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100 | l solution. For example, if demands and cost coef?cients are constant from period to period, then there is no incentive to build up inventory and therefore this will be the case. Consider a situation in which a ?rm produces two products, which we will call products 1 and 2. Table 16.1 gives descriptive data for these two products. In addition to the direct raw material costs associated with each product, we assume a $5,000 per week ?xed cost for labor and capital. Furthermore, there are 2,400 m |
101 | product, we assume a $5,000 per week ?xed cost for labor and capital. Furthermore, there are 2,400 minutes (5 days per week, 8 hours per day) of time available on workstations A to D. We assume that all these data are identical from week to week. Therefore, there is no reason to build inventory in one week to sell in a subsequent week. (If we can meet maximum demand this week with this week’s production, then the same thing is possible next week.) Thus, we can restrict our Table 16.1 Input Da |
102 | roduction, then the same thing is possible next week.) Thus, we can restrict our Table 16.1 Input Data for Single-Period AP Example Product 1 2 Selling price Raw material cost Maximum weekly sales Minutes per unit on workstation A Minutes per unit on workstation B Minutes per unit on workstation C Minutes per unit on workstation D $90 $45 100 15 15 15 15 $100 $40 50 10 30 5 5 Chapter 16 Aggregate and Workforce Planning 567 attention to a single week, and the only issue is the approp |
103 | Aggregate and Workforce Planning 567 attention to a single week, and the only issue is the appropriate amount of each product to produce. A Cost Approach. Let us begin by looking at this problem from a simple cost standpoint. Net pro?t per unit of product 1 sold is $45 ($90 - 45), while net pro?t per unit of product 2 sold is $60 ($100 - 40). This would seem to indicate that we should emphasize production of product 2. Ideally, we would like to produce 50 units of product 2 to meet maximum |
104 | ize production of product 2. Ideally, we would like to produce 50 units of product 2 to meet maximum demand, but we must check the capacity of the four workstations to make sure this is possible. Since workstation B requires the most time to make a unit of product 2 (30 minutes) among the four workstations, this is the potential constraint. Producing 50 units of product 2 on workstation B will require 30 minutes per unit × 50 units = 1,500 minutes This is less than the available 2,400 minutes o |
105 | equire 30 minutes per unit × 50 units = 1,500 minutes This is less than the available 2,400 minutes on workstation B, so producing 50 units of product 2 is feasible. Now we need to determine how many units of product 1 we can produce with the leftover capacity. The unused time on workstations A to D after subtracting the time to make 50 units of product 2 we compute as 2,400 - 10(50) = 1,900 minutes on workstation A 2,400 - 30(50) = 900 minutes on workstation B 2,400 - 5(50) = 2,150 minutes on |
106 | utes on workstation A 2,400 - 30(50) = 900 minutes on workstation B 2,400 - 5(50) = 2,150 minutes on workstation C 2,400 - 5(50) = 2,150 minutes on workstation D Since one unit of product 1 requires 15 minutes of time on each of the four workstations, we can compute the maximum possible production of product 1 at each workstation by dividing the unused time by 15. Since workstation B has the least remaining time, it is the potential bottleneck. The maximum production of product 1 on workstation |
107 | st remaining time, it is the potential bottleneck. The maximum production of product 1 on workstation B (after subtracting the time to produce 50 units of product 2) is 900 = 60 15 Thus, even though we can sell 100 units of product 1, we have capacity for only 60. The weekly pro?t from making 60 units of product 1 and 50 units of product 2 is $45 × 60 + $60 × 50 - $5,000 = $700 Is this the best we can do? A Bottleneck Approach. The preceding analysis is entirely premised on costs and considers |
108 | we can do? A Bottleneck Approach. The preceding analysis is entirely premised on costs and considers capacity only as an afterthought. A better method might be to look at cost and capacity, by computing a ratio representing pro?t per minute of bottleneck time used for each product. This requires that we ?rst identify the bottleneck, which we do by computing the minutes required on each workstation to satisfy maximum demand and 568 Part III Principles in Practice seeing which machine is mos |
109 | on to satisfy maximum demand and 568 Part III Principles in Practice seeing which machine is most overloaded.7 This yields 15(100) + 10(50) = 2,000 minutes on workstation A 15(100) + 30(50) = 3,000 minutes on workstation B 15(100) + 5(50) = 1,750 minutes on workstation C 15(100) + 5(50) = 1,750 minutes on workstation D Only workstation B requires more than the available 2,400 minutes, so we designate it the bottleneck. Hence, we would like to make the most pro?table use of our time on work |
110 | designate it the bottleneck. Hence, we would like to make the most pro?table use of our time on workstation B. To determine which of the two products does this, we compute the ratio of net pro?t to minutes on workstation B as $45 = $3 per minute spent processing product 1 15 $60 = $2 per minute spent processing product 2 30 This calculation indicates the reverse of our previous cost analysis. Each minute spent processing product 1 on workstation B nets us $3, as opposed to only $2 per minute s |
111 | ach minute spent processing product 1 on workstation B nets us $3, as opposed to only $2 per minute spent on product 2. Therefore, we should emphasize production of product 1, not product 2. If we produce 100 units of product 1 (the maximum amount allowed by the demand constraint), then since all workstations require 15 minutes per unit of one, the unused time on each workstation is 2,400 - 15(100) = 900 minutes Then since workstation B is the slowest operation for producing product 2, this is |
