Production Capacity Modeling of Alternative, Nonidentical, Flexible Machines

The International Journal of Flexible Manufacturing Systems, 14, 345 359, 2002 c 2002 Kluwer Academic Publishers Manufactured in The Netherlands Production Capacity Modeling of Alternative, Nonidentical, Flexible Machines GEORGE LIBEROPOULOS Department of Mechanical and Industrial Engineering, University of Thessaly, Pedion Areos, GR-38334 Volos, Greece glib@mieuthgr Abstract Analyzing the production capacity of a flexible manufacturing system consisting of a number of alternative, nonidentical, flexible machines, where each machine is capable of producing several different part types simultaneously (by flexibly allocating its production capacity among these part types), is not a trivial task The production capacity set of such a system is naturally expressed in terms of the machine-specific production rates of all part types In this paper we also express it in terms of the total production rates of all part types over all machines More specifically, we express the capacity set as the convex hull of a set of points corresponding to all possible assignments of machines to part types, where in each assignment each machine allocates all its capacity to only one part type First, we show that within each subset of assignments having a given number of machines assigned to each part type, there is a unique assignment that corresponds to an extreme point of the capacity set Then, we propose a procedure for generating all the extreme points and facets of the capacity set Numerical experience shows that when the number of part types is less than four, the size of the capacity set (measured in terms of the number of variables times the number of constraints) is smaller, if the capacity set is expressed in terms of the total production rates of all part types over all machines than if it is expressed in terms of the machine-specific production rates of all part types When the number of part types is four or more, however, the opposite is true Key Words: alternative, nonidentical machines, production capacity, production flow control 1 Introduction and motivation In the early 1980s, Kimemia and Gershwin (1983) formulated a paradigm for the optimal control of production flow in a flexible manufacturing system Their paradigm received considerable attention in the years that followed and spawned a large number of follow-on papers Some of the material in these papers is included in the books by Gershwin (1994), Sethi and Zhang (1994), and Maimon, Khmelnitsky, and Kogan (1998) Kimemia and Gershwin s paradigm concerns a flexible manufacturing system consisting of M machine groups Machine group m has a number of identical machines, where each machine is perfectly flexible and can allocate all or part of its production capacity to N different part types simultaneously The machines are unreliable, and the main difficulty the control system faces is to meet production requirements while the machines fail and are repaired at random times The system is modeled as a continuous-time, mixed-state (having both discrete and continuous elements) stochastic dynamic system The discrete constituent of the state, α, is the vector of machine states, where each machine may be either up or down The continuous portion of the state, x, is the vector of surpluses, ie, the cumulative differences between production and demand for all part types The objective is to minimize

346 LIBEROPOULOS these differences The production rate vector, u (the control), is constrained to be within a capacity set, (α), which is determined by the set of operational machines The production surplus vector satisfies the differential equation ẋ(t) = u(t) d, where d is a vector of constant demand rates A feedback control law that determines the current production rate as a function of the current machine state and current production surplus, u(x, α), is sought Kimemia and Gershwin (1983) formulated this problem as a stochastic optimal control problem and set out to solve it using dynamic programming They assumed that the machine state α(t) is a continuous-time Markov process and showed that the optimal control of the system is described by a set of stochastic partial differential (Bellman) equations involving the optimal cost-to-go or value function J (x, α) More specifically, they showed that the optimal feedback law u (x, α) is the solution to the linear program (LP), min x J (x, α) u subject to u (α) In other words, the optimal production rate vector u (x, α) is the control u in the production capacity set (α) that minimizes the inner product x J (x, α) u and thereby causes the steepest descent in J (x, α) Unfortunately, an exact solution of the Bellman equations for the J (x, α) is practically infeasible for systems producing more than one part type Caramanis and Liberopoulos (1992) proposed quadratic approximations of the value functions J (x, α) and used simulation and infinitesimal perturbation analysis to optimize the parameters of these functions (see also Liberopoulos and Caramanis, 1994, 1995) The simulation involves solving a linear program (LP) repeatedly in order to determine the optimal production rate vector u(x(t), α(t)), which changes dynamically in time, since x(t) and α(t) change dynamically in time This LP has the form, min c(x(t), α(t)) u, subject to u (α(t)), (1) where c(x(t), α(t)) is the gradient of the quadratic approximation of J (x(t), α(t)) In order to solve LP (1), (α) must be precisely defined To precisely define (α), consider the machine state α = (α 1,α 2,,α M ), where α m is the number of operational machines at machine group m Let ynm k be the rate at which machine group m performs operation k on type n parts (measured in parts per unit time) Let τnm k be the time required to complete the operation It then follows that ynm k τ nm k α m, for all m (2) n k The product ynm k τ nm k is the proportion of time in which all of the machines of machine group m perform operation k on part type n The left-hand side of inequality (2) is therefore the total amount of work performed on all part types on machine group m per unit time The inequality follows because the total amount of work per unit time performed on machine group m cannot exceed the time that the machines of machine group m are available By conservation of the flow, the rate at which any given operation at any given machine group is performed on part type n is the same as the rate at which any other operation at any other

