Datasets:

Lennard-Heuer

/

Into-into-500

like 0

0

C H 16 A P T E R Aggregate and Workforce Planning And I remember misinformation followed us like a plague, Nobody knew from time to time if the plans were changed. Paul Simon 16.1 Introduction A variety of manufacturing management decisions require information about what a plant will produce over the next year or beyond. Examples include the following: 1. Staf?ng. Recruiting and training new workers is a time-consuming process. Management needs a long-term production plan to decide h

1

ing new workers is a time-consuming process. Management needs a long-term production plan to decide how many and what type of workers to add and when to bring them online in order to meet production needs. Conversely, eliminating workers is costly and painful, but sometimes necessary. Anticipating reductions via a long-term plan makes it possible to use natural attrition, or other gentler methods, in place of layoffs to achieve at least part of the reductions. 2. Procurement. Contracts with sup

2

s, in place of layoffs to achieve at least part of the reductions. 2. Procurement. Contracts with suppliers are frequently set up well in advance of placing actual orders. For example, a ?rm might need an opportunity to “certify” the subcontractor for quality and other performance measures. Additionally, some procurement lead times are long (e.g., for high-technology components they may be 6 months or more). Therefore, decisions regarding contracts and long-lead-time orders must be made on the

3

ths or more). Therefore, decisions regarding contracts and long-lead-time orders must be made on the basis of a long-term production plan. 3. Subcontracting. Management must arrange contracts with subcontractors to manufacture entire components or to perform speci?c operations well in advance of actually sending out orders. Determining what types of subcontracting to use requires long-term projections of production requirements and a plan for in-house capacity modi?cations. 4. Marketing. Market

4

ctions of production requirements and a plan for in-house capacity modi?cations. 4. Marketing. Marketing personnel should make decisions on which products to promote on the basis of both a demand forecast and knowledge of which products have tight capacity and which do not. A long-term production plan incorporating planned capacity changes is needed for this. 553 554 Part III Principles in Practice The module in which we address the important question of what will be produced and when it w

5

ractice The module in which we address the important question of what will be produced and when it will be produced over the long range is the aggregate planning (AP) module. As Figure 13.9 illustrated, the AP module occupies a central position in the production planning and control (PPC) hierarchy. The reason, or course, is that so many important decisions, such as those listed, depend on a long-term production plan. Precisely because so many different decisions hinge on the long-range produc

6

ng-term production plan. Precisely because so many different decisions hinge on the long-range production plan, many different formulations of the AP module are possible. Which formulation is appropriate depends on what decision is being addressed. A model for determining the time of staf?ng additions may be very different from a model for deciding which products should be manufactured by outside subcontractors. Yet a different model might make sense if we want to address both issues simultaneo

7

e subcontractors. Yet a different model might make sense if we want to address both issues simultaneously. The staf?ng problem is of suf?cient importance to warrant its own module in the hierarchy of Figure 13.9, the workforce planning (WP) module. Although high-level workforce planning (projections of total staf?ng increases or decreases, institution of training policies) can be done using only a rough estimate of future production based on the demand forecast, low-level staf?ng decisions (tim

8

a rough estimate of future production based on the demand forecast, low-level staf?ng decisions (timing of hires or layoffs, scheduling usage of temporary hires, scheduling training) are often based on the more detailed production information contained in the aggregate plan. In the context of the PPC hierarchy in Figure 13.9, we can think of the AP module as either re?ning the output of the WP module or working in concert with the WP module. In any case, they are closely related. We highlight

9

module or working in concert with the WP module. In any case, they are closely related. We highlight this relationship by treating aggregate planning and workforce planning together in this chapter. As we mentioned in Chapter 13, linear programming is a particularly useful tool for formulating and solving many of the problems commonly faced in the aggregate planning and workforce planning modules. In this chapter, we will formulate several typical AP/WP problems as linear programs (LPs). We wil

10

s. In this chapter, we will formulate several typical AP/WP problems as linear programs (LPs). We will also demonstrate the use of linear programming (LP) as a solution tool in various examples. Our goal is not so much to provide speci?c solutions to particular AP programs, but rather to illustrate general problem-solving approaches. The reader should be able to combine and extend our solutions to cover situations not directly addressed here. Finally, while this chapter will not make an LP expe

11

to cover situations not directly addressed here. Finally, while this chapter will not make an LP expert out of readers, we do hope that they will become aware of how and where LP can be used in solving AP problems. If managers can recognize that particular problems are well suited to LP, they can easily obtain the technical support (consultants, internal experts) for carrying out the analysis and implementation. Unfortunately, far too few practicing managers make this connection; as a result, m

12

d implementation. Unfortunately, far too few practicing managers make this connection; as a result, many are hammering away at problems that are well suited to linear programming with manual spreadsheets and other ad hoc approaches. 16.2 Basic Aggregate Planning We start with a discussion of simple aggregate planning situations and work our way up to more complex cases. Throughout the chapter, we assume that we have a demand forecast available to us. This forecast is generated by the forecast

13

we assume that we have a demand forecast available to us. This forecast is generated by the forecasting module and gives estimates of periodic demand over the planning horizon. Typically, periods are given in months, although further into the future they can represent longer intervals. For instance, periods 1 to 12 might represent the next 12 months, while periods 13 to 16 might represent the four quarters following these 12 months. A typical planning horizon for an AP module is 1 to 3 years.

