Question

*2. In Example 4.3-2, suppose that TOYCO is studying the possibility of introducing a fourth toy: fire trucks. The assembly does not make use of operation 1. Its unit assembly times on operations 2 and 3 are 1 and 3 minutes, respectively. The revenue per unit is $4. Would you advise TOYCO to introduce the new product?

Answer 1

i E If the introduce fire truicksi Let, 44. represent the daily member of units assembled of fire trucks . The problem becomes, Maximize 2=3x, +262 + 5x2 +484 subject to, 2+249 + X2 5430 Coperations) 394 +223 +245460 (operation 2) 34 + 422 "+39035420 (operation 3) 21,22,23,24>0. Here 2,=0; x2 = 101.25. X3 = 227.5 ; Ky=5 is the optimal solution and Max Z = 1360: But toys can not be infractional value So, our final solution would be, x,20, H 2 2 101) xz 2227, Xy=5. ... and profit is, 221357 i Toyco Should introduce the nelle product.

The simplex method is given below:

Problem is Max Z = 3X1 + 2x2 + 5x3 + 4x4 subject to *1 + 2x2 + x3 S430 3x1 + 2x3 + x45 460 X1 + 4x + 3x5420 and X1, X2, X3, X4 20; The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate 1. As the constraint-1 is of type's' we should add slack variable 51 2. As the constraint-2 is of type's 'we should add slack variable S2 3. As the constraint-3 is of type's' we should add slack variable S3 After introducing slack variables Max Z = 3x1 + 2x2 + 5x3 + 4x4 +0S1 +0S2 +0S3 subject to *1 + 2x2 + x3 + Si =430 3x1 + 2x3 + x4 +52 = 460 *1 + 4x2 + 3x4 + S3 =420 and X1, X2, 43, 44, 51, 52, 5320 Iteration-1 3 25 0 0 0 MinRatio x Si S2 X3 430 = 430 400 – 230 - Z = 0 z; 2-C -51

Negative minimum 2; - Cis-and its column index is 3. So, the entering variable is Xz. Minimum ratio is 230 and its row index is 2. So, the leaving basis variable is 52. ... The pivot element is 2. Entering = x3, Departing = S2, Key Element = 2 + R2 (new) = R2 (old) = 2 + R, (new) = R, (old) - R (new) + R (new) = R3(old) Iteration-2 2 5 4 0 0 0 Min Ratio 49/ 200 7= 100 - 0.5 420 4201 0 420 = 105 Z = 1150 2; 2-C 7.5 4.5 2.5 -15 2.5 2. 5 0 0 0 Negative minimum Z;-C; is -2 and its column index is 2. So, the entering variable is xy- Minimum ratio is 100 and its row index is 1. So, the leaving basis variable is $1.

.. The pivot element is 2. Entering = x2, Departing = 51, Key Element = 2 + R (new) = R1 (old) = 2 + R (new) = R (old) + R3(new) = R3(old) - 4R(new) Iteration-3 5 4 0 0 0 Min Ratio X3 100 -0.25 1 -0.25 0.5 -0.25 0.5 2.39 = 460 Z = 1350 1 0 7 4 2 0 5 0 2 -2 2 2 2-6 Negative minimum Z; -;is -2 and its column index is 4. So, the entering variable is x4 Minimum ratio is 5 and its row index is 3. So, the leaving basis variable is 53.

Minimum ratio is 5 and its row index is 3. So, the leaving basis variable is Sz. .. The pivot element is 4. Entering = x 4, Departing = 53, Key Element - 4 + R3(new) = R3(old) = 4 + R (new) = R (old) +0.25R(new) + R (new) = R(old) - 0.5R2(new) 0 0 S2 Min Ratio CB 2 Iteration-4 B *2 x3 *4 Z = 1360 0.0625 3 X1 -0.125 1.25 0.5 XB 101.25 227.5 5 2 5 4 0 X2 X3 X4S 1 0 0 0.375 0 1 0 0.25 0 0 1 -0.5 2 54 . 0 0 0 0 0 -0.1875 0.375 5 -0.125 4 2 8 0.25 2.5 2.5 0.25 0.5 0.5 Z;-9 5 Since all Z; -C;20 Hence, optimal solution is arrived with value of variables as: X1 = 0, X2 = 101.25, x3 = 227.5,x4 = 5 Max Z = 1360

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