# Linear Programming Notes VII Sensitivity Analysis

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1 Linear Programming Notes VII Sensitivity Analysis 1 Introduction When you use a mathematical model to describe reality you must make approximations. The world is more complicated than the kinds of optimization problems that we are able to solve. Linearity assumptions usually are significant approximations. Another important approximation comes because you cannot be sure of the data that you put into the model. Your knowledge of the relevant technology may be imprecise, forcing you to approximate values in A, b, or c. Moreover, information may change. Sensitivity analysis is a systematic study of how sensitive (duh) solutions are to (small) changes in the data. The basic idea is to be able to give answers to questions of the form: 1. If the objective function changes, how does the solution change? 2. If resources available change, how does the solution change? 3. If a constraint is added to the problem, how does the solution change? One approach to these questions is to solve lots of linear programming problems. For example, if you think that the price of your primary output will be between \$100 and \$120 per unit, you can solve twenty different problems (one for each whole number between \$100 and \$120). 1 This method would work, but it is inelegant and (for large problems) would involve a large amount of computation time. (In fact, the computation time is cheap, and computing solutions to similar problems is a standard technique for studying sensitivity in practice.) The approach that I will describe in these notes takes full advantage of the structure of LP programming problems and their solution. It turns out that you can often figure out what happens in nearby linear programming problems just by thinking and by examining the information provided by the simplex algorithm. In this section, I will describe the sensitivity analysis information provided in Excel computations. I will also try to give an intuition for the results. 2 Intuition and Overview Throughout these notes you should imagine that you must solve a linear programming problem, but then you want to see how the answer changes if the problem is changed. In every case, the results assume that only one thing about the problem changes. That is, in sensitivity analysis you evaluate what happens when only one parameter of the problem changes. 1 OK, there are really 21 problems, but who is counting? 1

2 To fix ideas, you may think about a particular LP, say the familiar example: max 2x 1 + 4x 2 + 3x 3 + x 4 subject to 3x 1 + x 2 + x 3 + 4x 4 12 x 1 3x 2 + 2x 3 + 3x 4 7 2x 1 + x 2 + 3x 3 x 4 10 x 0 We know that the solution to this problem is x 0 = 42, x 1 = 0; x 2 = 10.4; x 3 = 0; x 4 = Changing Objective Function Suppose that you solve an LP and then wish to solve another problem with the same constraints but a slightly different objective function. (I will always make only one change in the problem at a time. So if I change the objective function, not only will I hold the constraints fixed, but I will change only one coefficient in the objective function.) When you change the objective function it turns out that there are two cases to consider. The first case is the change in a non-basic variable (a variable that takes on the value zero in the solution). In the example, the relevant non-basic variables are x 1 and x 3. What happens to your solution if the coefficient of a non-basic variable decreases? For example, suppose that the coefficient of x 1 in the objective function above was reduced from 2 to 1 (so that the objective function is: max x 1 + 4x 2 + 3x 3 + x 4 ). What has happened is this: You have taken a variable that you didn t want to use in the first place (you set x 1 = 0) and then made it less profitable (lowered its coefficient in the objective function). You are still not going to use it. The solution does not change. Observation If you lower the objective function coefficient of a non-basic variable, then the solution does not change. What if you raise the coefficient? Intuitively, raising it just a little bit should not matter, but raising the coefficient a lot might induce you to change the value of x in a way that makes x 1 > 0. So, for a non-basic variable, you should expect a solution to continue to be valid for a range of values for coefficients of nonbasic variables. The range should include all lower values for the coefficient and some higher values. If the coefficient increases enough (and putting the variable into the basis is feasible), then the solution changes. What happens to your solution if the coefficient of a basic variable (like x 2 or x 4 in the example) decreases? This situation differs from the previous one in that you are using the basis variable in the first place. The change makes the variable contribute less to profit. You should expect that a sufficiently large reduction makes you want to change your solution (and lower the value the associated variable). For example, if the coefficient of x 2 in the objective function in the example were 2 instead of 4 (so that the objective was max 2x 1 +2x 2 +3x 3 +x 4 ), 2