PRODUCTION CAPACITY MODELING 347 machine group is performed on the same part type, and is equal to the total rate at which operation k is performed on part type n, u n, that is, u n = m ynm k, for all k and n (3) The production capacity set (α) isdefined to be the set of all production rate vectors u = (u 1, u 2,,u N ) such that there exist feasible production rates ynm k satisfying (2) and (3) Note that the constraint set of LP (1) is expressed in terms of the machine-specific production rates ynm k, whereas the objective function of LP (1) is expressed in terms of the total production rates u n (α) is the projection of a polyhedral set into a lower dimensional subspace It is therefore convex and polyhedral This set is a representation of the production capacity of the system A question that arises naturally is, Would it be computationally more efficient for solving LP (1) to express (α) in terms of the variables u n instead of the variables ynm k? This question is particularly important in view of the fact that LP (1) needs to be solved repeatedly during each simulation run (since x(t) and α(t) change dynamically in time) and this provided the motivation for writing this article More specifically, our goal in this article is to determine the size of (α), when (α) is expressed in terms of the variables u n, and compare it with the size of (α), when (α) is expressed in terms of the variables ynm k Once the size of (α) is known, it can be used as a surrogate for the computational efficiency of solving LP (1), in the sense that the smaller the size of (α), the bigger the computational efficiency of solving LP (1) The size of (α) is measured in terms of the number of variables multiplied by the number of constraints The problem of modeling the capacity of alternative, nonidentical resources arises in almost all manufacturing environments where process technology is evolving, as well as in many other application areas Surprisingly, there is not much published research on this subject Leachman and Carmon (1992) proposed a procedure to generate the capacity set of alternative machine types assuming that processing times among alternative machine types are identical or proportional across the operations that they can perform Hung and Wang (1997) extended this procedure and applied it to the bin allocation planning problem (a material planning problem) in the semiconductor industry They called their procedure acceptable material set generation procedure Both procedures were devised within typical, finite-horizon, production or material planning problem formulations, but assumed that processing times among alternative machines are identical (or proportional) In this article we assume that processing times among alternative machines are different The remaining part of this article is organized as follows In Section 2 we formulate the problem of determining the size of (α) for a particular machine state α, where a given number of machines is up We therefore omit the dependence on α and call this set, for notational simplicity In Section 3 we propose an algorithm for determining the number of extreme points and the number of nonredundant facets of, as expressed in terms of the total production rates u n In Section 4 we report on the numerical experience for systems with up to four part types and four machines, and in Section 5 we draw conclusions

