_____________________________________________________________________________________
A Production Planning Model for an Unreliable Production Facility: Case of Finite Horizon and Single Demand _____________________________________________________________________________________
MOHSEN ELHAFSI The A. Gary Anderson Graduate School of Management University of California Riverside, CA 92521-0203, USA Phone: (909) 787-4557 Fax: (909) 787-3970 Email:
[email protected]
_____________________________________________________________________________________ Abstract We study a two-level inventory system that is subject to failures and repairs. The
objective is to minimize the expected total cost so as to determine the production plan for a single quantity demand. The expected total cost consists of the inventory carrying costs for finished and unfinished items, the backlog cost for not meeting the demand due-date, and the planning costs associated with the ordering schedule of unfinished items. The production plan consists of the optimal number of lot sizes, the optimal size for each lot, the optimal ordering schedule for unfinished items, and the optimal due-date to be assigned to the demand. To gain insight, we solve special cases and use their results to device an efficient solution approach for the main model. The models are solved to optimality and the solution is either obtained in closed form or through very efficient algorithms.
_____________________________________________________________________________________ Key words: Production Planning, Inventory Control, Lot Sizing, Stochastic, Optimization.
Electronic copy available at: http://ssrn.com/abstract=1298072
1. INTRODUCTION Most of the literature on inventory control and production planning has dealt with the assumption that the demand for a product will continue infinitely in the future either in a deterministic or in a stochastic fashion. This assumption does not always hold true. There are instances where the demand is a single quantity in time or lumped over a short period of time. Examples include seasonal demand such as in the retail fashion clothing industry, Christmas items, and commodity markets for produce used for canned food, to cite a few. Furthermore, with the development of E-Commerce, Contract Manufacturing has emerged in recent years as the preferred mode of production in the high-tech industry. Contract manufacturing is different from traditional supplier-manufacturer settings in that few large contractors dominate the marketplace. These contractors are able to support a large number of Original Equipment Manufacturers (EOMs) with products ranging from cellular phones to laser printers. In this sense, contractors accept contracts from OEMs based on a single quantity to be delivered by a certain due date. The contract may stipulate such things as the number and time of deliveries, the flexibility to change the original ordered quantity (upward or downward), due-date for the whole contracted quantity, and other aspects. In many manufacturing settings that emphasize part commonality, a standard part (e.g., drum for washing machines or shaft for auto parts) is produced in a single quantity to cover a short planning horizon. The part is then used in the final assembly of different finished products. For such cases, typically, some kind of Materials Requirement Planning (MRP) system is used to determine the standard part’s demand, by a Bill Of Materials (BOM) explosion from the finished products’ assembly schedules. The demand for the standard part is then offset in time to allow for a production lead-time and may be other lead-times such as transportation, maintenance and repair of equipment. These lead-times are usually assumed given. But, with the increasing pressure on manufacturing firms to compete on the basis of responsiveness, lead-times have to be continually monitored and shortened. Therefore, setups may have to be reduced to increase the productive capacity of the system, resulting in larger lot sizes of raw material, Work-In-Process (WIP), finished goods, and higher inventory costs. So it is clear that a trade-off operating cost must be achieved to balance setup and inventory costs while maintaining a competitive production lead-time.
2 Electronic copy available at: http://ssrn.com/abstract=1298072
Single quantity demand models are common. One shortcoming of these models is their inability to give information on the operating characteristic of the production plan of such demand (i.e., in what lot sizes or when should a lot be released for production). For instance, the newsvendor model only indicates the quantity to order/manufacture and does not give any details on how this quantity should be ordered/manufactured and when it should be delivered. Lot-sizing models can be treated either in continuous or discrete time. Schwarz (1972) presented the first finite-horizon continuous time model. He presented a model for determining optimal policies for finite-horizon Economic Order Quantity and Economic Manufacturing Quantity problems. He further showed that the optimal policy for the finite-horizon converges to the optimal policy for the infinite-horizon as the time horizon becomes longer. Discrete time finite-horizon models were initiated by Wagner and Whitin (1958). They divided the planning horizon into discrete periods of time and assumed that a fixed cost is incurred for each lot, and a holding cost assessed on inventory level at the end of a period. Their objective was to minimize the total setup and inventory costs over the planning horizon while satisfying the demand for each period. Currently, there is a sizeable literature in this area extending the basic model of Wagner and Whitin to include capacity constraints, multiple products and multiple stages. This literature includes the work of Baker et al. (1978), Florian and Klein (1971), Karmarkar et al. (1985), Karmarkar and Schrage (1981), Afentakis et al. (1984), McLaren (1976) and Zangwill (1969). Also, heuristic techniques for solving discrete time finite-horizon models were proposed by Gorham (1970), Silver and Meal (1973), Lambrecht and Vanderveken (1979) and Graves (1981) to cite a few. Although these models offer very good insights into the nature of the lot sizing problem and many of its aspects, they do not take into account the stochastic nature of the availability of the production facility. Incorporating failures and repairs explicitly in the model may reveal new insights about the capacity and batch sizing relationship. Finally, most of these models dealt with single-level lot sizing problems. Incorporating raw material costs and lead-time delivery adds accuracy to the model and new insights about the relationship between raw material costs, finished goods costs and lot sizing. In this paper, we address the above issues. We use a continuous-time modeling approach and, explicitly, incorporate the stochastic nature of the production facility to model production capacity shortages through failures and repairs as well as time spent in setups. We consider the cost of holding inventory of finished and unfinished items, backlog costs for not meeting the demand due-date, and the cost of the ordering schedule of 3
unfinished items. In Section 2, we present the notation and some preliminary results that will be used throughout the paper. In Section 3, we give the mathematical formulation of the main model. In Section 4, we study special cases of the main model. In Section 5, we propose a solution approach to the main model. Section 6 concludes the paper. 2. NOTATION AND PRELIMINARY The following notation will be used throughout the rest of the paper. Other symbols will be introduced as need. Q:
Size of the demand.
n:
Number of lots the quantity Q is split into (decision variable.)
qi :
Size of the ith lot to be processed (decision variable.)
d:
Delivery due-date for the Quantity Q.
xi :
Planned lead-time between the receipt of Lots i and i+1 (decision variable.)
ti :
Total processing time of Lot i (random variable.)
τi :
Total repair time while processing Lot i (random variable.)
K:
Setup/ordering cost per lot ($).
S:
Setup time before a lot is processed.
hu : Inventory cost rate of unfinished items ($/unit/time unit). hf : Inventory cost rate of finished items ($/unit/time unit.) b:
Backlog penalty for not meeting the due-date, d ($/unit/time unit.)
p:
Rescheduling/productivity loss penalty ($/time unit.)
