Presentation on theme: "SENSITIVITY ANALYSIS. luminous lamps produces three types of lamps A, B And C. These lamps are processed on three machines X, Y and Z. the full technology."— Presentation transcript:
luminous lamps produces three types of lamps A, B And C. These lamps are processed on three machines X, Y and Z. the full technology and input restrictions are given in the following table.
The Linear Programming model for the above problem can be written as Maximize Z = 12 x1 + 3 x2 + x3 subject to 10 x1 + 2 x2 + x3 /< 100 7 x1 + 3 x2 + 2x3 /< 77 2 x1 + 4 x2 + x3 /< 80 x1, x2, x3 >/ 0 Solving the above problem by the simplex method x1 = 73/ 8, x2 = 35/ 8, x3 = 0 and the optimal value of the objective function is 981/8
SENSITIVITY ANALYSIS First we find out whether a previously determined optimal solution remains optimal if the value of problem is changed. We will observe the optimum solution for following conditions. 1.Changes in Objective Function co-efficients A.Variables not included in the solution (i.e X3) (i.e. changes in the non-basic variable) B.Variables included in the solution (i.e. X1, X2) (i.e. Changes in basic variable) 2.Changes in the bi values : (RHS side of constraints) RHS Ranging. (i.e. changes in available resources)
Changes in objective function co-efficient Variable not included in the solution. Or Changes in non basic variable i.e. X3 As given in the optimum solution X3 is not in the solution. The Z value for X3 is 11/16 and its profit contribution is (i.e. product C) is also 1 Rs. This implies that a profit margin of at least 11/16 must be increased for this product if the firm would decide to include this product. So the increased margin of 11/16 means that total profit margin of the product should more than 11/16 + 1 = 27/16. The same thing can be treated like if the current solution is changed to 27/16 with the introduction of product C than the solution optimal beyond that it is no longer optimum. The information is useful to the manager as he/she can make decision of pricing without interfering optimal solution.
Variable included in the solution or changes in basic variable of obj fn For the given problem X1, X2 are the basic variables. Consider whether the change of profit per unit of a product, causes a change in the optimal solution of the problem. Over a certain range, a change positive or negative into unit profit would not cause a change in the optimal solution. To determine this, divide final Z raw values by corresponding raw values in the table. i.e. for product X1 it should be values in Z raw / values of X1 raw.
The least positive quotient (i.e Z / X1) is 5 implies that how much profit could decrease without changing solution. And least negative is 3 implies that maximum increase in the profit that would not cause a change in solution. So the range of profit would be (12-5=)7 to 15 (=12-(-3)) for product 1. in case the profit per product 1 falls below 7 or becomes higher than 15 rs. the optimum solution would change. In a similar way profit range for product 3 is (3- 3/5) and (3 -(-15/7). i.e. 12/5 to 36/7. That means that the variation in the profit per product 2 between 2.4 to 5.14 Would not change the optimum solution.
Change in the bi values or available resources or RHS Ranging In this case aim is to find that whether the change in available resources would make optimum solution unchanged or not. Or if so than in what range we can change the available resources. So the question is how many resources(i.e machine hours) in particular department may be increased or decreased that the profit become unchanged. To determine this we divide value of solution column with the respective slack variables. i.e. for our problem if we want to determine how many machine hourse of M1 can be changed without affecting final solution than we have to divide solution column / S1 raw. (the corrosponding slack variable for M1).
From the values least positive number provides answer to the question how many hours can be decreased and least negative number gives the how may hours can be increased in addition to the available hours. So for M1 least positive number is 354/11 and least negative number is -10. So number of hours should be 100- 354/11 to 100 – (-10). i.e. 746/11 (67.81) to 110 hours. It can be concluded that between 67.81 to 110 hours of the machine 1 the profit become unchanged. Similarly for machine M2 the range of number of hours would be 70 to 870/9 ( i.e. 96.67) hours.