DUALITY AND SENSITIVITY ANALYSIS

DUALITY AND SENSITIVITY ANALYSIS Understanding Duality No learning of Linear Programming is complete unless we learn the concept of Duality in linear programming. It is impossible to separate the linear programming problem and its dual. Several algorithms to solve special linear programming problems are centered around the concept of duality. Even though the emphasis of this book is to solve LP problems and understand to interpret the solutions, we deviate from the main focus of applications to understand duality. We will however retain the methodology of solving LPs to understand the concepts and applications. We explain duality using an example. In Illustration 1.1, we formulated and solved the LP problem to maximize the revenue for the bakery. The problem is to Maximize 32X 1 + 25X 2 5X 4X 59 4X 3X 46 X1, X2 0. The optimum solution obtained using the solver is X 1 = 7, X 2 = 6 with Z = 374 We consider another LP given below: Illustration 1.18 Minimize 59Y 1 + 46Y 2 5Y 4Y 32 4Y 3Y 25 Y1, Y2 0. We solve this LP using the solver. The final output is given in Table 1.16 Table 1.16 Solution for Illustration 1.18 y1 y2 4 3 374 32 >= 32

25 >= 25 4 >= 0 3 >= 0 The optimum solution is Y 1 = 4, Y 2 = 3 with objective function value W = 374. Now considering the maximization problem (called P1) and the minimization problem (called P2) we make the following observations: 1. P1 and P2 had the same value of the objective function at the optimum 2. The objective function coefficient coefficients of P1 are the RHS values of P2 and vice versa. 3. The number of variables in P1 and the number of constraints in P2 are equal and vice versa. The obvious question is Is there a relationship between P1 and P2? The relationship is established using the discussion that follows: 1. Consider P1 (the maximization problem). If there were no constraints the objective function value is. Let us try to get upper estimates for the value of Z. 2. We multiply the second constraint by 9 to get 36X 1 + 27X 2 414. Since X 1 and X 2 are 0, 32X 1 + 25X 2 36X 1 + 27X 2 414. Therefore Z * 414. 3. We multiply the first constraint by 7 to get 35X 1 + 28X 2 413. Since X 1 and X 2 are 0, 32X 1 + 25X 2 35X 1 + 28X 2 413. Therefore Z * 413 and this is a better upper estimate for Z * 4. We add the two constraints to get 9X 1 + 7X 2 105. This inequality holds because X 1 and X 2 are 0, We multiply this constraint by 4 to get 36X 1 + 28X 2 420. Since X 1 and X 2 are 0, 32X 1 + 25X 2 36X 1 + 28X 2 420. Therefore 420 is an upper estimate of Z * but we ignore this because our current best estimate is 413. 5. We multiply the constraint 4X 1 + 3X 2 46 by 25/3 to get 33.33X 1 + 25X 2 383.33. Based on the above discussions, 383.33 is a better upper estimate for Z *. 6. We multiply the constraint 5X 1 + 4X 2 59 by 32/5 to get 32X 1 + 25.6X 2 377.6. Based on the above discussions, 377.6 is a better upper estimate for Z *. 7. We multiply the constraint 9X 1 + 7X 2 105 by 25/7 to get 32.14X 1 + 25X 2 375. Now 375 is a better upper estimate for Z *. Here we have added the two constraints and multiplied by 25/7. This is the same as multiplying the constraints by 25/7 and adding them. 8. We can multiply the first constraint by a and the second by b and add them. If on addition the coefficients of X 1 and X 2 are 59 and 46 respectively, the RHS value (which is 59a + 46b) is an upper estimate of Z. We should have a, b 0, otherwise the sign of the inequality will change and we cannot add the inequalities. 9. If we want the best estimate of Z * (as small an upper estimate as possible) we need to define a and b such that 59a + 46b is as small as possible.

