- Email: [email protected]
Optimal time varying lot-sizing models under inflationary conditions
EUROPEAN JOURNAL OF OPERATIONAL RESEARCH ELSEVIER
European Journal of Operational Research 89 (1996) 313-325
Theory and Methodology
Optimal time varying lot-sizing models under inflationary conditions M . A . H a r i g a a, *, M . B e n - D a y a
b
a College of Engineering, King Saucl University, P.O. Box 800, Riyadh 11421, Saudi Arabia b Systems Engineering Department, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia Received August 1993; revised August 1994
Abstract
This paper deals with the inventory replenishment problem over a fixed planning horizon for items with linearly time-varying demand and under inflationary conditions. We develop models and optimal solution procedures with and without shortages. We do not put any restriction on the length of the replenishment cycles making the proposed methods the first optimal solution procedure for this problem. Using four examples, we illustrate the proposed solution procedures and study the effect of changing the inflation and discount rates on the optimal replenishment schedules. Keywords: Inventory; Inflation; S h o r t a g e s
1. Introduction
The classical E O Q formula assumes that all relevant costs are constant. It also assumes a constant demand rate over an infinite planning horizon. The latter assumption is only valid during the saturation phase of the product life cycle and for finite periods of time. Furthermore, many items such as electronics components, fashionable clothes, and domestic goods observe periods of rapid sales increase after gaining consumers' acceptance. At the same time sales for other products may drop drastically due to the introduction of new competitive products or to the change in consumers' preferences. As to the first assumption, since many countries experience high annual inflation rates, it is no longer reasonable to ignore the effect of inflation and time value of money. Clearly, there is need for the reformulation of the optimal inventory management policies taking into account these different factors. The inventory problem with linear trend in demand has been the subject of several studies since the exact procedure of Donaldson [7]. Heuristic methods have been introduced by Silver [17], Mitra et al. [13], and Phelps [15] among others. Recently, Hariga [10] has developed a new optimal iterative procedure which can be applied to both linearly increasing and decreasing demands. However, all these
* Corresponding author.
0377-2217/96/$15.00 © 1996 Elsevier Science B.V. All rights reserved SSDI 0 3 7 7 - 2 2 1 7 ( 9 4 ) 0 0 2 5 6 - 8
314
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
exact and approximate procedures have not recognized the time value of money and inflation in determining the replenishment schedule. Among the first authors considering inventory problems with inflation, Buzacott [4] has studied the E O Q model under different pricing policies. Following the work of Buzacott, several attempts (Bierman and Thomas [3], Misra [11,12], Aggrawal [1], Chandra and Bahner [5], Gurnani [9], and Banks et al. [2]) have been made to extend his approach to various inflationary situations. However, one common characteristic of all these studies is the assumption that the market demand is time-invariant. In a recent paper, Datta and Pal [6] have studied the effect of inflation and time-value of money on an inventory model with linear trend in demand and shortages. They have based their model on the restrictive assumption of equal replenishment cycle in order to minimize the present value of the total inventory. In this paper, we follow the discounted cash flow approach to develop two time varying lot sizing models for the inventory replenishment problem with Iinear trend in demand taking into account the effects of inflation and time value of money. We first introduce a model with infinite shortage cost. We then reformulate and propose a solution procedure for this model under the assumption that shortages are permitted and are fully backordered. In both models, we do not put any restriction on the length of the replenishment cycles. We do not consider internal and external inflation rates but rather use one inflation rate for all costs. In [5], Chandra and Bahner have pointed out that it is always possible to assume that these two rates are equal and to combine the internal and external elements of each type of cost, if appropriate. Moreover, the introduction of these two rates would not alter the analysis presented in this paper, but would rather add to the complexity of the derived mathematical expressions. The proposed models extend the ones in [1-5,9,11,12] to the case of time varying demand. Our models also relax the restrictive assumptions of equal replenishment cycles and equal positive inventory periods in each cycle considered by Datta and Pal. We extend the model proposed by Hariga [10] by considering the effect of inflation and time value of money. This makes our models the first optimal solution procedures for the inventory replenishment problem with finite and infinite shortage cost and linear trend in demand under inflationary conditions. To the best of our knowledge this problem has not been addressed so far. This paper is organized as follows. In the next section, we state our assumptions and notations. The optimal replenishment model without shortages and its solution procedure are presented in Section 3. Four numerical examples are also presented to illustrate the solution procedure. In Section 4, we develop and solve the model with finite shortage cost. Finally, Section 5 concludes the paper.
2. Assumptions and notations We make the following assumptions: A single item is held in stock over a known and finite horizon which is H years long. Replenishment occurs instantaneously at an infinite rate. Shortages are not allowed. No inventory is carried at time zero and at the end of the time horizon. The demand rate is known and varies linearly with time, namely D ( t ) = a + bt, where a >~ 0, and b is an arbitrary constant. 6. The problem parameters a, b, and H must satisfy D ( H ) -~ a + b H >~ O. The following notation will be used: A0: Fixed replenishment cost per lot at time zero. Co: Unit cost at time zero. h0: Unit inventory holding cost per unit time excluding the cost of capital. 1. 2. 3. 4. 5.