348 LIBEROPOULOS 2 Problem formulation In the rest of this article, to simplify matters, we concentrate on a flexible manufacturing system consisting of M machine groups, where each machine group has only one machine This is without loss of generality because one can always view multiple identical machines as a single machine with a given capacity The machines are different from each other, but every machine can allocate all or part of its production capacity to N different part types We assume that each part type needs only one operation that can be performed on any of the machines In short, we consider a flexible manufacturing system consisting of M alternative, nonidentical, flexible machines, where each machine can allocate all or part of its production capacity to N different part types simultaneously Such a system was considered in Maimon and Gershwin (1988) Note that when parts visit several machines in series, the sequencing and mixing of part types should be taken into account when defining the production capacity set, as was pointed out by Lasserre (1992) For ease of exposition, we concentrate on a particular machine state, namely the state where all machines are up, that is, we assume that α m = 1, for m = 1, 2,,M This is without loss of generality because any other state in which some of the machines are up and the rest are down is equivalent to the state where all machines are up in a system with fewer machines For the state where all machines are up, the production capacity set, henceforth referred to as for notational simplicity, is defined to be the set of all production rate vectors u = (u 1, u 2,,u N ) such that there exist feasible machine-specific production rates y nm satisfying, N y nm τ nm 1, for m = 1, 2,,M, (4) n=1 u n = M y nm, for n = 1, 2,,N (5) m=1 Our goal is to determine the size of, as expressed in terms of the total production rates u n, and to compare it to the size of, as expressed in terms of machine-specific production rates y nm The size of, as expressed in terms of the production rates y nm,is (NM)M = NM 2, where NM is the number of the y nm variables and M is the number of constraints (4) The size of, as expressed in terms of the total production rates u n,isnf, where N is the number of the u n variables and F is the number of nonredundant facets of, as expressed in terms of u n We need to find F To illustrate this, consider a simple example of two machines and two part types, that is, M = 2 and N = 2, with parameters, τ 11 = 1/2,τ 21 = 1, for the first machine, and τ 12 = 1,τ 22 = 2/3, for the second machine Then, can be expressed in terms of machinespecific production rates y nm by the constraints 1/2y 11 + y 21 1 and y 12 + 2/3y 22 1 In this case, the size of is 2 2 2 = 8 can also be expressed in terms of the total production rates u n by the constraints 1/5u 1 + 2/5u 2 1 and 1/3u 1 + 2/9u 2 1 In this case, the number of nonredundant facets of is 2, and therefore the size of is 2 2 = 4 For a graphical representation of these constraints see Figure 1

PRODUCTION CAPACITY MODELING 349 Figure 1, as expressed in terms of the total production rates u n, for a case where M = 2, N = 2 3 Extreme points and facets of Ω The production capacity set, as expressed in terms of the total production rates u n,isa convex polyhedron One way to describe a convex polyhedron is to characterize it as the convex hull of a finite set of points To find a set of points, expressed in terms of the rates u n, whose convex hull is, we consider all possible assignments of M machines to N part types, such that in each assignment, each machine allocates all its capacity to only one part type The number of such assignments is N M Let S ={1, 2,,N M } be a set of indexes to all such assignments, and let s be an element of S, that is, s S For every s S, let znm s be a variable that is equal to one, if machine m is assigned to part type n in assignment s, and zero, otherwise Because for any s S, each machine is assigned to only one part type, it follows that N n=1 zs nm = 1, for m = 1, 2,,M and all s S Also, for every s S, let pn s be the maximum total production rate of part type n in assignment s, that is, pn s = M m=1 zs nm /τ nm, for n = 1, 2,,N, and let p s = (p1 s, ps 2,,ps N ) Then, the production capacity set, as expressed in terms of the rates u n, is the convex hull of points 0 and p s, for all s S Note, however, that not all points p s are extreme points (or vertices) of To obtain a nonredundant polyhedral description of, we need to identify only those points p s that are extreme points of Before setting out to identify the extreme points of, as expressed in terms of the total production rates u n, let us first try to count them To count the number of extreme points of, as expressed in terms of the total production rates u n, consider all partitions of S into subsets S r, such that each subset S r contains all assignments that have a given number of machines assigned to each part type To define each subset S r S, let in r be the number of machines that are assigned to part type n in every s S r, and let i r = (i1 r, ir 2,,ir N ) Clearly, the ir n satisfy N in r = M and 0 ir n M, for n = 1, 2,,N n=1 With the above definition in mind, S r can be defined in terms of the in r as follows: { S r = s : M znm s = ir n }, n = 1, 2,,N m=1