14

our quarters following these 12 months. A typical planning horizon for an AP module is 1 to 3 years. Chapter 16 16.2.1 555 Aggregate and Workforce Planning A Simple Model Our ?rst scenario represents the simplest possible AP module. We consider this case not because it leads to a practical model, but because it illustrates the basic issues, provides a basis for considering more realistic situations, and showcases how linear programming can support the aggregate planning process. Although

15

ituations, and showcases how linear programming can support the aggregate planning process. Although our discussion does not presume any background in linear programming, the reader interested in how and why LP works is advised to consult Appendix 16A, which provides an elementary overview of this important technique. For modeling purposes, we consider the situation where there is only a single product, and the entire plant can be treated as a single resource. In every period, we have a demand

16

product, and the entire plant can be treated as a single resource. In every period, we have a demand forecast and a capacity constraint. For simplicity, we assume that demands represent customer orders that are due at the end of the period, and we neglect randomness and yield loss. It is obvious under these simplifying assumptions that if demand is less than capacity in every period, the optimal solution is to simply produce amounts equal to demand in every period. This solution will meet all d

17

solution is to simply produce amounts equal to demand in every period. This solution will meet all demand just-in-time and therefore will not build up any inventory between periods. However, if demand exceeds capacity in some periods, then we must work ahead (i.e., produce more than we need in some previous period). If demand cannot be met even by working ahead, we want our model to tell us this. To model this situation in the form of a linear program, we introduce the following notation: t =

18

s. To model this situation in the form of a linear program, we introduce the following notation: t = an index of time periods, where t = 1, . . . , t¯, so t¯ is planning horizon for problem dt = demand in period t, in physical units, standard containers, or some other appropriate quantity (assumed due at end of period) ct = capacity in period t, in same units used for dt r = pro?t per unit of product sold (not including inventory carrying cost) h = cost to hold one unit of inventory for one per

19

oduct sold (not including inventory carrying cost) h = cost to hold one unit of inventory for one period X t = quantity produced during period t (assumed available to satisfy demand at end of period t) St = quantity sold during period t (we assume that units produced in t are available for sale in t and thereafter) It = inventory at end of period t (after demand has been met); we assume I0 is given as data In this notation, X t , St , and It are decision variables. That is, the computer program

20

ven as data In this notation, X t , St , and It are decision variables. That is, the computer program solving the LP is free to choose their values so as to minimize the objective, provided the constraints are satis?ed. The other variables—dt , ct , r , h—are constants, which must be estimated for the actual system and supplied as data. Throughout this chapter, we use the convention of representing variables with capital letters and constants with lowercase letters. We can represent the problem

21

ing variables with capital letters and constants with lowercase letters. We can represent the problem of maximizing net pro?t minus inventory carrying cost subject to capacity and demand constraints as Maximize t¯  r St - h It (16.1) t=1 Subject to: St = dt t = 1, . . . , t¯ (16.2) 556 Part III Principles in Practice X t = ct t = 1, . . . , t¯ (16.3) It = It-1 + X t - St t = 1, . . . , t¯ (16.4) X t , St , It = 0 t = 1, . . . , t¯ (16.5) The objective function computes net

22

. . . , t¯ (16.4) X t , St , It = 0 t = 1, . . . , t¯ (16.5) The objective function computes net pro?t by multiplying unit pro?t r by sales St in each period t, and subtracting the inventory carrying cost h times remaining inventory It at the end of period t, and summing over all periods in the planning horizon. Constraints (16.2) limit sales to demand. If possible, the computer will make all these constraints tight, since increasing the St values increases the objective function. The only

23

ll these constraints tight, since increasing the St values increases the objective function. The only reason that these constraints will not be tight in the optimal solution is that capacity constraints (16.3) will not permit it.1 Constraints (16.4), which are of a form common to almost all multiperiod aggregate planning models, are known as balance constraints. Physically, all they represent is conservation of material; the inventory at the end of period t(It ) is equal to the inventory at the

24

conservation of material; the inventory at the end of period t(It ) is equal to the inventory at the end of period t - 1(It-1 ) plus what was produced during period t(X t ) minus the amount sold in period t (St ). These constraints are what force the computer to choose values for X t , St , and It that are consistent with our verbal de?nitions of them. Constraints (16.5) are simple non-negativity constraints, which rule out negative production or inventory levels. Many, but not all (e.g., not

25

ty constraints, which rule out negative production or inventory levels. Many, but not all (e.g., not Solver in Excel), computer packages for solving LPs automatically force decision variables to be non-negative unless the user speci?es otherwise. 16.2.2 An LP Example To make the above formulation concrete and to illustrate the mechanics of solving it via linear programming, we now consider a simple example. The Excel spreadsheet shown in Figure 16.1 contains the unit pro?t r of $10, the one-p

26

imple example. The Excel spreadsheet shown in Figure 16.1 contains the unit pro?t r of $10, the one-period unit holding cost h of $1, the initial inventory I0 of 0, and capacity and demand data ct and dt for the next 6 months. We will make use of the rest of the spreadsheet in Figure 16.1 momentarily. For now, we can express LP (16.1)–(16.5) for this speci?c case as Maximize 10(S1 + S2 + S3 + S4 + S5 + S6 ) - 1(I1 + I2 + I3 + I4 + I5 + I6 ) (16.6) Subject to: Demand constraints S1 = 80 (16.

27

4 + S5 + S6 ) - 1(I1 + I2 + I3 + I4 + I5 + I6 ) (16.6) Subject to: Demand constraints S1 = 80 (16.7) S2 = 100 (16.8) S3 = 120 (16.9) S4 = 140 (16.10) S5 = 90 (16.11) S6 = 140 (16.12) 1 If we want to consider demand as inviolable, we could remove constraints (16.2) and replace S with d t t in the objective and constraints (16.4). The problem with this, however, is that if demand is capacityinfeasible, the computer will just come back with a message saying “infeasible,” which doesn’

28

capacityinfeasible, the computer will just come back with a message saying “infeasible,” which doesn’t tell us why. The formulation here will be feasible regardless of demand; it simply won’t make sales equal to demand if there is not enough capacity, and thus we will know what demand we are incapable of meeting from the solution. Chapter 16 Figure 16.1 Input spreadsheet for linear programming example. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