L:
Delivery lead-time of unfinished items.
λ:
Failure rate of the production facility.
1 µ : Expected repair time of the production facility. r:
Processing rate of the production facility (units/time unit.)
a + = max(a,0) and a − = max(−a,0) for any real value a.
We also define the following two vectors.
q = ( q1 , q2 , " , qn ) . x = ( x1 , x2 , " , xn −1 ) . Using the above notation, we have
4
n
∑q i =1
i
=Q.
(1)
In this paper, capacity limitation/loss is accounted for explicitly through failures and repairs. We assume that failures occur according to a Poisson process. The assumption on the failure distribution represents systems with many unreliable sources or components, each susceptible to breakdown (see Barlow and Proschan 1965). The repair time distribution is assumed to be general for the moment, and repair times are assumed to be independent and identically distributed (i.i.d.) random variables. Let I u (qi ) ( I f (qi ) ) denote the expected inventory of unfinished goods (finished goods) while processing a lot of size qi . Clearly, we have I u (qi ) = I f (qi ) . Sivazlian (1992) showed that I u (qi ) =
qi2 1 + 2r µ
∫
qi
0
M (u)du .
(2)
M (u ) represents the expected number of failures, while processing a lot of size u. It is given by the following integral equation. u
M (u ) = F ( u r ) + ∫ M (u − x )dF ( x r ) .
(3)
o
F(.) denotes the cumulative distribution function (cdf) of the time between failures. In the case of Poisson arrival of failures, Equation (3) reduces to M (u ) = λ u r . Substituting in (2), we obtain I u (qi ) = I f (qi ) =
qi2 . 2R
(4)
Here, R = r (1 + λ µ ) represents the expected production rate of the production facility, taking into account failures and repairs. Notice that R < r . In the next section, we derive the cost components of the main model and give its mathematical formulation. 3. THE MAIN MODEL
As indicated in the introduction, we are interested in determining an optimal production plan for the single quantity Q. Let n represent the number of lots required to produce the quantity Q. Raw material or unfinished items require a delivery lead-time L and an ordering cost K. We assume that the delivery lead-time remains constant even if
5
more than one order are placed at the same time, or before receiving a pending order. A setup time S is incurred before the production of a lot is initiated. This setup time could be the result of cleaning and calibration activities that may take a non-negligible amount of time. Since the delivery of unfinished items takes L time units, the setup can be performed either S time units before receiving a lot of unfinished items or as soon as a lot has been received at the facility. In the first case, one saves the inventory costs associated with holding inventory of unfinished items while the facility is being set up. In the second case, an extra cost of hu qi S will be incurred for each lot of size qi . For mathematical convenience, we assume that the setup time starts only when unfinished items are presents at the facility. This assumption will not alter the optimal production plan. Indeed, the total inventory costs of holding unfinished items at the facility while it is being set up is equal to
∑
n i =1
hu qi S = huQS , which is a constant and therefore, has no bearing on the optimal
solution. This is usually the case in practice, also, since production managers will usually not allow a setup to be initiated before all raw materials and components are physically present at the production facility. We assume that the whole quantity is due by a certain due-date d. In many applications, a due-date is imposed on the completion time of the quantity Q. Such applications include MRP for part components and seasonal demand for finished goods. To avoid triviality, we assume that d ≥ L + S + Q r . In other words, the due-date must be at least equal to the processing time of the quantity Q, without random interruptions plus the setup and delivery lead-time of unfinished items. We are now ready to give the problem formulation. We first develop the cost components of the objective function. 3.1
Work-In-Process Cost
The work-in-process cost, denoted by WC, consists of the inventory holding costs associated with unfinished as well as finished items while a lot is being processed. Using the preliminary results of Section 2, the expected work-in-process cost is given by n
n
i =1
i =1
WC = ∑ hu I u (qi ) + hf I f ( qi ) = ( hu + hf ) ∑
6
qi2 . 2R
(5)
3.2
Planning Cost for Unfinished Items
Because of the lead-time delivery, L, of unfinished items and because of the randomness of the processing times, we need to plan when to receive lots of unfinished items from a vendor or from an upstream facility, or equivalently, when to place orders for unfinished items. Let xi denote the planned lead-time between the receipt of Lot i, and the receipt of Lot i+1. The xi ’s are decision variables that will determine our ordering schedule. Figure 1 shows the relationship between the processing times of the lots and their receipt times. Without loss of generality, we assume that Lot 1 is ordered at time zero, hence received at time L. Let σ i denote the time processing of Lot i starts. Let Ci denote the completion time of Lot i. Then, we can show that σ i and Ci obey the following recursions
i −1
j =1
σ i = max L + ∑ x j , Ci −1 , and Ci = ti + σ i , i = 1, " , n − 1 . Here, ti = S + qi r + τ i denotes the total processing time of Lot i, τ i = ∑ k =0 i Yk denotes the N (q )
total repair time while processing Lot i, N (qi ) denotes a Poisson random variable counting the number of failures while processing Lot i of size qi , and Yk denotes the repair time after each failure. The Yk ’s are i.i.d. random variables. Throughout the rest of the paper, we use the convention Y0 = 0 . In general, we can write i −1
i −2
σ i = L + max ∑ x j , ∑ x j + j =1
j =1
i −1
i −1
1
i −1
∑ t ," , ∑ x + ∑ t , ∑ t
j =i −1
j
i i −2 i −1 Ci = L + max ∑ x j + ∑ t j ,∑ x j + j =i j =1 j =1
j =1
j
j =2
j
j =1
j
, i = 1, " , n − 1 .
