the following tree (for each subproblem any omitted variable equals 0):

10 Note that since optimal objective function value for any candidate solution associated with a branch must be an integer, SP 3 can at best yield a z-value of 10, so we need not branch on SP 3. Thus the optimal solution is z = 10, x 1 = x 2 = 0, x 3 = x 4 = 1. From the Review Section (pages ): 3. Let z i = 1 if gymnast i enters both events z i = 0 otherwise x i = 1 if gymnast i enters only BB x i = 0 otherwise y i = 1 if gymnast i enters just FE y i = 0 Then the appropriate IP is max z = 16.7z z z x x y y 6 s.t. z 1 + z z 6 = 3 x 1 + x x 6 = 1 y 1 + y y 6 = 1 x 1 + y 1 + z 1 1 x 2 + y 2 + z 2 1 x 3 + y 3 + z 3 1 x 4 + y 4 + z 4 1 x 5 + y 5 + z 5 1 x 6 + y 6 + z 6 1 All variables equal 0 or 1.

11 4. Let x ij = 1 if students from district i are sent to school j x ij = 0 otherwise Then the appropriate IP is Min z = 110x x x x x x x x x x 52 s.t. 110x x x x x x x x x x x x x x x x x x x x 51 or 0 8x 11-10x 21-10x x x 51 30x x x x x x x x x x 52 or 0 8x 12-10x 22-10x x x 52 x 11 + x 12 = 1, x 21 + x 22 = 1, x 31 + x 32 = 1 x 41 + x 42 = 1, x 51 + x 52 = 1 All variables = 0 or 1 5. Let x 1 = 1 if RS is signed x 1 = 0 otherwise x 2 = 1 if BS is signed x 2 = 0 otherwise x 3 = 1 if DE is signed x 3 = 0 otherwise x 4 = 1 if ST is signed x 4 = 0 otherwise x 5 = 1 if TS is signed x 5 = 0 otherwise max z = 6x 1 + 5x 2 + 3x 3 + 3x 4 + 2x 5 s.t. 6x 1 + 4x 2 + 3x 3 + 2x 4 + 2x 5 12 (1) x 2 + x 3 + x 4 2 (2) x 1 + x 2 + x 3 + x 5 2 (3) x 1 + x 2 1 (4) All variables = 0 or 1 7.

12 8. z = 0, x 1 = 0, x 2 = 0 is the optimal solution to the LP relaxation. Since this solution is feasible for the IP, it is optimal for the IP SP 1 z = 40/3 t = 1 x 1 = 20/3 x 2 = 10/ x 2 3 x SP 2 SP 3 t = 2 z = 13 t = 3 x 1 =5 x 2 = From the tree we find the optimal solution to be z = 13, x 1 = 5, x 2 = 4. Note that since SP 1 has z = 13 1/3, the z =

13 13 value for SP 3 implies that SP 3 yields an optimal solution. Thus SP 2 need not be considered.