29

g example. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 557 Aggregate and Workforce Planning B C D E F G H Constants: r h I_O t c_t d_t 10 1 0 1 100 80 2 100 100 3 100 120 4 120 140 5 120 90 6 120 140 Total 660 670 Variables: t X_t S_t I_t 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 5 0 0 0 6 0 0 0 Total 0 0 0 Objective: Net Profit: $0 Constraints: S_1 S_2 S_3 S_4 S_5 S_6 X_1 X_2 X_3 X_4 X_5 X_6 I_1-I_0-X_1+S_1 I_2-I_1-X

30

t Profit: $0 Constraints: S_1 S_2 S_3 S_4 S_5 S_6 X_1 X_2 X_3 X_4 X_5 X_6 I_1-I_0-X_1+S_1 I_2-I_1-X_2+S_2 I_3-I_2-X_3+S_3 I_4-I_3-X_4+S_4 I_5-I_4-X_5+S_5 I_6-I_5-X_6+S_6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 r*(S_1+S_2+S_3+S_4+S_5+S_6) - h*(I_1+I_2+I_3+I_4+I_5+I_6) <= <= <= <= <= <= <= <= <= <= <= <= = = = = = = d_1 d_2 d_3 d_4 d_5 d_6 c_1 c_2 c_3 c_4 c_5 c_6 80 100 120 140 90 140 100 100 100 120 120 120 0 0 0 0 0 0 Note: X_t, S_t and I_t must be >= 0 Capacity constraints X 1 = 100 (16.

31

00 120 120 120 0 0 0 0 0 0 Note: X_t, S_t and I_t must be >= 0 Capacity constraints X 1 = 100 (16.13) X 2 = 100 (16.14) X 3 = 100 (16.15) X 4 = 120 (16.16) X 5 = 120 (16.17) X 6 = 120 (16.18) Inventory balance constraints I1 - X 1 + S1 = 0 (16.19) I2 - I1 - X 2 + S2 = 0 (16.20) I3 - I2 - X 3 + S3 = 0 (16.21) I4 - I3 - X 4 + S4 = 0 (16.22) I5 - I4 - X 5 + S5 = 0 (16.23) I6 - I5 - X 6 + S6 = 0 (16.24) Non-negativity constraints X 1, X 2, X 3, X 4, X 5, X 6 = 0 (16.25)

32

- I5 - X 6 + S6 = 0 (16.24) Non-negativity constraints X 1, X 2, X 3, X 4, X 5, X 6 = 0 (16.25) S1 , S2 , S3 , S4 , S5 , S6 = 0 (16.26) I1 , I2 , I3 , I4 , I5 , I6 = 0 (16.27) 558 Part III Principles in Practice Some linear programming packages allow entry of a problem formulation in a format almost identical to (16.6) to (16.27) via a text editor. While this is certainly convenient for very small problems, it can become prohibitively tedious for large ones. Because of this, the OM

33

for very small problems, it can become prohibitively tedious for large ones. Because of this, the OM research community has done considerable work to develop modeling languages that provide user-friendly interfaces for describing large-scale optimization problems (see Fourer, Gay, and Kernighan 1993 for an excellent example of a modeling language). Conveniently for us, LP is becoming so prevalent that our spreadsheet package, Microsoft Excel, has an LP tool, called the Solver, built right into

34

nt that our spreadsheet package, Microsoft Excel, has an LP tool, called the Solver, built right into it. We can represent and solve formulations (16.6) to (16.27) right in the spreadsheet shown in Figure 16.1. The following technical note provides details on how to do this. Technical Note: Using the Excel LP Solver Although the reader should consult the Excel documentation for details about the release in use, we will provide a brief overview of the LP solver in Excel 2007. The ?rst step is t

35

e release in use, we will provide a brief overview of the LP solver in Excel 2007. The ?rst step is to establish cells for the decision variables (B11:G13 in Figure 16.1). We have initially entered zeros for these, but we can set them to be anything we like; thus, we could start by setting X t = dt , which would be closer to an optimal solution than zeros. The spreadsheet is a good place to play what-if games with the data. However, eventually we will turn over the problem of ?nding optimal val

36

what-if games with the data. However, eventually we will turn over the problem of ?nding optimal values for the decision variables to the LP solver. Notice that for convenience we have also entered a column that totals X t , St , and It . For example, cell H11 contains a formula to sum cells B11:G11. This allows us to write the objective function more compactly. Once we have speci?ed decision variables, we construct an objective function in cell B16. We do this by writing a formula that multip

37

ariables, we construct an objective function in cell B16. We do this by writing a formula that multiplies r (cell B2) by total sales (cell H12) and then subtracts the product of h (cell B3) and total inventory (cell H13). Since all the decision variables are zero at present, this formula also returns a zero; that is, the net pro?t on no production with no initial inventory is zero. Next we need to specify the constraints (16.7) to (16.27). To do this, we need to develop formulas that compute th

38

to specify the constraints (16.7) to (16.27). To do this, we need to develop formulas that compute the left-hand side of each constraint. For constraints (16.7) to (16.18) we really do not need to do this, since the left-hand sides are only X t and St and we already have cells for these in the variables portion of the spreadsheet. However, for clarity, we will copy them to cells B19:B30. We will not do the same for the non-negativity constraints (16.25) to (16.27), since it is a simple matter t

39

not do the same for the non-negativity constraints (16.25) to (16.27), since it is a simple matter to choose all the decision variables and force them to be greater than or equal to zero in the Excel Solver menu. Constraints (16.19) to (16.24) require us to do work, since the left-hand sides are formulas of multiple variables. For instance, cell B31 contains a formula to compute I1 - I0 - X 1 + S1 (that is, B13 - B4 - B11 + B12). We have given these cells names to remind us of what they repres

40

S1 (that is, B13 - B4 - B11 + B12). We have given these cells names to remind us of what they represent, although any names could be used, since they are not necessary for the computation. We have also copied the values Figure 16.2 Speci?cation of objectives and constraints in Excel. Solver Parameters Set Target Cell: $B$16 Equal To: Max By Changing Cells: Min $B$11:$G$13 Solve Value of: 0 Close Guess Subject to the Constraints: $B$11:$G$13 >= 0 $B$19:$B$30 <= $D$19:$D$30 $B$31:$B$36