4 2.4 Relationship to the Dual The objective function coefficients correspond to the right-hand side constants of resource constraints in the dual. The primal s right-hand side constants correspond to objective function coefficients in the dual. Hence the exercise of changing the objective function s coefficients is really the same as changing the resource constraints in the dual. It is extremely useful to become comfortable switching back and forth between primal and dual relationships. 3 Understanding Sensitivity Information Provided by Excel Excel permits you to create a sensitivity report with any solved LP. The report contains two tables, one associated with the variables and the other associated with the constraints. In reading these notes, keep the information in the sensitivity tables associated with the first simplex algorithm example nearby. 3.1 Sensitivity Information on Changing (or Adjustable) Cells The top table in the sensitivity report refers to the variables in the problem. The first column (Cell) tells you the location of the variable in your spreadsheet; the second column tells you its name (if you named the variable); the third column tells you the final value; the fourth column is called the reduced cost; the fifth column tells you the coefficient in the problem; the final two columns are labeled allowable increase and allowable decrease. Reduced cost, allowable increase, and allowable decrease are new terms. They need definitions. The allowable increases and decreases are easier. I will discuss them first. The allowable increase is the amount by which you can increase the coefficient of the objective function without causing the optimal basis to change. The allowable decrease is the amount by which you can decrease the coefficient of the objective function without causing the optimal basis to change. Take the first row of the table for the example. This row describes the variable x 1. The coefficient of x 1 in the objective function is 2. The allowable increase is 9, the allowable decrease is 1.00E+30, which means 10 30, which really means. This means that provided that the coefficient of x 1 in the objective function is less than 11 = = original value + allowable increase, the basis does not change. Moreover, since x 1 is a non-basic variable, when the basis stays the same, the value of the problem stays the same too. The information in this line confirms the intuition provided earlier and adds something new. What is confirmed is that if you lower the objective coefficient of a non-basic variable, then your solution does not change. (This means that the allowable decrease will always be infinite for a non-basic variable.) The example also demonstrates your value will stay the same. 4

5 that increasing the coefficient of a non-basic variable may lead to a change in basis. In the example, if you increase the coefficient of x 1 from 2 to anything greater than 9 (that is, if you add more than the allowable increase of 7 to the coefficient), then you change the solution. The sensitivity table does not tell you how the solution changes, but common sense suggests that x 1 will take on a positive value. Notice that the line associated with the other non-basic variable of the example, x 3, is remarkably similar. The objective function coefficient is different (3 rather than 2), but the allowable increase and decrease are the same as in the row for x 1. It is a coincidence that the allowable increases are the same. It is no coincidence that the allowable decrease is the same. We can conclude that the solution of the problem does not change as long as the coefficient of x 3 in the objective function is less than or equal to 10. Consider now the basic variables. For x 2 the allowable increase is infinite while the allowable decrease is 2.69 (it is to be exact). This means that if the solution won t change if you increase the coefficient of x 2, but it will change if you decrease the coefficient enough (that is, by more than 2.7). The fact that your solution does not change no matter how much you increase x 2 s coefficient means that there is no way to make x 2 > 10.4 and still satisfy the constraints of the problem. The fact that your solution does change when you increase x 2 s coefficient by enough means that there is a feasible basis in which x 2 takes on a value lower than (You knew that. Examine the original basis for the problem.) The range for x 4 is different. Line four of the sensitivity table says that the solution of the problem does not change provided that the coefficient of x 4 in the objective function stays between 16 (allowable increase 15 plus objective function coefficient 1) and -4 (objective function coefficient minus allowable decrease). That is, if you make x 4 sufficiently more attractive, then your solution will change to permit you to use more x 4. If you make x 4 sufficiently less attractive the solution will also change. This time to use less x 4. Even when the solution of the problem does not change, when you change the coefficient of a basic variable the value of the problem will change. It will change in a predictable way. Specifically, you can use the table to tell you the solution of the LP when you take the original constraints and replace the original objective function by max 2x 1 + 6x 2 + 3x 3 + x 4 (that is, you change the coefficient of x 2 from 4 to 6), then the solution to the problem remains the same. The value of the solution changes because now you multiply the 10.4 units of x 2 by 6 instead of 4. The objective function therefore goes up by The reduced cost of a variable is the smallest change in the objective function coefficient needed to arrive at a solution in which the variable takes on a positive value when you solve the problem. This is a mouthful. Fortunately, reduced costs are redundant information. The reduced cost is the negative of the allowable increase for non-basic variables (that is, if you change the coefficient of x 1 by 7, then you arrive at a problem in which x 1 takes on a positive 5