112 | 00) = 900 minutes Then since workstation B is the slowest operation for producing product 2, this is what limits the amount we can produce. Each unit of product 2 requires 30 minutes on B; thus, we can produce 900 = 30 30 units of product 2. The net pro?t from producing 100 units of product 1 and 30 units of product 2 is $45 × 100 + $60 × 30 - $5,000 = $1,300 This is clearly better than the $700 we got from using our original analysis and, it turns out, is the best we can do. But will this meth |
113 | we got from using our original analysis and, it turns out, is the best we can do. But will this method always work? A Linear Programming Approach. To answer the question of whether the previous “bottleneck ratio” method will always determine the optimal product mix, we consider a slightly modi?ed version of the previous example, with data shown in Table 16.2. The only changes in these data relative to the previous example are that the processing time of product 2 on workstation B has been incr |
114 | tive to the previous example are that the processing time of product 2 on workstation B has been increased from 30 to 35 minutes and the processing times for products 1 and 2 on workstation D have been increased from 15 and 5 to 25 and 14, respectively. 7 The alert reader should be suspicious at this point, since we know that the identity of the “bottleneck” can depend on the product mix in a multiproduct case. Chapter 16 569 Aggregate and Workforce Planning Table 16.2 Input Data for Modi |
115 | ltiproduct case. Chapter 16 569 Aggregate and Workforce Planning Table 16.2 Input Data for Modi?ed Single-Period AP Example Product 1 2 Selling price Raw material cost Maximum weekly sales Minutes per unit on workstation A Minutes per unit on workstation B Minutes per unit on workstation C Minutes per unit on workstation D $90 $45 100 15 15 15 25 $100 $40 50 10 35 5 14 To execute our ratio-based approach on this modi?ed problem, we ?rst check for the bottleneck by computing the min |
116 | r ratio-based approach on this modi?ed problem, we ?rst check for the bottleneck by computing the minutes required on each workstation to meet maximum demand levels: 15(100) + 10(50) = 2,000 minutes on workstation A 15(100) + 35(50) = 3,250 minutes on workstation B 15(100) + 5(50) = 1,750 minutes on workstation C 25(100) + 14(50) = 3,200 minutes on workstation D Workstation B is still the most heavily loaded resource, but now workstation D also exceeds the available 2,400 minutes. If we designa |
117 | eavily loaded resource, but now workstation D also exceeds the available 2,400 minutes. If we designate workstation B as the bottleneck, then the ratio of net pro?t to minute of time on the bottleneck is $45 = $3.00 per minute spent processing product 1 15 $60 = $1.71 per minute spent processing product 2 35 which, as before, indicates that we should produce as much product 1 as possible. However, now it is workstation D that is slowest for product 1. The maximum amount that can be produced on |
118 | r, now it is workstation D that is slowest for product 1. The maximum amount that can be produced on D in 2,400 minutes is 2,400 = 96 25 Since 96 units of product 1 use up all available time on workstation D, we cannot produce any product 2. The net pro?t from this mix, therefore, is $45 × 96 - $5,000 = -$680 This doesn’t look very good—we are losing money. Moreover, while we used workstation B as our bottleneck for the purpose of computing our ratios, it was workstation D that determined how m |
119 | as our bottleneck for the purpose of computing our ratios, it was workstation D that determined how much product we could produce. Therefore, perhaps we should have designated workstation D as our bottleneck. If we do this, the ratio of net pro?t to minute of time on the bottleneck is $45 = $1.80 per minute spent processing product 1 25 $60 = $4.29 per minute spent processing product 2 14 570 Part III Principles in Practice This indicates that it is more pro?table to emphasize production o |
120 | Part III Principles in Practice This indicates that it is more pro?table to emphasize production of product 2. Since workstation B is slowest for product 2, we check its capacity to see how much product 2 we can produce, and we ?nd 2,400 = 68.57 35 Since this is greater than maximum demand, we should produce the maximum amount of product 2, which is 50 units. Now we compute the unused time on each machine as 2,400 - 10(50) = 1,900 minutes on workstation A 2,400 - 35(50) = 650 minutes on work |
121 | each machine as 2,400 - 10(50) = 1,900 minutes on workstation A 2,400 - 35(50) = 650 minutes on workstation B 2,400 - 5(50) = 2,150 minutes on workstation C 2,400 - 14(50) = 1,700 minutes on workstation D Dividing the unused time by the minutes required to produce one unit of product 1 on each workstation gives us the maximum production of product 1 on each to be 1,900 = 126.67 units on workstation A 15 650 = 43.33 units on workstation B 15 2,150 = 143.33 units on workstation C 15 1,700 = 68 u |
122 | tion A 15 650 = 43.33 units on workstation B 15 2,150 = 143.33 units on workstation C 15 1,700 = 68 units on workstation D 25 Thus, workstation B limits production of product 1 to 43 units, so total net pro?t for this solution is $45 × 43 + $60 × 50 - $5,000 = -$65 This is better, but we are still losing money. Is this the best we can do? Finally, let’s bring out our big gun (not really that big, since it is included in popular spreadsheet programs) and solve the problem with a linear programmi |
123 | , since it is included in popular spreadsheet programs) and solve the problem with a linear programming package. Letting X 1 (X 2 ) represent the quantity of product 1 (2) produced, we formulate a linear programming model to maximize pro?t subject to the demand and capacity constraints as Maximize 45X 1 + 60X 2 - 5,000 (16.33) Subject to: X 1 = 100 (16.34) X 2 = 50 (16.35) 15X 1 + 10X 2 = 2,400 (16.36) 15X 1 + 35X 2 = 2,400 (16.37) 15X 1 + 5X 2 = 2,400 (16.38) 25X 1 + 14X 2 = 2,400 |
124 | 2,400 (16.36) 15X 1 + 35X 2 = 2,400 (16.37) 15X 1 + 5X 2 = 2,400 (16.38) 25X 1 + 14X 2 = 2,400 (16.39) Chapter 16 571 Aggregate and Workforce Planning Problem (16.33)–16.39) is trivial for any LP package. Ours (Excel) reports the solution to this problem to be Optimal objective = $557.94 X 1* = 75.79 X 2* = 36.09 Even if we round this solution down (which will certainly still be capacity-feasible, since we are reducing production amounts) to integer values X 1* = 75 X 2* = 36 we ge |