41

0 Close Guess Subject to the Constraints: $B$11:$G$13 >= 0 $B$19:$B$30 <= $D$19:$D$30 $B$31:$B$36 = 0 Options Add Change Reset All Delete Help Chapter 16 559 Aggregate and Workforce Planning Figure 16.3 Add Constraint Add Constraint dialog box in Excel. Cell Reference: Constraint: $B$19:$B$30 OK =D$19:$D$30 <= Cancel Add Figure 16.4 Solver Options Setting Excel to use linear programming. Max Time: 100 Iterations: 100 Precision: 0.000001 Tolerance: 5 Convergence: 0

42

programming. Max Time: 100 Iterations: 100 Precision: 0.000001 Tolerance: 5 Convergence: 0.001 seconds Help OK Cancel Load Model... % Save Model... Help Assume Linear Model Use Automatic Scaling Assume Non-Negative Estimates Derivatives Show Iteration Results Search Tangent Forward Newton Quadratic Central Conjugate of the right-hand sides of the constraints into cells D19:D36 and labeled them in column E for clarity. This is not strictly necessary, but does make it ea

43

D19:D36 and labeled them in column E for clarity. This is not strictly necessary, but does make it easier to specify constraints in the Excel Solver, since whole blocks of constraints can be speci?ed (for example, B19:B30 = D19:D30). The equality and inequality symbols in column C are also unnecessary, but make the formulation easier to read. To use the Excel LP Solver, we choose Formula/Solver from the menu. In the dialog box that comes up (see Figure 16.2), we specify the cells containing the

44

from the menu. In the dialog box that comes up (see Figure 16.2), we specify the cells containing the objective, choose to maximize or minimize, and specify the cells containing decision variables (this can be done by pointing with the mouse). Then we add constraints by choosing Add from the constraints section of the form. Another dialog box (see Figure 16.3) comes up in which we ?ll in the cell containing the left-hand side of the constraint, choose the relationship (=, =, or =), and ?ll in t

45

l containing the left-hand side of the constraint, choose the relationship (=, =, or =), and ?ll in the right-hand side. Note that the actual constraint is not shown explicitly in the spreadsheet; it is entered only in the Solver menu. However, the right-hand side of the constraint can be another cell in the spreadsheet or a constant. By specifying a range of cells for the right-hand side and a constant for the left-hand side, we can add a whole set of constraints in a single command. For insta

46

constant for the left-hand side, we can add a whole set of constraints in a single command. For instance, the range B11:G13 represents all the decision variables, so if we use this range as the left-hand side, a = symbol, and a zero for the right-hand side, we will represent all the non-negativity constraints (16.25) to (16.27). By choosing the Add button after each constraint we enter, we can add all the model constraints. When we are done, we choose the OK button, which returns us to the orig

47

dd all the model constraints. When we are done, we choose the OK button, which returns us to the original form. We have the option to edit or delete constraints at any time. Finally, before running the model, we must tell Excel that we want it to use the LP solution algorithm.2 We do this by choosing the Options button to bring up another dialog box (see Figure 16.4) and choosing the Assume Linear Model option. This form also allows 2 Excel can also solve nonlinear optimization problems and wil

48

ar Model option. This form also allows 2 Excel can also solve nonlinear optimization problems and will apply the nonlinear algorithm as a default. Since LP is much more ef?cient, we de?nitely want to choose it as long as our model meets the requirements. All the formulations in this chapter are linear and therefore can use LP. 560 Part III Principles in Practice us to limit the time the model will run and to specify certain tolerances. If the model does not converge to an answer, the most

49

el will run and to specify certain tolerances. If the model does not converge to an answer, the most likely reason is an error in one of the constraints. However, sometimes increasing the search time or reducing tolerances will ?x the problem when the solver cannot ?nd a solution. The reader should consult the Excel manual for more detailed documentation on this and other features, as well as information on upgrades that may have occurred since this writing. Choosing the OK button returns us to

50

formation on upgrades that may have occurred since this writing. Choosing the OK button returns us to the original form. Once we have done all this, we are ready to run the model by choosing the Solve button. The program will pause to set up the problem in the proper format and then will go through a sequence of trial solutions (although not for long in such a small problem as this). Basically, LP works by ?rst ?nding a feasible solution—one that satis?es all the constraints—and then generatin

51

LP works by ?rst ?nding a feasible solution—one that satis?es all the constraints—and then generating a succession of new solutions, each better than the last. When no further improvement is possible, it stops and the solution is optimal: It maximizes or minimizes the objective function. Appendix 16A provides background on how this process works. The algorithm will stop with one of three answers: 1. Could not ?nd a feasible solution. This probably means that the problem is infeasible; that is,

52

s: 1. Could not ?nd a feasible solution. This probably means that the problem is infeasible; that is, there is no solution that satis?es all the constraints. This could be due to a typing error (e.g., a plus sign was incorrectly typed as a minus sign) or a real infeasibility (e.g., it is not possible to meet demand with capacity). Notice that by clever formulation, one can avoid having the algorithm terminate with this depressing message when real infeasibilities exist. For instance, in formula

53

ithm terminate with this depressing message when real infeasibilities exist. For instance, in formulation (16.6) to (16.27), we did not force sales to be equal to demand. Since cumulative demand exceeds cumulative capacity, it is obvious that this would not have been feasible. By setting separate sales and production variables, we let the computer tell us where demand cannot be met. Many variations on this trick are possible. 2. Does not converge. This means either that the algorithm could not

54

ons on this trick are possible. 2. Does not converge. This means either that the algorithm could not ?nd an optimal solution within the allotted time (so increasing the time or decreasing the tolerances under the Options menu might help) or that the algorithm is able to continue ?nding better and better solutions inde?nitely. This second possibility can occur when the problem is unbounded: The objective can be driven to in?nity by letting some variables grow positive or negative without bound.