(6)
1 i i t , " , x t , + ∑ ∑ ∑ j j j ∑ t j , i = 1, " , n − 1 . j =i −1 j =1 j =2 j =1 i
(7)
In this model, in addition to the inventory cost of unfinished and finished items while processing a lot, we have two extra costs. One is incurred when a lot is received before its processing begins (hence, it is stored as inventory of unfinished items.) The other is incurred when a lot is received after its planned start time (therefore, causing rescheduling and expediting costs or loss of productivity at the production facility.) Let $p per time unit be the penalty that the production facility incurs due to late arrival of a lot of unfinished items. For Lot i, the expected cost due to receiving Lot i before its planned start time is given by +
i −1 qi hu E σ i − L − ∑ j =1 x j = qi hu E max {(ti −1 − xi −1 ) ," , (ti −1 − xi −1 ) + " + (t1 − x1 )} . +
7
Lot 1
Lot 2
Lot 3
Lot n
x3
x2
x1
L
Lot 4 xn-1
time
..... t1
t2
t3
tn-1
tn
Figure 1. Sample Ordering Scheme The expected cost due to late arrival at the facility is given by −
i −1 pE Ci −1 − L − ∑ j =1 x j = pE max {( ti −1 − xi −1 ) ," , ( ti −1 − xi −1 ) + " + ( t1 − x1 )} . −
Let Gi ( x , q ) = max {(ti − xi ) , (ti − xi ) + (ti −1 − xi −1 ) ," , (ti − xi ) + " + (t1 − x1 )} . G0 ( x , q ) = 0 . Hence, the expected planning cost for unfinished items, denoted by PC, is given by n −1
(
PC = ∑ hu qi E [Gi −1 ( x , q )] + pE [Gi −1 ( x , q )] i =1
+
−
).
(8)
3.3. Holding Cost for Finished Items
Using (7), one can show that the completion time of the entire quantity Q is given by
Cn , the completion time of the last lot. This is given by. n −1
Cn = L + ∑ xi + Gn −1 ( x , q ) + tn . i =1
Since the whole quantity Q has to be delivered at once (i.e., no partial deliveries of lots), finished lots have to be stored in inventory waiting for the last lot to be completed. The holding time for Lot i is equal to Cn − Ci . As a result, the holding cost for finished items, denoted by HC, is given by HC = ∑ i =1 hf qi E [Cn − Ci ] . n
But, notice that
8
n −1
n −1
n −1 hf qi E [Cn − Ci ] = ∑ hf qi E ∑ j =i (C j +1 − C j ) ∑ i =1 i =1 n −1
= hf ∑ qi ∑ j =i E C j +1 − C j n −1
i =1
By definition, we have j
C j = L + ∑ xk + Gj ( x , q) . k =1
Hence,
C j +1 − C j = x j +1 + G j +1 ( x , q ) − G j ( x , q )
= t j +1 + max {0, G j ( x , q )} − G j ( x , q ) = t j +1 + (G j ( x , q ) )
−
Also, E[t j ] = S + q j r + M ( q j ) µ . Which, in the case of Poisson arrival of failures, reduces to E[t j ] = S + q j R . Substituting in the expression of HC, we obtain n −1
HC = hf S ∑ (n − i )qi + i =1
hf R
n −1 n −1
n −1 n −1
i =1 j =i
i =1 j =i
∑∑ qi q j +1 + hf ∑∑ qi E G j ( x , q ) . −
(9)
3.4. Due-Date Cost
The due-date cost, denoted by DC, is the cost incurred either for not meeting the duedate delivery of the quantity Q or for completing it before it due-date d. It is given by DC = hf QE [Cn − d ] + bQE [Cn − d ] . −
+
(10)
3.5. The Mathematical Model
The expected total cost, denoted by C (n, q, x ) , is the sum of WC, PC, HC, DC, and the total ordering cost, given by nK (since an ordering cost of $K is incurred with each lot). The model can then be given as follows.
9
(
n qi2 n −1 Minimize C n q x nK h h ( , , ) = + + ( u f ) ∑ 2R + ∑ hu qi E [Gi −1 ( x, q)]+ + pu E [Gi −1 ( x, q)]− i =1 i =1 n −1 n n −1 n −1 n −1 − 1 hf − + h f S ∑ ( n − i )qi + ∑∑ qi q j +1 + h f ∑∑ qi E G j ( x, q) R i =1 j =i i =1 i =1 j = i − + + h f QE [Cn − d ] + bQE [Cn − d ] n s.t. ∑ qi = Q i =1 q i ≥ 0, i = 1, 2," , n xi ≥ 0, i = 1, 2," , n
) (11)
The decision variables are: the lot sizes, qi , the number of lots, n, and the planned leadtimes, xi . It is clear that the model is very complicated and very difficult to solve directly. Indeed, this is a mixed nonlinear integer-programming problem. To gain some insights and develop an efficient solution approach to solve it, we study special instances of the model. In addition to their practical relevance, these special cases will serve as building blocks for developing an efficient solution approach to the main model. 4. SPECIAL CASES OF THE MAIN MODEL
In the next section, we study the simplest instance of Problem (11). 4.1. Model I
Model I is obtained by making the following assumptions: A1. Delivery of unfinished items is instantaneous (i.e., the delivery time L is zero). A2. The setup time, S, at the production facility is negligible (i.e., S is zero.) A3. There is no due date on the Quantity Q (i.e., d = Cn .) A4. Partial delivery of finished goods is allowed (i.e., once a lot is completed, it is
shipped either to a customer or to the next production stage or final assembly stage). Assumptions A1 and A2 imply that there is no need to plan for delivery of unfinished items. In other words, we can let xi = ti , for all i, and all we have to do is wait until a lot is finished to order and receive unfinished items for the production of the next lot. As a consequence, PC = 0 . Assumption 3 implies that DC = 0 . Assumption A4 implies that only the work-in-
10
process inventory cost, WC, is involved. The mathematical model, in this case, can be written as follows. n qi2 Minimize C( n, q ) = nK + ( hu + hf ) ∑ i =1 2 R n s.t. ∑ qi = Q i =1 qi ≥ 0, i = 1,2," , n
(12)
This model is very similar to the finite-horizon demand model studied by Schwarz (1972). To solve this problem, we will proceed as in Schwarz (1972). First, using Lagrange multipliers, it is not difficult to show that, at the optimal solution, all qi ’s are equal (a similar proof is shown below.) In other words, unfinished items should be ordered in equal lot sizes, qi = Q n for all i. Hence, the problem becomes one of finding the optimal number of orders n*. The expected total cost is then written as follows. C (n, Q ) = nK + ( hu + hf )
Q2 . 2Rn
This cost function has the property that its optimal cost value, C * (Q ) , as a function of Q, is the lower envelop of the set of functions C (n, Q ) , n = 1,2," . It is not difficult to show that C * (Q ) = C (n, Q ) for n and Q satisfying the following double inequality. 2n(n − 1) KR 2n(n + 1) KR ≤Q ≤ . (hu + hf ) (hu + hf )
(13)
As a consequence, the optimal number of orders n* satisfies 2n( n + 1) KR Q≤ . ( hu + hf )
(14)
Where x denotes the smallest integer less or equal to x. Written in a different way, condition (14) becomes *
n (n
*
(h + 1) ≥
u
+ hf ) Q 2 2 KR
.