350 LIBEROPOULOS The number of elements of each subset S r is equal to the number of ways of partitioning a population of M machines into N subpopulations of sizes in r, n = 1, 2,,N, and is given by M!/ N n=1 ir n! To illustrate this, let us revisit the example at the end of Section 2 The capacity constraints for this example are shown in Figure 1, where m denotes the production capacity set of machine m and is expressed in terms of the machine-specific production rates y nm In the example, S is partitioned into three subsets, S 1 ={1}, S 2 ={3}, and S 3 ={2, 4} Subset S 1 contains the only assignment where both (all) machines are assigned to part type 1, that is, i 1 = (2, 0) Subset S 2 contains the only assignment where both (all) machines are assigned to part type 2, that is, i 2 = (0, 2) Subset S 3 contains the only two assignments where one machine is assigned to part type 1 and the other machine is assigned to part type 2, that is, i 3 = (1, 1) We now have the following result Proposition 1 In every subset S r S there is a unique assignment, s r, such that ps r is an extreme point of Proof: Consider a particular subset of assignments, say subset S r1, and consider a particular assignment in this subset, say s 1, that is, s 1 S r1 Suppose that a different assignment, say s 2, can be obtained from assignment s 1 by changing the individual assignment of only one machine, say machine m 1, from part type n 1 to some other part type, say part type n 2 Clearly, s 2 belongs to a different subset, say S r2, that is, s 2 S r2 S r1 Subsets S r1 and S r2 are related by i r 2 n 1 = i r 1 n 1 1, i r 2 n 2 = i r 1 n 2 + 1, and i r 2 n = ir 1 n, for all n n 1, n 2 Now suppose that a new assignment, say s 1, such that s 1 s 1, can be obtained from assignment s 2 by changing the individual assignment of only one machine, say machine m 2, where m 2 m 1, from part type n 2 back to part type n 1 Clearly, s 1 S r 1 Point p s 1 may be either an interior point (see Figure 2(a)), or an exterior point (see Figure 2(b)) with respect to the supporting hyperplane defined by the edge connecting points p s 1 and p s 2 Inthefirst case, p s 1 cannot be an exterior point of In the second case, p s 1 cannot be an extreme point of Therefore, only one of the points s 1 and s 1 may be an extreme point of This argument Figure 2 Projection of production rate vectors p s 1, p s 2, and p s 1 on the u n1 u n2 plane