55

bjective can be driven to in?nity by letting some variables grow positive or negative without bound. Usually this is the result of a failure to properly constrain a decision variable. For instance, in the above model, if we forgot to specify that all decision variables must be non-negative, then the model will be able to make the objective arbitrarily large by choosing negative values of It , t = 1, . . . , 6. Of course, we do not generate revenue via negative inventory levels, so it is importa

56

= 1, . . . , 6. Of course, we do not generate revenue via negative inventory levels, so it is important that non-negativity constraints be included to rule out this nonsensical behavior.3 3. Found a solution. This is the outcome we want. When it occurs, the program will write the optimal values of the decision variables, objective value, and constraints into the spreadsheet. Figure 16.5 shows the spreadsheet as modi?ed by the LP algorithm. The program also offers three reports—Answer, 3 We will

57

e spreadsheet as modi?ed by the LP algorithm. The program also offers three reports—Answer, 3 We will show how to modify the formulation to allow for backordering, which is like allowing negative inventory positions, without this inappropriately affecting the objective function, later in this chapter. Chapter 16 Figure 16.5 Output spreadsheet for LP example. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 561 Aggregate and Workforce

58

4 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 561 Aggregate and Workforce Planning B C D E F G H Constants: r h I_0 t c_t d_t 10 1 0 1 100 80 2 100 100 3 100 120 4 120 140 5 120 90 6 120 140 Total 660 670 Variables: t X_t S_t I_t 1 100 80 20 2 100 100 20 3 100 120 0 4 120 120 0 5 110 90 20 6 120 140 0 Total 650 650 60 Objective: Net Profit: $6,440 Constraints: S_1 S_2 S_3 S_4 S_5 S_6 X_1 X_2 X_3 X_4 X_5 X_6 I_1-I_0-X_1+S_1 I_2-I_1-X_2+S_2 I_3-

59

40 Constraints: S_1 S_2 S_3 S_4 S_5 S_6 X_1 X_2 X_3 X_4 X_5 X_6 I_1-I_0-X_1+S_1 I_2-I_1-X_2+S_2 I_3-I_2-X_3+S_3 I_4-I_3-X_4+S_4 I_5-I_4-X_5+S_5 I_6-I_5-X_6+S_6 80 100 120 120 90 140 100 100 100 120 110 120 0 0 0 0 0 0 r*(S_1+S_2+S_3+S_4+S_5+S_6) - h*(I_1+I_2+I_3+I_4+I_5+I_6) <= <= <= <= <= <= <= <= <= <= <= <= = = = = = = 80 100 120 140 90 140 100 100 100 120 120 120 0 0 0 0 0 0 d_1 d_2 d_3 d_4 d_5 d_6 c_1 c_2 c_3 c_4 c_5 c_6 Note: X_t, S_t and I_t must be >= 0 Sensitivity, and Limits—w

60

3 d_4 d_5 d_6 c_1 c_2 c_3 c_4 c_5 c_6 Note: X_t, S_t and I_t must be >= 0 Sensitivity, and Limits—which write information about the solution into other spreadsheets. For instance, highlighting the Answer report generates a spreadsheet with the information shown in Figures 16.6 and 16.7. Figure 16.8 contains some of the information contained in the report generated by choosing Sensitivity. Now that we have generated a solution, let us interpret it. Both Figure 16.5—the ?nal spreadsheet—and Fig

61

that we have generated a solution, let us interpret it. Both Figure 16.5—the ?nal spreadsheet—and Figure 16.6 show the optimal decision variables. From these we see that it is not optimal to produce at full capacity in every period. Speci?cally, the solution calls for producing only 110 units in month 5 when capacity is 120. This might seem odd given that demand exceeds capacity. However, if we look more carefully, we see that cumulative demand for periods 1 to 4 is 440 units, while cumulative

62

look more carefully, we see that cumulative demand for periods 1 to 4 is 440 units, while cumulative capacity for those periods is only 420 units. Thus, even when we run ?at out for the ?rst 4 months, we will fall short of meeting demand by 20 units. Demand in the ?nal 2 months is only 230 units, while capacity is 240 units. Since our model does not permit backordering, it does not make sense to produce more than 230 units in months 5 and 6. Any extra units cannot be used to make up a previous

63

produce more than 230 units in months 5 and 6. Any extra units cannot be used to make up a previous shortfall. Figure 16.7 gives more details on the constraints by showing which ones are binding or tight (i.e., equal to the right-hand side) and which ones are nonbinding or slack, and by how much. Most interesting are the constraints on sales, given in (16.7) to (16.12), and capacity, in (16.13) to (16.18). As we have already noted, the capacity constraint on X 5 is nonbinding. Since we produce

64

to (16.18). As we have already noted, the capacity constraint on X 5 is nonbinding. Since we produce only 110 units in month 5 and have capacity for 120, this constraint is slack by 10 units. This means that if we changed this constraint 562 Part III Principles in Practice Figure 16.6 Figure 16.7 Optimal values report for LP example. Optimal constraint status for LP example. Microsoft Excel 12.0 Answer Report Worksheet: [BasicCap.xls]Figure 16.6 Report Created: 8/29/2007 3:11:48 PM M

65

Excel 12.0 Answer Report Worksheet: [BasicCap.xls]Figure 16.6 Report Created: 8/29/2007 3:11:48 PM Microsoft Excel 12.0 Answer Report Worksheet: [BasicCap.xls]Figure 16.7 Report Created: 8/29/2007 3:11:48 PM Target Cell (Max) Cell Name $B$16 Net_Profit Original Value Final Value $0.00 $6,440.00 Adjustable Cells Cell Name $B$11 X_1 $C$11 X_2 $D$11 X_3 $E$11 X_4 $F$11 X_5 $G$11 X_6 $B$12 S_1 $C$12 S_2 $D$12 S_3 $E$12 S_4 $F$12 S_5 $G$12 S_6 $B$13 I_1 $C$13 I_2 $D$13 I_3 $E$13 I_4 $F$13 I_5 $G

