SCHOOL OF BUSINESS, ECONOMICS AND MANAGEMENT. BF360 Operations Research

Transcription

1 SCHOOL OF BUSINESS, ECONOMICS AND MANAGEMENT BF360 Operations Research Unit 3 Moses Mwale moses.mwale@ictar.ac.zm

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3 BF360 Operations Research Contents Unit 3: Sensitivity and Duality Sensitivity Analysis Example. Giapetto s Problem Graphical Introduction to Sensitivity Analysis... 5 Graphical Analysis of the Effect of a Change in an Objective Function Coefficient... 5 Graphical Analysis of the Effect of a Change in a Right-Hand Side on the LP s Optimal Solution... 7 Shadow Prices Reduced Costs Dual Problem... 12

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5 BF360 Operations Research Unit 3: Sensitivity and Duality The most important topic of linear programming is of course solving linear programs. We have just covered the topic in the previous lectures. The second most important topics in linear programming are sensitivity analysis and duality. This lecture covers at least the rudiments of Sensitivity Analysis. 3.1 Sensitivity Analysis What and Why is Sensitivity Analysis? When one uses a mathematical model to describe reality one must make approximations. The world is more complicated than the kind of optimization problems that we are able to solve. Indeed, it may well be that the shortest model that explains the universe is the universe itself. Linearity assumptions usually are significant approximations. Another important approximation comes because one cannot be sure of the data one puts into the model. One s knowledge of the relevant technology may be imprecise, forcing one to approximate the parameters A, b and c in the LP max s. t. Moreover, information may change. z = c x Ax b x 0 Sensitivity Analysis is a systematic study of how sensitive the solutions of the LP 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 c changes in its parameter c i, how does the solution change? 2. If the resources available change, i.e., the constraint vector b change in its parameter b i, how does the solution change? 3. If a new constraint is added to the problem, how does the solution change? We shall give answers to the questions 1 and 2. Question 1 is related to the concept of reduced cost, a.k.a. the opportunity cost. Question 2 is related to the concept of shadow price, a.k.a. the marginal price. The question 3 will be completely ignored in these lectures. 3

6 4 Unit 3: Sensitivity and Duality One approach to these questions is to solve lots and lots of LPs: One LP to each change in the parameters. Consider the following classical problem: Example. Giapetto s Problem Giapetto s Woodcarving Inc. manufactures two types of wooden toys: soldiers and trains. A soldier sells for K27 and uses K10 worth of raw materials. Each soldier that is manufactured increases Giapetto s variable labor and overhead costs by K14. A train sells for K21 and uses K9 worth of raw materials. Each train built increases Giapetto s variable labor and overhead costs by K10. The manufacture of wooden soldiers and trains requires two types of skilled labor: carpentry and finishing. A soldier requires 2 hours of finishing labor and 1 hour of carpentry labor. A train requires 1 hour of finishing and 1 hour of carpentry labor. Each week, Giapetto can obtain all the needed raw material but only 100 finishing hours and 80 carpentry hours. Demand for trains is unlimited, but at most 40 soldier are bought each week. Giapetto wants to maximize weekly profits (revenues - costs). The LP for Giapetto s is max z = 3x 1 + 2x 2 (objective function) s. t. 2x 1 + x (finishing constraint) x 1 + x 2 80 (carpentry constraint) x 1 40 (demand for soldiers) x 1, x 2 0 (sign constraints) For example, in Giapetto s problem there might be uncertainty in what is the actual market demand for soldiers. It was assumed to be 40, but it could be anything between 30 and 50. We could then solve the Giapetto s LP separately for market demands 30, 31,, 49, 50. So, we would solve 20 different LPs (21, actually, but who s counting). If it is also assumed that the the profit for soldiers might not be exactly K3 but could be anything between K2.5 and K3.5, then we could also solve the LP separately for profits K2.5, K2.6,, K3.4, K3.5. Combining this with the different LPs we got from the uncertainty in the market demand we would then have = 200 different LPs to solve (well, = 231 if you count correctly). This checking the scenarios method would work, and it is indeed widely used in practice. This method has only two problems: (1) It is inelegant, and (2) it would involve a large amount of calculations. These problems are, however, not critical. Indeed, solving hundreds of LPs is not that time-consuming with modern computers and efficient algorithms like the Simplex. As for the inelegance of the scenario-based method: Who cares about elegance these days? Nevertheless, we shall try to be at least a little bit elegant in this chapter.