PRODUCTION CAPACITY MODELING 351 can be repeated to show that in subset S r1 S there is a unique assignment, s1 S r 1, such that p s 1 is an extreme point of Proposition 1 states that in every subset S r, there is a unique assignment, s r, which corresponds to an extreme point of All other assignments in S r correspond to interior points of To illustrate Proposition 1 consider again the example of Figure 1 In this example, subset S 3 contains the only two assignments (s = 2 and s = 4) where one machine is assigned to part type 1 and the other machine is assigned to part type 2, that is, i 3 = (1, 1) Of these two assignments only one corresponds to an extreme point of, namely assignment s = 2 The following result is a consequence of Proposition 1 Corollary 1 The number of extreme points of (excluding the origin 0) is ( M+N 1 N 1 ) Proof: By Proposition 1, the number of extreme points of is equal to the number of subsets S r that partition S This number is equal to the number of ways in which M identical items (corresponding to the M machines) can be placed in N in-line compartments (corresponding to the N part types) that are separated with N 1 partitions, where each compartment may contain all, some, or none of the items Viewed differently, this is equal to the number of combinations of N 1 partitions and M items in a row Corollary gives the number of extreme points of, as expressed in terms of the rates u n For example, if N = 2, the number of extreme points of is equal to M + 1 (see Figures 1 and 11) If N = 3, the number of extreme points of is equal to (M + 2)(M + 1)/2 (see Figures 4 7) What remains to be done is to generate these points and the facets of that they define An efficient way to screen out many of the points p s that are not extreme points of is given by Proposition 2 that follows To state Proposition 2 we first need the following definitions Consider a particular assignment s Of all machines that are assigned to part type n 1 in assignment s, let m s n 1 n 2 be that machine which, when switched from part type n 1 to part type n 2, yields the largest relative gain in production capacity, that is, m s n 1 n 2 [ ] = arg max τn1 m/τ n2 m (6) m:zn s 1 m =1 Let σ (s, n 1, n 2 ) be the assignment that is obtained from assignment s by changing the individual assignment of machine m s n 1 n 2 from part type n 1 to part type n 2, where m s n 1 n 2 is given by (6) Let P s be the set of all part types that have at least one machine assigned to them in assignment s, that is, P s ={n 1 : zn s 1 m = 1, for some m} We now have the following result Proposition 2 If m σ (s,n 1,n 2 ) n 2 n 1 m s n 1 n 2 or equivalently σ (σ (s, n 1, n 2 ), n 2, n 1 ) s, for any n 1 P s and any n 2 {1, 2,,N}, then p s is not an extreme point of Proof: The proof is similar to that of Proposition 1 Assignment σ (s, n 1, n 2 ) is obtained from assignment s if the individual assignment of machine m s n 1 n 2 is changed from part

352 LIBEROPOULOS Figure 3 Projection of production rate vectors p s, p σ (s,n 1,n 2 ), and p σ (σ (s,n 1,n 2 ),n 2,n 1 ) on the u n1 u n2 plane type n 1 to part type n 2 so that the largest relative gain in production capacity is obtained Assignment σ (σ (s, n 1, n 2 ), n 2, n 1 ) is obtained from assignment σ (s, n 1, n 2 ) if the individual assignment of machine m σ (s,n 1,n 2 ) n 2 n 1 is changed from part type n 2 back to part type n 1 so that the largest relative gain in production capacity is obtained If m σ (s,n 1,n 2 ) n 2 n 1 m s n 1 n 2, then σ (σ (s, n 1, n 2 ), n 2, n 1 ) s and clearly p σ (σ (s,n 1,n 2 ),n 2,n 1 ) is an exterior point with respect to the supporting hyperplane defined by the edge connecting points p s and p σ (s,n 1,n 2 ) (see Figure 3) This means that point p s is not an extreme point of Proposition 2 provides a sufficient but not necessary condition for screening out points p s that are not extreme points of In other words, if a point p s satisfies this condition, it can be eliminated from the list of candidate extreme points of If it does not satisfy this condition, however, it is not necessarily an extreme point of Numerical experience, however, has shown that in a typical case there are only a few points p s that do not satisfy this condition and are not extreme points of Figure 4 shows an example of the production capacity set that results after eliminating all points that satisfy the condition stated in Proposition 2, for a case where M = 4 and N = 3 Note that two of the points that are not eliminated, namely A and B, are interior points of Figure 4, as expressed in terms of the total production rates u n, for a case where M = 4, N = 3