66

C$12 S_2 $D$12 S_3 $E$12 S_4 $F$12 S_5 $G$12 S_6 $B$13 I_1 $C$13 I_2 $D$13 I_3 $E$13 I_4 $F$13 I_5 $G$13 I_6 Original Value Final Value 0 100 0 100 0 100 0 120 0 110 0 120 0 80 0 100 0 120 0 120 0 90 0 140 0 20 0 20 0 0 0 0 0 20 0 0 Constraints Cell Name Cell Value Formula 0 $B$31=0 $B$31 I_1-I_0-X_1+S_1 0 $B$32=0 $B$32 I_2-I_1-X_2+S_2 0 $B$33=0 $B$33 I_3-I_2-X_3+S_3 0 $B$34=0 $B$34 I_4-I_3-X_4+S_4 0 $B$35=0 $B$35 I_5-I_4-X_5+S_5 0 $B$36=0 $B$36 I_6-I_5-X_6+S_6 80 $B$19<=$D$19 $B$19 S_1 100 $

67

4+S_4 0 $B$35=0 $B$35 I_5-I_4-X_5+S_5 0 $B$36=0 $B$36 I_6-I_5-X_6+S_6 80 $B$19<=$D$19 $B$19 S_1 100 $B$20<=$D$20 $B$20 S_2 120 $B$21<=$D$21 $B$21 S_3 120 $B$22<=$D$22 $B$22 S_4 90 $B$23<=$D$23 $B$23 S_5 140 $B$24<=$D$24 $B$24 S_6 100 $B$25<=$D$25 $B$25 X_1 100 $B$26<=$D$26 $B$26 X_2 100 $B$27<=$D$27 $B$27 X_3 120 $B$28<=$D$28 $B$28 X_4 110 $B$29<=$D$29 $B$29 X_5 120 $B$30<=$D$30 $B$30 X_6 100 $B$11>=0 $B$11 X_1 100 $C$11>=0 $C$11 X_2 100 $D$11>=0 $D$11 X_3 120 $E$11>=0 $E$11 X_4 110 $F$11>=0 $F

68

$11>=0 $B$11 X_1 100 $C$11>=0 $C$11 X_2 100 $D$11>=0 $D$11 X_3 120 $E$11>=0 $E$11 X_4 110 $F$11>=0 $F$11 X_5 120 $G$11>=0 $G$11 X_6 80 $B$12>=0 $B$12 S_1 100 $C$12>=0 $C$12 S_2 120 $D$12>=0 $D$12 S_3 120 $E$12>=0 $E$12 S_4 90 $F$12>=0 $F$12 S_5 140 $G$12>=0 $G$12 S_6 20 $B$13>=0 $B$13 I_1 20 $C$13>=0 $C$13 I_2 0 $D$13>=0 $D$13 I_3 0 $E$13>=0 $E$13 I_4 20 $F$13>=0 $F$13 I_5 0 $G$13>=0 $G$13 I_6 Status Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Binding Binding Bindin

69

Status Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Binding Binding Binding Not Binding Binding Binding Binding Binding Binding Binding Not Binding Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Not Binding Binding Binding Not Binding Binding Slack 0 0 0 0 0 0 0 0 0 20 0 0 0 0 0 0 10 0 100 100 100 120 110 120 80 100 120 120 90 140 20 20 0 0 20 0 by a

70

0 0 0 0 0 0 0 20 0 0 0 0 0 0 10 0 100 100 100 120 110 120 80 100 120 120 90 140 20 20 0 0 20 0 by a little (e.g., reduced capacity in month 5 from 120 to 119 units), it would not change the optimal solution at all. In this same vein, all sales constraints are tight except that for S4 . Since sales are limited to 140, but optimal sales are 120, this constraint has slackness of 20 units. Again, if we were to change this sales constraint by a little (e.g., limit sales to 141 units), the optimal s

71

f we were to change this sales constraint by a little (e.g., limit sales to 141 units), the optimal solution would remain the same. In contrast with these slack constraints, consider a binding constraint. For instance, consider the capacity constraint on X 1 , which is the seventh one shown in Figure 16.7. Since the model chooses production equal to capacity in month 1, this constraint is tight. If we were to change this constraint by increasing or decreasing capacity, the solution would change

72

If we were to change this constraint by increasing or decreasing capacity, the solution would change. If we relax the constraint by increasing capacity, say, to 101 units, then we will be able to satisfy an additional unit of demand and therefore the net pro?t will increase. Since we will produce the extra item in month 1, hold it for 3 months to month 4 at a cost of $1 per month, and then sell it for $10, the overall increase in the objective from this change will be $10 - 3 = $7. Conversely,

73

it for $10, the overall increase in the objective from this change will be $10 - 3 = $7. Conversely, if we tighten the constraint by decreasing capacity, say to 99 units, then we will be able to carry only 19 units from Chapter 16 Figure 16.8 Sensitivity analysis for LP example. 563 Aggregate and Workforce Planning Microsoft Excel 12.0 Sensitivity Report Worksheet: [BasicCap.xls]Figure 16.8 Report Created: 8/29/2007 3:11:48 PM Adjustable Cells Cell Name $B$11 X_1 $C$11 X_2 $D$11 X_3 $E$1

