This preview has intentionally blurred parts. Sign up to view the full document

View Full Document

Unformatted Document Excerpt

Latus John Professor Saffarian MGMT 3413 A. ( Material) 6x1 + 4x2 =48 6x + 4(0) = 48 6(0) + 4x = 48 6x = 48 4x = 48 X1 = 8 x2 = 12 (Labor) 4x + 8x = 80 4x + 8(0) = 80 4(0) + 8x = 80 4x = 80 8x = 80 X 1= 20 x2 = 10 2(6x +4x = 48) 1(4x + 8x = 80) 12x + 8x = 96 4x + 8x = 80 8x = 16 X1 = 2 4(2) + 8x2 = 80 8 + 8x2 = 80 8x2 = 72 X2 = 9 Z= 4(2) + 3(9) Z= 35 1. X1=2 X2=9 Z=35 2. No slack 3. No surplus 4. No redundant Constraints B. Maximize Z=2x1+10x2 Durability 10x+4x=40 Strength x+6x=24 Time x+2x=14 10x1+4(0) = 40 X1=4 10(0) +4x2=40 X2=10 X1+6(0) = 24 X1=24 1(0) +6x2=24 X2=4 X1+2(0) =14 X1=14 1(0) +2x2=14 X2=7 D-S [1.5(10x1+4x2=40) x1+6x2=24] = 14x1=36 X1=2.57 X2=3.57 D-T [10x1+4x2=40- 2(x1+2x2=14)]= 8x1=12 X1=1.5 X2=6.25 S-T [x1+6x2=24 (x1+2x2=14)= 4x2=10 X2= 2.5 X1=9 Z= 2(2.57)+10(3.57)= 40.84 Z= 2(1.5)+10(6.25)= 65.5 (optimum) Z= 2(9)+10(2.5)= 43 1. X1=1.5 X2=6.25 Z=65.5 2. No slack 3. The Strength Constraint has a surplus of 15 4. No redundant constraints C. Maximize Z=6a+3b Material 20a+6b=600 Machinery 25a+20b=1,000 Labor 20a+30b=1,200 M-Mac [20a+6b=600- .8(25a+20b=1,000)] -10b=-200 b=20 a=24 John Latus Professor Saffarian MGMT 3413 M-L [20a+6b=600- 20a+30b=1,200] -24b=-600 b=25 a=22.5 Mac-L [.8(25a+20b=1,000)- 20a+20b=1,200] -4b=-400 b=100 a=-40 Z=6(24)+3(20)= 204 (optimum) Z=6(22.5)+3(25)= 210 (not optimum because it goes over the machinery hours limit) Z=6(-40)+3(100)=60 1. A=24 B=20 Z=204 2. Yes the Labor Constrain has a slack 120 of 3. No surplus 4. No redundant constraints Problem 2 Minimize Z=1.80S+2.20T Potassium 5S+8T=200 Carbohydrate 15S+6T=240 Protein 4S+12T=180 T T=10 S=40(t=0) T=25(s=0) S=16(t=0) T=40(s=0) S=45(t=0) T=15(s=0) T=20 S=8 S=34.28 T=3.57 S=11.53 T=11.15 P-C 3(5S+8T=200) 15S+6T=240 18T=360 P-Pr 1.5(5S+8T=200) 4S+12T=180 3.5S=120 C-Pr 2(15S+6T=240) 4S+12T=180 26S=300 1. 2. 3. 4. Minimize D E F S=8 T=20 Z=58.4 No slack 3 constrain has a surplus of 92 and the fourth constraint has a surplus of 10 The third constrain is redundant, it does not fit into the feasible area Z=2X1+3X2 4x+2x=20 2x+6x=18 1x+2x=12 X1=4.2 X1=2.67 X1=18 X2=1.6 X2=4.67 X2= -3 D-E 3(4x1+2x2=20) 2x1+6x2=18 10x1=42 D-F 4x1+2x2=20 1x1+2x2=12 3x1=8 E-F 2x1+6x2=18 3(x1+2x2=12) x1=18 1. X1=4.2 X2=1.6 Z=13.2 2. Yes constraint F has a slack of 4.6 3. No surplus 4. No redundant constraints John Latus Professor Saffarian MGMT 3413 Problem #4 (R) Raisin cost $1.50lb (P) Peanuts cost $.60lb (D) Deluxe bag use R 2/3 P 1/3 (S) Standard bag use R P Deluxe sell for $2.90 Standard sells for $2.55 Cost of Deluxe 2/3R + 1/3P= $1.20 Cost of Standard 1/2R + 1/2P= $1.05 Profit on Deluxe $1.70 Profit on Standard $1.50 Maximize Z= 1.70D+1.50S Raisins 2/3D+1/2S=90 Peanuts 1/3D+1/2S=60 D [less than or equal to] 110 S [less than or equal to] 110 2/3d+1/2s=90 1/3D+12S=60 D=60(s=0) S=45(d=0) D=20(s=0) S=30(d=0) 1/3D=30 D=90 S=60 1. 90 bags of Deluxe and 60 bags of Standard 2. $243 ... View Full Document

End of Preview

Sign up now to access the rest of the document