7 BF360 Operations Research 3.2 Graphical Introduction to Sensitivity Analysis Sensitivity analysis is concerned with how changes in an LP s parameters affect the optimal solution. Reconsider the Giapetto problem of Section 3.1: max z = 3x 1 + 2x 2 s. t. where 2x 1 + x (Finishing constraint) x 1 + x 2 80 (Carpentry constraint) x 1 40 (Demand constraint) x 1, x 2 0 x 1 = number of soldiers produced per week x 2 = number of trains produced per week The optimal solution to this problem is z = 180, x 1 = 20, x 2 = 60 (point B in Figure 1), and it has x 1, x 2, and s 3 (the slack variable for the demand constraint) as basic variables. How would changes in the problem s objective function coefficients or right-hand sides change this optimal solution? Graphical Analysis of the Effect of a Change in an Objective Function Coefficient If the contribution to profit of a soldier were to increase sufficiently, then it seems reasonable that it would be optimal for Giapetto to produce more soldiers (that is, s 3 would become non-basic). Similarly, if the contribution to profit of a soldier were to decrease sufficiently, then it would become optimal for Giapetto to produce only trains (x 1 would now be nonbasic). We now show how to determine the values of the contribution to profit for soldiers for which the current optimal basis will remain optimal. Let c 1 be the contribution to profit by each soldier. For what values of c 1 does the current basis remain optimal? 5

8 6 Unit 3: Sensitivity and Duality Figure 1: Analysis of Range of Values for Which c 1 Remains Optimal in Giapetto Problem Currently, c 1 = 3, and each isoprofit line has the form 3x 1 + 2x 2 = constant, or x 2 = 3x constant 2 and each isoprofit line has a slope of 3. From Figure 1, we see that if a 2 change in c 1 causes the isoprofit lines to be flatter than the carpentry constraint, then the optimal solution will change from the current optimal solution (point B) to a new optimal solution (point A). If the profit for each soldier is c 1, the slope of each isoprofit line will be c 1 2. Because the slope of the carpentry constraint is 1, the isoprofit lines will be flatter than the carpentry constraint if c 1 > 1, and the current 2

9 BF360 Operations Research basis will no longer be optimal. The new optimal solution will be (0, 80), point A in Figure 1. If the isoprofit lines are steeper than the finishing constraint, then the optimal solution will change from point B to point C. The slope of the finishing constraint is -2. If c 1 2 < 2 or c 1 > 4, then the current basis is no longer optimal and point C, (40, 20), will be optimal. In summary, we have shown that (if all other parameters remain unchanged) the current basis remains optimal for 2 c 1 4, and Giapetto should still manufacture 20 soldiers and 60 trains. Of course, even if 2 c 1 4, Giapetto s profit will change. For instance, if c 1 = 4, then Giapetto s profit will now be 4(20) + 2(60) = $200 instead of $180. Graphical Analysis of the Effect of a Change in a Right-Hand Side on the LP s Optimal Solution A graphical analysis can also be used to determine whether a change in the right-hand side of a constraint will make the current basis no longer optimal. Let b 1 be the number of available finishing hours. Currently, b 1 = 100. For what values of b 1 does the current basis remain optimal? From Figure 2, we see that a change in b 1 shifts the finishing constraint parallel to its current position. The current optimal solution (point B in Figure 2) is where the carpentry and finishing constraints are binding. Figure 2: Range of Values on Finishing Hours for Which Current Basis Remains Optimal in Giapetto Problem 7