125 | acity-feasible, since we are reducing production amounts) to integer values X 1* = 75 X 2* = 36 we get an objective of $45 × 75 + $60 × 36 - $5,000 = $535 So making as much product 1 as possible and making as much product 2 as possible both result in negative pro?t. But making a mix of the two products generates positive pro?t! The moral of this exercise is that even simple product mix problems can be subtle. No trick that chooses a dominant product or identi?es the bottleneck before knowing th |
126 | can be subtle. No trick that chooses a dominant product or identi?es the bottleneck before knowing the product mix can ?nd the optimal solution in general. While such tricks can work for speci?c problems, they can result in extremely bad solutions in others. The only method guaranteed to solve these problems optimally is an exact algorithm such as those used in linear programming packages. Given the speed, power, and user-friendliness of modern LP packages, one should have a very good reason to |
127 | the speed, power, and user-friendliness of modern LP packages, one should have a very good reason to forsake LP for an approximate method. 16.3.3 Extensions to the Basic Model A host of variations on the basic problem given in formulation (16.28)–(16.32) are possible. We discuss a few of these next; the reader is asked to think of others in the problems at chapter’s end. Other Resource Constraints. Formulation (16.28)–(16.32) contains capacity constraints for the workstations, but not for ot |
128 | aints. Formulation (16.28)–(16.32) contains capacity constraints for the workstations, but not for other resources, such as people, raw materials, and transport devices. In some systems, these may be important determinants of overall capacity and therefore should be included in the AP module. Generically, if we let bi j = units of resource j required per unit of product i k jt = number of units of resource j available in period t X it = amount of product i produced in period t we can express th |
129 | of resource j available in period t X it = amount of product i produced in period t we can express the capacity constraint on resource j in period t as m i=1 bi j X it = k jt (16.40) 572 Part III Principles in Practice Notice that bi j and k jt are the nonworkstation analogs to ai j and c jt in formulation (16.28)–(16.32). As a speci?c example, suppose an inspector must check products 1, 2, and 3, which require 1, 2, and 1.5 hours, respectively, per unit to inspect. If the inspector is |
130 | , 2, and 3, which require 1, 2, and 1.5 hours, respectively, per unit to inspect. If the inspector is available a total of 160 hours per month, then the constraint on this person’s time in month t can be represented as X 1t + 2X 2t + 1.5X 3t = 160 If this constraint is binding in the optimal solution, it means that inspector time is a bottleneck and perhaps something should be reorganized to remove this bottleneck. (The plant could provide help for the inspector, simplify the inspection procedu |
131 | ove this bottleneck. (The plant could provide help for the inspector, simplify the inspection procedure to speed it up, or use quality-at-the-source inspections by the workstation operators to eliminate the need for the extra inspection step.) As a second example, suppose a ?rm makes four different models of circuit board, all of which require one unit of a particular component. The component contains leadingedge technology and is in short supply. If kt represents the total number of these comp |
132 | ntains leadingedge technology and is in short supply. If kt represents the total number of these components that can be made available in period t, then the constraint represented by component availability in each period t can be expressed as X 1t + X 2t + X 3t + X 4t = kt Many other resource constraints can be represented in analogous fashion. Utilization Matching. As our discussion so far shows, it is straightforward to model capacity constraints in LP formulations of AP problems. However, we |
133 | s, it is straightforward to model capacity constraints in LP formulations of AP problems. However, we must be careful about how we use these constraints in actual practice, for two reasons. 1. Low-level complexity. An AP module will necessarily gloss over details that can cause inef?ciency in the short term. For instance, in the product mix example of the previous section, we assumed that it was possible to run the four machines 2,400 minutes per week. However, from our Factory Physics discussi |
134 | possible to run the four machines 2,400 minutes per week. However, from our Factory Physics discussions of Part II, we know that it is virtually impossible to avoid some idle time on machines. Any source of randomness (machine failures, setups, errors in the scheduling process, etc.) can diminish utilization. While we cannot incorporate these directly in the AP model, we can account for their aggregate effect on utilization. 2. Production control decisions. As we noted in Chapter 13, it may be |
135 | ggregate effect on utilization. 2. Production control decisions. As we noted in Chapter 13, it may be economically attractive to set the production quota below full average capacity, in order to achieve predictable customer service without excessive overtime costs. If the quota-setting module indicates that we should run at less than full utilization, we should include this fact in the aggregate planning module in order to maintain consistency. These considerations may make it attractive to pla |
136 | planning module in order to maintain consistency. These considerations may make it attractive to plan for production levels below full capacity. Although the decision of how close to capacity to run can be tricky, the mechanics of reducing capacity in the AP model are simple. If the c jt parameters represent practical estimates of realistic full capacity of workstation j in period t, adjusted for setups, worker breaks, machine failures, and other reasonable detractors, then we can simply de?at |