(15)
The optimal order quantity, q*, is then given by q * = Q n* .
11
OBSERVATION 1. Based on Equation (15), for a large quantity Q, the optimal number of
lots, n*, is large as well. Hence, n* ( n* + 1) ≈ n*2 ≈ ( hu + hf ) Q 2 2 KR . Written differently, we have q * = Q n* ≈
2 KR . This is exactly the EOQ formula with R as the demand rate. ( hu + hf )
4.2. Model II
Here, we modify Model I by dropping Assumption 4 and keep Assumptions A1, A2, and A3. In other words, finished goods have to be held in inventory until the last lot is processed, then the whole quantity Q is shipped out. This is the case of the retail fashion clothing industry where the demand for the whole season is prepared then shipped to the retail stores right before the selling season starts. Compared to Model I, the expected total cost includes HC now. The additional cost HC is given by (9). Notice that under Assumptions A1, A2, and A3, Gi ( x , q ) = 0 . Hence, E (Gi ( x , q ) ) = 0 . Also, since S = 0 , HC −
reduces to the following HC =
hf R
n −1
n
∑ ∑ qq i =1 j =i +1
i
j
.
(16)
The mathematical model can then be written as follows. n qi2 hf Minimize C n q nK h h ( , ) = + + + ( ) u f ∑ R i =1 2 R n s.t. ∑ qi = Q i =1 qi ≥ 0, i = 1,2," , n
n −1
n
∑ ∑ qq i =1 j =i +1
i
j
(17)
To solve this model, we first characterize the optimal solution, using the following Lemma. LEMMA 1.
At the optimal solution, all qi ’s are equal (i.e., qi = Q n , i = 1,2," , n .)
For convenience, all proofs are deferred to the appendix. Using Lemma 1, we have C (n, Q ) = nK + hu
Q2 Q2 + hf . 2Rn 2R
(18)
We notice that this objective function has the same structure as the one in Model I. Hence, n* is the smallest integer satisfying the following expression. n* ( n* + 1) ≥
huQ 2 . 2 KR
(19)
12
OBSERVATION 2. In this model, we notice that when optimizing for the number of lots, n,
only the inventory cost of unfinished items matters. The cost of holding inventory of finished goods, hf
Q2 , is a sunk cost. This is because inventory of finished goods keeps 2R
building up at a rate R until the whole quantity Q is processed no matter how Q is split. Hence, inventory holding costs associated with finished items will be the same regardless of the lot sizing policy.
OBSERVATION 3. Here also, we have q * = Q n* ≈
2 KR as Q gets large enough. hu
4.3. Model III
In many applications the production facility has to be set up before the production of a new lot is initiated. For instance, cleaning and calibration activities may take a nonnegligible amount of time. In this model, we keep the same assumptions as in Model II, but assume that the setup time S is significant. In other words, we drop Assumption A2 and keep Assumptions A1 and A3. Here again, since the unfinished items can be replenished instantaneously ( L = 0 ), there is no need to plan for delivery of unfinished items (i.e., xi = ti for all i.) In this case the mathematical model can then be written as follows. n h qi2 n −1 Minimize C n q nK h h ( , ) = + + + hf S ∑ i =1 (n − i )qi + f ( ) u f ∑ R i =1 2 R n s.t. ∑ qi = Q i =1 qi ≥ 0, i = 1,2," , n
n −1
n
∑ ∑ qq i =1 j =i +1
i
j
(20)
Notice that the only difference between Models II and III is the term hf S ∑ i =1 (n − i )qi , n −1
which is due to the setup time S. We now show, through Lemma 2, that the optimal qi ’s must not be equal. LEMMA 2.
qi =
The optimal lot sizes are given by Q n − 2i + 1 hf − h SR . 2 n u
(21)
In addition, n* is bounded above and satisfies following expression
13
n * (n * −1) hu Q ≤ . 2 hf SR
(22)
Observation 4. Intuitively, this result is expected, since the higher the number of lots the
longer the time spent in setups, and the higher the inventory cost will be for early processed lots. Hence, at the optimal solution, we expect qi < qi +1 , i = 1,2, " , n − 1 . It is interesting to note that this solution follows a rule similar to the Shortest Processing Time rule, where small jobs are processed first. Also, notice that when S = 0 , we retrieve the solution of Model II. DEFINITION 1.
A function
f (n)
defined on
]+
is convex in its argument if
f ( n + 2) − 2 f (n + 1) + f (n ) ≥ 0 .
Definition 1 states that for a function defined on the set of positive integers ] + to be convex it is sufficient that its second difference be greater or equal to zero. The second difference is the equivalent of the second derivative is the continuous case. Strict convexity corresponds to strict inequality in Definition 1. LEMMA 3.
The objective function C (n, q * ) , obtained after substituting the optimal qi* ’s, is
strictly convex in n. As a result of Lemma 3, one calculates an upper bound on n according to (22), then evaluates C (n, q * )
for increasing values of n and stops when
C (n + 1, q * ) − C (n, q * ) > 0 . The procedure is illustrated below. PROCEDURE 1 STEP 1.
Calculate n , the upper bound on the value of n*, using (22). Let n = 1 , q* = Q , and evaluate C ( n, q * ) .
STEP 2.
For n = 2, " , n
(
)
Calculate q* = q1* , " , qn* using (21).
(
)
(
)
If C n, q * − C n − 1, q * > 0 Let n* = n − 1 and got to Step 3. 14
n = n*
satisfying
End If. End For. Let n* = n . STEP 3.
The optimal solution to Model III is given by, n* and q*.
Notice that the approach used in models I and II does not work here since the cost functions C ( n, Q ) are not strictly increasing in n anymore. In other words, the lower envelope of the C ( n, Q ) curves is difficult to obtain in closed form. Nevertheless, the above algorithm is very fast since it only requires n evaluations and comparisons of the
(
)
C n, q * ’s, at most.
