Sensitivity analysis involves 'what if?' questions. This technique is used to determine how different values of an independent variable will impact a particular dependent variable under a given set of assumptions. This technique is used within specific boundaries that will depend on one or more input variables, such as the effect that changes in interest rates will have on a bond's price.
Sensitivity analysis is a way to predict the outcome of a decision if a situation turns out to be different compared to the key prediction(s). By creating a given set of scenarios, the analyst can determine how changes in one variable(s) will impact the target variable.
Every commercial linear-programming system provides this elementary sensitivity analysis, since the calculations are easy to perform using the tableau associated with an optimal solution. There are two variations in the data that invariably are reported: objective function and right-hand-side ranges.
Further, associated with each range is information concerning how the basis would change if the range were exceeded. Sensitivity analysis deals with making individual changes in the co-efficient of the objective function and the right hand sides of the constraints. It is the study of how changes in the co-efficient of a linear programming problem affect the optimal solution.
We can answer questions such as,
For example, a company produces two products x1 and x2 with the use of three different materials 1, 2 and 3. The availability of materials 1, 2 and 3 are 175, 50 and 150 respectively. The profit for selling per unit of product x1 is Rs. 40 and that of x2 is Rs. 30. The raw material requirements for the products are shown by equations, as given below.
zmax = 40x1 + 30x2
Subject to constraints
4x1 + 5x2≤ 175 ....................(i)
2x2≤ 50 ....................(ii)
6x1 + 3x2≤ 150 ....................(iii)
where x1, x2≥ 0
The optimal solution is
x1 = Rs. 12.50
x2 = Rs. 25.00
zmax = 40 * 12.50 + 30 * 25.00
= Rs. 1250.00
The problem is solved using TORA software and the output screen showing sensitivity analysis is given in Table.
Change in objective function co-efficients and effect on optimal solution
Sensitivity Analysis Table Output
Referring to the current objective co-efficient, if the values of the objective function coefficient decrease by 16 (Min. obj. co-efficient) and increase by 20 (Max. obj. coefficient) there will not be any change in the optimal values of x1 = 12.50 and x2 = 25.00. But there will be a change in the optimal solution, i.e. zmax
Note: This applies only when there is a change in any one of the co-efficients of variables i.e., x1 or x2. Simultaneous changes in values of the co-efficients need to apply for 100 Percent Rule for objective function co-efficients.
For x1, Allowable decrease = Current value - Min. Obj. co-efficient
= 40 – 24
= Rs. 16 ------------------ (i)
Allowable increase = Max. Obj. co-efficient – Current value
= 60 – 40
= Rs. 20.00 ---------------- (ii)
Similarly, For x2, Allowable decrease = Rs. 10.00 ---------------- (iii)
Allowable increase = Rs. 20.00 --------------- (iv)
For example, if co-efficient of x1 is increased to 48, then the increase is 48 – 40 = Rs. 8.00. From (ii), the allowable increase is 20, thus the increase in x1 coefficient is 8/20 = 0.40 or 40%.
If co-efficient of x2 is decreased to 27, then the decrease is 30 - 27 = Rs. 3.00.
From (iii), the allowable decrease is 10, thus the decrease in x2 co-efficient is 3/10 = 0.30 or 30%.
Therefore, the percentage of increase in x1 and the percentage of decrease in x2 is 40 and 30 respectively. i.e. 40% + 30% = 70%
For all the objective function co-efficients that are changed, sum the percentage of the allowable increase and allowable decrease. If the sum of the percentages is less than or equal to 100%, the optimal solution does not change, i.e., x1 and x2 values will not change.
But zmax will change, i.e.,
= 48(12.50) + 27(25)
= Rs. 1275.00
If the sum of the percentages of increase and decrease is greater than 100%, a different optimal solution exists. A revised problem must be solved in order to determine the new optimal values.
Change in the right-hand side constraints values and effect on optimal solution
Suppose an additional 40 kgs of material 3 is available, the right-hand side constraint increases from 150 to 190 kgs.
Now, if the problem is solved, we get the optimal values as
x1 = 23.61, x2 = 16.11 and zmax = 1427.78
From this, we can infer that an additional resources of 40 kgs increases the profit by
= 1427.78 – 1250 = Rs. 177.78
Therefore, for one kg or one unit increase, the profit will increase by
= 177.78 / 40
= Rs. 4.44
Dual price is the improvement in the value of the optimal solution per unit increase in the right-hand side of a constraint. Hence, the dual price of material 3 is Rs 4.44 per kg. Increase in material 2 will simply increase the unused material 2 rather than increase in objective function. We cannot increase the RHS constraint values or the resources. If the limit increases, there will be a change in the optimal values.
The limit values are given in Table 2.10, i.e., Min RHS and Max RHS values. For example, for material 3, the dual price Rs. 4.44 applies only to the limit range 150 kgs to 262.50 kgs.
Where there are simultaneous changes in more than one constraint RHS values, the 100 per cent Rule must be applied.
Reduced cost: The reduced cost associated with the nonnegativity constraint for each variable is the shadow price of that constraint (i.e., the corresponding change in the objective function per unit increase in the lower bound of the variable).
Shadow price: The shadow price associated with a particular constraint is the change in the optimal value of the objective function per unit increase in the right-hand-side value for that constraint, all other problem data remaining unchanged.
If the activity's reduced cost per unit is positive, then its unit cost of consumed resources is higher than its unit profit, and the activity should be discarded. This means that the value of its associated variable in the optimum solution should be zero.
Alternatively, an activity that is economically attractive will have a zero reduced cost in the optimum solution signifying equilibrium between the output (unit profit) and the input (unit cost of consumed resources).
In the problem, both x1 and x2 assume positive values in the optimum solution and hence have zero reduced cost.
Considering one more variable x3 with profit Rs. 50
zmax = 40x1 + 30x2 + 50x3
Subject to constraints,
4x1 + 5x2 + 6x3≤ 175 ....................(i)
2x2 + 1x3≤ 50 ....................(ii)
6x1 + 3x2 + 3x3≤ 150 ....................(iii)
where x1, x2, x3≥ 0
The sensitivity analysis of the problem is shown in the computer output below in Table.
The reduced cost indicates how much the objective function co-efficient for a particular variable would have to improve before that decision function assumes a positive value in the optimal solution.
The reduced cost of Rs.12.50 for decision variable x2 tells us that the profit contribution would have to increase to at least 30 + 12.50 = 42.50 before x3 could assume a positive value in the optimal solution.
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