A lot sizing heuristic for deteriorating items with shortages in growing and declining markets

Computers Ops Res. Vol.24, No. I I, pp. 1075-1083, 1997

Pergamon

P l h 8030go-.0548(96)00012-9

1997 Elsevier Science Ltd All rights reserved. Printed in Great Britain 0305-0548/97 $17.00+ 0.00

A LOT SLZING HEURISTIC FOR DETERIORATING ITEMS WITH S H O R T A G E S IN G R O W I N G A N D D E C L I N I N G M A R K E T S Moncer Harigal":~ and All AI-Alyan§ Industrial Engineering Program. College of Engineering, King Saud University, P.O. Box 800, Riyadh 11421, Saudi Arabia

(Received December 1995; in revised form Januat3., 1997) Scope and Purpose--The purpose of this article is to present a new heuristic procedure for the inventory problem of deteriorating items with time varying demand and shortages. In contrast to Wee's model, recently published in this journal, our model is not developed under the restrictive assumption of equal replenishment cycles. A comparative work shows that our heuristic procedure is outperforming the one proposed by Wee in both cost and computation time performances. Abstract--In this article, we consider the inventory replenishment problem with shortages over a fixed time horizon for a product deteriorating at a constant rate, We do not put restrictions on the length of the replenishment cycles and the form of the demand function making our model a general one. For linearly and exponentially time varying demands, the results of a comparative work show that our heuristic is economically and computationally more efficient than the equal cycle approach. © 1997 Elsevier Science Ltd 1. I N T R O D U C T I O N

In a recent article, published in the Journal of Computers and Operations Research, Wee [ I ] considered the inventory problem of an item deteriorating at a constant rate in an exponentially declining market over a fixed time horizon. In his model, Wee allowed shortages to be back-ordered, except for the last cycle during which shortages are prohibited. He further restricted the replenishment intervals to be of equal length. Under these assumptions, he derived the operating characteristics of the system and presented the optimization procedure that generates the optimal replenishment frequency and service level (fraction of time with no shortages). Goswami and Chaudhuri [2] attempted to solve the same inventory problem with a different assumption on the pattern of the demand rate which was linearly increasing. Their model was recently modified and generalized to any type of demand functions by Hariga [31. Hariga also showed that the equation giving the percentage service level for a fixed replenishment frequency is independent of the demand function parameters in contrast to the ones derived by Goswami and Chaudhuri equation (1 I) and Wee equation (32). Moreover, the resulting model of Hariga, because of its generality, is more lucid and easier to follow than the ones of [ 1] and [2]. In this article, we present a new heuristic procedure that is economically and computationally more efficient than the optimization procedure of Wee. We do not restrict the replenishment cycles to be equal since severe cost penalty may be incurred by such a restrictive assumption. Intuitively, as demand decreases over time one tends to increase the duration between consecutive orders to reduce the number of replenishments. The converse is also true for increasing demand to reduce the holding cost. Moreover, our procedure requires little computational effort to adjust the generated schedule as a result of a change in the time horizon. In the Section 2, we present the underlying assumptions and notation to be used throughout the article. In Section 3, we review briefly the equal replenishment cycles procedure. Then, we develop our heuristic procedure for a general continuous time-varying demand and illustrate the model with the two special 1"To whom all correspondence should be addressed (email: f45m036 @ksu.edu.sa). -1:M. A. Hariga is an Associate Professor of Industrial Engineering at King Saud University, Saudi Arabia, Dr. Hariga received his M. Eng. in Industrial Engineering from the National School of Engineering in Tunis, Tunisia and his Ph.D. in Operations Research and Industrial Engineering from Coroell University, Ithaca, NY. His current research interests include mathematical modelling of inventory systems, scheduling, and production planning. He has research papers published or forthcoming in the

Journal of Operational Research Steles.. HE Transactions, Naval Research Logistics, Computers and Industrial Engineering, Computers and Operations Research, European Journal of Operational Research, International Journal of Production Economics, International Journal of Production Research, International Journal of Operations and Production Management, etc. § All Al-Alyan is a master student in the Industrial Engineering department at King Saud University. He is currently working as operations manager in Arabian Medical Products Manufacturing Co., Riyadh, Saudi Arabia. 1075