10 8 Unit 3: Sensitivity and Duality If we change the value of b 1, then as long as the point where the finishing and carpentry constraints are binding remains feasible, the optimal solution will still occur where the finishing and carpentry constraints intersect. From Figure 2, we see that if b 1 > 120, then the point where the finishing and carpentry constraints are both binding will lie on the portion of the carpentry constraint below point D. Note that at point D, 2(40) + 40 = 120 finishing hours are used. In this region, x 1 > 40, and the demand constraint for soldiers is not satisfied. Thus, for b 1 > 120, the current basis will no longer be optimal. Similarly, if b 1 < 80, then the carpentry and finishing constraints will be binding at an infeasible point having x 1 < 0, and the current basis will no longer be optimal. Note that at point A, = 80 finishing hours are used. Thus (if all other parameters remain unchanged), the current basis remains optimal if 80 b Note that although for 80 b 1 120, the current basis remains optimal, the values of the decision variables and the objective function value change. For example, if 80 b 1 100, then the optimal solution will change from point B to some other point on the line segment AB. Similarly, if 100 b 1 120, then the optimal solution will change from point B to some other point on the line BD.

11 BF360 Operations Research As long as the current basis remains optimal, it is a routine matter to determine how a change in the right-hand side of a constraint changes the values of the decision variables. To illustrate the idea, let b 1 = number of available finishing hours. If we change b 1 to Δ, then we know that the current basis remains optimal for 20 Δ 20. Note that as b 1 changes (as long as 20 Δ 20), the optimal solution to the LP is still the point where the finishing-hour and carpentry-hour constraints are binding. Thus, if b 1 = Δ, we can find the new values of the decision variables by solving 2x 1 + x 2 = Δ and x 1 + x 2 = 80 This yields x 1 = 20+ Δ and x 2 = 60 - Δ. Thus, an increase in the number of available finishing hours results in an increase in the number of soldiers produced and a decrease in the number of trains produced. If b 2 (the number of available carpentry hours) equals 80+ Δ, then it can be shown that the current basis remains optimal for 20 Δ 20. If we change the value of b 2 (keeping 20 Δ 20), then the optimal solution to the LP is still the point where the finishing and carpentry constraints are binding. Thus, if b 2 = 80+ Δ, the optimal solution to the LP is the solution to 2x 1 + x 2 = 100 and x 1 + x 2 = 80 + Δ This yields x 1 = 20+ Δ and x 2 = 60-2Δ, which shows that an increase in the amount of available carpentry hours decreases the number of soldiers produced and increases the number of trains produced. Suppose b 3, the demand for soldiers, is changed to 40 + Δ. Then it can be shown that the current basis remains optimal for Δ 20. For Δ in this range, the optimal solution to the LP will still occur where the finishing and carpentry constraints are binding. Thus, the optimal solution will be the solution to 2x 1 + x 2 = 100 and x 1 + x 2 = 80 Of course, this yields x 1= 20 and x 2 = 60, which illustrates an important fact. Consider a constraint with positive slack (or positive excess) in an LP s optimal solution; if we change the right-hand side of this constraint in the range where the current basis remains optimal, then the optimal solution to the LP is unchanged. 9

12 10 Unit 3: Sensitivity and Duality Exercise 1. Show that if the contribution to profit for trains is between $1.50 and $3, the current basis remains optimal. If the contribution to profit for trains is $2.50, then what would be the new optimal solution? 2. Show that if available carpentry hours remain between 60 and 100, the current basis remains optimal. If between 60 and 100 carpentry hours are available, would Giapetto still produce 20 soldiers and 60 trains? 3. Show that if the weekly demand for soldiers is at least 20, then the current basis remains optimal, and Giapetto should still produce 20 soldiers and 60 trains. Shadow Prices There are two central concepts in sensitivity analysis. They are so important that LP solvers will typically print their values in their standard reports. These are the shadow prices for constraints and reduced costs for decision variables. In this subsection we consider the shadow prices, and show where they are represented in the LP solver reports. Definition. The Shadow Price of a constraint is the amount that the objective function value would change if the said constraint is changed by one unit given that the optimal BVs don t change. Remarks. Note the clause given that the optimal BVs don t change. This means that the shadow price is valid for small changes in the constraints. If the optimal corner changes when a constraint is changed, then the interpretation of the shadow price is no longer valid. It is valid, however, for all changes that are small enough, i.e., below some critical threshold. Shadow prices are sometimes called Marginal Prices. This is actually a much more informative name than the nebulous shadow price. Indeed, suppose you have a constraint that limits the amount of labor available to 40 hours per week. Then the shadow price will tell you how much you would be willing to pay for an additional hour of labor. If your shadow price is K10 for the labor constraint, for instance, you should pay no more than K10 an hour for additional labor. Labor costs of less than K10 per hour will increase the objective value; labor costs of more than K10 per hour will decrease the objective value. Labor costs of exactly K10 will cause the objective function value to remain the same.