PRODUCTION CAPACITY MODELING 353 Adefinite but computationally more demanding way of testing if a point p s is an extreme point of is to solve the LP: max w s subject to M M y nm = w s znm s /τ nm( = w s pn) s, for n = 1, 2,,N, m=1 m=1 N y nm τ nm 1, n=1 for m = 1, 2,,M, (7) where the znm s, n = 1, 2,,N, m = 1, 2,,M, are given 0 and 1 coefficients for each p s Let w s, ynm, n = 1, 2,,N, m = 1, 2,,M, be the optimal solution of LP (7) The optimal solution w s is the maximum scalar by which vector p s may be multiplied without violating the capacity constraints (4) and consequently the production capacity set Clearly, if w s = 1, then p s is an extreme point of Ifw s > 1, then p s is an interior point of A graphical representation of LP (7) is shown in Figure 1 for a case where M = 2 and N = 2, and consequently S ={1, 2, 3, 4} Note that point p 4 is not an extreme point of because there exists a scalar w 4, such that w 4 > 1 and w 4 p 4 Points p 1, p 2, and p 3,on the other hand, are all extreme points of Before proceeding any further with our analysis, the following parenthetical comment is in order Certainly, the solution w s = 1 and y nm = znm s /τ nm, for all n and m, is a feasible solution of LP (7) Call this solution χ s for short One way of testing if point p s is an extreme point of, which would be more efficient than solving LP (7) from scratch, would be to examine the sign of the reduced cost coefficients of LP (7) corresponding to solution χ s More specifically, if none of these reduced cost coefficients is negative, then solution χ s is the optimal solution, and therefore p s is an extreme point of Otherwise, χ s is not the optimal solution, and therefore p s is not an extreme point of Unfortunately, however, these reduced cost coefficients cannot be computed because not all of the M + N basic variables corresponding to solution χ s are known Instead, only M +1 of the basic variables are known, namely w s and those y nm for which the corresponding znm s variables are equal to one The remaining N 1 basic variables, whose value is zero, cannot be determined Therefore, in order to test whether a point p s is definitely an extreme point of or not, LP (7) must be solved from scratch Based on our previous analysis, we propose the following procedure for generating the production capacity set, as expressed in terms of the total production rates u n 31 Procedure for generating 1 Construct a directed graph G = (S, E), where the set of nodes S is the set of indexes to all assignments and the set of arcs E is the set of ordered pairs (s,σ(s, n 1, n 2 )), for all n 1 P s, n 2 {1, 2,,N}, and s S 2 Drop the nodes (and the arcs that are attached to them) for which the sufficient condition stated in Proposition 2 holds The remaining graph is no longer a directed graph because

354 LIBEROPOULOS both (s,σ(s, n 1, n 2 )) and (σ (s, n 1, n 2 ), s) are arcs of the graph, for all n 1 P s, n 2 {1, 2,,N}, and s S 3 Drop the nodes (and the arcs that are attached to them) for which the solution to LP (7), w s, is such that w s > 1 4 The dropped nodes represent all the interior points of Call the resulting graph G G represents the extreme points of and the edges between them 5 Identify the facets of from graph G For systems that produce N part types, the facets of are (N 1)-dimensional and are defined as the convex hull of at least N points or, alternatively, of at least N(N 2)-dimensional faces 4 Numerical results We used the procedure described in the previous section to construct the production capacity set for cases of up to N = 4 part types and M = 4 machines Figures 4 9 show the resulting graph G for these cases Tables 1 3 show the number of facets of, as expressed in terms of the total production rates u n, that are in the positive orthant, for the cases where N = 2, 3, and 4, shown in Figures 5 10 They also show the size of, as expressed in terms of the rates y nm and u n, respectively, for the same cases, where the following notation is used: C number of extreme points of (excluding point 0), as expressed in terms of the total production rates u n ; F( j) number of facets of, as expressed in terms of the rates u n that are defined by j (N 2)-dimensional hyperplanes and are in the positive orthant; F total number of facets of, as expressed in terms of the rates u n, that are in the positive orthant; Figure 5 Graph G for a case where N = 3, M = 2