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
315
The inflation rate. The discount rate representing the value of money. Discount rate net of inflation, that is R = r - i. R: Number of replenishment periods during the time horizon H. n: Total time elapsed up to and including the j-th replenishment period, j = 1 , . . . , n, where t/ t o = O a n d tn = H . Length of the j-th replenishment period, j -- 1 , . . . , n, where T/= tj - ty_ 1T]: I(t): Inventory level at time t, 0 ~
r:
3. A time-varying lot sizing model without shortages
3.1. Preliminaries The inventory level at any time t during the replenishment cycle is solution of the following differential equation with boundary condition I(tj) = O.
(dl(t)/dt)=-D(t),
tj_l<_t<_tj,
t0=0,
tn=H,
or
I(t) = fttJD(u) du,
ti_ 1 < t <_tj.
Next, we determine the expression of the present value of the various costs involved for a cycle. The present value of the inventory holding cost incurred during the j-th cycle is H C ( t j _ l , t j ) = h o f ti e ( i - r ) t l ( t ) d t = h o f t' e - R t ( f t ' D ( u ) d u ) a t . tj_ 1 tj_ 1 \ t Integrating by parts, we obtain HC(t/_l, tj)= -~
ftlJl(e-Rt~-l--e--Rt)D(t)._ dt.
(1)
Note that as R --+ O, it can be shown that (1) reduces to
hof tj ( t - t j _ l ) D ( t )
dt
~j- 1 which corresponds to the case of no inflation (Hariga [10]). The present value of the total variable cost incurred during the j-th cycle is
-1 + Coe-Rt-,Q(ti_l,
+ nc(tj_,
tj)
M.A. Hariga, M. Ben-Daya ~European Journal of Operational Research 89 (1996) 313-325
316
where Q(tj_ 1, tj) is the lot received at time tj_ 1 to satisfy the demand of the j-th cycle, that is,
The present value of the total cost over the time horizon is n
n
h0
K ( n ) = A 0 ~ e -R`J-a + C o Y'~ e-ms-'ftts D ( t ) dt + j=l
j=l
j-I
-
R j=l
/,
e -Rtj-' + C E e-RtJ-i tj D ( t ) dt -
=A 0 j=l
j=l
n
Y'. fts (e_ms_~_ e _ m ) D ( t ) dt
j-1
tj-1
- fo "e -RtD(t)
at
(2)
where C = C O+ ho/R. In the next lemma, we show that under a certain condition, it is optimal to place only one order during the planning horizon. L e m m a 1. I f the discount rate net of inflation is negative and less than the inventory carrying cost rate
(excluding the cost of capital invested in inventories), then the present value of the total cost is minimized by placing only one order during the planning horizon. Proof. Note that for negative discount rate net of inflation, C > 0 if RC o < - h o. Since h 0 = ioC o where i 0 is the inventory carrying cost rate excluding the cost of capital, the above inequality becomes R < - i 0. In this case it is easy to see that = A 0 e - m l + C(e -m' - 1)
K(2) -K(1)
(t) dt,
which is clearly a positive number. H e n c e K(2) > K(1).
[]
For fixed n, leaving out the constant term, the problem becomes:
tj-1
j=l
subject to
tj>tj_l,
t0=0,
t,=H.
In order to determine tj, j = 1 , . . . , n - 1, we proceed as follows. Taking the derivative of the objective function of the above optimization problem with respect to tj and equating to zero, we get
Re-l%(Ao+CftS+xD(t) t ""tj
dt} = C D ( t j ) ( e - R t i - l - e - R t J ) ,
j= 1,2 ..... n-1,
(3)
]
which can be written as
O(ti)[(eR
- 1)/R]
-Ao/C= f"+'D(t) dt. tj
Observe that as R --* 0, (4) becomes
D ( t i )-T i = l t~S + i D ( t ) d- t , ts
j=l
.... ,n-l,
(4)
M.A. Hariga, M. Ben-Daya/ European Journal of Operational Research 89 (1996) 313-325
317
which corresponds to the optimality conditions in the case of no inflation (Hariga [10]). Now since D ( t ) = a + bt, it follows that
ftJ+lD(t) d t =
1 2 T j + I D ( t i ) + ~bTj+ 1.
tj
Substituting in (4), we get 1 2 = Ty+lD(ti) + gbTj+l,
O ( t y ) [ ( e RTj - 1 ) / R ] - A o / C
j = 1,...,n-
1,
(5)
12)]
or
A o / C + ~bTy+ 1 + D(tj)
'
(6)
j = 1 . . . . . n - 1.
An alternative formula can be obtained by solving (5) as a second order polynomial for Ty+l, leading to Tj+I= ~
(tj.)Z+2b D ( t j )
-R-
-
-D(tj)
,
j= 1..... n-
1.
(7)
Note that it is enough to know Tn to determine all other T/s using (6). Alternatively, if T 1 is known, one can determine all other T/s using (7). It has been shown in [10] that when R = 0 and b > 0,
(8)
> Tn.