1076

Moncer Hariga and Ali AI-Alyan

cases of linear and exponential demand trends. In Section 5, a numerical comparative study is carded out to show the superiority of our heuristic procedure. Finally, the last section concludes the article. 2. ASSUMPTIONS AND NOTATION The assumptions and notation used in this article are slightly different to that considered by Wee and are given here for the sake of completeness. (a) A single item with a constant rate of deterioration 0 is considered. (b) Deterioration of the units is considered only after they have been received into inventory. (c) There is no replacement or repair of the deteriorated units during the period under consideration. (d) Demand is a known continuous function of time, D(t). (e) The rate of replenishment is infinite. (f) Carrying cost, (72, applies to good units only. (g) The system operates only for a prescribed period of H units of time. (h) Shortages are allowed except for the last cycle and are fully backordered at cost C 3 per unit per unit of time. (i) The order quantity, inventory level and demand are treated as continuous variables, and the number of replenishments, n, is treated as discrete variable. (j) The ordering cost is C~ per replenishment. (k) C is the cost of deteriorated unit. (1) T~is the time of the j-th replenishment, j = 1, 2 ..... n with T~=0. (m) ti is the time at which the inventory in the j-th cycle drops to zero, j = 1, 2 ..... n with t,=H. (n) rj is the percentage service level of the j-th cycle (fraction of time with no shortages in the j-th cycle), j = 1, 2 ..... n - I. Then, Tj, tj, and r~ are related by the following relation. ti=rjT)+(l

-

rj)Tj_ t j= 1,2..... n - 1

(I)

3. EQUAL REPLENISHMENT CYCLES HEURISTIC Using the assumptions listed in the previous section, Hariga [31 obtained the following mathematical relation for the total inventory cost t ,_, r,[, W(n,t,T)=nCt +CaO-' ~, (e ~ ' - r ' ' - 1)D(u)du+C3 E (Tj+~-u)D(u)du, J=J rj j= ,j

(2)

where C4=C2+ 0C. For linearly time varying demand, i.e., D(u)=a+bu, where a>0 and a+b H>-O, (2) becomes; W( n,t,T) =nC~ + C40- i ~, { (e ~',- r~ _ I )( OD(Q - b ) + O(b - a O)(t~- Tj) j=l

OZB ( t ~ - T~)} + C3 n~l - 2 ~ J., {(T~+j-tj)2(3a+b(T~+,+2tj))}.

(3)

When the trend in the demand is exponential i.e., D(u)=ae ~, where a>0, the total inventory cost can be written as; W(n,t,T) =nCi

+aC40-1

~ OehT; j=l ~. b(b+O)

b

+ - b+O

+aC3~1

-

.

b

'

if0

+b~0

(4)

and, W(n,t,T)=nCi +aC40-' J=~,ebr~{ tj - Tj =0

.aC,

b2

eh',(T~.l -- tj) "~ if O+b b J (5)

A lot sizing heuristic for deteriorating items

1077

Comparing (4) and relation (24) of Wee [1], our expression of the total inventory cost is much mathematically simpler and given in a more compact form. Next, for a given value of n and using the assumption of equal replenishment cycles, we have; H T~= (j - 1) - j = 1,2 ..... n, n

(6)

H t j = ( r j + j - 1 ) - j = 1,2 ..... n - 1. n

(7)

and,

Moreover, after setting the derivative of W(n,t,T) in relation (2) with respect to tj equal to zero, we obtain;

C4O-'(e~', -r,'- I ) - C3(Tj÷~- t ) = 0 j = 1,2 ..... n - 1. Next, using (6) and (7), this equation can be transformed to;

C40-I(e*/~/n - 1) - C3(l - r) H = 0 j = 1,2 ..... n - 1. n It is clear now that rj=r for j = l, 2 ..... n - 1 (see also Hariga [3]). Letting the left hand side of the last equation equal E(r) and replacing r/by r, we get;

E(r)=C40-I(ee"t" - l) - C3(i - r) _H =0 n

(8)

Observe that, in contrast to relations (11) of Goswami and Chaudhuri [2] and (32) of Wee [l], the optimal value of the service level is independent of the demand parameters. Moreover, (8) can be easily solved by the dichotomous search method on the interval [0, l] since E(0)< 0, E ( l ) > 0 and E ' ( r ) > 0. Then, the optimal replenishment schedule can be obtained by determining a list of minimum W(n,r*) for a range of values of n. The values of n and r yielding the minimum W* are taken to be the optimal ones.