13 BF360 Operations Research Reduced Costs Definition. Remark. Let us then consider the reduced costs. Remember that the shadow prices were associated to the constraints, or if you like Simplex language to the slacks. The reduced costs are associated to the decision variables. The Reduced Cost u i for an NBV decision variable x i is the amount the objective value would decrease if x i would be forced to be 1, and thus a BV given that the change from x i = 0 to x i = 1 is small. i.e for any non-basic variable, the reduced cost for the variable is the amount by which the non-basic variable s objective function coefficient must be improved before that variable will become a basic variable in some optimal solution to the LP Here are some interpretations and remarks of reduced costs that should help you to understand the concept: Exercise The clause given that the change from x i = 0 to x i = 1 is small is a similar clause that the clause given that the optimal BVs don t change was in Definition of shadow price. Indeed, it may be, e.g., that forcing x i 1 will make the LP infeasible. Remember: In sensitivity analysis we are talking about small changes whatever that means. The analysis may, and most often will, fail for big changes. Decision variables that are BVs do not have reduced costs, or, if you like, their reduced costs are zero. The reduced cost is also known as Opportunity Cost. Indeed, suppose we are given the forced opportunity (there are no problems only opportunities) to produce one unit of x i that we would not otherwise manufacture at all. This opportunity would cost us, since our optimized objective would decrease to a suboptimal value. Indeed, we have now one more constraint the forced opportunity in our optimization problem. So, the optimal solution can only get worse. The decrease of the objective value is the opportunity cost. The reduced cost u i of x i is the amount by which the objective coefficient c i for x i needs to change before x i will become non-zero. Radioco manufactures two types of radios. The only scarce resource that is needed to produce radios is labor. At present, the company has two laborers. Laborer 1 is willing to work up to 40 hours per week and is paid $5 per hour. Laborer 2 will work up to 50 hours per week for $6 per hour. The price as well as the resources required to build each type of radio are given in Table 1. 11

14 12 Unit 3: Sensitivity and Duality Letting x i be the number of Type i radios produced each week, Radioco should solve the following LP: max z = 3x 1 + 2x 2 s. t. 2x 1 + 2x x 1 + 2x 2 50 x 2 0 a) For what values of the price of a Type 1 radio would the current basis remain optimal? b) For what values of the price of a Type 2 radio would the current basis remain optimal? c) If laborer 1 were willing to work only 30 hours per week, then would the current basis remain optimal? Find the new optimal solution to the LP. d) If laborer 2 were willing to work up to 60 hours per week, then would the current basis remain optimal? Find the new optimal solution to the LP. e) Find the shadow price of each constraint. 3.3 Dual Problem Associated with any LP there is another LP, called the dual and then the original LP is called the primal. The relationship between the primal and dual is important because it gives interesting economic insights. Also, it is important because it gives a connection between the shadow prices and the reduced costs. In general, if the primal LP is a maximization problem, the dual is a minimization problem and vice versa. Also, the constraints of the primal LP are the coefficients of the objective of the dual problem and vice versa. If the constraints of the primal LP are of type then the constraints of the dual LP are of type and vice versa.