10. We therefore define a and b to minimize 59a + 46b is minimized. This leads us to another LP problem (where the variables are defines as Y 1 and Y 2 instead of a and b) which is P2. Minimize 59Y 1 + 46Y 2 5Y 4Y 32 4Y 3Y 25 Y1, Y2 0. Therefore problem P2 is born out of P1. The problem P2 is called the dual of the given problem P1 (called the primal). 1. Do the primal and dual have the same value of the objective function at the optimum? Yes. If both have optimum solutions their objective function value is the same. This result is called the main duality theorem. If the primal and dual have optimal solutions, the objective function value at the optimum for both is the same. 2. Can we read the solution of the dual from the solution of the primal? Yes. When the LP is written into the solver, we click options and click linear model and non negative. After we solve the LP, the solver dialogue box shows option for three reports. We can click on them and press OK. The reports are available as active sheets. If we go to the sensitivity report, we see a table under constraints. This part is shown in Table 1.17 Table 1.17- Sensitivity analysis report Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$6 x1 59 4 59 2.333333333 1.5 $B$7 x1 46 3 46 1.2 1.75 $B$8 x1 7 0 0 7 1E+30 $B$9 x1 6 0 0 6 1E+30 The shadow prices against the two constraints are 4 and 3 respectively. These are the values of Y 1 and Y 2 at the optimum. (We will see the relationship between the shadow prices and the dual solution later). 3. Does the algebraic method show the dual solution? YES. The algebraic method terminated for our LP problem with the equations

X 1 = 7 + 3X 3 4X 4 X 2 = 6-4X 3 + 5X 4 and Z = 374-4X 3 3X 4. We terminated by saying that the revenue function cannot be maximized further by increasing X 3 or X 4 (by increasing the non utilization or by keeping some resources idle). At the optimum the solution is X 1 = 7 and X 2 =6. The coefficients of X 3 and X 4 in the objective function at the optimum (in the algebraic method) represent the dual solution (with a negative sign). The dual solution is Y 1 = 4 and Y 2 =3 with W = 374 (It is customary to use W to represent the objective function of the dual). 4. Since the primal and dual have the same value of the objective function, how do we define the dual variables? The optimum solution to the dual is Y 1 = 4 and Y 2 =3 with W = 59 x 4 + 46 x 3 = 374. The values Y 1 = 4 and Y 2 =3 represent the worth of the 59 units of the first resource and 46 units of the second resource at the optimum. The dual solution represents the worth of the resources at the optimum. 5. What is shadow price and why is it related to the resources (primal constraints)? We have already seen that the dual solution represents the worth of one unit of each resource at the optimum. Let us explain it further using an example. Here we increase the first resource of the primal by 1 to 60. Illustration 1.19 Solve the LP - Maximize 32X 1 + 25X 2 5X 4X 60 4X 3X 46 X1, X2 0. The optimum solution to the primal (using solver) is given in Table 1.18 Table 1.18 Solver Solution x1 x2 4 10 obj fn 378 60 <= 60

46 <= 46 4 >= 0 10 >= 0 The optimum solution is X 1 = 4, X 2 = 10 with Z = 378. We continue to produce both the items but with different quantities. The objective function value has increased by 4 to 378. This shows that if we can increase the RHS of the first resource by 1, the profit increases by 4 (as long as we continue to produce the same products as before). The worth of the additional resource at the optimum is the extent to which the revenue goes up on increase of the resource. This is called the shadow price of the resource at the optimum or the worth of an additional amount of the resource at the optimum or the marginal value of the resource at the optimum. This is represented by the optimum solution of the dual. We restore the RHS of the first constraint to 59 and decrease the second resource by 1 to 45. Illustration 1.20 Solve the LP - Maximize 32X 1 + 25X 2 5X 4X 59 4X 3X 45 X1, X2 0. The optimum solution to the primal (using solver) is given in Table 1.19 Table 1.19 Optimum solution from solver x1 x2 3 11 obj fn 371 59 <= 59 45 <= 45 3 >= 0 11 >= 0 The optimum solution is X 1 = 3, X 2 = 11 with Z = 371. We continue to produce both the items but with different quantities. The objective function value has decreased by 3 to 371. This shows that if we decrease the RHS of the first resource by 1, the profit decreases by 3 (as long as we continue to produce the same products as before). The worth of the additional resource at the optimum is the extent to which the