137 | or setups, worker breaks, machine failures, and other reasonable detractors, then we can simply de?ate capacity by multiplying these by a constant factor. For instance, if either historical experience or the quote-setting module indicates that it is reasonable to run at a fraction Chapter 16 573 Aggregate and Workforce Planning q of full capacity, then we can replace constraints (16.30) in LP (16.28)–(16.32) by m ai j X it = qc jt for all j, t i=t The result will be that a binding ca |
138 | LP (16.28)–(16.32) by m ai j X it = qc jt for all j, t i=t The result will be that a binding capacity constraint will occur whenever a workstation is loaded to 100q percent of capacity in a period. Backorders. In LP (16.28)–(16.32), we forced inventory to remain positive at all times. Implicitly, we were assuming that demands had to be met from inventory or lost; no backlogging of unmet demand was allowed. However, in many realistic situations, demand is not lost when not met on time. Cus |
139 | mand was allowed. However, in many realistic situations, demand is not lost when not met on time. Customers expect to receive their orders even if they are late. Moreover, it is important to remember that aggregate planning is a longterm planning function. Just because the model says a particular order will be late, that does not mean that this must be so in practice. If the model predicts that an order due 9 months from now will be backlogged, there may be ample time to renegotiate the due dat |
140 | an order due 9 months from now will be backlogged, there may be ample time to renegotiate the due date. For that matter, the demand may really be only a forecast, to which a ?rm customer due date has not yet been attached. With this in mind, it makes sense to think of the aggregate planning module as a tool for reconciling projected demands with available capacity. By using it to identify problems that are far in the future, we can address them while there is still time to do something about th |
141 | ms that are far in the future, we can address them while there is still time to do something about them. We can easily modify LP (16.28)–(16.32) to permit backordering as follows: Maximize t¯ ri Sit - h i Iit+ - pit- (16.41) t=1 Subject to: d it = Sit = d¯it for all i, t (16.42) for all j, t (16.43) Iit = Iit-1 + X it - Sit for all i, t (16.44) Iit = Iit+ - Iit- for all i, t (16.45) X it , Sit , Iit+ , Iit- = 0 for all i, t (16.46) m ai j X it = c jt i=1 The main chang |
142 | .45) X it , Sit , Iit+ , Iit- = 0 for all i, t (16.46) m ai j X it = c jt i=1 The main change was to rede?ne the inventory variable Iit as the difference Iit+ - Iit- , where Iit+ represents the inventory of product i carried from period t to t + 1 and Iit- represents the number of backorders carried from period t to t + 1. Both Iit+ and Iit- must be non-negative. However, Iit can be either positive or negative, and so we refer to it as the inventory position of product i in period t. A |
143 | her positive or negative, and so we refer to it as the inventory position of product i in period t. A positive inventory position indicates on-hand inventory, while a negative inventory position indicates outstanding backorders. The coef?cient pi is the backorder analog to the holding cost h i and represents the penalty to carry one unit of product i on backorder for one period of time. Because both Iit- and Iit+ appear in the objective with negative coef?cients, the LP solver will never make b |
144 | both Iit- and Iit+ appear in the objective with negative coef?cients, the LP solver will never make both of them positive for the same period. This simply means that we won’t both carry inventory and incur a backorder penalty in the same period. In terms of modeling, the most troublesome parameters in this formulation are the backorder penalty coef?cients pi . What is the cost of being late by one period on one unit 574 Part III Principles in Practice of product i? For that matter, why sho |
145 | ne period on one unit 574 Part III Principles in Practice of product i? For that matter, why should the lateness penalty be linear in the number of periods late or the number of units that are late? Clearly, asking someone in the organization for these numbers is out of the question. Therefore, one should view this type of model as a tool for generating various long-term production plans. By increasing or decreasing the pi coef?cients relative to the h i coef?cients, the analyst can increa |
146 | y increasing or decreasing the pi coef?cients relative to the h i coef?cients, the analyst can increase or decrease the relative penalty associated with backlogging. High pi values tend to force the model to build up inventory to meet surges in demand, while low pi values tend to allow the model to be late on satisfying some demands that occur during peak periods. By generating both types of plans, the user can get an idea of what options are feasible and select among them. To accomplish this, |
147 | ns, the user can get an idea of what options are feasible and select among them. To accomplish this, we need not get overly ?ne with the selection of cost coef?cients. We could set them with the simple equations h i = api (16.47) pi = ß (16.48) where a represents the one-period interest rate, suitably in?ated to penalize uncompetitive cycle times caused by excess inventory, and pi represents the raw materials cost of one unit of product i, so that api represents the interest lost on the mon |
148 | the raw materials cost of one unit of product i, so that api represents the interest lost on the money tied up by holding one unit of product i in inventory. Analogously, ß represents a (somewhat arti?cial) cost per period of delay on any product. The assumption here is that the true cost of being late (expediting costs, lost customer goodwill, lost future orders, etc.) is independent of the cost or price of the product. If equations (16.47) and (16.48) are valid, then the user can ?x a and ge |