15 BF360 Operations Research Let us now give the formal definition of the dual. We assume that the primal LP is in standard form. Since all LPs can be transformed into a standard form this assumption does not restrict the generality of the duality. The assumption is made only for the sake of convenience. Definition The dual of the standard form LP is max z = c 1 x 1 + c 2 x c n x n s. t a 11 x 1 + a 12 x a 1n x n b 1 ; a 21 x 1 + a 22 x a 2n x n b 2 ; a m1 x 1 + a m2 x a mn x n b m ; x 1, x 2, ; x n 0: min w = b 1 y 1 + b 2 y b m y m s. t a 11 y 1 + a 21 y a m1 y m c 1 ; a 11 y 1 + a 22 y a m2 x m c 2 ; a 1n y 1 + a 2n y a mn y m c n ; y 1, y 2,, y m 0 In matrix form the duality can be written as: The dual of the LP max z = c x s.t. Ax b x 0 is min w = b y s.t. A y c y 0 13

16 14 Unit 3: Sensitivity and Duality Example: Consider the LP Example: The Dakota Problem The Dakota problem is max z = 60x x x 3 s.t. 8x 1 + 6x 2 + x x 1 + 2x x x x x 3 8 x 1, x 2, x 3 0 where x 1 = number of desks manufactured x 2 = number of tables manufactured x 3 = number of chairs manufactured (Lumber constraint) (Finishing constraint) (Carpentry constraint) Remark. Then, reading down, we find the Dakota dual to be min w = 48y y 2 + 8y 3 s.t. 8y 1 + 4y 2 + 2y y 1 + 2y y 3 30 y y y 3 20 y 1, y 2, y 3 0 Let us discuss briefly about concept of duality in general and the duality of Definition in particular. In general, dual is a transformation with the following property: Transforming twice you get back. This is the abstract definition of duality.

17 BF360 Operations Research Example. Looking at Definition one sees the dual is LP itself. So, it can be transformed into a standard form, and the one can construct the dual of the dual. When one does so one gets back to the original primal LP, i.e., the dual of the dual is primal. So, the dual of Definition deserves its name. We have already seen one duality between LPs before: A minimization problem is in duality with a maximization problem with the transform where the objective function is multiplied by -1. The usefulness of this simple duality was that we only need to consider maximization problems, and the solution of the minimization problem is -1 times the solution of the corresponding maximization problem in this simple duality. Also, the optimal decisions in the maximization and minimization problems are the same. The duality of Definition is more complicated than the simple multiply by -1 duality of the previous point. This makes the duality of Definition in some sense more useful than the simple multiply by - 1 duality. Indeed, since the transformation is more complicated, our change of perspective is more radical, and thus this transformation gives us better intuition of the original problem. Let us find a dual of an LP that is not in standard form. Consider the LP min z = 50x x x 3 s. t: 2x 1 + 3x 2 + 4x x x x x 1 + x 2 + x 3 = 1 x 1, x 2, x 3 0 The LP of Example above is not in standard form. So, before constructing its dual, we transform it into standard form. This is not necessary. Sometimes we can be clever, and find the dual without first transforming the primal into standard form. But we don t feel clever now. So, here is the standard form: max z = 50x 1 20x 2 30x 3 s: t: 2x 1 3x 2 4x x x x x 1 + x 2 + x 3 1 x 1 x 2 x 3 1 x 1, x 2, x

18 16 Unit 3: Sensitivity and Duality Now we are ready to present the dual: min w = 11y y 2 + y 3 y 4 s: t: 2y y 2 + y 3 y y y 2 + y 3 y y y 2 + y 3 y 4 30 y 1, y 2, y 3, y 4 0 (we used variable -w in the dual because there was variable -z in the standard form primal). Note now that th dual LP is in dual standard form : It is a minimization problem with only inequalities of type. The original primal LP was a minimization problem. So, it is natural to express the dual LP as a maximization problem. Also, inequalities of type are more natural to maximization problems than the opposite type inequalities. So, let us transform the dual LP into a maximization problem with type inequalities. In fact, let us transform the dual LP into a standard form. We obtain max w = 11y 1 111y 2 y 3 + y 4 s. t: 2y 1 12y 2 y 3 + y y 1 13y 2 y 3 + y y 1 14y 2 y 3 + y 4 30 y 1, y 2, y 3, y Economic Interpretation of Dual Let us recall again the Dakota Furniture s problem (without the market demand constraint that turned out to be irrelevant anyway): The Dakota problem is max z = 60x x x 3 s.t. 8x 1 + 6x 2 + x 3 48 (Lumber constraint) where 4x 1 + 2x x 3 20 (Finishing constraint) 2x x x 3 8 (Carpentry constraint) x 1, x 2, x 3 0 x 1 = number of desks manufactured x 2 = number of tables manufactured x 3 = number of chairs manufactured