TI> T2> ""
This is not true in general here. The following lemma identifies a case where (8) still holds. 1.emma 2. I f b > 0, R < 0 and C > 0, then T 1 > Z 2 > . . . > T n. Proof. Using the mean value theorem,. ftt/+lD(t) dt = T~+ 1D(0) where tj < 0 < tj+ 1. Since b > 0, it follows • . J a . that D ( t ) < D(O). Multiplying both sldes by Tj+I, we obtain
rj+lO(tj) <
rj+lV(0
) .
(9)
Also, for R < 0 it can be shown that > (e
(10)
- O/R.
From (5) we have O ( t j ) [ ( e RTj - 1 ) / R ] = A o / C
+ Ti+ID(0 ).
Using (9) and (10), we obtain [ e RTj - -
D(tj)Tj>D(tj)[
1
-~
Hence (Tj - Tj+I)D(tj) > A o / C
Ao
Ao
=--~-+ Tj+ID(O)>~-+D(tj)Tj+
> 0 or Tj > Tj+p
[]
1.
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
318
3.2. Solution procedure Recall that the problem is to find the number of orders and their replenishment epochs tl, t2,... , t~_ 1 which solve the following problem. Minimize
j ~ l e -nt~-I A°+CftJtj_ 1D(t) dt
subject to
t1 > t j_l,
t0=0, t~=H.
We propose a solution procedure for finding the optimal n and In particular, we do not consider the usual [6] restriction that method rely on the development of the previous section. For enough to find T~ and then use (6) to determine the length determined as follows. Let
T. = T = f o ( T ),
1I(
t/s without any additional assumptions. the length of the cycles are equal. Our fixed n, it is clear, from (6), that it is of the other intervals. Now Tn can be
1 2
Ao/C + ~bfj_l(T )
T,,_j=fi(T ) = ~ln 1 +R fj_l(T) + D(H) -bE~-ifk(T)
)] ' j = 1.... ,n - 1.
Then T. is the solution of the equation n--1
F(T)=
E fj(T) -H= j=0
0.
(al)
It can be shown by induction that the function fj(T), j = 1. . . . , n - 1, is increasing in T. Observe that
f l ( T ) is increasing in T. Therefore standard search procedures can be used to solve (11) for Tn. Once Tn is found one can use (6) to find all T/s and consequently all t/s. So far we assumed that n is known. To find the optimal value of n, we repeat the scheme of finding the length of the replenishment periods for increasing values of n (n = 1, 2 . . . ). We stop when the total variable cost starts increasing. We can now outline the complete procedure for finding the optimal (T1, T 2 , . . . , T~).
Algorithm Step 1. Start with n = 1. Step 2. Find the optimal replenishment schedule (n, T 1, T 2 .... , Tn) by solving equation (11) for Tn and then using (6) to determine the remaining T/s.
Step 3. Compute the total variable cost using (2). Step 4. Increase n by 1 and repeat Step 2 and 3. Step 5. Repeat Steps 2 - 4 until the cost increases for the first time. The optimal value of n is the one corresponding to the minimal cost. This value of n and the associated T/s form the optimal replenishment schedule. Note that a similar procedure can be used for decreasing market demands (b < 0) using the recursive equation (7) instead of (6). A more efficient way of searching for the optimal value of n may be to start close to the optimal n using some approximation. Then we find the optimal value of n above or below this value. A simple formula for an approximate value of n is
n=[ l °rnl
M,4. Hariga, M. Ben-Daya ~European Journal of Operational Research 89 (1996) 313-325
319
Table 1 Computational results for R = 0.2, with A 0 = 2, C O= 1, h 0 = 0.32, H = 10
D(t)= 10+ t
D(t)= 1 0 - t
(n * = 14, K * = 82.12)
(n * = 7, K * = 44.75)
TI = 0.831 Tz = 0.804 T3 = 0.780 T4 0.758 T5 = 0.740 T6 = 0.723 T7 = 0.708
T1 = 0.910 T2 = 0.964 T3 = 1.035 T4 = 1.131 T5 = 1.278 T6 = 1.547 T7 = 3.135
=
Ts = 0.695 T9 = 0.683 T10 = 0.672 Tll = 0.663 Tt2 = 0.655 7"i3= 0.647 T14 = 0.641
where T * is the economic order interval using an equivalent constant demand equal to the total demand over (0, H ) divided by H and a unit holding cost equal to (h 0 + RC0); [x] is the smallest integer greater t h a n or e q u a l to x. B e i n g t h e s u m o f a h o l d i n g c o s t w h i c h is d e c r e a s i n g i n n a n d a n o r d e r i n g c o s t w h i c h is i n c r e a s i n g i n n , t h e t o t a l c o s t f u n c t i o n is c o n v e x i n n . T h e r e f o r e t h e m i n i m u m o f t h e c o s t f u n c t i o n is g l o b a l a n d c a n b e determined efficiently.