Model III is interesting in the sense that it shows that earlier lots have sizes that are smaller than later lots and that the lot size increases with the processing rank of the lot. But, some manufacturers are reluctant to using varying lot sizes and prefer equal lot sizes throughout the production process. So let us study the impact of using equal lot sizes instead of the optimal lot sizes given by (21). To do so, we first derive the additional cost incurred (which we refer to as the cost gap and denote by ∆C ) when equal lot sizes are used. A straightforward calculation leads to the following expression. ∆C (h f , hu , S , R, Q ) =
h 2f hu
S 2R
hf (n * −1)n * 2 5(n * +1) − ( 3(n*) − 11n * +4 ) hu 6
(23)
To measure the impact of using equal lot sizes rather than the optimal lot sizes, we study the cost gap with respect to its parameters. Notice that ∆C , implicitly, depends on the quantity Q through n*. Also, notice that ∆C becomes zero when S is zero. Indeed, in this case we retrieve Model II. Table 1 shows a sensitivity analysis of ∆C to the parameters h f , hu , S, R, and Q. We computed ∆C for different values of each of the parameters. The
latter were doubled, tripled, quadrupled, and so forth. The first column shows the multiplier used for each of the parameters. The entries in the table show the relative percentage gap ( ∆C C (n*, q*) ) × 100% . It is striking to see that the relative percentage gap is very small (maximum of 0.68%). This result reveals an inherent robustness of the model with respect to the choice of the lot size. To further test this robustness, we reversed the order of the optimal quantities in (21) starting with the larger lots first. This resulted in a percentage cost gap of only 0.44%.
15
Table 1. Percentage gap when equal lot sizes are used hf hu Q S R K 1 0.21 0.23 0.23 0.23 0.23 0.23 2 0.19 0.16 0.31 0.31 0.30 0.17 3 0.20 0.15 0.37 0.43 0.42 0.12 4 0.23 0.11 0.38 0.43 0.43 0.08 5 0.20 0.14 0.41 0.68 0.33 0.08 6 0.18 0.08 0.44 0.48 0.47 0.08 An important parameter in this model is the setup time. To see its impact on the average cost, we plotted, in Figure 2, the expected total cost, C (n*, q*) , as a function of the setup time S. Notice in Figure 2, that when we double the setup time, the cost increases only by 1.2%.
1.07
Total Average Cost
1.06 1.05 1.04 1.03 1.02 1.01 1 1
2
3
4
5
6
7
8
9
10
11
12
Setup Time (in S time units)
Figure 2. Setup time effect. So far, we have not imposed a due-date on the completion time of the quantity Q. In the next section, we extend Model III by dropping Assumption A3 and keeping Assumption A1 only. 4.4. Model IV
Here, we extend Model III by implicitly imposing the due-date, d, through the objective function. This will allow us to optimize the due-date as well. This is useful in case we want to quote a customer a reliable due-date for the delivery of the quantity Q.
16
To avoid triviality, we assume that d ≥ S + Q r (since L is still zero.) In other words, the due-date must be at least equal to the processing time of the quantity Q, without random interruptions if it were produced in one lot. As stated in the main model, we assume that any quantity processed after the due-date is completely backlogged and incurs a backlog penalty of $b per unit of finished items per unit of time. Hence, compared to Model III, we have DC as additional cost. DC = hf QE [Cn − d ] + bQE [Cn − d ] . −
+
Since L = 0 , we have xi = ti , for all i. Hence, n −1 n −2 Cn = L + max ∑ x j , ∑ x j + j =1 j =1
1 n −1 n −1 t , " , x t , + ∑ ∑ ∑ j j j ∑ t j + tn j = n −1 j =1 j =2 j =1 n −1
n −1 n n −1 n −1 = max ∑ t j , ∑ t j ," , ∑ t j + tn = ∑ t j j =1 j =1 j =1 j =1
DC can then be written as follows. −
n n DC = hf QE ∑ t j − d + bQE ∑ t j − d j =1 j =1
+
−
n n Q Q = hf QE + nS + ∑ τ j − d + bQE + nS + ∑ τ j − d j =1 j =1 r r
+
As before, τ i = ∑ k =0 i Yk represents the total repair time while processing Lot i. Notice that N (q )
∑
n
τ = ∑ i =1 ∑ k =0 Yk = ∑ k =0 Yk . n
N ( qi )
N (Q )
i =1 i
Hence, −
+
N (Q ) N (Q ) Q Q DC = hf QE + nS + ∑ k =0 Yk − d + bQE + nS + ∑ k =0 Yk − d r r .
This indicates that DC does not depend on the qi ’s and only depends on the number of lot sizes, n. In other words, with respect to the due-date cost, it does not matter how we partition the quantity Q. As a consequence, the optimal lot sizes are still calculated according to Lemma 2. If we let ψ (⋅) denote the cdf of the random variable follows.
17
∑
N (Q ) k =0
Yk , we can write the model as
n h n −1 n qi2 n −1 + hf S ∑ i =1 ( n − i )qi + f ∑ ∑ qi q j Minimize C (n, q ) = nK + ( hu + hf ) ∑ R i =1 j =i +1 i =1 2 R Q ∞ d − − nS Q Q + hf Q ∫ r d − − nS − u dψ (u ) + bQ ∫ Q u + + nS − d dψ (u ) d − − nS 0 r r r n s.t. ∑ qi = Q i =1 qi ≥ 0, i = 1,2," , n
LEMMA 4.
(24)
The objective function, C (n, q * ) , obtained after substituting the optimal qi* ’s, is
strictly convex in n. As a result of Lemma 4, the optimal solution can be obtained using the same procedure used to solve Model III. To illustrate Models IV, we assume that the repair times follow an exponential distribution with parameter µ and probability density function (pdf) ϕY ( y ) = µ e − µ y , y>0. In this case, the cdf ψ (t ) of the total repair time during the processing of the quantity Q, can be shown to have the following expression. t 0 Let nk = n − 1 and got to Step 3.
19
End If. End For. Let nk = n , and q k = ( q1 , " , qnk ) using (21). Let C(nk , q k , d k ) denote the optimal cost. 2. If C (nk , q k , d k ) − C (nk −1 , q k −1 , d k −1 ) < ε Go to Step 3. Else Set k=k+1, Compute d k using (26), and go to Step 2. End If The optimal solution has been reached.
STEP 3.
n* = n k , q* = q k , and d* = d k . Notice in Procedure 2, the computation of the expected total cost, C(n,q,d), and condition (26) involve numerical integration for general repair time distributions. Closed form expression of the distribution of the random variable ξ = ∑ k =0 Yk can be obtained by N (Q )
approximating it by simple distributions such as Erlang or Gamma (See Rosling (1989) for details.) We now turn our attention to the main module. We will use some of the results derived above to devise an efficient heuristic solution approach. 5. MAIN MODEL: SOLUTION APPROACH
Here, we go back to our original assumptions and drop Assumptions A1, A2, A3, and A4 that we used for the special case models. Before we proceed with the solution approach we need to develop few preliminary results Noticing that
∑( n −1 i =1
hu qi E [Gi −1 ( x, q )] + ( ih f qi + p ) E [Gi −1 ( x, q) ] +
−
)
n −1
(
= ∑ hu qi E [Gi −1 ( x, q) ] + pE [Gi −1 ( x, q) ] i =1
+
n −1 n −1
+ h f ∑∑ qi E G j ( x, q )
−
−
)
(27)
i =1 j = i
and letting pi = ihf qi + p , the complete mathematical model can then be written as follows.