revenue decreases on decrease of the resource. This is called the shadow price of the resource at the optimum. This is represented by the optimum solution of the dual. 6. How do I write the dual for a given primal? The relationship between the primal and the dual are given in Table 1.20 Table 1.20 Primal dual relationships Primal Dual Maximization Minimization Number of variables Number of constraints Number of constraints Number of variables Objective function values RHS values RHS values Objective function values Coefficient matrix for Transpose of the constraint coefficient constraints matrix variable constraint variable constraint Unrestricted variable Constraint is an equation constraint variable constraint variable equation Unrestricted variable Illustration 1.21 Consider the LP Minimize 59Y 1 + 46Y 2 + 37Y 3 5Y 4Y 3Y 32 3 4Y 3Y 8Y 26 3 2Y 5Y 9Y 60 3 3Y 2Y 3Y 15 3 Y 0, Y unrestricted, Y 0. 3 Write the dual to this LP and solve the dual? Using the information in Table 1.20 (treating the minimization problem as primal) we get the dual as Maximize 32X 1 +26X 2 + 60X 3 +15X 4 5X 4X 2X 3X 59 3 4 4X 3X 5X 2X 46 3 4 3X 8X 9X 3X 37 3 4 X unrestricted, X 0, X 0, X 0 3 4

We solve the primal and the dual optimally using the solver. Since we have a variable and an unrestricted variable, we go to options and leave out the non negativity restriction there. We write Y 1 0 and Y 2 0 as explicit constraints. We don t have a restriction on Y 2 and we do not restrict it. There is no constraint for Y 2 unrestricted. Table 1.21 gives the solver output. Table 1.21 Solver output y1 y2 y3 8-2 0 obj 380 32 = 32 26 >= 26 6 <= 60 20 >= 15 8 >= 0 0 <= 0 The optimum solution is Y 1 = 8, Y 2 = -2 and Y 3 = 0 with Z = 380 We now solve the dual. The solution to the dual is given in Table 1.22 Table 1.22 Solution to the dual x1 x2 x3 x4 7 6 0 0 obj fn 380 59 <= 59 46 = 46 69 >= 37 6 >= 0 0 <= 0 0 >= 0 The solution is X 1 = 7, X 2 = 6 with W = 380. 7. It was mentioned that the dual value for a resource is the increase in profit for unit increase in the resource. If we increase the resource by a large value will the objective function value increase at the rate of the dual?

Illustration 1.22 Maximize 32X 1 + 25X 2 5X 4X 59 4X 3X 45 X1, X2 0 We know that the dual solution is Y 1 =4, Y 2 = 3. We have already seen that if the RHS of the first constraint becomes 60, the solution to the primal is X 1 =4, X 2 = 10 with Z = 378. When RHS = 62, the solution is X 1 = 0, X 2 = 15.333 with Z = 383.33. We observe that the objective function has not increased by 12 when the RHS of the first constraint is increased by 3. When RHS = 61.3, the solution is X 1 = 0.1, X 2 = 15.2with Z = 383.2 We observe that the objective function has increased by 9.2 when the RHS of the first constraint is increased by 3.2. The linear increase continues because we continue to produce X 1 and X 2. At RHS = 61.333, we produce X 1 = 0, X 2 = 15.333 with Z = 383.33. The limiting value up to which the linearity holds is 2.333. This is given in Table 6.2 as the allowable increase and allowable decrease for both the resources. As we keep increasing (decreasing) the value of the RHS (resource) the combination of products made changes. At that point, the shadow price changes and the proportional increase (decrease) in the objective function also changes. 8. Is the first dual variable related to the first constraint or the primal (resource)? Yes. We have as many dual variables as the number of constraints in the primal. For every primal constraint there is a dual variable. The dual variable represents the value of the corresponsing resource at the optimum. 9. Can a dual variable take zero value? Yes. We explain this through an example