3.3. N u m e r i c a l examples
We illustrate the above procedure using the following examples for both growing and declining markets. The value of the parameters for the first example are:
D ( t ) = 10 + t , H =
10, A 0 = 2, h 0 = 0 . 3 2 , C O = 1.
F o r t h e s e c o n d e x a m p l e t h e d e m a n d is g i v e n b y D ( t ) = 10 - t a n d t h e r e m a i n i n g p a r a m e t e r s as t h e s a m e a s f o r t h e f i r s t e x a m p l e . B o t h p r o b l e m s a r e s o l v e d f o r R = - 0 . 2 a n d R = 0.2. T a b l e s 1 a n d 2 summarize the computational results using the algorithm discussed above. For the case of increasing demand
w i t h n o e f f e c t o f i n f l a t i o n ( R = 0), t h e o p t i m a l r e p l e n i s h m e n t
s c h e d u l e h a s 11 o r d e r s a n d a t o t a l c o s t o f 193.84. T h e o p t i m a l i n v e n t o r y p o l i c y f o r t h e s e c o n d p r o b l e m w i t h o u t i n f l a t i o n is t o p l a c e 6 o r d e r s d u r i n g t h e p l a n n i n g h o r i z o n w i t h a t o t a l c o s t o f 72.92. We run some experiments to study the effect of varying the discount rate net of inflation on the o p t i m a l n u m b e r of o r d e r s a n d total cost for t h e two e x a m p l e s . T h e r e s u l t s i n d i c a t e t h a t as R gets s m a l l e r t h e s o l u t i o n g e n e r a t e d b y o u r a l g o r i t h m c o n v e r g e s t o t h e n o i n f l a t i o n s o l u t i o n ( R = 0) a s e x p e c t e d . W e also observed that, for both examples, the optimal number of orders increases with R and the total cost decreases.
Table 2 Computational results for R = -0.2, with A 0 = 2, C o = 1, h 0 = 0.32, H = 10 D ( t ) = 10+ t (n * = 6, K * = 611.438)
D(t)= 10- t (n * = 3, K ~"= 137.21)
T~ = 2.069 T2 = 1.831 T3 = 1.673 T4 = 1.557 T5 = 1.470 Z 6 = 1.400
T1 = 2.129 T2 = 2.580 T3 = 5.291
M.A. Hariga, M. Ben-Daya/European Journal of OperationalResearch 89 (1996)313-325
320
4. A time-varying lot sizing model with shortages In this section, we develop an optimal time-varying lot-sizing model with the relaxed assumption that shortages are allowed and are completely backordered. We also assume that an extra order is placed at the end of the time horizon (t = H ) just to cover up the shortages during the last replenishment cycle. Moreover, each unsatisfied unit of demand results in a cost of ~'0 per unit of time. The lot-size received at the beginning of each replenishment cycle is partly used to satisfy the accumulated backorders in the previous cycle. The amount of inventory left after clearing the backorder position is depleted at time sy (sy > ty_l). The length of time until the inventory level reduces to zero is Iy that is, Iy = s y - ty_l. The demand from time sy to ty is satisfied when the lot size of ( j + 1)st replenishment is received. The length of time over which shortages are incurred is Sy, that is, Sy = ty - sy, These times are illustrated in Fig. 1 in case of increasing demand.
4.1. Preliminaries Consider the j-th replenishment cycle ( j = 1, 2 . . . . . n). It is clear from Fig. 1 that the inventory level, and the shortage level S(t), are given by
I(t),
I ( t ) = ftSJD(t) dt,
tj_x <-t ~ s j,
S(t)
si
sy
dt,
The present value of the inventory holding cost as to time zero is HC(ty_ 1, sy) = h ° f sy (e - R g - ' - e - R t ) D ( t ) dt,
R Jty_l
j = 1, 2 , . . . n.
Inventorylevel
,o= o
=
T2 v
Fig. 1. Variation of the inventorylevel as function of time for increasing demand.
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
321
The present value of the shortage cost is ftt - R t
SC(sj, t j ) = .,oj~t e
S(t) dt
or, after an integration by parts, ~T 0
sc(sj, tj)=
ftJ(e-gt-e-gtOD(t )dt, st
j = 1, 2 . . . . . n.
The present value of the purchase cost as to time zero is
=Cofst e-gtt-~D(t)
P(sj_ 1, sj)
dt,
j=l,
2,...,n,
sj-1
where s o -- 0. Therefore, the present value of the total cost of the system during the j-th cycle is st
h.
s~
gJ=A°e-Rtt-l+C°e-Rtj-lfst - lD(t) dt + R ~t-1 (e -R%' - - e - R t ) D ( t ) dt
Jt•
f't(e-"t-e-"'OD(t) dt. S]
The present value of the total cost during the entire planning horizon is n
K(n) = E Kj+KH,
(13)
y=l where K H = A 0 e -RH +Coe-RHfs~D(t ) d t
(14)
represents the present value of the replenishment and purchasing costs of the extra order placed at time H. Substituting K i and rearranging terms, the present value of the total cost can be written as
g ( n ) = A o e - R H + ~__ e -mj-1 A o + -RJtt_D(t) dt j=l
~e-roD(t)
-~re-Rt~[t'D(t)R Jsi dt + ~r--P-°% R .e
D ( t ) dt
(15)
where h = h o + CoR and ~r = ~'o - CoR. For fixed n, the partial derivative of K(n) with respect to each sj gives
- - eh h+~"
-Rtt-~ +
'rr e_Rt j = e-RSt, h+~"
j = 1, 2 . . . . . n.
