2. Linear Programming – Graphical Method 2) Maximise: Z=3X 1 + 2X 2 s.t. 2X 1 + X 2 2 3X 1 + 4X 2 12 X 1 , X 2 0 Equation When X 1 = 0 When X 2 = 0 2X 1 + X 2 = 2 0 + X 2 = 2 X 2 = 2 Point is (0,2) 2 X 1 + 0 = 2 X 1 = 1 Point is (1,0) 3X 1 + 4 X 2 = 12 0 + 4 X 2 = 12 X 2 = 3 Point is (0,3) 3X 1 + 0 = 12 X 1 = 4 Point is (4,0) 1 2 1 2 3X 1 + 4 X 2 ≥ 12 2X 1 + X 2 ≤ 2 X 1 X 2 ≤ Shaded area towards the origin Conclusion: Since there is no common area, there is feasible solution for the given problem ≥ Shaded area away from the origin 3 3 4
3. Identify the common area for the following examples a) 2 X + Y 3 Y – 3 X 1 X, Y 0 Equation When X = 0 When Y = 0 2X + Y = 3 0 + Y = 3 Y = 3 Point is (0, 3) 2 X + 0 = 3 X = 1.5 Point is (1.5, 0) Y - 3 X = 1 Y - 3 (0) = 1 Y = 1 Point is (0, 1) 0 – 3 X = 1 X = -1/3 Point is (-1/3, 0) 1 2 1 2 2X + Y ≥ 3 Y -3X ≤ 1 X 1 X 2 ≤ Shaded area towards the origin ≥ Shaded area away from the origin 3 3 4
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5. Linear Programming – Graphical Method 3) Maximise: Z=3X 1 + 8X 2 s.t. X 1 + X 2 = 200 X 1 80 X 2 60 X 1 ,X 2 0 Equation When X 1 = 0 When X 2 = 0 X 1 + X 2 = 200 0 + X 2 = 200 X 2 = 200 Point is (0,200) X 1 + 0 = 200 X 1 = 200 Point is (200,0) X 1 = 80 Point is (80,0) 50 100 150 50 100 X 1 + X 2 = 200 X 1 ≤ 80 A B X 1 X 2 ≤ Shaded area towards the origin Point Co-Ordinate Z max = 3 X 1 + 8 X 2 A 0, 200 3 x 0 + 8 x 200 = 1600 B 80, 120 3 x 80 + 8 x 120 = 1200 X 2 = 60 Point is (0, 60) 200 150 200 = Shaded area only on the line ≥ Shaded area away from the origin X 2 ≥ 60 (0, 200) (80, 120) Conclusion: X 1 = 0 and X 2 = 200 will give Z max = 1600
6. Linear Programming – Graphical Method 4) Maximise: Z=3X 1 + 2X 2 s.t. -2X 1 + X 2 1 X 1 2 X 1 + X 2 3 X 1 ,X 2 0 1 2 3 1 2 X 1 + X 2 ≤ 3 X 1 ≤ 2 A X 1 X 2 ≤ Shaded area towards the origin Conclusion: X 1 = 2 and X 2 = 1 will give Z max = 8 4 3 4 - 2 X 1 + X 2 ≤ 1 - 1 Equation When X 1 = 0 When X 2 = 0 - 2X 1 + X 2 = 1 0 + X 2 = 1 X 2 = 1 Point is (0,1) - 2 X 1 + 0 = 1 X 1 = - 0.5 Point is (-0.5,0) X 1 = 2 Point is (2,0) X 1 + X 2 = 3 Point is (0, 3) Point is (3, 0) B C D E To find C -2 X 1 + X 2 = 1 ----- (1) X 1 + X 2 = 3 ----- (2) (1)-(2) -3 X 1 = - 2 X 1 = 2/3 Substitute X 1 value in Eq (2) 2/3 + X 2 = 3 Therefore X 2 = 7/3 Point Co-Ordinate Z max = 3 X 1 + 2 X 2 A 0, 0 3 x 0 + 2 x 0 = 0 B 0, 1 3 x 0 + 2 x 1 = 2 C 2/3, 7/3 3 x 2/3 + 2 x 7/3 = 20/3 D 2, 1 3 x 2 + 2 x 1 = 8 E 2, 0 3 x 2 + 2 x 0 = 6