Illustration 1.23 Maximize 32X 1 + 25X 2 5X 4X 59 4X 3X 46 3X 2X 30 X1, X2 0. The optimum solution to the primal is given in Table 1.23 Table 1.23 Optimum solution from the solver X1 X2 1 13.5 369.5 59 <= 59 44.5 <= 46 30 <= 30 The solution is given by X 1 = 1, X 2 = 13.5 and Z = 369.5. We store the sensitivity result for this solution. Table 1.24 shows the sensitivity results. Table 1.24 Sensitivity results Microsoft Excel 12.0 Sensitivity Report Worksheet: [Book1]Sheet1 Report Created: 24-05-2010 22:19:38 Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$2 X1 1 0 32 5.5 0.75 $C$2 X2 13.5 0 25 0.6 3.666666667 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$6 X1 59 5.5 59 3 1E+30 $B$7 X1 44.5 0 46 1E+30 1.5

$B$8 X1 30 1.499999999 30 3 1E+30 From the shadow prices, we find that the dual solution is Y 1 = 5.5, Y 2 = 0 and Y 3 = 1.5 with W = 59 x 5.5 + 30 x 1.5 = 369.5. We observe that the objective function value of the primal and dual at the optimum is 369.5. Here we observe that Y 2 = 0 indicating that the value of the second dual variable is zero. From Table 6.8 we observe that the primal optimal solution X 1 = 1, X 2 = 13.5 completely utilizes resource 1 and 3 (in the primal) but uses only 44.5 out of the available 46 units of the second resource. We also observe that Y 2 (the dual variable corresponding to this resource) is zero. At the optimum, when the primal resource is utilized fully, the dual variable has a positive value. When a resource is not utilized fully the corresponding dual variable is zero. The dual variable is the marginal value of the resource at the optimum. It also represents the price at which an additional resource can be bought at the optimum. It is obvious that when a resource is not utilized fully, increasing it will not increase the revenue. Therefore the shadow price or marginal value of such a resource is zero. Illustration 1.24 Consider a small factory that makes two types of furniture tables and chairs. They use two resources wood and steel. The amount of steel used per unit is 4 kg and 5 kg for a table and a chair. The amount of wood required per table and chair is 6 and 5 kg. We have 60 kg of steel and 80 kg of wood. The selling price of table and chair is Rs 80 and 70 respectively. Find the number of tables and chairs to be made to maximize revenues. Let X 1 and X 2 be the number of tables and chairs produced. The corresponding LP problem is to Maximize 80X 1 + 70X 2 4X 1 + 5X 2 60 6X 1 + 5X 2 80 X 1, X 2 0 The optimum solution to the LP is X 1 = 10, X 2 = 4 with revenue = 1080. Worth of the resource at the optimum Consider the optimum solution X 1 = 10, X 2 = 4. We observe that we have consumed 60 kg of steel and 80 kg of wood to make 10 tables and 4 chairs. In the example, at the optimum we have consumed all the resources that we had, converted them into produces that have fetched us the revenue. It is therefore not incorrect to consider that the worth if the resources put together is the total revenue that we have earned by converting them into products. Let us try to compute the worth of the resources at the optimum using the example