149 | st or price of the product. If equations (16.47) and (16.48) are valid, then the user can ?x a and generate many different production plans by varying the single parameter ß. Overtime. The previous representations of capacity assume each workstation is available a ?xed amount of time in each period. Of course, in many systems there is the possibility of increasing the time via the use of overtime. Although we will treat overtime in greater detail in our upcoming discussion of workforce planning |
150 | e. Although we will treat overtime in greater detail in our upcoming discussion of workforce planning, it makes sense to note quickly that it is a simple matter to represent the option of overtime in a product mix model, even when labor is not being considered explicitly. To do this, let l j = cost of 1 hour of overtime at workstation j; a cost parameter O jt = overtime at workstation j in period t in hours; a decision variable We can modify LP (16.41)–(16.46) to allow overtime at each worksta |
151 | od t in hours; a decision variable We can modify LP (16.41)–(16.46) to allow overtime at each workstation as follows: Maximize n t¯ {ri Sit - h i Iit+ - pi Iit- - l j O jt } t=1 (16.49) j=1 Subject to: d it = Sit = d¯it for all i, t (16.50) ai j X it = c jt + O jt for all j, t (16.51) Iit = Iit-1 + X it - Sit for all i, t (16.52) Iit = Iit+ - Iit- for all i, t (16.53) X it , Sit , Iit+ , Iit- O jt = 0 for all i, j, t (16.54) m i=1 Chapter 16 575 Aggregate and Workf |
152 | Sit , Iit+ , Iit- O jt = 0 for all i, j, t (16.54) m i=1 Chapter 16 575 Aggregate and Workforce Planning The two changes we have made to LP (16.41)–(16.46) were to ¯ 1. Subtract the cost of overtime at stations 1, . . . , n, which is tt=1 nj=1 l j O jt , from the objective function. 2. Add the hours of overtime scheduled at station j in period t, denoted by O jt , to the capacity of this resource c jt in constraints (16.51). It is natural to include both backlogging and overtime |
153 | of this resource c jt in constraints (16.51). It is natural to include both backlogging and overtime in the same model, since these are both ways of addressing capacity problems. In LP (16.49)–(16.54), the computer has the option of being late in meeting demand (backlogging) or increasing capacity via overtime. The speci?c combination it chooses depends on the relative cost of backordering (pi ) and overtime (l j ). By varying these cost coef?cients, the user can generate a range of production |
154 | and overtime (l j ). By varying these cost coef?cients, the user can generate a range of production plans. Yield Loss. In systems where product is scrapped at various points in the line due to quality problems, we must release extra material into the system to compensate for these losses. The result is that workstations upstream from points of yield loss are more heavily utilized than if there were no yield loss (because they must produce the extra material that will ultimately be scrapped). |
155 | were no yield loss (because they must produce the extra material that will ultimately be scrapped). Therefore, to assess accurately the feasibility of a particular demand pro?le relative to capacity, we must consider yield loss in the aggregate planning module in systems where scrap is an issue. We illustrate the basic effect of yield loss in Figure 16.9. In this simple line, a, ß, and ? represent the fraction of product that is lost to scrap at workstations A, B, and C, respectively. If we re |
156 | ent the fraction of product that is lost to scrap at workstations A, B, and C, respectively. If we require d units of product to come out of station C, then, on average, we will have to release d/(1 - ? ) units into station C. To get d/(1 - ? ) units out of station B, we will have to release d/[(1 - ß)(1 - ? )] units into B on average. Finally, to get the needed d/[(1 - ß)(1 - ? )] out of B, we will have to release d/[(1 - a)(1 - ß)(1 - ? )] units into A. We can generalize the speci?c example o |
157 | will have to release d/[(1 - a)(1 - ß)(1 - ? )] units into A. We can generalize the speci?c example of Figure 16.9 by de?ning yi j = cumulative yield from station j onward (including station j) for product i If we want to get d units of product i out of the end of the line on average, then we must release d yi j (16.55) units of i into station j. These values can easily be computed in the manner used for the example in Figure 16.9 and updated in a spreadsheet or database as a function of the |
158 | er used for the example in Figure 16.9 and updated in a spreadsheet or database as a function of the estimated yield loss at each station. Using equation (16.55) to adjust the production amounts X it in the manner illustrated in Figure 16.9, we can modify the LP formulation (16.28)–(16.32) to consider yield loss Figure 16.9 Yield loss in a three-station line. 1–? A ? 1–? B ? 1–? C ? 576 Part III Principles in Practice as follows: Maximize t¯ ri Sit - h i Iit (16.56) t=1 S |
159 | 576 Part III Principles in Practice as follows: Maximize t¯ ri Sit - h i Iit (16.56) t=1 Subject to: d it = Sit = d¯it for all i, t (16.57) m ai j X it for all j, t (16.58) Iit = Iit-1 + X it - Sit for all i, t (16.59) X it , Sit , Iit = 0 for all i, t (16.60) i=1 yi j = c jt As one would expect, the net effect of this change is to reduce the effective capacity of workstations, particularly those at the beginning of the line. By altering the yi j values (or better ye |
160 | kstations, particularly those at the beginning of the line. By altering the yi j values (or better yet, the individual yields that make up the yi j values), the planner can get a feel for the sensitivity of the system to improvements in yields. Again as one would intuitively expect, the impact of reducing the scrap rate toward the end of the line is frequently much larger than that of reducing scrap toward the beginning of the line. Obviously, scrapping product late in the process is very costl |
161 | crap toward the beginning of the line. Obviously, scrapping product late in the process is very costly and should be avoided wherever possible. If better process control and quality assurance in the front of the line can reduce scrap later, this is probably a sound policy. An aggregate planning module like that given in LP (16.56)–(16.60) is one way to get a sense of the economic and logistic impact of such a policy. 16.4 Workforce Planning In systems where the workload is subject to variatio |