20
(
)
n qi2 n −1 + − + ∑ hu qi E [Gi −1 ( x, q )] + pi E [Gi −1 ( x, q )] Minimize C (n, q, x) = nK + ( hu + h f ) ∑ i =1 2 R i =1 n − n − n −1 1 1 hf − + + h f S ∑ ( n − i )qi + ∑∑ qi q j +1 + h f QE [Cn − d ] + bQE [Cn − d ] R i =1 j =i i =1 n s.t. ∑ qi = Q i =1 q i ≥ 0, i = 1, 2," , n xi ≥ 0, i = 1, 2," , n
(28)
We will assume that the qi ' s are known for the moment. Also, throughout this section, we will confine ourselves to continuous probability distributions for the repair time. Discrete versions can be solved for through dynamic programming or enumeration. Before, we proceed with the analysis, let hi = hu qi , i = 1,2," , n − 1 . Also, let ϕ i (⋅) and ψ i (⋅) denote the pdf and cdf, respectively, of ti , the total processing time of Lot i. Using this notation, we first address the following optimization problem. Minimize
n −1
(
f ( x) = ∑ hi E [Gi ( x, q )] + pi E [Gi ( x, q ) ] i =1
+
−
)
(29)
s.t. xi ≥ 0, i = 1, 2," , n − 1
Notice that
E Gi ( x , q ) = +
E Gi ( x , q ) =
∑
∫ )
b1
∑
∫ )
d1
all ( a1 ,b1 ,", ai ,bi
−
all ( c1 , d1 ,", ci , di
a1
c1
ϕ1 (u1 ) " ∫
bi −1
ϕ1 (u1 ) " ∫
di −1
ai −1
ci −1
ϕi −1 (ui −1 )∫
bi
ai
ϕi −1 (ui −1 )∫
di
ci
(ui − ai ) ϕi (ui )dui " du1
(30)
( di − ui ) ϕi (ui )dui " du1
(31)
The vectors ( a1 , b1 ," , ai , bi ) and ( c1 , d1 ," , ci , di ) are generated according to a binary tree. For example, when i=2 the binary tree is generated according to Figure 4. The “+” (“−“) branch of the tree indicates that the term at the end of that branch should be included in the expression of E [Gi ( x , q )]
+
( E [Gi ( x , q )] ). Each level of the tree, from top to bottom, −
corresponds to an index i. For instance, level 1 corresponds to E [G1 ( x , q )] and E [G1 ( x , q )] , +
−
level 2 corresponds to E [G2 ( x , q )] and E [G2 ( x , q )] , and so on. As an example, for n=3, the +
−
objective function in (29) can be written as
21
∞
x1
( u1 − x1 )ϕ1 (u1 )du1 + p ∫0 ( x1 − u1 )ϕ1 (u1 )du1 x
f ( x) = h2 ∫
1
{ p {ψ ( x ) ∫ 1
1
∞
∞
∞
2
1
1
( u2 − x2 )ϕ 2 (u2 )du2 + ∫x ∫x + x −u ( u1 + u2 − x1 − x2 )ϕ 2 (u2 )ϕ1 (u1 )du2 du1 x
h3 ψ 1 ( x1 ) ∫
x2
0
x1 + x2
( x2 − u2 )ϕ 2 (u2 )du2 + ∫x
2
∫
x1 + x2 − u1
0
1
1
−
( x1 + x2 − u1 − u2 )ϕ 2 (u2 )ϕ1 (u1 )du2 du1
}
+
(0, x1)
(x1,∞)
−
+
(0, X1, 0, x2)
}
−
(0, x1, x2, ∞)
(x1, x1+ x2,0, x1+ x2-u1)
+ (x1, ∞, x1+ x2-u1, ∞)
Figure 4.: Binary tree for the case n=3. In Elhafsi (2001), we studied Problem (29). We showed that f(x) is strictly convex in x. Hence, Problem (29) has a unique solution. The solution procedure is based on the following recursive formulas for the evaluation of f(x). G1 ( x, q ) = ( t1 − x1 ) and in general, Gi ( x, q ) = ( ti − xi ) + max ( 0, Gi −1 ( x, q) ) .
Letting φi (⋅) and Φi (⋅) denote the pdf and cdf, respectively, of the random variable Gi ( x , q ) , we have
φ1 (u1 ) = ϕ1 ( u1 + x1 ) , and in general,
φi (u ) = ϕ i ( u + xi ) Φ i −1 (0) +
u + xi
∫ ϕ (u + x i
i
− v ) φi −1 (v)dv .
(32)
0
Therefore, ∞
∞ u + xi
0
0
E [Gi ( x, q) ] = Φ i −1 (0) ∫ uϕ i ( u + xi ) du + ∫ +
0
E [Gi ( x, q) ] = −Φ i −1 (0) ∫ uϕ i ( u + xi ) du − −
− xi
∫
uϕ i ( u + xi − v ) φi −1 (v)dvdu
(33)
0
0 u + xi
∫ ∫
− xi
uϕ i ( u + xi − v ) φi −1 (v) dvdu
(34)
0
(33) and (34) have the advantage of being recursive and involve only double integration, while direct evaluation of f(x) involves multiple integrals that have to be generated in a
22
combinatorial fashion for a given n. Furthermore, closed form expression of f(x) quickly becomes extremely difficult to derive as n increases. The reader may realize that solving Problem (28) to optimality is an extremely difficult task. Fortunately, one can develop an efficient solution approach to the problem at hand based on the following observations. OBSERVATION 5.
The optimal solution of Problem (29) is not very sensitive to the choice of
the lot sizes, qi . OBSERVATION 6.
The optimal solution of Problem (29) slightly favors increasing lot sizes
compared to equal lot sizes. OBSERVATION 7.