74

.8 Report Created: 8/29/2007 3:11:48 PM Adjustable Cells Cell Name $B$11 X_1 $C$11 X_2 $D$11 X_3 $E$11 X_4 $F$11 X_5 $G$11 X_6 $B$12 S_1 $C$12 S_2 $D$12 S_3 $E$12 S_4 $F$12 S_5 $G$12 S_6 $B$13 I_1 $C$13 I_2 $D$13 I_3 $E$13 I_4 $F$13 I_5 $G$13 I_6 Final Reduced Objective Cost Coefficient Value 0 0 100 0 0 100 0 0 100 0 0 120 0 0 110 0 0 120 0 10 80 0 10 100 0 10 120 0 10 120 0 10 90 0 10 140 0 -1 20 0 -1 20 0 -1 0 -11 -1 0 0 -1 20 -2 -1 0 Allowable Increase 1E+30 1E+30 1E+30 1E+30 1 1E+30 1E+3

75

1 20 0 -1 20 0 -1 0 -11 -1 0 0 -1 20 -2 -1 0 Allowable Increase 1E+30 1E+30 1E+30 1E+30 1 1E+30 1E+30 1E+30 1E+30 1 1E+30 1E+30 3 2 1 11 1 2 Allowable Decrease 7 8 9 10 9 1 3 2 1 7 10 9 7 7 7 1E+30 9 1E+30 Allowable Increase 20 20 20 20 110 20 0 0 0 1E+30 10 10 20 20 20 20 1E+30 20 Allowable Decrease 0 0 0 120 10 10 20 20 20 20 90 20 0 0 0 120 10 10 Constraints Final Shadow Price Cell Name Value 7 0 $B$31 I_1-I_0-X_1+S_1 8 0 $B$32 I_2-I_1-X_2+S_2 9 0 $B$33 I_3-I_2-X_3+S_3 10 0 $B$34 I_4-I_

76

Value 7 0 $B$31 I_1-I_0-X_1+S_1 8 0 $B$32 I_2-I_1-X_2+S_2 9 0 $B$33 I_3-I_2-X_3+S_3 10 0 $B$34 I_4-I_3-X_4+S_4 0 0 $B$35 I_5-I_4-X_5+S_5 1 0 $B$36 I_6-I_5-X_6+S_6 3 80 $B$19 S_1 2 100 $B$20 S_2 1 120 $B$21 S_3 0 120 $B$22 S_4 10 90 $B$23 S_5 9 140 $B$24 S_6 7 100 $B$25 X_1 8 100 $B$26 X_2 9 100 $B$27 X_3 10 120 $B$28 X_4 0 110 $B$29 X_5 1 120 $B$30 X_6 Constraint R.H. Side 0 0 0 0 0 0 80 100 120 140 90 140 100 100 100 120 120 120 month 1 to month 3 and will therefore lose one unit of demand i

77

0 140 90 140 100 100 100 120 120 120 month 1 to month 3 and will therefore lose one unit of demand in month 3. The loss in net pro?t from this unit will be $8 ($10 - $2 for 2 months’ holding). The sensitivity data generated by the LP algorithm shown in Figure 16.8 gives us more direct information on the sensitivity of the ?nal solution to changes in the constraints. This report has a line for every constraint in the model and reports three important pieces of information:4 1. The shadow price

78

very constraint in the model and reports three important pieces of information:4 1. The shadow price represents the amount the optimal objective will be increased by a unit increase in the right-hand side of the constraint. 2. The allowable increase represents the amount by which the right-hand side can be increased before the shadow price no longer applies. 3. The allowable decrease represents the amount by which the right-hand side can be decreased before the shadow price no longer applies. A

79

the amount by which the right-hand side can be decreased before the shadow price no longer applies. Appendix 16A gives a geometric explanation of how these numbers are computed. 4 The report also contains sensitivity information about the coef?cients in the objective function. See Appendix 16A for a discussion of this. 564 Part III Principles in Practice To see how these data are interpreted, consider the information in Figure 16.8 on the seventh line of the constraint section for the capa

80

d, consider the information in Figure 16.8 on the seventh line of the constraint section for the capacity constraint X 1 = 100. The shadow price is $7, which means that if the constraint is changed to X 1 = 101, net pro?t will increase by $7, precisely as we computed above. The allowable increase is 20 units, which means that each unit capacity increase in period 1 up to a total of 20 units increases net pro?t by $7. Therefore, an increase in capacity from 100 to 120 will increase net pro?t by

81

eases net pro?t by $7. Therefore, an increase in capacity from 100 to 120 will increase net pro?t by 20 × 7 = $140. Above 20 units, we will have satis?ed all the lost demand in month 4, and therefore further increases will not improve pro?t. Thus, this constraint will become nonbinding once the right-hand side exceeds 120. Notice that the allowable decrease is zero for this constraint. What this means is that the shadow price of $7 is not valid for decreases in the right-hand side. As we comput

82

means is that the shadow price of $7 is not valid for decreases in the right-hand side. As we computed above, the decrease in net pro?t from a unit decrease in the capacity in month 1 is $8. In general, we can determine only the effect of changes outside the allowable increase or decrease range by actually changing the constraints and rerunning the LP solver. The above examples are illustrative of the following general behavior of linear programming models: 1. Changing the right-hand sides of

83

of the following general behavior of linear programming models: 1. Changing the right-hand sides of nonbinding constraints by a small amount does not affect the optimal solution. The shadow price of a nonbinding constraint is always zero. 2. Increasing the right-hand side of a binding constraint will increase the objective by an amount equal to the shadow price times the size of the increase, provided that the increase is smaller than the allowable increase. 3. Decreasing the right-hand side o

84

rovided that the increase is smaller than the allowable increase. 3. Decreasing the right-hand side of a binding constraint will decrease the objective by an amount equal to the shadow price times the size of the decrease, provided that the decrease is smaller than the allowable decrease. 4. Changes in the right-hand sides beyond the allowable increase or decrease range have an indeterminate effect and must be evaluated by resolving the modi?ed model. 5. All these sensitivity results apply to c