6 value in the solution). This is the same as saying that the allowable increase in the coefficient is 7. The reduced cost of a basic variable is always zero (because you need not change the objective function at all to make the variable positive). Neglecting rare cases in which a basis variable takes on the value 0 in a solution, you can figure out reduced costs from the other information in the table: If the final value is positive, then the reduced cost is zero. If the final value is zero, then the reduced cost is negative one times the allowable increase. Remarkably, the reduced cost of a variable is also the amount of slack in the dual constraint associated with the variable. With this interpretation, complementary slackness implies that if a variable that takes on a positive value in the solution, then its reduced cost is zero. 3.2 Sensitivity Information on Constraints The second sensitivity table discusses the constraints. The cell column identifies the location of the left-hand side of a constraint; the name column gives its name (if any); the final value is the value of the left-hand side when you plug in the final values for the variables; the shadow price is the dual variable associated with the constraint; the constraint R.H. side is the right hand side of the constraint; allowable increase tells you by how much you can increase the right-hand side of the constraint without changing the basis; the allowable decrease tells you by how much you can decrease the right-hand side of the constraint without changing the basis. Complementary Slackness guarantees a relationship between the columns in the constraint table. The difference between the Constraint Right-Hand Side column and the Final Value column is the slack. (So, from the table, the slack for the three constraints is 0 (= 12 12), 37 (= 7 ( 30)), and 0 (= 10 10), respectively. We know from Complementary Slackness that if there is slack in the constraint then the associated dual variable is zero. Hence CS tells us that the second dual variable must be zero. Like the case of changes in the variables, you can figure out information on allowable changes from other information in the table. The allowable increase and decrease of non-binding variables can be computed knowing final value and right-hand side constant. If a constraint is not binding, then adding more of the resource is not going to change your solution. Hence the allowable increase of a resource is infinite for a non-binding constraint. (A nearly equivalent, and also true, statement is that the allowable increase of a resource is infinite for a constraint with slack.) In the example, this explains why the allowable increase of the second constraint is infinite. One other quantity is also no surprise. The allowable decrease of a non-binding constraint is equal to the slack in the constraint. Hence the allowable decrease in the second constraint is 37. This means that if you decrease the right-hand side of the second constraint from its original value (7) to anything greater than 30 you do not change the optimal basis. In fact, the only part of the solution that changes when you do this is that the value of the slack variable for this constraint changes. In this paragraph, the point is only this: If you solve an LP and find that a constraint is not binding, 6

8 4 Example In this section I will present another formulation example and discuss the solution and sensitivity results. Imagine a furniture company that makes tables and chairs. A table requires 40 board feet of wood and a chair requires 30 board feet of wood. Wood costs \$1 per board foot and 40,000 board feet of wood are available. It takes 2 hours of skilled labor to make an unfinished table or an unfinished chair. Three more hours of labor will turn an unfinished table into a finished table; two more hours of skilled labor will turn an unfinished chair into a finished chair. There are 6000 hours of skilled labor available. (Assume that you do not need to pay for this labor.) The prices of output are given in the table below: Product Price Unfinished Table \$70 Finished Table \$140 Unfinished Chair \$60 Finished Chair \$110 We want to formulate an LP that describes the production plans that the firm can use to maximize its profits. The relevant variables are the number of finished and unfinished tables, I will call them T F and T U, and the number of finished and unfinished chairs, C F and C U. The revenue is (using the table): 70T U + 140T F + 60C U + 110C F,, while the cost is 40T U + 40T F + 30C U + 30C F (because lumber costs \$1 per board foot). The constraints are: 1. 40T U + 40T F + 30C U + 30C F T U + 5T F + 2C U + 4C F The first constraint says that the amount of lumber used is no more than what is available. The second constraint states that the amount of labor used is no more than what is available. Excel finds the answer to the problem to be to construct only finished chairs ( I m not sure what it means to make a sell 1 3 chair, but let s assume that this is possible). The profit is \$106, Here are some sensitivity questions. 1. What would happen if the price of unfinished chairs went up? Currently they sell for \$60. Because the allowable increase in the coefficient is \$50, it would not be profitable to produce them even if they sold for the same amount as finished chairs. If the price of unfinished chairs went down, then certainly you wouldn t change your solution. 8

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