PRODUCTION CAPACITY MODELING 355 Figure 6 Graph G for a case where N = 3, M = 3 Figure 7 Graph G for a case where N = 3, M = 4 S 1 size of (variables constraints), as expressed in terms of the machine-specific production rates y nm ; S 2 size of (variables constraints), as expressed in terms of the total production rates u n, where S 1 = (NM)M = NM 2 and S 2 = NF Based on the results reported in Tables 1 3, we make the following observations When N = 2, the structure of, as expressed in terms of the total production rates u n,is simple Namely, has M one-dimensional facets, each of which is defined as the convex hull of two zero-dimensional points In fact, can be constructed trivially as follows: Start from the point where all machines are assigned to one part type, say part type 2, that is, (u 1, u 2 ) = (0, M m=1 1/τ 2m) Then, lay the machine-specific production capacity sets in

356 LIBEROPOULOS Figure 8 Graph G for a case where N = 4, M = 2 Figure 9 Graph G for a case where N = 4, M = 3 order of decreasing-slope (see Figure 11) Clearly, in this case, S 1 = 2M 2, S 2 = 2M, and therefore S 1 S 2 When N > 2, however, the number of facets of, as expressed in terms of the total production rates u n, is greater than M, for M > 1 More specifically, has M facets that are defined as the convex hull of exactly N (N 2)-dimensional faces, but it also has other facets that are defined as the convex hull of more than N (N 2)-dimensional faces For example, when N = 3 and M = 2, consists of two triangles and one parallelogram (in R 3, a triangle is a two-dimensional facet defined as the convex hull of three one-dimensional faces (edges), and a parallelogram is a two-dimensional facet defined as the convex hull of four one-dimensional faces (edges)) (see Figure 5)

PRODUCTION CAPACITY MODELING 357 Table 1 Number of facets of, as expressed in terms of the rates u n, and size of, as expressed in terms of the rates y nm and u n, respectively, for N = 2 M C F(2) F S 1 S 2 1 2 1 1 2 2 2 3 2 2 8 4 3 4 3 3 18 6 4 5 4 4 32 8 M M + 1 M M 2M 2 2M Table 2 Number of facets of, as expressed in terms of the rates u n, and size of, as expressed in terms of the rates y nm and u n, respectively, for N = 3 M C F(3) F(4) F S 1 S 2 1 3 1 0 1 3 3 2 6 2 1 3 12 9 3 10 3 3 6 27 18 4 15 4 6 10 48 30 M (M+1)(M+2) 2 M (M 1)M 2 M(M+1) 2 3M 2 3M(M+1) 2 Table 3 Number of facets of, as expressed in terms of the rates u n, and size of, as expressed in terms of the rates y nm and u n, respectively, for N = 4 M C F(4) F(5) F(6) F S 1 S 2 1 4 1 0 0 1 4 4 2 10 2 2 0 4 16 16 3 20 3 6 1 10 36 40 4 35 4 12 4 20 64 80 M (M+1)(M+2)(M+3) 2 M (M 1)M (M 2)(M 1)M 6 M(M+1)(M+2) 6 4M 2 4M(M+1)(M+2) 6 The results in Tables 1 3 reveal a pattern, which, if extrapolated to the general case of N part types and M machines, leads to the following conjecture Conjecture 1 The total number of facets of in the positive orthant, F, as expressed in terms of total production rates u n, is ( M+N 2 N 1 )

358 LIBEROPOULOS Figure 10 Graph G for a case where N = 4, M = 4 Figure 11, as expressed in terms of the total production rates u n, for a case where N = 2 By Conjecture 1, the size of, as expressed in terms of total production rates u n,is ( ) M + N 2 N(M + N 2)! S 2 = N = N 1 (N 1)!(M 1)! Given that that the size of expressed in terms of the rates y nm is S 1 = NM 2,we obtain the following result Corollary 2 S 1 S 2, for N < 4, and S 1 S 2, for N 4 Corollary 2 states that when the number of part types is less than four, the size of is smaller, if is expressed in terms of the total production rates of all part types over all machines, u n, than if it is expressed in terms of the machine-specific production rates of all part types, y nm When the number of part types is four or more, however, the opposite is true To elaborate this, suppose that N = 3, M = 10 Then, by Conjecture 1, F = 55 and S 2 = 3 55 = 165, whereas S 1 = (3 10) 10 = 300, that is, S 1 is approximately