162 | pact of such a policy. 16.4 Workforce Planning In systems where the workload is subject to variation, because of either a changing workforce size or overtime load, it may make sense to consider the aggregate planning (AP) and workforce planning (WP) modules in tandem. Questions of how and when to resize the labor pool or whether to use overtime instead of workforce additions can be posed in the context of a linear programming formulation to support both modules. 16.4.1 An LP Model To illust |
163 | e context of a linear programming formulation to support both modules. 16.4.1 An LP Model To illustrate how an LP model can help address the workforce-resizing and overtime allocation questions, we will consider a simple single-product model. In systems where product routings and processing times are either almost identical, so that products can be aggregated into a single product, or entirely separate, so that routings can be analyzed separately, the single-product model can be reasonable. I |
164 | separate, so that routings can be analyzed separately, the single-product model can be reasonable. In a system where bottleneck identi?cation is complicated by different processing times and interconnected routings, a planner would most likely need an explicit multiproduct model. This involves a straightforward integration of a product mix model, like those we discussed earlier, with a workforce-planning model like that presented next. Chapter 16 Aggregate and Workforce Planning 577 We in |
165 | e-planning model like that presented next. Chapter 16 Aggregate and Workforce Planning 577 We introduce the following notation, paralleling that which we have used up to now, with a few additions to address the workforce issues. 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¯t = maximum demand in period t d t = minimum sales allowed in period t a j = time required on |
166 | izon d¯t = maximum demand in period t d t = minimum sales allowed in period t a j = time required on workstation j to produce one unit of product b = number of worker-hours required to produce one unit of product c jt = capacity of workstation j in period t r = net pro?t per unit of product sold h = cost to hold one unit of product for one period l = cost of regular time in dollars per worker-hour l = cost of overtime in dollars per worker-hour e = cost to increase workforce by one worker-hou |
167 | ur l = cost of overtime in dollars per worker-hour e = cost to increase workforce by one worker-hour per period e = cost to decrease workforce by one worker-hour per period X t = amount produced in period t St = amount sold in period t It = inventory at end of t (I0 is given as data) Wt = workforce in period t in worker-hours of regular time (W0 is given as data) Ht = increase (hires) in workforce from period t - 1 to t in worker-hours Ft = decrease (?res) in workforce from period t - 1 to t |
168 | force from period t - 1 to t in worker-hours Ft = decrease (?res) in workforce from period t - 1 to t in worker-hours Ot = overtime in period t in hours We now have several new parameters and decision variables for representing the workforce considerations. First, we need b, the labor content of one unit of product, in order to relate workforce requirements to production needs. Once the model has used this parameter to determine the number of labor hours required in a given month, it has two op |
169 | s used this parameter to determine the number of labor hours required in a given month, it has two options for meeting this requirement. Either it can schedule overtime, using the variable Ot and incurring cost at rate lt , or it can resize the workforce, using variables Ht and Ft and incurring a cost of e (e ) for every worker added (laid off). To model this planning problem as an LP, we will need to make the assumption that the cost of worker additions or deletions is linear in the number o |
170 | need to make the assumption that the cost of worker additions or deletions is linear in the number of workers added or deleted; that is, it costs twice as much to add (delete) two workers as it does to add (delete) one. Here we are assuming that e is an estimate of the hiring, training, out?tting, and lost productivity costs associated with bringing on a new worker. Similarly, e represents the severance pay, unemployment costs, and so on associated with letting a worker go. Of course, in real |
171 | severance pay, unemployment costs, and so on associated with letting a worker go. Of course, in reality, these workforce-related costs may not be linear. The training cost per worker may be less for a group than for an individual, since a single instructor can train many workers for roughly the same cost as a single one. On the other hand, the plant disruption and productivity falloff from introducing many new workers may be much more severe than those from introducing a single worker. Althoug |
172 | oducing many new workers may be much more severe than those from introducing a single worker. Although one can use more sophisticated models to consider such sources of nonlinearity, we will stick 578 Part III Principles in Practice with an LP model, keeping in mind that we are capturing general effects rather than elaborate details. Given that the AP and WP modules are used for long-term general planning purposes and rely on speculative forecasted data (e.g., of future demand), this is pr |
173 | eneral planning purposes and rely on speculative forecasted data (e.g., of future demand), this is probably a reasonable choice for most applications. We can write the LP formulation of the problem to maximize net pro?t, including labor, overtime, holding, and hiring/?ring costs, subject to constraints on sales and capacity, as Maximize t¯ {r St - h It - lWt - l Ot - eHt - e Ft } (16.61) d t = St = d¯t for all t (16.62) a j X t = c jt for all j, t (16.63) It = It-1 + X t - St for |
174 | d t = St = d¯t for all t (16.62) a j X t = c jt for all j, t (16.63) It = It-1 + X t - St for all t (16.64) Wt = Wt-1 + Ht - Ft for all t (16.65) bX t = Wt + Ot for all t (16.66) X t , St , It , Ot , Wt , Ht , Ft = 0 for all t (16.67) t=1 Subject to: The objective function in formulation (16.61) computes pro?t as the difference between net revenue and inventory carrying costs, wages (regular and overtime), and workforce increase/decrease costs. Constraints (16.62) are the u |