4. A NEW HEURISTIC PROCEDURE The new heuristic that we propose for solving the inventory problem with deterioration, shortages, and time-varying demand relaxes the restrictive assumptions of equal replenishment cycles and constant service level during each cycle. In fact, as it is shown in the numerical comparisons, severe cost penalty may be incurred by such restrictions. Moreover, in contrast to the equal replenishment cycles heuristic, our approach requires little computational effort to adjust the replenishment schedule when the planning horizon changes. After partially differentiating the total inventory cost (2) with respect to tj and Ti, the following necessary optimality conditions can be obtained;

0

(.4

,,.,

o

C4 O- i(eeO,-r) _ l)+tj Ti÷t= -~

(10)

Our heuristic is based on the observation that once t~ =t is fixed, all the remaining tj (j=2,3,...) and (j=2,3,...) can be determined form (9) and (10). In fact, for a given t~ =t, (10) will be used to determine /'2 (recall T~=0). Then, t 2 will be obtained by solving (9). By repeating this process, other t/s and T/s can be determined consecutively. Therefore, we need only to find a good starting value for t~ to generate the rest of the replenishment schedule. In our heuristic, we suggested to start with a value for tj that is obtained from the following equation. Sum of holding, shortages, and deterioration costs during the first cycle = ordering cost. Mathematically, this can be written as:

1078

Moncer Hariga and Ali AI-Alyan T~(I I)

tl

C40 -~ ~ (e ~'- l)D(u)du+C3

~ (T2(tO-u)D(uJdu=C ,,

0

h

where T2(tO is obtained from (10), T2(t~)= C--A4O-J(e ~''-T'>- l)+tl, ¢3

Therefore, t t is the solution to the following equation II

T2i/I)

0

h

F(tl)=C40 -I ~ (e * ' - l)D(u)du+C 3

(T2(h)-u)D(u)du-Cl=O,

(11)

which is obtained from the above equation by simply moving Ct to the left hand side. For increasing demand, there exists a unique solution to (11) since F(0)= - Cj, F(oo) > 0, and F' (t t) > 0. However, for decreasing demand, if F(y)<0, where y is the solution to T2(y) - x = 0 and x is the solution to D(t)=0, then only one order should be placed during the entire planning horizon. Moreover, if the solution to (1 I) is larger than H, i.e., t~>H, then, similarly, the replenishment frequency is equal to one. However, if t,
h(t) = t e ~D(u)du - g(tj_ ,,Tj) = 0 j = 2,3 .... 0

(12)

where g(tj _ l,T) is the right hand side of (9) and it is a constant term for given tj_ ~and T~. It can be easily verified that h ' ( t ) > 0 and h(T)<0. Moreover, for increasing demand, it can be shown that h(oo)>0 which implies the existence of a unique solution to (12). For decreasing demand, suppose that t,(k= 1,2,3 .... j) and T,(k=2,3 ..... j + l ) were determined using (9) and (10) and we are in (j+l)-st iteration of the algorithm. If h(y)~0, where y is the solution to Tj+,(y)-x=0, then the algorithm will terminate, and otherwise will search for tj. ] in the interval [Ti÷t, y]. Finally, for both increasing and decreasing demands, the relations (9) and (10) will be used successively until, tj. ~, the first timing of shortages will exceed H. At this stage, we let n=j+ 1 and decide between two replenishment schedules that are obtained by linear proportional adjustment of the timings of the original schedule. The first replenishment schedule has n orders placed during the planning horizon with;

T~I)=Tk

k=2,3 ..... n

t~t)=t~ H k= 1,2..... n, t,

whereas in the second one the number of orders to be replenished is (n - 1) and;

T~?':T, _H_ k=2,3 ..... n - 1 t,,_ t

t~?~=tt, tt__ k=1,2 ..... n - 1. tn- I

Then, the replenishment schedule with the least cost is the one to be selected. This solution procedure is next outlined in an algorithmic form for both increasing and decreasing demands.

Step 1 Step 1.1: If demand is a decreasing function of time and F(y)-0, then let n= 1 and stop. Step 1.2: In other cases (increasing demand, decreasing demand with F(y)>0), find t t and T2 by solving (11).