The objective function in Problem (28) is strictly convex in the number
of lot sizes, n. Observation 5 is based on an extensive numerical experiment where different sets of lot sizes where generated randomly from a Normal distribution with variance equal three times the value of the mean to allow for enough variability among the lot sizes. The maximum difference between the optimal values was insignificant in all cases. What we noticed is that the planned lead-times, xi , will always adjust to accommodate any change in the lot sizes. In other words, if the lot size increases the corresponding xi will shift accordingly to minimize the total cost in (29). Observation 6 is based on a result we developed in Elhafsi (2001) where it is better to have processing times with narrower distributions (i.e., with smaller variance) for early lots. Indeed, there is a direct relationship between the lot size, qi , and the variance of the distribution of the processing time of that lot, which is given by 2 λ qi r µ 2 . Observation 6 is in agreement with the optimal solution of Models III and IV, which states that we should produce lots with smaller sizes first according to (21). Observation 7 is very difficult to prove theoretically. But, numerical experiments showed that it is always the case. Intuitively, we would expect such a result. Using Observations 5, 6, and 7, we can substitute the lot sizes qi according to (21) in the main model (28) and solve for n, the number of lot sizes and xi ’s, the planned lead-
23
times. As a result, we will use a procedure similar to the one use for solving Models III and IV. The procedure is given below. PROCEDURE 3 STEP 1.
Calculate n , the upper bound on the value of n*, using (22). Let n = 1 , q* = Q , and evaluate C ( n, q*, x ) .
STEP 2.
For n = 2, " , n
(
)
Calculate q* = q1* , " , qn* using (21).
(
)
(
)
If C n, q * , x n − C n − 1, q * , x n −1 > 0 Let n* = n − 1 , x * = x n −1 , and got to Step 3. End If. End For. Let n* = n and x * = x n . STEP 3.
The optimal solution to Problem (28) is given by, n*, x* and q*.
A few comments are in order here. To compute the term h f QE [Cn − d ] + bQE [Cn − d ] , let −
+
Θ(⋅) denote the cdf of Cn , the completion time of the Quantity Q. Then, we have h f QE [Cn − d ] + bQE [Cn − d ] = h f Q ∫ ( d − u ) d Θ(u ) + bQ ∫ ( u − d ) d Θ(u ) . −
+
d
0
0
d
Θ(⋅) can be shown to satisfy the following expression. dΘ(u ) = ∫
∞
− xn −1
ϕ n (u − L − ∑ i =1 xi − v)φn −1 (v ) dv , u ≥ 0 . n −1
(35)
Here, ϕ n (⋅) is the pdf of the total processing time, tn , of the last lot, and φn −1 (⋅) is the pdf of Gn −1 ( x , q ) . φn −1 (⋅) is generated recursively according to (32). Procedure 3 is based on the assumption that the due-date d is given. If the due-date becomes a decision variable, a similar argument to Lemma 5 shows that the optimal duedate satisfies the following condition. Θ (d *) =
b b , or d* = Θ −1 b+ h b + hf f
.
(36)
24
d* can be easily computed using a one dimensional line search procedure such as a bisection search. The following procedure can be used to determine the optimal due-date in conjunction with the optimal lot sizes and optimal planned lead-times. PROCEDURE 4 STEP 1.
Calculate n , the upper bound on the value of n*, using (22). Set k = 1 , and let d k = Q R + S + L (Initial guess.) Set ε as the tolerance.
STEP 2.
(
)
1. Let n = 1 , q = Q , and evaluate C n, q, x k , d k . For n = 2, " , n Calculate q = ( q1 , " , qn ) using (21).
(
)
(
)
If C n, q, x k , d k − C n − 1, q, x k , d k > 0 Let nk = n − 1 and got to Step 3. End If. End For. Let nk = n , and q k = ( q1 , " , qnk ) using (21). Let C (nk , q k , x k , d k ) denote the optimal cost. 3. If C (nk , q k , x k , d k ) − C( nk −1 , q k −1 , x k −1 , d k −1 ) < ε Go to Step 3. Else Set k=k+1, Compute d k using (36), and go to Step 2. End If STEP 3.
The optimal solution has been reached. n* = n k , q* = q k , x * = x k , and d* = d k .
A Numerical Example
To illustrate the main model, we again assume that the repair times are exponentially distributed. In this case, the pdf of the total processing times, ti , is given by
25
0 − λqi r dt ϕi (t ) = e e −( λqi r + µ (t − ai ))
t < ai
( λqi r ) ( µ (t − ai ) ) I1 ( 2 ( λqi r ) µ (t − ai ) ) dt,
t = ai ai < t < ∞
Here, ai = S + qi r . The data for the example is given in Table 2. Table 2. Example Data. Parameter Value 1000 units Q 2 units/hour r λ 0.05/hour µ 0.05/hour 17 hours S hu $0.075/unit/hour hf $0.425/unit/hour p b L d
$200/hour $4.25/unit/hour 40 hours 1200 hours
Applying Procedure 3, we obtain the following solution. Table 3. Example solution. Decision variable Value 4 n* (105.5, 201.8, 298.2, 394.5) q* (223, 325, 460) x* $80,855 Expected Total Cost If the due-date d is a decision variable, then by applying Procedure 4, we obtain Table 4. Example solution when the due-date is a decision variable. Decision variable Value 4 n* (105.5, 201.8, 298.2, 394.5) q* (223, 325, 460) x* 1484 hours d* $37,303 Expected Total Cost
26
Notice the difference in expected total cost when d becomes a decision variable. According to (36), one can assign an optimal due-date d* which will be met with a probability of b (b + hf ) . The latter is known as the service level in the literature. In our numerical case it is 91%. In other words, if we assign a due-date of 1484 hours to the demand Q, we are sure 91% that the assigned due-date will be met. 6. CONCLUSION
In this paper, we studied a two level inventory production system that is prone to random failures and repairs. Our goal was to determine a production plan for a single quantity type of demand. We proposed a main model with the objective to minimize the expected total cost consisting of inventory carrying costs for both finished and unfinished items, backlog cost for not meeting the demand’s due-date, and cost of planning for the receipt of unfinished items. To gain insight, we solved instances of the main model that arise from different manufacturing situations. These special cases were either solved analytically or by means of very efficient algorithms. Some of the results indicate that it is not always optimal to produce in equal lot sizes for certain manufacturing situations. Other results confirm the robustness of inventory models. Despite the complexity of the main model, a very efficient solution approach was devised to solve it. This solution approach was based on some of the observations and procedures developed with the special cases. This work can take several future directions. For instance some of the special cases can be generalized to the multiple product case. Others can be extended to the multistage production case. Yet another direction would be to consider multi-level inventory systems such as in Material Requirement Planning.
27
REFERENCES
[1]
P. Afentakis, B. Gavish, U.S. Karmarkar, Exact Solutions to the Lot-Sizing Problem in Multi-Stage Assembly Systems, Management Science 30 (1984) 222-239.