175 | s, wages (regular and overtime), and workforce increase/decrease costs. Constraints (16.62) are the usual bounds on sales. Constraints (16.63) are capacity constraints for each workstation. Constraints (16.64) are the usual inventory balance equations. Constraints (16.65) and (16.66) are new to this formulation. Constraints (16.65) de?ne the variables Wt , t = 1, . . . , t¯, to represent the size of the workforce in period t in units of worker-hours. Constraints (16.66) constrain the worker-hou |
176 | e of the workforce in period t in units of worker-hours. Constraints (16.66) constrain the worker-hours required to produce X t , given by bX t , to be less than or equal to the sum of regular time plus overtime, namely, Wt + Ot . Finally, constraints (16.67) ensure that production, sales, inventory, overtime, workforce size, and labor increases/decreases are all non-negative. The fact that It = 0 implies no backlogging, but we could easily modify this model to account for backlogging in a mann |
177 | 0 implies no backlogging, but we could easily modify this model to account for backlogging in a manner like that used in LP(16.41)–(16.46). 16.4.2 A Combined AP/WP Example To make LP (16.61)–(16.67) concrete and to give a ?avor for the manner in which modeling, analysis, and decision making interact, we consider the example presented in the spreadsheet of Figure 16.10. This represents an AP problem for a single product with unit net revenue of $1,000 over a 12-month planning horizon. We assu |
178 | roblem for a single product with unit net revenue of $1,000 over a 12-month planning horizon. We assume that each worker works 168 hours per month and that there are 15 workers in the system at the beginning of the planning horizon. Hence, the total number of labor hours available at the start of the problem is W0 = 15 × 168 = 2,520 There is no inventory in the system at the start, so I0 = 0. The cost parameters are estimated as follows. Monthly holding cost is $10 per unit. Regular-time labor |
179 | e cost parameters are estimated as follows. Monthly holding cost is $10 per unit. Regular-time labor (with bene?ts) costs $35 per hour. Overtime is paid at time-and-a-half, Chapter 16 579 Aggregate and Workforce Planning Figure 16.10 Initial spreadsheet for workforce planning 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 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 B C D E F G H I J |
180 | 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 B C D E F G H I J K L M Parameters: r h l l' e e' b l_0 W_0 t d_t 1000 10 35 52.5 15 9 12 0 2520 1 200 2 220 3 230 4 300 5 400 6 450 7 320 8 180 9 170 10 170 11 160 12 180 Decision Variables: t Xt Wt Ht Ft It Ot 1 0.00 0.00 0.00 0.00 0.00 0.00 2 0.00 0.00 0.00 0.00 0.00 0.00 3 0.00 0.00 0.00 0.00 0.00 0.00 4 0.00 0.00 0.00 0.00 0.00 0.00 5 0.00 0.00 0.00 0.00 0.00 0.00 6 0.00 0.00 0.00 0.0 |
181 | .00 0.00 0.00 4 0.00 0.00 0.00 0.00 0.00 0.00 5 0.00 0.00 0.00 0.00 0.00 0.00 6 0.00 0.00 0.00 0.00 0.00 0.00 7 0.00 0.00 0.00 0.00 0.00 0.00 8 0.00 0.00 0.00 0.00 0.00 0.00 9 0.00 0.00 0.00 0.00 0.00 0.00 10 0.00 0.00 0.00 0.00 0.00 0.00 11 0.00 0.00 0.00 0.00 0.00 0.00 12 0.00 0.00 0.00 0.00 0.00 0.00 Objective: Profit: Constraints: I1-I0-X1 I2-I1-X2 I3-I2-X3 I4-I3-X4 I5-I4-X5 I6-I5-X6 I7-I6-X7 I8-I7-X8 I9-I8-X9 I10-I9-X10 I11-I10-X11 I12-I11-X12 W1-W0-H1+F1 W2-W1-H2+F2 W3-W2-H3+F3 |
182 | X6 I7-I6-X7 I8-I7-X8 I9-I8-X9 I10-I9-X10 I11-I10-X11 I12-I11-X12 W1-W0-H1+F1 W2-W1-H2+F2 W3-W2-H3+F3 W4-W3-H4+F4 W5-W4-H5+F5 W6-W5-H6+F6 W7-W6-H7+F7 W8-W7-H8+F8 W9-W8-H9+F9 W10-W9-H10+F10 W11-W10-H11+F11 W12-W11-H12+F12 bX1-W1-O1 bX2-W2-O2 bX3-W3-O3 bX4-W4-O4 bX5-W5-O5 bX6-W6-O6 bX7-W7-O7 bX8-W8-O8 bX9-W9-O9 bX10-W10-O10 bX11-W11-O11 bX12-W12-O12 $2,980,600.00 = -d_1 0.00 -200 = -d_2 0.00 -220 = -d_3 0.00 -230 = -d_4 0.00 -300 = -d_5 0.00 -400 = -d_6 0.00 -450 = -d_7 0.00 -320 = -d_8 0.00 -18 |
183 | = -d_3 0.00 -230 = -d_4 0.00 -300 = -d_5 0.00 -400 = -d_6 0.00 -450 = -d_7 0.00 -320 = -d_8 0.00 -180 = -d_9 0.00 -170 = -d_10 0.00 -170 = -d_11 0.00 -160 = -d_12 0.00 -180 = -2520.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 = 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 <= 0.00 0 Note: All decision variables must be >= 0 580 Part III Principles in Practice which is e |
184 | 0.00 0 Note: All decision variables must be >= 0 580 Part III Principles in Practice which is equal to $52.50 per hour. It costs roughly $2,500 to hire and train a new worker. Since this worker will account for 168 hours per month, the cost in terms of dollars per worker-hour is $2,500 = $14.88 ˜ $15 per hour 168 Since this number is only a rough approximation, we will round to an even $15. Similarly, we estimate the cost to lay off a worker to be about $1,500, so the cost per hour of red |
185 | . Similarly, we estimate the cost to lay off a worker to be about $1,500, so the cost per hour of reduction in the monthly workforce is $1,500 = $8.93 ˜ $9 per hour 168 Again, we will use the rounded value of $9, since data are rough. Notice that the projected demands (dt ) in the spreadsheet have a seasonal pattern to them, building to a peak in months 5 and 6, and tapering off thereafter. We will assume that backordering is not an option and that demands must be met, so the main issue will be |
186 | ill assume that backordering is not an option and that demands must be met, so the main issue will be how to do this. Let us begin by expressing LP (16.61)–(16.67) in concrete terms for this problem. Because we are assuming that demands are met, we set St = dt , which eliminates the need for separate sales variables St and sales constraints (16.62). Furthermore, to keep things simple, we will assume that the only capacity constraints are those posed by labor (i.e., it requires 12 hours of labor |