In the next lemma, we derive the optimality conditions of the s/s in case of no effect of inflation. Lemma 3. In case of no effect of inflation, (16) becomes
ho( sj - tj_ 1) - 7%( tj - si) = O.
(16)
MM. Hariga, M. Ben-Daya ~European Journal of Operational Research 89 (1996) 313-325
322
Proof. Recall that h = h o + CoR and 7r = ~'0 - CoR. T h e n (16) can be rewritten as
ho(e-Rtt-1 _ e-R*O C ° ( e - R t j - I -- e - R t o +
+ 7r°
R
(e-Rt~ _ e-Rs) R = O.
Clearly, as R goes to zero, the last e q u a t i o n b e c o m e s ho(s i - t j _ a) -7ro(s j - t ) = N o t e that (16) can also be rewritten as h
7r
h+~"
h+~"
--eRT:+--=eRSt,
j=l,
O.
2 . . . . . n,
(17)
or 'W
1+ ~-(1-
e - R S 0 = eRlt,
j=1,2
. . . . . n.
[]
(lS)
L e m m a 4. A feasible replenishment schedule that minimizes K(n) exists if
- h o / C o < R < zro/C o.
(19)
Proof. T h e p r o o f is trivial by noting that a r e p l e n i s h m e n t schedule g e n e r a t e d by solving (17) is feasible if 1 < eRSt < e RT~ for positive R , and e RT~ < e Rst < 1 for negative R. Observe that condition (19) simply states that h and ~- have to be positive for any value of R.
[]
Next, setting the partial derivation of K(n) with respect to each tj equal to zero yields
-RAo-hf*t+lD(t) dt +rrf"D(t) d t = 0 , tj
j= 1,2 ..... n-
1.
(20)
sj
Clearly as R goes to zero, (20) b e c o m e s
hf*J+lD(t) d t = r r f t J D ( t ) dt, .t.
d~
ti
j=1,2 .... ,n-
1.
sj
L e m m a 5. I f shortages are not allowed, then (20) reduces to (4). Proof. If shortages are not permitted, t h e n ~-o --+ ~ implies, using (16), that sj ~ tj. Next, observe that (20) can be rewritten as
-Rao-hfS~+'D(t)
dt q- (h + r : ) f " D ( t )
st
dt=0.
(21)
st
Moreover, using (17), we have h + 7r = h ( e R ~ -
1)/(e Rs, - 1).
(22)
Substituting (22) into (21), we obtain
-*t4o - h fS'+'D(t) dt + h(e~, "- 1) ft'D(t) dt//(e R(tt-sj) st
-
-
1) = 0,
sj
which, as sj ~ tj and si+ 1 ~ tj+ 1, reduces to egTJ -- 1
-RAo-hlt'+'D(t)-a. " dt + h - - D ( t j ) = 0 . ti
R
(23)
M.A. Hariga, M. Ben-Daya /European Journal of Operational Research 89 (1996) 313-325 F i n a l l y , (4) c a n b e o b t a i n e d Now substituting
after dividing (23) by h.
323
[].
D ( t ) = a + bt i n t o (20), w e g e t
rr . l_~bSy2 rrD(t,)Sy + hZj+l[ D(ty) + ½bZj+, ] + R A 0 = 0 ' _
j = 1, 2 . . . . , n .
(24)
S o l v i n g ( 2 4 ) as a s e c o n d p o l y n o m i a l f o r Sy, w e o b t a i n
Sy=(rrD(tj)+_x/A)/(rrb), depending
on the signs of b and
j=1,2
(25)
..... n-l,
(rrD(tj) - v ~ ) , w h e r e
A = T r 2 D 2 ( t j ) - 2 r r b { h l j + l [ D ( t j ) + ½bIj+l] + R A 0 } . 4.2. Solution procedure N o t e t h a t if T n is k n o w n , t h e n S n c a n b e e a s i l y o b t a i n e d f r o m (17). T h e n , I n c a n b e d e t e r m i n e d b y s u b s t i t u t i n g t h e v a l u e o f S n i n t o (18). T h i s i n t u r n l e a d s t o S n _ 1 b y u s i n g (25). T h e r e f o r e , b y r e p e a t i n g t h i s p r o c e s s all t h e T / s , I/s, a n d S / s c a n b e d e t e r m i n e d . H o w e v e r , t h e v a l u e o f T n s h o u l d l e a d t o a f e a s i b l e s c h e d u l e ; t h a t is all T / s m u s t s u m t o H . A o n e - d i m e n s i o n a l s e a r c h t e c h n i q u e b a s e d o n t h e bisection method can be used for this purpose. The optimal number of replenishments n can be found u s i n g a s c h e m e s i m i l a r t o t h a t d i s c u s s e d i n S e c t i o n 3.2.