19 BF360 Operations Research Then, reading down, we find the Dakota dual to be min w = 48y y 2 + 8y 3 s.t. 8y 1 + 4y 2 + 2y 3 60 (desk) 6y 1 + 2y y 3 30 (table) y y y 3 20 (chair) y 1, y 2, y 3 0 We have given the constraints the names (desk), (table), and (chair). Those were the decision variables x 1, x 2 and x 3 in the primal LP. By symmetry, or duality, we could say that y 1 is associated with lumber, y 2 with finishing, and y 3 with carpentry. What is going on here? It is instructive to represent the data of the Dakota s problem in a table where we try to avoid taking Dakota s point of view: Now, the table above can be read either horizontally of vertically. You should already know how the read the table above horizontally. That is the Dakota s point of view. But what does it mean to read the table vertically? Here is the explanation, that is also the economic interpretation of the dual LP: Suppose you are an entrepreneur who wants to purchase all of Dakota s resources maybe you are a competing furniture manufacturer, or maybe you need the resources to produce soldiers and trains like Giapetto. Then you must determine the price you are willing to pay for a unit of each of Dakota s resources. But what are the Dakota s resources? Well they are lumber, finishing hours, and carpentry hours, that Dakota uses to make its products. So, the decision variables for the entrepreneur who wants to buy Dakota s resources are: y 1 = price to pay for one unit of lumber y 2 = price to pay for one hour of finishing labor y 3 = price to pay for one hour of carpentry labor Now we argue that the resource prices y 1, y 2, y 3 should be determined by solving the Dakota dual problem. 17

20 18 Unit 3: Sensitivity and Duality First note that you are buying all of Dakota s resources. Also, note that this is a minimization problem: You want to pay as little as possible. So, the objective function is min w = 48y y 2 + 8y 3 : Indeed, Dakota has 48 units of lumber, 20 hours of finishing labor, and 8 hours of carpentry labor. Now we have the decision variables and the objective. How about constraints? In setting the resource prices y 1, y 2, and y 3, what kind of constraints do you face? You must set the resource prices high enough so that Dakota would sell them to you. Now Dakota can either use the resources itself, or sell them to you. How is Dakota using its resources? Dakota manufactures desks, tables, and chair. Take desks first. With 8 units of lumber, 4 hours of finishing labor, and 2 hours of carpentry labor Dakota can make a desk that will sell for 60. So, you have to offer more than 60 for this particular combination of resources. So, you have the constraint 8y 1 + 4y 2 + 2y 3 60 But this is just the first constraint in the Dakota dual, denoted by (desk). Similar reasoning shows that you must pay at least 30 for the resources Dakota uses to produce one table. So, you get the second constraint, denoted by (table), of the Dakota dual: 6y 1 + 2y y 3 30 Similarly, you must offer more than 20 for the resources the Dakota can use itself to produce one chair. That way you get the last constraint, labeled as (chair), of the Dakota dual: y y y 3 20 We have just interpreted economically the dual of a maximization problem. Let us then change our point of view to the opposite and interpret economically the dual of a minimization problem. Exercise Find the duals of the following LPs: 1 max z = 2x 1 + x 2 s.t. x 1 + x 2 1 x 1 + x 2 3 x 1 2x 2 4 x 1, x 2 0

21 BF360 Operations Research 2 min w = y 1 y 2 s.t. 2y 1 + y 2 4 y 1 + y 2 1 y 1 + 2y 2 3 y 1, y max z = 4x 1 x 2 + 2x 3 s.t. x 1 + x 2 5 2x 1 + x 2 7 2x 2 + x 3 6 x 1 + x 3 = 4 x 1 0, x 2, x 3 urs 4 min w = 4y 1 + 2y 2 y 3 s.t. y 1 + 2y 2 6 y 1 y 2 + 2y 3 = 8 y 1, y 2 0, y 3 urs 19