85

ect and must be evaluated by resolving the modi?ed model. 5. All these sensitivity results apply to changes in one right-hand side variable at a time. If multiple changes are made, the effects are not necessarily additive. Generally, multiple-variable sensitivity analysis must be done by resolving the model under the multiple changes. 16.3 Product Mix Planning Now that we have set up the basic framework for formulating and solving aggregate planning problems, we can examine some commonly enco

86

framework for formulating and solving aggregate planning problems, we can examine some commonly encountered situations. The ?rst realistic aggregate planning issue we will consider is that of product mix planning. To do this, we need to extend the model of the previous section to consider multiple products explicitly. As mentioned previously, allowing multiple products raises the possibility of a “?oating bottleneck.” That is, if the different products require different amounts of processing t

87

a “?oating bottleneck.” That is, if the different products require different amounts of processing time on the various workstations, then the workstation that is most heavily loaded during a period may well depend on the mix of products run during that period. If ?exibility in the mix is possible, we can use the AP module to adjust the mix in accordance with available capacity. And if the mix is essentially ?xed, we can use the AP module to identify bottlenecks. Chapter 16 16.3.1 565 Aggr

88

s essentially ?xed, we can use the AP module to identify bottlenecks. Chapter 16 16.3.1 565 Aggregate and Workforce Planning Basic Model We start with a direct extension of the previous single-product model in which demands are assumed ?xed and the objective is to minimize the inventory carrying cost of meeting these demands. To do this, we introduce the following notation: i = an index of product, i = 1, . . . , m, so m represents total number of products j = an index of workstation, j =

89

product, i = 1, . . . , m, so m represents total number of products j = an index of workstation, j = 1, . . . , n, so n represents total number of workstations t = an index of period, t = 1, . . . , t¯, so t¯ represents planning horizon d¯it = maximum demand for product i in period t d it = minimum sales5 allowed of product i in period t ai j = time required on workstation j to produce one unit of product i c jt = capacity of workstation j in period t in units consistent with those used to de?

90

of product i c jt = capacity of workstation j in period t in units consistent with those used to de?ne ai j ri = net pro?t from one unit of product i h i = cost6 to hold one unit of product i for one period t X it = amount of product i produced in period t Sit = amount of product i sold in period t Iit = inventory of product i at end of period t (Ii0 is given as data) Again, X it , Sit , and Iit are decision variables, while the other symbols are constants representing input data. We can give

91

t are decision variables, while the other symbols are constants representing input data. We can give a linear program formulation of the problem to maximize net pro?t minus inventory carrying cost subject to upper and lower bounds on sales and capacity constraints as Maximize Subject to: t¯ t=1 m i=1 ri Sit - h i Iit (16.28) d it = Sit = d¯it m i=1 ai j X it = c jt for all i, t (16.29) for all j, t (16.30) Iit = Iit-1 + X it - Sit for all i, t (16.31) X it , Sit , Iit = 0 for

92

) for all j, t (16.30) Iit = Iit-1 + X it - Sit for all i, t (16.31) X it , Sit , Iit = 0 for all i, t (16.32) In comparison to the previous single-product model, we have adjusted constraints (16.29) to include lower, as well as upper, bounds on sales. For instance, the ?rm may have long-term contracts that obligate it to produce certain minimum amounts of certain products. Conversely, the market for some products may be limited. To maximize pro?t, the computer has incentive to set pro

93

the market for some products may be limited. To maximize pro?t, the computer has incentive to set production so that all these constraints will be tight at their upper limits. However, this may not be possible due to capacity constraints (16.30). Notice that unlike in the previous formulation, we now have capacity constraints for each workstation in each period. By noting which of these constraints are tight, we can identify those resources that limit production. Constraints (16.31) are the mul

94

nts are tight, we can identify those resources that limit production. Constraints (16.31) are the multiproduct version of the balance equations, and constraints (16.32) are the usual non-negativity constraints. 5 This might represent ?rm commitments that we do not want the computer program to violate. 6 It is common to set h i equal to the raw materials cost of product i times a one-period interest rate to represent the opportunity cost of the money tied up in inventory; but it may make sense

95

rest rate to represent the opportunity cost of the money tied up in inventory; but it may make sense to use higher values to penalize inventory that causes long, uncompetitive cycle times. 566 Part III Principles in Practice We can use LP (16.28)–(16.32) to obtain several pieces of information, including 1. Demand feasibility. We can determine whether a set of demands is capacity-feasible. If the constraint Sit = d¯it is tight, then the upper bound on demand d¯it is feasible. If not, then

96

If the constraint Sit = d¯it is tight, then the upper bound on demand d¯it is feasible. If not, then it is capacity-infeasible. If demands given by the lower bounds on demand d it are capacity-infeasible, then the computer program will return a “could not ?nd a feasible solution” message and the user must make changes (e.g., reduce demands or increase capacity) in order to get a solution. 2. Bottleneck locations. Constraints (16.30) restrict production on each workstation in each period. By not

97

tleneck locations. Constraints (16.30) restrict production on each workstation in each period. By noting which of these constraints are binding, we can determine which workstations limit capacity in which periods. A workstation that is consistently binding in many periods is a clear bottleneck and requires close management attention. 3. Product mix. If we are unable, for capacity reasons, to attain all the upper bounds on demand, then the computer will reduce sales below their maximum for some

98

ain all the upper bounds on demand, then the computer will reduce sales below their maximum for some products. It will try to maximize revenue by producing those products with high net pro?t, but because of the capacity constraints, this is not a simple matter, as we will see in the following example. 16.3.2 A Simple Example Let us consider a simple product mix example that shows why one needs a formal optimization method instead of a simpler ad hoc approach for these problems. We simplify ma

99

a formal optimization method instead of a simpler ad hoc approach for these problems. We simplify matters by assuming a planning horizon of only one period. While this is certainly not a realistic assumption in general, in situations where we know in advance that we will never carry inventory from one period to the next, solving separate one-period problems for each period will yield the optimal solution. For example, if demands and cost coef?cients are constant from period to period, then the