4.3. Numerical examples T h e s o l u t i o n p r o c e d u r e o u t l i n e d a b o v e is n o w i l l u s t r a t e d u s i n g t h e s a m e f o u r e x a m p l e s s o l v e d i n S e c t i o n 3. W e f u r t h e r a s s u m e a u n i t s h o r t a g e o f 0.76 p e r u n i t o f t i m e ; t h a t is, ~ 0 = 0.76. T h e r e f o r e , t h e f e a s i b l e v a l u e s f o r R a r e i n t h e i n t e r v a l ( - 0 . 3 2 , 0.76). T a b l e s 3 a n d 4 s u m m a r i z e t h e r e s u l t s o b t a i n e d b y solving the four example problems. I n c a s e o f n o - i n f l a t i o n ( R = 0), t h e s e t w o e x a m p l e p r o b l e m s ( D ( t ) = 10 + t, a n d D ( t ) = 10 - t ) w e r e s o l v e d u s i n g H a r i g a ' s o p t i m a l s o l u t i o n p r o c e d u r e [10]. F o r i n c r e a s i n g d e m a n d t h e o p t i m a l s o l u t i o n is t o p l a c e 9 o r d e r s a t a c o s t o f 188.69. F o r d e c r e a s i n g d e m a n d t h e o p t i m a l r e p l e n i s h m e n t s c h e d u l e h a s 5 o r d e r s w i t h a t o t a l c o s t o f 71.82.
Table 3 Computational results for R = 0.2, with A 0 = 2, C o = 1, h 0 = 0.32, ~'0 = 0.76, H = 10
D(t) = 10 + t
D(t) = 1 0 - t
( N * = 10, K * = 75.63 (74.18))
( N * = 6, K * = 40.76 (40.33))
T 1 = 1.183 T2 = 1.124 T 3 = 1.075 T4 = 1.033 Ts = 0.998 T6 = 0.967 T7 = 0.939 Ts = 0.915 T9 = 0.893 Ta0 = 0,873
TI = T2 = T3 = T4 = T5 = T6 =
11 = 0.578 I 2 = 0.551 I 3 = 0.528 /4 = 0.509 I s = 0.493 16 = 0.478 /7 = 0.465 I 8 = 0.454 19 = 0.443 110 = 0.434
1.220 1.300 1.406 1.560 1.830 2.684
11 = 12 = 13 = 14 = 15 = 16 =
0.595 0.632 0.679 0.748 0.865 1.213
M.A. Hariga, M. Ben-Daya ~European Journal of Operational Research 89 (1996) 313-325
324
Table 4 Computational results for R = -0.2, with D(t) = 10 + t (N * = 6, K * = 621.25 (585.64)) T1 = T2 = T3 = T4 = T5 = T6 =
1.976 1.804 1.681 1.585 1.509 1.445
Ii = I2 = 13 = I4 = Is = 16 =
A 2 = 2 , C O = 1, h 0 =
0.32, ~'0 = 0.76, H = 10 D(t) = 1 0 - t (N * = 3, K * = 151.28 (135.99))
1.791 1.633 1.519 1.432 1.362 1.304
7"i = 2.181 T2 = 2.660 T3 = 5.159
11 = 1.980 I 2 = 2.426 13 = 4.788
T h e c o m p a r i s o n o f T a b l e 1 with T a b l e 3 r e v e a l s t h a t t h e fact o f allowing s h o r t a g e s d e c r e a s e d t h e n u m b e r o f r e p l e n i s h m e n t s a n d t h e t o t a l cost for positive d i s c o u n t r a t e n e t o f inflation. By c o m p a r i n g T a b l e s 2 a n d 4, n o t i c e t h e u n e x p e c t e d i n c r e a s e in t h e t o t a l cost. This is d u e to t h e cost o f t h e e x t r a o r d e r p l a c e d at t i m e H . A s a m a t t e r o f fact, t h e t o t a l cost excluding this e x t r a cost, i n d i c a t e d b e t w e e n p a r e n t h e s i s in T a b l e s 3 a n d 4, is l o w e r t h a n t h e cost w i t h o u t s h o r t a g e s . It c a n b e s e e n f r o m (13) a n d (14) that, for R < 0 a n d l a r g e H , t h e cost o f t h e e x t r a o r d e r , K H, h a s a significant effect o n t h e p r e s e n t value o f t o t a l cost. U s i n g t h e s e e x a m p l e , we s t u d i e d t h e effect o f c h a n g i n g t h e d i s c o u n t r a t e n e t o f inflation on t h e o p t i m a l n u m b e r o f o r d e r s a n d t o t a l costs. T h e results o b t a i n e d i n d i c a t e t h a t m o r e o r d e r s a r e p l a c e d for s m a l l e r v a l u e s o f R. M o r e o v e r , as it is e x p e c t e d t h e t o t a l cost d e c r e a s e s as t h e d i s c o u n t r a t e n e t o f inflation decreases.