[2]
K.R. Baker, P.S. Dixon, M.J. Magazine, A. Silver, An Algorithm for the Dynamic LotSize Problem with Time-Varying Production Capacity Constraints, Management Science 16 (1978) 1710-1720.
[3]
R.E. Barlow, F. Proschan, Mathematical Theory and Reliability, John Wiley & Sons, New York, 1965.
[4]
M. Elhafsi, Optimal Lead-Time Planning in Serial Production Systems with Earliness and Tardiness Costs, IIE Transactions, Forthcoming March (2002).
[5]
M. Florian, M. Klein, Deterministic Production Planning with Concave Costs and capacity Constraints, Management Science 18 (1971) 12-20.
[6]
T. Gorham, Dynamic Order Quantities, Production and Inventory Management, 2nd Quarter, (1970).
[7]
S.C., Graves, Multi-Stage Lot-Sizing; An Interactive Procedure. In Multi-Level Production/Inventory Control Systems: Theory and Practices, Schwarz L. B., TIMS, North-Holland, New York, 1981.
[8]
U.S. Karmarkar, S. Kekre, S. Kekre, Lot Sizing in Multi-Item, Multi-Machine Job Shops, IIE Transactions 17 (1985) 290-298.
[9]
U.S. Karmarkar, L. Shrage, The Deterministic Dynamic Product Cycling Problem, Operations Research 33 (1985) 326-345.
[10] M. Lambrecht, H. Vanderveken, Heuristic Procedure for the Single Operations, MultiItem Loading Problem, IIE Transactions December 1979. [11] B.J. McLaren, A study of Multiple Level Lot Sizing Techniques for Material Requirements Planning Systems, Ph.D. Dissertation, Purdue University, 1976. [12] K. Rosling, Optimal Inventory Policies for Assembly Systems under Random Demands, Operations Research 37 (1989) 565-600. [13] L.B. Schwarz, Economic Order Quantities for Products with Finite Demand Horizons. IIE Transactions 4 (1972) 234-237. [14] E.A. Silver, M.C. Meal, Heuristic for Selecting Lot-Sizing Quantities for the Case of a Deterministic
Time-Varying
Demand
Rate
and
Discrete
Opportunities
Replenishment, Production and Inventory Management 2nd Quarter (1973). 28
for
[15] B.D. Sivazlian, The Filtered Counting Process and its Applications to Stochastic Manufacturing Systems, Department of Industrial and Systems Engineering, University of Florida, Research Report 92-13, 1992. [16] H.M., Wagner, T.M. Whitin, Dynamic Version of the Economic Lot Size Model, Management Science 5 (1958) 89-96. [17] W. Zangwill, A Backlogging Model and Multi-Echelon Model of a Dynamic Economic Lot-Size Production System-A Network Approach, Management Science 15 (1969) 506-527.
29
APPENDIX Proof of Lemma 1:
Fixing n and using Lagrange multiplier η for the equality constraint in (17), we obtain the following Lagrangian function. n
L(η , q ) = nK + ( hu + hf ) ∑ i =1
qi2 hf + 2R R
n −1
n
∑ ∑ qq i =1 j =i +1
i
j
n + η ∑ qi − Q . i =1
Taking the partial derivatives with respect to the qi ’s and η , and setting the resulting equations to zero gives the following system of (n+1) equations in (n+1) unknowns.
hf q2 + " + hf qn + η R = 0 (hu + hf )q1 + hf q1 + (hu + hf )q2 + " + hf qn + η R = 0 # hf q1 + hf q2 + " + (hu + hf )qn + η R = 0 q1 + q2 + " + qn =Q Adding the first n equations together and solving for η gives
η=−
(hu + nhf )Q nR
.
Now, notice that the n × n system (after substituting for η , and excluding the last equation) is symmetric and diagonally dominant. Therefore, its determinant is nonzero. This means that the system has a unique solution. It is not difficult to check that the only solution to this system is the one with all qi ’s equal. ■ Proof of Lemma 2:
Proceeding as in Lemma 1, when taking the partial derivatives of the Lagrangian function, L(η , q ) , and setting them to zero, the only difference in the (n + 1) × (n + 1) system of linear equations resides in the right hand side. After solving for η , we obtain the following n × n system. (hu + hf )q1 + + hf qn = ( hu + hf ) Q n − (n − 1)hf SR 2 hf q2 + " hf q1 + (hu + hf )q2 + " + hf qn = ( hu + hf ) Q n − (n − 3)hf SR 2 # hf q1 + hf q2 + " + (hu + hf )qn = ( hu + hf ) Q n − hf SR 2 The solution of the above system is unique and is given by
30
qi =
Q n − 2i + 1 hf − h SR , i = 1,2," , n . n 2 u
Further, since the qi ’s must be nonnegative, then the smallest of the qi ’s ( q1 ) must be nonnegative. This translates to n(n − 1) hu Q ≤ .■ 2 hf SR Proof of Lemma 3:
The proof is a straightforward application of Definition 1, hence omitted. ■ Proof of Lemma 4:
The proof is a straightforward application of Definition 1, hence omitted. ■ Proof of Lemma 5:
Let g (d ) = hf Q ∫
d−
0
Q − nS r
∞ Q Q d − r − nS − u dψ (u ) + bQ ∫d − Q −nS u + r + nS − d dψ (u) . r
Taking the derivative with respect to d and setting it to zero gives condition (26). Taking the second derivative gives Q (b + hf )dψ d − − n * S ≥ 0 . r Hence, g (d ) is strictly convex in d and d* is the unique solution of equation (26). ■
31
Lot 1 L
Lot 2 x1
Lot 3
Lot 4
Lot n
x3
x2
xn-1
time
..... t1
t2
t3
Figure 1. Sample Ordering Scheme
32
tn-1
tn
1.07
Total Average Cos
1.06 1.05 1.04 1.03 1.02 1.01 1 1
2
3
4
5
6
7
8
9
Setup Time (in S time units)
Figure 2. Setup time effect.
33
10
11
12
1.3 Expected Total Cost
1.2 1.1 1 0.9 0.8 0.7 0.6 0.5 0.4 18
20
22
24
26
28
30
32
Due-Date
Figure 3. Due-Date effect on the expected total cost.
34
−
+
(0, x1)
− (0, x1, 0, x2)
(x1,∞) + (0, x1, x2, ∞)
− (x1, x1+ x2,0, x1+ x2-u1)
+ (x1, ∞, x1+ x2-u1, ∞)
Figure 4.: Binary tree for the case n=3.
35