Illustration 1.25 Consider the data given in Illustration 1.24. Use the optimum solution X 1 = 10 and X 2 = 4 (if required) and try to formulate an LP problem to determine the value (or worth) of the resources at the optimum? Let Y 1 and Y 2 be the worth of 1 kg of steel and wood at the optimum. The worth of a table is 4Y 1 + 6Y 2 and the worth of a chair is 5Y 1 + 5Y 2. Now 4Y 1 + 6Y 2 should be 80 and 5Y 1 + 5Y 2 70 (The worth of a table should not be less than the selling price of 80; otherwise we would not convert the raw material into products and sell them). If the value of the unit resource represents the cost price of the resource (at the optimum), we would minimize the cost price which is to minimize 60Y 1 + 80Y 2. The corresponding LP that determines the value of the resources at the optimum is to Minimize 60Y 1 + 80Y 2 4Y 1 + 6Y 2 80 5Y 1 + 5Y 2 70 Y 1, Y 2 0 The optimum solution to this LP is Y 1 = 2, Y 2 = 12 with total worth = 1080. It is to be noted that expectedly, the total revenue is equal to the total worth of the resource at the optimum. Shadow price and the marginal value of the resources When we computed the value of the resource, we assumed that Y 1 and Y 2 represented the cost price or market price of a unit of the resource. This price is the shadow price of the resource and represents the price of unit resource at the optimum. In our example, we were able to convert 60 kg of steel and 80 kg of wood to 10 tables and 4 chairs having total revenue (worth) of Rs 1080. Let us solve the problem assuming that we have 61 kg of steel and 80 kg of wood. Illustration 1.26 Solve the problem in Illustration 1.24 considering 61 kg of steel and interpret the solution? What happens if we wish to have 70 kg and 90 kg of steel? The corresponding LP would be Maximize 80X 1 + 70X 2 4X 1 + 5X 2 61 6X 1 + 5X 2 80 X 1, X 2 0 The optimum solution is given by X 1 = 9.5, X 2 = 4.6 with revenue = 1082. We observe that a unit increase of steel results in an increase of revenue of Rs 2. If we were to buy the extra

resource (at the optimum) from the market, its worth is Rs 2 because it can fetch us additional revenue of Rs 2. This is the same as Y 1 = 2 in example 2. The value of Y 1 is called the marginal value of resource 1 at the optimum which is the additional worth of a unit of the resource. The reader would observe that the optimum solution now has X 1 = 9.5 and X 2 = 4.6. The reader can ask a question as to whether we can make and sell a fractional number of tables and chairs. The answer is that we are solving LP problem that allows fractional values for the variables. In fact the proportionality assumption makes us believe that 0.5 tables is worth Rs 40 just as half a kg of the first resource is worth Re 1 at the optimum! When we solve for 70 kg of steel, the optimum solution is given by X 1 = 5, X 2 = 10 with revenue = 1100. The additional 10 kg of steel fetched us additional revenue of Rs 20 at the rate of Rs 2/kg for the additional 10kgs. We solve for 70 kg of steel and 90 kg of wood. The optimum solution is given by X 1 = 10 and X 2 = 6 with revenue = 1220. The additional revenue is 1220 1080 = 140. The extra 10 kg of steel is worth 20 and the extra 10 kg of wood is worth 120 (at the rate of 12 per kg for 10 kg note that Y 2 =10, indicating that the worth of 1 kg of wood at the optimum is Rs 10). We observe that the additional cost of the resource is equal to the additional revenue. We solve for 90 kg of steel and 80 kg of wood. The optimum solution is given by X 2 = 16 with revenue = 1120. If we assume that the marginal value of steel is Rs 2, the additional cost is Rs 60 while the additional revenue is only Rs 40. Here, the additional cost does not get reflected as additional revenue. We observe that the proportionality works till we increase the steel quantity to 80 after which we do not produce tables. The proportionality will hold only till the same set of products is made (at different quantities). When the set of products made changes, the extra cost and the extra revenue are not equal. Illustration 1.27 Consider the data in Illustration 1.24. In addition it is observed that it takes 8 hours to make a table and 5 hours to make a chair. The company works 5 days a week and 16 hours a day. Assume that the products have to be made in a week. We have a third constraint which relates the time required to the time available. The problem is to Maximize 80X 1 + 70X 2 4X 1 + 5X 2 60 6X 1 + 5X 2 80 8X 1 + 5X 2 80 X 1, X 2 0