5. Conclusion I n this p a p e r w e d e v e l o p e d two o p t i m a l s o l u t i o n p r o c e d u r e s for a d e t e r m i n i s t i c i n v e n t o r y p r o b l e m with a l i n e a r t r e n d in d e m a n d a n d t a k i n g into a c c o u n t i n f l a t i o n a n d t i m e v a l u e o f m o n e y . T h e first s o l u t i o n p r o c e d u r e g e n e r a t e s o p t i m a l r e p l e n i s h m e n t s c h e d u l e s w h e n s h o r t a g e s a r e n o t allowed. T h e second solution method provides optimal ordering policies when shortages are permitted and are c o m p l e t e l y b a c k o r d e r e d . B o t h s o l u t i o n p r o c e d u r e s d o n o t p u t any r e s t r i c t i o n o n t h e l e n g t h o f t h e r e p l e n i s h m e n t cycles. T o t h e b e s t o f o u r k n o w l e d g e t h e s e a r e t h e first o p t i m a l p r o c e d u r e s for this t y p e o f inventory problems. T h e d e v e l o p m e n t o f a p p r o x i m a t e s o l u t i o n p r o c e d u r e s for t h e i n v e n t o r y r e p l e n i s h m e n t p r o b l e m with l i n e a r t r e n d in d e m a n d u n d e r i n f l a t i o n a r y c o n d i t i o n s is a p o s s i b l e t o p i c for f u t u r e r e s e a r c h . T h e s i m p l e s t e x t e n s i o n a l o n g this line o f r e s e a r c h is to use t h e E O Q m o d e l with t h e d e m a n d p e r u n i t o f t i m e e q u a l to t h e t o t a l d e m a n d over t h e interval (0, H ) d i v i d e d by H , h = h o + CoR, a n d ¢r = ~'0 - CoR. I n this case, t h e n u m b e r o f o r d e r s p l a c e d over t h e p l a n n i n g h o r i z o n is given by n = v/H 2h'n'(a + ½ b H ) / ( 2 A o ( h
+ ~r)),
a n d t h e l e n g t h o f t h e r e p l e n i s h m e n t cycles a r e Tj=T~
j=
1, 2 , . . . , n -
1,
T,,=H-(n-1)T,
where T= ~2A0(h + rr)/(h'n'(a + ½bH)) .
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
325
Another approximate solution procedure can be obtained by extending Silver's Heuristic [17] to accommodate for inflation and time value money.
References [1] Aggrawal, S.C., "Purchase inventory decision models for inflationary conditions", Interfaces 11 (1981) 18-23. [2] Banks, J., Mangiameli, P.M., and Schwarzbach, H., "Static inventory models and inflationary increases", The Engineering Economist 26/2 (1982) 91-112. [3] Bierman, H., Thomas, J., "Inventory decision under inflationary conditions", Decision Sciences 8 (1977) 151-155. [4] Buzacott, J.A., "Economic order quantity with inflation", Operational Research Quarterly 26 (1975) 553-558. [5] Chandra, M.J., and Bahner, M.L., "The effects of inflation and the time value of money on some inventory systems", Intematioual Journal of Production Research 23/4 (1985) 723-730. [6] Datta, T.K., and Pal, A.K., "Effects of inflation and time-value of money on an inventory model with linear time-dependent demand rate and shortages", European Journal of Operational Research 52 (1991) 326-333. [7] Donaldson, W.A., "Inventory replenishment policy for a linear trend in demand: An analytical approach", Operational Research Quarterly 28 (1970) 663-670. [8] Goyal, S.K., "A heuristic for replenishment of trended inventories considering shortages", Journal of the OperationalResearch Society 39 (1988) 885-887. [9] Guruani, C., "Economic analysis of inventory systems", International Journal of Production Research 21/2 (1983) 261-277. [10] Hariga, M., "The inventory replenishment problem with a linear trend in demand", Computers and Industrial Engineering 24/2 (1993) 143-150. [11] Misra, R.B., " A study of inflationary effects on inventory systems", Logistics Spectrum 9/3 (1975). [12] Misra, R.B., " A note on optimal inventory management under inflation", Naval Research Logistics Quarterly 26 (1979) 161-165. [13] Mitra, A., Cox, J.F., and Jesse, Jr., R.R., " A note on determining order quantities with a linear trend in demand", Journal of the Operational Research Society 35 (1984) 141-144. [14] Murdeshwar, T.M., "Inventory replenishment policy for linearly increasing demand considering shortages-an optimal solution", Journal of the Operational Research Society 39 (1988) 687-692. [15] Phelps, R.I., "Optimal inventory rule for a linear trend in demand with constant replenishment period", Journal of the Operational Research Society 31 (1980) 439-442. [16] Ritchie, E., "The EOQ for linear increasing demand; a simple optimal solution", Journal of the Operational Research Society 35 (1984) 949-952. [17] Silver, E.A., " A simple inventory replenishment decision rule for a linear trend in demand", Journal of the Operational Research Society 30 (1979) 71-75.