Step 1.3: If t~-->H, let n= 1 and stop, Otherwise let j = 1 and goto Step 2. Step 2 Step 2.1: If demand is a decreasing function of time then: Step 2.1.1: If Tj÷t-->Hand h(y)-<0 then

A lot sizing heuristic for deterioratingitems

1079

• n=j

• F o r k = l ton H

t, ,--t, I.

r,.-r,-

H t,,

• Compute W(n,t,T) using (2) and Stop. Step 2.1.2: If Ti. f
tj

T,.-T, H k= 1,2,...j

tj

Step 2.1.2.3: If W(]+ I,t,T)0, search for tj÷~in the interval [Tj÷~,x] using (9) and goto Step 3. Step 2.2: If demand is increasing, determine tj.j by (9) and goto Step 3:

Step 3: Step 3.1: If tj÷l
• Compute T~+I using (10). • Goto Step 2.

Step 3.2: If tj÷~>--Hthen Step 3.2.1: Compute W(j+ l,t,T) where H tk.--t k - tj+l H Tk.--T,-tj+l

k=

1,2.... j+ 1

k=l,2 .... j + l

Step 3.2.2: Compute WU,t,T) where H t, ~---tk-

k= 1,2.... d

r,-r, n-

k= 1,2.... d

tj

tj

Step 3.2.3: If W(j+ I,t,T)
1080

Moncer Hariga and All AI-Alyan

In this case;

F(tt)=

C40-la

01;

I e(h~.a~ elnl b(b+0) b+0 b + --

C3

[ebr:-eh"(b(T2-tO+ l)]-Ci b+O-~O (13)

e b'' 1 ) a C40-ta t I - ~ - + ~ +C3~[ehr:-eb"(b(T2-tO+l)]-Cib+O=O Moreover, substituting D(t)=a e ~' into (9) or (12) we get b (l-e-~r~-"-°)]+T~ t t~=~+bln [ I+C--:C3 b+O

b+0--~0

tj= C3 e ~ C4 b (ehr'--eb" ')+Tj

(14)

b+O=O

Note that the condition F(y)-<0 in Step 1.1. of the algorithm becomes

C4a b(b + 0)

Cl--<0

(15)

for decreasing exponential demand since y=x=~. In this case, only one order should be placed during the entire planning horizon. Moreover, it can be shown that the condition h(y)-<0 in Steps 2.1.1. and 2.2.2. is equivalent to;

t j - T j > ~ In

1 - bC4

C3 b+O

(16)

and should be used only when b+ 0<0 since in the other cases (b>0, b<0 and b+ 0>0) a solution to (9) will always exist. Next, our heuristic procedure for the case of exponential trend in demand is illustrated numerically using the data of the following three example problems with example 2 being the one used by Wee [ 1]. Example 1: Example 2: Example 3:

a=10, a=500, a= 10,

b=0.98, b=-0.98, b= - 0.08,

Ct=250, C~=250, C~=250,

C2=40, C2=40, C2=40,

C=200, C=200, C=200,

C3=80, C3=80, C3=80,

0=0.08, 0=0.08, 0=0.08,

H=4. H=4. H=4.

The results for these three example problems are reported in Table 1. Observe that our heuristic generated lower cost replenishment schedules in each of the three cases. Table I. Results of the three example problems. Example Our heuristic Wee's Heuristic Our heuristic Wee's heuristic Percentage saving by our heuristic

I

2

3

II 14 5196.6 6357.4 22.34

10 II 5115.6 5732.6 12.06

3 4 1629.1 1656.7 1.70

k

T,

I 2 3 4 5 6 7 8 9 10 II

0 0.931 1.569 2.054 2.445 2.773 3.055 3.302 3.522 3.722 3.903

t, 0.542 1.304 1.853 2.283 2.638 2.939 3.200 3.432 3.640 3.828 4.0

7",

0 0.204 0.427 0.675 0.954 1.272 1.642 2.084 2.635 3.365

t, 0.120 0.335 0.573 0.839 1.140 1.489 1.90 I 2AO6 3.062 4.0

T,

0 1.483 3.044

0.862 2.390 4.0

A lot sizing heuristic for deteriorating items

1081

4.2. Linear trend in demand Letting D(t)=a+b t in the derived mathematical expressions of the general model, we obtain;

2b 2"]

C3

F(t O=C40 -~ (e ~' - l )( OD(tO - b)+ O(b - aO)fi - 0 ~ t j ] + -6- (T2 - h):(3a+b(T2 + 2tO) - C,, (17) and,

h(t)=eS'( O D ( t ) - b ) - e ~ [ 2C~02 ~ 4 (~ - tj_t)(D(T)+ D(tj_ t))+ OD(T~) - b ] .