PRODUCTION CAPACITY MODELING 359 twice as big as S 2 In this case it is preferable to express in terms of the total production rates u n Now suppose that N = 10, M = 10 Then, by Conjecture 1, F = 48,620 and S 2 = 10 48,620 = 486,200, whereas S 1 = (10 10) 10 = 1000, that is, S 2 is approximately 500 times larger than S 1! In this case it is preferable to express in terms the machine-specific production rates y nm 5 Conclusions The production capacity set,, ofm alternative, nonidentical, flexible machines, each capable of producing N different part types, was expressed in terms of the total production rates over all machines u n, as the convex hull of the set of points corresponding to all possible assignments of machines to part types First, it was shown that within each subset of assignments having the same number of machines assigned to each part type, there is a unique assignment that corresponds to an extreme point of Then, a procedure for generating all the extreme points and facets of was proposed The topology and geometry of, as expressed in terms of the total production rates u n, is not trivial, as the two-part type case might have misled us to believe Numerical experience showed that if the number of part types is less than four, the size of is smaller, if is expressed in terms of the total production rates u n than in terms of the production rates y nm If the number of part types is four or more, however, the size of is larger if is expressed in terms of the rates u n than in terms of the rates y nm References Caramanis, M and Liberopoulos, G, Perturbation Analysis for the Design of Flexible Manufacturing System Flow Controllers, Operations Research, Vol 40, No 6, pp 1107 1125 (November December 1992) Gershwin, SB, Manufacturing Systems Engineering, Prentice-Hall, Englewood Cliffs, NJ (1994) Hung, Y-F and Wang, Q-Z, A New Formulation Technique for Alternative Material Planning An Approach for Semiconductor Bin Allocation Planning, Computers and Industrial Engineering, Vol 32, No 2, pp 281 297 (April 1997) Kimemia, J and Gershwin, S B, An Algorithm for the Computer Control of a Flexible Manufacturing System, IIE Transactions, Vol 15, No 4, pp 353 362 (December 1983) Lasserre, J B, New Capacity Sets for the Hedging Point Strategy, International Journal of Production Research, Vol 30, No 12, pp 2941 2949 (December 1992) Leachman, R C and Carmon, T F, On Capacity Modeling for Production Planning with Alternative Machine Types, IIE Transactions, Vol 24, No 4, pp 62 72 (September 1992) Liberopoulos, G and Caramanis, M, Infinitesimal Perturbation Analysis for Second Derivative Estimation and Design of Manufacturing Flow Controllers, Journal of Optimization Theory and Applications, Vol 81, No 2, pp 297 327 (May 1994) Liberopoulos, G and Caramanis, M, Dynamics and Design of a Class of Parameterized Manufacturing Flow Controllers, IEEE Transactions on Automatic Control, Vol 40, No 6, pp 1018 1028 (June 1995) Maimon, O Z and Gershwin, S B, Dynamic Scheduling and Routing for Flexible Manufacturing Systems that Have Unreliable Machines, Operations Research, Vol 36, No 2, pp 279 292 (March April 1988) Maimon, O Z, Khmelnitsky, E, and Kogan, K, Optimal Flow Control in Manufacturing Systems: Production, Planning and Scheduling, Kluwer Academic Publishers, Dordrecht, The Netherlands (1998) Sethi, S P and Zhang, Q, Hierarchical Decision Making in Stochastic Manufacturing Systems, Birkhauser Boston, Cambridge, MA (1994)