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9. Minimise: Z = 4X 1 + 2X 2 s.t. X 1 + 2 X 2 ≥ 2 3 X 1 + 2 X 2 ≥ 3 4 X 1 + 3 X 2 ≥ 6 X 1 , X 2 ≥ 0 Equation When X 1 = 0 When X 2 = 0 X 1 + 2 X 2 = 2 0 + 2 X 2 = 2 X 2 = 1 Point is (0, 1) X 1 + 0 = 2 X 1 = 2 Point is (2, 0) 3X 1 + 2 X 2 = 3 0 + 2 X 2 = 3 X 2 = 1.5 Point is (0, 1.5) 3X 1 + 0 = 3 X 1 = 1 Point is (1,0) 4X 1 + 3 X 2 = 6 0 + 3 X 2 = 6 X 2 = 2 Point is (0, 2) 4X 1 + 0 = 6 X 1 = 1.5 Point is (1.5, 0) 1 1 2 X 2 2 3 X 1 ≥ Shaded area away from the origin 3X 1 + 2 X 2 ≥ 3 4X 1 + 3 X 2 ≥ 6 B A C X 1 + 2 X 2 ≥ 2 To find B X 1 + 2X 2 = 2 ----- (1) 4 X 1 + 3 X 2 = 6 ---- (2) (1) X 4 4 X 1 + 8 X 2 = 8 (Sub) -5 X 2 = - 2 X 2 = 2/5 = 0.4 Substitute X 2 value in Eq (1) X 1 + 2 (0.4) = 2 Therefore X 1 = 1.2 Point Co-Ordinate Z max = 4 X 1 + 2 X 2 A 2, 0 4 x 2 + 2 x 0 = 8 B 1.2, 0.4 4 x 1.2 + 2 x 0.4 = 5.6 C 0, 2 4 x 0 + 2 x 2 = 4 Conclusion: X 1 = 0 and X 2 = 2 will give Z min = 4
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11. Equation When X 1 = 0 When X 2 = 0 X 1 = 20 Point is (20, 0) X 2 = 70 Point is (70, 0) X 1 + X 2 = 120 Point is (0, 120) Point is (120, 0) Max Z = -150 X 1 + 100 X 2 + 2800000 S.t. (i) X 1 ≥ 20 ; (ii) X 2 ≥ 70 and X 2 ≤ 140 (iii) X 1 + X 2 ≤ 140 ; (iv) X 1 ≤ 60 (v) X 1 + X 2 ≥ 120 X 1 , X 2 ≥ 0 X 1 = 60 Point is (60, 0) X 2 = 140 Point is (140, 0) X 1 + X 2 = 140 Point is (0, 140) Point is (140, 0) 20 X 1 X 2 40 60 80 100 120 140 20 40 60 80 100 120 140 ≤ Shaded area towards the origin ≥ Shaded area away from the origin X 1 ≥ 20 X 2 ≥ 70 X 1 ≤ 60 X 2 ≤ 140 X 1 + X 2 ≥ 120 X 1 + X 2 ≤ 140 A B C D E Point Co-Ordinate Z max = - 150 X 1 - 100 X 2 +280000 A 20, 100 -150 x 20 - 100 x 100+ 280000 = 267000 B 20, 120 -150 x 20 - 100 x 120 + 280000 = 265000 C 60, 80 -150 x 60 - 100 x 80 + 280000 = 263000 D 60, 70 -150 x 60 - 100 x 70 + 280000 = 264000 E 50, 70 -150 x 50 - 100 x 70 + 280000 = 265500 Conclusion: X 1 = 20 , X 2 = 100 and X 3 = 80 will give Z max = 267000 X 1 = 20 ; X 2 = 100 ; Therefore X 3 = 200-(X 1 +X 2 ) = 200 – (20+100) = 80