(I 8)

The heuristic is next illustrated by considering the following three examples with linearly time varying demand. Example 4 is the one solved by Goswami and Chaudhuri [2] and Hariga [3]. Example 4: Example 5: Example 6:

a=20, a=0, a=20,

b=2,

Ca =90,

b=2, b= - 0.5,

C~=90, Ct =90,

6",--5, C3= 1.5, 6"_,=5, C3= 1.5, C,_=5, C3= 1.5,

C=0.5, C=0.5, C=0.5,

0=0.01, O=0.01, 0=0.01,

H= 11. H= 11. H= I 1.

The results of the three examples are shown in Table 2 from which one can notice the significant cost saving achieved by our heuristic. 5. C O M P A R A T I V E

STUDY

To show the superiority of our heuristic in terms of cost and computation time performances, we solved the 49 example problems of Wee for declining exponential demand. The results of these problems are reported in Table 3. The measure of computation performance used in the comparison is the ratio of CPU time of the equal cycle heuristic to the CPU time of our heuristic. The results indicated that our procedure produced lower cost solution than equal cycle heuristic for all 49 problems. The percentage of cost saving by our heuristic varies from 5.84% to 20.20% whereas the computation time ranges from 1.3 to 4.1. The average saving of these cases is 13.65% and our heuristic is on the average 2.7 times faster in execution than the equal cycles approach. The two heuristics were also compared in case of exponentially increasing demand using another 49 problems that differ from the previous ones by the values of the b parameter. The results of this comparison are shown in Table 4. Once again, it can be observed that our heuristic outperformed the equal cycle method economically and computationaily. The percentage cost saving of our heuristic varies from 10.49% to 24.97% with an average value of 19.82%. The computation time ratio changes from 4.3 to 19.1 and our heuristic is 9.4 faster on the average. The computation superiority of our heuristic can be explained by the fact that the equal cycle method has to evaluate several replenishment schedules with different number of orders (n= 1,2,...). However, in our procedure, the replenishment frequency is determined when the first tj exceeds H. Problems with linearly time varying demand were also solved to evaluate the superiority of the cost and computation time performances of our heuristic. The problems solved consist of 49 cases generated from example 4 by keeping a=20 and changing the remaining parameters one at a time by 10%. Table 2. Results of examples 4.5. and 6. Example Our heuristic Wee's heuristic Our heuristic Wee's heuristic Percentage saving by our heuristic k I 2 3 4 5 6

4 6 7 944.725 1133.67 20.00

0 2.494 4.738 6.809 8.753 10.595

0.574 3.011 5.215 7.257 9.177 Il

5 3 5 560.889 794.498 41.65

0 6.519 10.281

1.496 7.384 II

6 4 5 735.302 831.426 13.07

0 3.251 6.626 10.149

0.748 4.027 7.436 II

1082

Moncer Hariga and All AI-Alyan Table 3. Results of the 49 example pcoblems of Wee (Exponential decline) C

200

220

240

260

12.16 3.0

12.06 3.1

11.92 3.4

11.72 3.0

11.52 2.0

200

225

250

275

300

325

12.39 3.9

11.98 3.4

11.98 3. I

12.06 3.1

11.60

1.8

11.66 2.0

11.95 2.4

28

32

36

40

44

48

52

I 1.90 3. I

I 1.06 2.6

12.26 3.0

12.06 3.1

I 1.62 2.0

I I. I I 2.0

12. I I 3.4

56

64

72

80

88

96

104

10.64 2.5

11.84 2.5

12.01 3.0

12.06 3. I

11.97 3.0

11.82 1.9

11.68 3.6

H

2.8

3.2

3.6

4.0

4.4

4.8

5.2

% cost saving of our heuristic Computation time ratio

5.84 1.3

7.25 2.6

9.87 3.1

12.06 3. I

13.86 3.5

16.23 3.4

20.20 3.6

0

0.056

0.064

0.72

0.0g

0.088

0.096

0.104

% cost saving of our heuristic Computation time ratio C~ % cost saving of our heuristic Computation time ratio