187 | sume that the only capacity constraints are those posed by labor (i.e., it requires 12 hours of labor to produce each unit of product). No other machine or resource constraints need be considered. Thus we can omit constraints (16.63). Under these assumptions, the resulting LP formulation is Maximize 1,000(d1 + · · · + d12 ) - 10(I1 + · · · + I12 ) -35(W1 + · · · + W12 ) - 52.5(O1 + · · · + O12 ) -15(H1 + · · · + H12 ) - 9(F1 + · · · + F12 ) (16.68) Subject to: I1 - I0 - X 1 = -d1 (16.69) I |
188 | ) -15(H1 + · · · + H12 ) - 9(F1 + · · · + F12 ) (16.68) Subject to: I1 - I0 - X 1 = -d1 (16.69) I2 - I1 - X 2 = -d2 (16.70) I3 - I2 - X 3 = -d3 (16.71) I4 - I3 - X 4 = -d4 (16.72) I5 - I4 - X 5 = -d5 (16.73) I6 - I5 - X 6 = -d6 (16.74) I7 - I6 - X 7 = -d7 (16.75) I8 - I7 - X 8 = -d8 (16.76) I9 - I8 - X 9 = -d9 (16.77) I10 - I9 - X 10 = -d10 (16.78) I11 - I10 - X 11 = -d11 (16.79) I12 - I11 - X 12 = -d12 (16.80) Chapter 16 581 Aggregate and Workforce Planning W1 - H |
189 | (16.79) I12 - I11 - X 12 = -d12 (16.80) Chapter 16 581 Aggregate and Workforce Planning W1 - H1 + F1 = 2,520 (16.81) W2 - W1 - H2 + F2 = 0 (16.82) W3 - W2 - H3 + F3 = 0 (16.83) W4 - W3 - H4 + F4 = 0 (16.84) W5 - W4 - H5 + F5 = 0 (16.85) W6 - W5 - H6 + F6 = 0 (16.86) W7 - W6 - H7 + F7 = 0 (16.87) W8 - W7 - H8 + F8 = 0 (16.88) W9 - W8 - H9 + F9 = 0 (16.89) W10 - W9 - H10 + F10 = 0 (16.90) W11 - W10 - H11 + F11 = 0 (16.91) W12 - W11 - H12 + F12 = 0 (16.92) 12X 1 - W1 |
190 | F10 = 0 (16.90) W11 - W10 - H11 + F11 = 0 (16.91) W12 - W11 - H12 + F12 = 0 (16.92) 12X 1 - W1 - O1 = 0 (16.93) 12X 2 - W2 - O2 = 0 (16.94) 12X 3 - W3 - O3 = 0 (16.95) 12X 4 - W4 - O4 = 0 (16.96) 12X 5 - W5 - O5 = 0 (16.97) 12X 6 - W6 - O6 = 0 (16.98) 12X 7 - W7 - O7 = 0 (16.99) 12X 8 - W8 - O8 = 0 (16.100) 12X 9 - W9 - O9 = 0 (16.101) 12X 10 - W10 - O10 = 0 (16.102) 12X 11 - W11 - O11 = 0 (16.103) 12X 12 - W12 - O12 = 0 (16.104) X t , It , Ot , Wt , Ht , Ft = 0 |
191 | 2X 11 - W11 - O11 = 0 (16.103) 12X 12 - W12 - O12 = 0 (16.104) X t , It , Ot , Wt , Ht , Ft = 0 t = 1, . . . , 12 (16.105) Objective (16.68) is identical to objective (16.61), except that the St variables have been replaced with dt constants.8 Constraints (16.69)–(16.80) are the usual balance constraints. For instance, constraint (16.69) simply states that I1 = I0 + X 1 - d1 That is, inventory at the end of month 1 equals inventory at the end of month 0 (i.e., the beginning of the proble |
192 | ntory at the end of month 1 equals inventory at the end of month 0 (i.e., the beginning of the problem) plus production during month 1, minus sales (demand) in month 1. We have arranged these constraints so that all decision variables are on the 8 Since the d values are ?xed, the ?rst term in the objective function is not a function of our decision t variables and could be left out without affecting the solution. We have kept it in so that our model reports a sensible pro?t function. 582 Pa |
193 | cting the solution. We have kept it in so that our model reports a sensible pro?t function. 582 Part III Principles in Practice left-hand side of the equality and constants (dt ) are on the right-hand side. This is often a convenient modeling convention, as we will see in our analysis. Constraints (16.81) to (16.92) are the labor balance equations given in constraints (16.65) of our general formulation. For instance, constraint (16.81) represents the relation W1 = W0 + H1 - F1 so that the |
194 | formulation. For instance, constraint (16.81) represents the relation W1 = W0 + H1 - F1 so that the workforce at the end of month 1 (in units of worker-hours) is equal to the workforce at the end of month 0, plus any additions in month 1, minus any subtractions in month 1. Constraints (16.93) to (16.104) ensure that the labor content of the production plan does not exceed available labor, which can include overtime. For instance, constraint (16.93) can be written as 12X 1 = W1 + O1 In the spre |
195 | can include overtime. For instance, constraint (16.93) can be written as 12X 1 = W1 + O1 In the spreadsheet shown in Figure 16.10, we have entered the decision variables X t , Wt , Ht , Ft , It , and Ot into cells B16:M21. Using these variables and the various coef?cients from the top of the spreadsheet, we express objective (16.68) as a formula in cell B24. Notice that this formula reports a value equal to the unit pro?t times total demand, or 1,000(200 + 220 + 230 + 300 + 400 + 450 + 320 + 1 |
196 | alue equal to the unit pro?t times total demand, or 1,000(200 + 220 + 230 + 300 + 400 + 450 + 320 + 180 + 170 + 170 + 160 + 180) = $2,980,000 because all other terms in the objective are zero when the decision variables are set at zero. We enter formulas for the left-hand sides of constraints (16.69) to (16.80) in cells B27:B38, the left-hand sides of constraints (16.81) to (16.92) in cells B39:B50, and the left-hand sides of constraints (16.93) to (16.104) in cells B51:B62. Notice that many of |
197 | B50, and the left-hand sides of constraints (16.93) to (16.104) in cells B51:B62. Notice that many of these constraints are not satis?ed when all decision variables are equal to zero. This is hardly surprising, since we cannot expect to earn revenues from sales of product we have not made. A convenient aspect of using a spreadsheet for solving LP models is that it provides us with a mechanism for playing with the model to gain insight into its behavior. For instance, in the spreadsheet of Figur |
198 | r playing with the model to gain insight into its behavior. For instance, in the spreadsheet of Figure 16.11 we try a chase solution where we set production equal to demand (X t = dt ) and leave Wt = W0 in every period. Although this satis?es the inventory balance constraints in cells B27:B38, and the workforce balance constraints in cells B39:B50, it violates the labor content constraints in cells B52:B57. The reason, of course, is that the current workforce is not suf?cient to meet demand wit |
199 | lls B52:B57. The reason, of course, is that the current workforce is not suf?cient to meet demand without using overtime. We could try adding overtime by adjusting the Ot variables in cells B21:M21. However, searching around for an optimal solution can be dif?cult, particularly in large models. Therefore, we will let the LP solver in the software do the work for us. Using the procedure we described earlier, we specify constraints (16.69) to (16.105) in our model and turn it loose. The result is |