The optimum solution to the LP is given by X 1 = 5, X 2 = 8 with revenue = 960. It is observed that 60 units of steel is consumed, 70 units of wood is consumed and 80 hours of time is consumed. It is observed that the second resource is not fully consumed. The sensitivity report from the excel solver shows that the shadow prices are 8, 0 and 6 for the three resources. The worth of the resources are 60 x 8 + 80 x 0 + 80 X 6 = 960 which is equal to the revenue. It is observed that the shadow price of resource 2 is zero. This is because this resource is not fully consumed. Obviously the marginal value (cost) of this resource is zero because we need not pay extra to buy a unit of this resource since it is available with us. Shadow prices are positive for resources that are scarce (fully consumed). Shadow price is zero when a resource is not fully consumed. Does it mean that the wood used has no value? The answer is No. It is acceptable to say that the marginal value (or the implied cost of procuring an extra kg) is zero since the resource is available with us. When the shadow price is zero we should not assume that the worth is zero. The 70 kg of wood consumed has a value which is not visible at the optimum. Only the marginal value at the optimum has to be interpreted as zero. The total revenue of 960 is distributed between the value of the other two resources at the optimum Sensitivity Analysis Sensitivity analysis deals with the sensitivity of the solution to an LP to changes in problem parameters. The parameters that can change are: 1. Change in Right hand side (RHS) values in the constraints 2. Change in objective function values 3. Changes in coefficients of constraints. In addition two more aspects are included in sensitivity analysis. These are 1. Adding a variable (product) 2. Adding a constraint. In the days when LPs were solved by hand or when solvers were not available freely, it was necessary to compute the effect of the above changes from the optimal solution. Sensitivity analysis methods essentially did that and obtained the optimum solution after the changes from the given optimum. So long as the changes were not on the coefficients of the constraints corresponding to the decision variables in the optimum solution, it is possible to carry out sensitivity analysis from the present optimum solution.

Since we are using the excel solver, we do not emphasize much on sensitivity analysis because it is possible to solve the problem from the beginning using the solver for every change made. We need to make changes in the template and modify the constraints or the objective in the solver. Illustration 1.28 Maximize 32X 1 + 25X 2 5X 4X 59 4X 3X 46 X1, X2 0. Table 1.25 shows the results of modifying some of the parameters. Table 1.25 Results after modifying some parameters No. Parameter Original Changed value Solution value 1. Objective coeff 1 32 33.333 X 1 = 7, X 2 = 6 with Z = 383.33 2. Objective coeff 1 32 33.5 X 1 = 11.5 X 2 = 0 with Z = 385.25 3. Objective coeff 1 32 31.25 X 1 = 7, X 2 = 6 with Z = 368.75 X 1 = 0, X 2 = 14.75 with Z = 368.75 4. Objective coeff 1 32 31 X 1 = 0, X 2 = 14.75 with Z = 368.75 5. Objective coeff 2 25 25.6 X 1 = 7, X 2 = 6 with Z = 377.6 X 1 = 0, X 2 = 14.75with Z = 377.6 6. Objective coeff 2 25 26 X 1 = 0, X 2 = 14.75with Z = 383.5 7. Objective coeff 2 25 24 X 1 = 7, X 2 = 6 with Z = 368 X 1 = 11.5, X 2 = 0 with Z = 368 8. RHS 1 59 60 X 1 = 4, X 2 = 10 with Z = 378 9. RHS1 59 61.333 X 1 =0, X 2 = 15.333 with Z = 383.333 10. RHS1 59 57.5 X 1 = 11.5, X 2 = 0 with Z = 368 11. RHS2 46 47 X 1 = 11, X 2 = 1 with Z = 377 12. RHS2 46 47.2 X 1 = 11.8, X 2 = 0 with Z = 377.6 13. RHS2 46 44.25 X 1 = 0, X 2 = 14.75 with Z = 368.75 From Table 1.25 we observe the following:

1. When the objective function coefficient of the first variable (C 1 ) increases up to 33.33, we produce X 1 and X 2. When it increases to 33.5 we produce only X 1. The limiting value for this coefficient is an increase of 1.5 at/after which the products made changes. 2. The lower limiting value for C 1 is a reduction of 0.75 at/after which we make only X 2. At the limiting value we can have two (or more) solutions. We have shown two solutions under item 3. In fact at all limiting values we have alternate optima. 3. When the objective function coefficient of the first variable (C 2 ) can be increased up to 25.6, we produce X 1 and X 2. When it increases to 25.6 we produce only X 2. The limiting value for this coefficient is an increase of 0.6 at/after which the products made changes. At the limiting value we can have two (or more) solutions. We have shown two solutions under item 5. In fact at all limiting values we have alternate optima. 4. The lower limiting value for C 2 is a reduction of 1 at/after which we make only X 1. At the limiting value we can have two (or more) solutions. We have shown two solutions under item 7. In fact at all limiting values we have alternate optima. 5. We observe that when the RHS of the first constraint goes up to 61.33 we start producing only X 2 and if it reduces to 57.5 we make only X 1. The limiting values are 2.333 and 1.5 respectively. 6. We observe that when the RHS of the second constraint goes up to 47.2 we start producing only X 1 and if it reduces to 44.25 we make only X 2. The limiting values are 1.2 and 1.75 respectively. The observations in Table 1.25 are based on solving the LP problem at least ten times using the solver. It is not a difficult task but one of the reports that can be obtained from the solution to the original problem (sensitivity analysis report) gives all the above information. After we solve the problem, we get a message keep solver solution and we can retain the solver solution. On the right hand side we have the following three options: 1. Answer 2. Sensitivity 3. Limits We can click on the Sensitivity and the solver creates a sheet called Sensitivity report 1. The sheet looks as shown in Table 1.26 Table 1.26 Sensitivity Report Microsoft Excel 12.0 Sensitivity Report Worksheet: [Book1]Sheet1 Report Created: 5/31/2010 3:09:56 PM

Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$2 x1 6.999999998 0 32 1.3333333340.749999999 $C$2 x2 6.000000002 0 25 0.599999999 1 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$6 x1 59 4.000000003 59 2.333333332 1.5 $B$7 x1 46 2.999999996 46 1.2 1.749999999 The allowable increase and decrease for the coefficients in the objective function and for the constraints can be read from Table 1.26. The shadow prices indicate the optimum solution to the dual. Thus the sensitivity analysis report provides certain basic information about the solution to the dual. It also tells us the extent to which the solution is sensitive to changes in coefficients in the objective function and to RHS values. The limiting values where there is a change in the products made is known when the original problem is solved. Illustration 1.29 Consider the data given in Illustration 1.24. The sensitivity analysis report given when the problem is solved using Excel solver is reproduced in Table 1.27 Interpret the solver report to explain the sensitivity of the LP parameters. Table 1.27 Solver Report Variable Values Obj fn Increase Decrease X 1 10 80 4 24 X 2 4 70 30 3.333 Constraint Value Shadow Price Increase Decrease 1 60 2 20 6.6666 2 80 12 10 20 The table shows that the optimum solution to the given problem is X 1 = 10, X 2 = 4. The shadow prices for the two resources are 2 and 12 respectively. When the objective function coefficient for X 1 can be increased to 84 and decreased to 56, we will continue to produce both items. When this value goes out of this range, we will not produce tables. Similarly when the revenue from chair is within 100 and 66.666, we will produce chairs. When the revenue goes beyond these values, we will produce a different combination of items.

When the availability of resource 1 is between 80 and 53.333, the revenue increase (or decrease) will be at the rate of Rs 2 per kg. Similarly, till the availability of the second resource is between 90 and 60, the increase or decrease in revenue will be Rs 12 for unit increase or decrease of the resource. What happens when the objective function coefficients are 50 and 50 and the availability of resources is 60 and 50? The optimum solution to the LP is X 2 = 10 and revenue = 500. The problem becomes a new LP which is quite unrelated to the old LP that we are solving. It only shows that some linear relationships among the variables are not valid at the new values.