C: % cost saving of our heuristic Computation time ratio C~ % cost saving of our heuristic Compotation time ratio

140

160

180

12.28 3.1

12.23 2.6

175

% cost saving of our heuristic Computation time ratio

12.30 1.7

12.25 3.0

12.16 3.0

12.06 3.1

11.91 3.0

11.70 3. I

11.48 1.9

b

-0.686

-0.784

-0.882

4).98

- 1.078

- I. 176

- 1.274

5.85 2.4

7.90 4. I

10.01 3. I

12.06 3. I

13.37 1.8

17.17 2.5

18.00 2.5

% cost .saving of our heuristic Computation time ratio Base column

Similarly, 49 problems were obtained from example 5 and another 49 cases from example 6. The overall results of this comparative work are given in Table 5. Based on these results, the same conclusion about the superiority of our heuristic can be drawn for linearly time varying demands.

Table 4. Results of the 49 example problems with exponential growing market C % cost saving of our heuristic Computation time ratio C~ % cost saving of our heuristic Computation time ratio C: % cost saving of our heuristic Computation time ratio C~ % cost saving of our heuristic Computation time ratio H % cost saving of our heuristic Computation time ratio 0 % cost saving of our heuristic Computation time ratio b % cost saving of our heuristic Compotatiou time ratio Base column

140

160

180

200

220

240

260

17.04 8.5

17.0q 8.0

17.22 8.4

17.28 10.8

17.34 8.6

17.38 9. I

15.51 11.6

175

200

225

250

275

300

325

16.79 1I. 1

16.99 9.9

17.14 9.4

17.28 10.8

17.37 7.6

17.52 7.6

17.67 6.8

28

32

36

40

44

48

52

16.68 9.5

16.84 7.4

17.06 8.5

17.28 10.8

17.47 11.3

17.63 9.5

17.84 9.9

56

64

72

g0

88

96

19.07 9.8

18.26 10.3

17.71 10.3

17.28 10.8

16.86 11.3

16.61 8.8

16.32 9. I

2.8

3.2

3.6

4.0

4.4

4.8

5.2

I 1.43 4.3

13.30 4.8

15.15 7.3

17.28 10.8

19.51 11.2

21.88 14.3

24.34 19.1

0.72

0.08

17.22 8.9

17.28 10.8

0.056 17.04 8.2 0.686 10.49 5. I

0.064 17.09 8.0 0.784 12.63 6.3

0.882 14.87 8.2

0.98 17.28 10.8

0.088 17.34 8.4 1.078 19.74 10.8

0.096 17.38 10.7 I. 176 22.31 14.6

104

0.104 17.51 8.8 1.274 24.97 13.8

A lot sizing heuristic for deteriorating items

1083

Table 5. Overall results of linearly time varying demand problems % cosl saving of our Compuuaiontime heudstic ratio Average Maximum Average Maximum

Demand

Linearly increasing with a=20 Linearly increasing with a=0 Linearly decreasing

23.47 49.00 15.63

36.87 70.23 27.73

1.5 1.3 1.0

3.7 3.7 2.2

6. C O N C L U S I O N

A new heuristic procedure for the inventory problem of deteriorating items with time varying demand and shortages is developed in this article. In contrast to Wee's model, we do not put a restriction on the replenishment cycles. Besides its cost and computation superiority, our heuristic requires little computational effort to adjust its replenishment schedule as a result of a change in the time horizon. In our comparative work, we also tested another heuristic procedure with a different value for h. In this heuristic, t~ is based on the economic order quantity formula and is given by;

ll----

•/

2ClCsH H

•

C4(C3+C4) J D(u)du o

However, me results for this heuristic were not encouraging because it was dominated by the one presented in this article. REFERENCES I.

Wee, H. M., A deterministic lot-size inventory model for deteriorating items with shortages on a declining market. Computers

and Operations Research, 1995, 22, 345-356. :2. 3.

Goswami, A. and Chaudhuri, K. S., An EOQ model for deteriorating items with shortages and linear trend in demand. Journal of Operational Research Society, 1991, 42, 1105-1110. Hariga, M. A., An EOQ model for deteriorating items with shortages and time-varying demand. Journal of Operational Research Socieo,, 1995, 46, 398-404.