Question

(b) A LP model and its solution outputs and sensitivity report are as below. LP Model: Maximise R 7X+5Y+10 Z Subjective to: CI: 2X +Y+32 <= 50 C3: X 3 C6: X, Y, Z)so

Answer 1

Solution:

1:

The optimum objective value will decrease from 101.33 to 89.67

Explanation:

Max Z-7x1 +5x2-10x3 subject to 21 2 3*3s 3x45x3 545 *324 and xj *x320, The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate 1 As the constraint-1 is of type' we should add slack variable S1 2. As the constraint-2 is of type's' we should add slack variable S2 3 As the constraint-3 is of type" we should subtract surplus variable S and add artificial variable 1 4. As the constraint-4 is of type" 'e should subtract surplus variable s, and add artificial variable A After introducing slack,surplus,artificial variables Max Z = 7x1 + 5x2-10x3-osl-os, +0S3 + 05, . M41 . M42 subject to 2x1x2 3x3t S1 50 3x1 + 42 5x3 -45 Iteration-1 10 MinRatio 50 = 16.6667 45 Z,- C r- 5


Negative minimum Z, - C, is-M-10 and its column index is 3. So, the entering variable is x Minimum ratio is 4 and its row index is 4. So, the leaving basis variable is A2 . The pivot element is 1 Enteringx3, DepartingA, Key Element1 + R4(new)-R4(old) +R (new)-R(old)-3R,(new) t R2(new)R2(old)- 5R4(new) + R3(new- R(old) Iteration 2 10 MinRatio x3 38 38 38 25 25 10 10 10 Z, - C M-5↑ Negative minimum Z, - C, is-M- 5 and its column index is 2. So, the entering variable is x2.


Minimum ratio is 2 and its row index is 3. So, the leaving basis variable is A The pivot element is 1 Enteringx2, DepartingAKey Element -1 +Rg(new)- R (old) + R,(new)-R,(old) - Rz(new) + Rj(new)-R2(old) - 4R (new) Iteration-3 10 MinRatio S, 36 36 17 3.4 17 10 Z-50 2, - C 10 T Negative minimum 2,-c, is-10 and its column index is 7. So, the entering variable is S4. Minimum ratio is 3.4 and its row index is 2. So, the leaving basis variable is S2 The pivot element is 5


Entering S Departing S, Key Elements + R (new)R(old) 5 + Rj(new)R(old) - 3R2(new) +R(new)-R (old) 10 MinRatic 129 129 129 17 5 17 ' = 5.6667 → 37 5 37 T12.3333 10 Z 84 10 Z,-C Negative minimum z-c, is-1 and its column index is 1 So, the entering variable is x1 Minimum ratio is 5.6667 and its row index is 2. So, the leaving basis variable is S4


. The pivot element is Enteringx, DepartingS, Key Element - + RI(new) Ri(old) R(new) + R(new)-R(old) + R4(new) = R4(old)-782(new) 10 *1 MinRatic 74 17 X- 10 269 13 10 Since all 0 Hence, optimal solution is arrived with value of variables as 269 Max Z

2:
The Value of Optimal objective function changes and becomes 104.4

The Optimum values of X, Y and Z are as follows,

X = 0
Y = 2
Z = 8.4

Explanation:

Problem is Max Z-6x1 6X2 + 11x3 subject to x324 and 11-x2-x3 0; The problem is converted to canonical fom by adding slack, surplus and artificial variables as appropiate 1. As the constraint-1 is of type 'we should add slack variable s 2. As the constraint-2 is of type' we should add slack variable s2 3. As the constraint-3 is of type'we should subtract surplus variable S and add artificial variable A 4. As the constraint-4 is of type'2 we should subtract surplus variable S4 and add artificial variable A2 After introducing slack,surplus,artificial variables subject to 2x12 +3x3 S - 50 Iteration-1 MinRatio 50 50 16.6667 50 10 50 6M M-11 t M-6


Negative minimum z,- C, is -M- 11 and its column index is 3. So, the entering variable is xj Minimum ratio is 4 and its row index is 4. So, the leaving basis variable is The pivot element is 1 Enteringx3,Departing, Key Element 1 + R (new)-R(old)- 5R(new) + R (new)-R (old) Iteration-2 MinRatio 38 38 30 22M+44 M-6T Negative minimum Z,- C, is -M-6 and its column index is 2. So, the entering variable is x


Minimum ratio is 2 and its row index is 3. So, the leaving basis variable is A : The pivot element is 1 Entering x2, Departing A1, Key Element-1 + R (new) -R (old) + Ri (new) = R1(old) . R3(new) + R,(new) = R2(old)-4A3(new) t R (new)-R4(old) Iteration-3 MinRatio 36 36 = 4.4- Z = 56 Negative minimum z- C, is 11 and its column index is 7. So, the entering variable is s, Minimum ratio is 4.4 and its row index is 2. So, the leaving basis variable is S2. . The pivot element is 5.


Entering S, Departing S, KeyElement s + Ri(new)-R(old) 3R (new) + R (new) R old) t R(new) R(old)+R (new) Iteration-4 MinRatio 114 42 522 14 Zi-C Since all Z,- C0 Hence, optimal solution is arrived with value of variables as 42 *0,22, 522