European Journal of Operational Research 89 (1996) 313-325
Theory and Methodology
Optimal time varying lot-sizing models under inflationary conditions M . A . H a r i g a a, *, M . B e n - D a y a
b
a College of Engineering, King Saucl University, P.O. Box 800, Riyadh 11421, Saudi Arabia b Systems Engineering Department, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia Received August 1993; revised August 1994
Abstract
This paper deals with the inventory replenishment problem over a fixed planning horizon for items with linearly time-varying demand and under inflationary conditions. We develop models and optimal solution procedures with and without shortages. We do not put any restriction on the length of the replenishment cycles making the proposed methods the first optimal solution procedure for this problem. Using four examples, we illustrate the proposed solution procedures and study the effect of changing the inflation and discount rates on the optimal replenishment schedules. Keywords: Inventory; Inflation; S h o r t a g e s
1. Introduction
The classical E O Q formula assumes that all relevant costs are constant. It also assumes a constant demand rate over an infinite planning horizon. The latter assumption is only valid during the saturation phase of the product life cycle and for finite periods of time. Furthermore, many items such as electronics components, fashionable clothes, and domestic goods observe periods of rapid sales increase after gaining consumers' acceptance. At the same time sales for other products may drop drastically due to the introduction of new competitive products or to the change in consumers' preferences. As to the first assumption, since many countries experience high annual inflation rates, it is no longer reasonable to ignore the effect of inflation and time value of money. Clearly, there is need for the reformulation of the optimal inventory management policies taking into account these different factors. The inventory problem with linear trend in demand has been the subject of several studies since the exact procedure of Donaldson [7]. Heuristic methods have been introduced by Silver [17], Mitra et al. [13], and Phelps [15] among others. Recently, Hariga [10] has developed a new optimal iterative procedure which can be applied to both linearly increasing and decreasing demands. However, all these
* Corresponding author.
0377-2217/96/$15.00 © 1996 Elsevier Science B.V. All rights reserved SSDI 0 3 7 7 - 2 2 1 7 ( 9 4 ) 0 0 2 5 6 - 8
314
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
exact and approximate procedures have not recognized the time value of money and inflation in determining the replenishment schedule. Among the first authors considering inventory problems with inflation, Buzacott [4] has studied the E O Q model under different pricing policies. Following the work of Buzacott, several attempts (Bierman and Thomas [3], Misra [11,12], Aggrawal [1], Chandra and Bahner [5], Gurnani [9], and Banks et al. [2]) have been made to extend his approach to various inflationary situations. However, one common characteristic of all these studies is the assumption that the market demand is time-invariant. In a recent paper, Datta and Pal [6] have studied the effect of inflation and time-value of money on an inventory model with linear trend in demand and shortages. They have based their model on the restrictive assumption of equal replenishment cycle in order to minimize the present value of the total inventory. In this paper, we follow the discounted cash flow approach to develop two time varying lot sizing models for the inventory replenishment problem with Iinear trend in demand taking into account the effects of inflation and time value of money. We first introduce a model with infinite shortage cost. We then reformulate and propose a solution procedure for this model under the assumption that shortages are permitted and are fully backordered. In both models, we do not put any restriction on the length of the replenishment cycles. We do not consider internal and external inflation rates but rather use one inflation rate for all costs. In [5], Chandra and Bahner have pointed out that it is always possible to assume that these two rates are equal and to combine the internal and external elements of each type of cost, if appropriate. Moreover, the introduction of these two rates would not alter the analysis presented in this paper, but would rather add to the complexity of the derived mathematical expressions. The proposed models extend the ones in [1-5,9,11,12] to the case of time varying demand. Our models also relax the restrictive assumptions of equal replenishment cycles and equal positive inventory periods in each cycle considered by Datta and Pal. We extend the model proposed by Hariga [10] by considering the effect of inflation and time value of money. This makes our models the first optimal solution procedures for the inventory replenishment problem with finite and infinite shortage cost and linear trend in demand under inflationary conditions. To the best of our knowledge this problem has not been addressed so far. This paper is organized as follows. In the next section, we state our assumptions and notations. The optimal replenishment model without shortages and its solution procedure are presented in Section 3. Four numerical examples are also presented to illustrate the solution procedure. In Section 4, we develop and solve the model with finite shortage cost. Finally, Section 5 concludes the paper.
2. Assumptions and notations We make the following assumptions: A single item is held in stock over a known and finite horizon which is H years long. Replenishment occurs instantaneously at an infinite rate. Shortages are not allowed. No inventory is carried at time zero and at the end of the time horizon. The demand rate is known and varies linearly with time, namely D ( t ) = a + bt, where a >~ 0, and b is an arbitrary constant. 6. The problem parameters a, b, and H must satisfy D ( H ) -~ a + b H >~ O. The following notation will be used: A0: Fixed replenishment cost per lot at time zero. Co: Unit cost at time zero. h0: Unit inventory holding cost per unit time excluding the cost of capital. 1. 2. 3. 4. 5.
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
315
The inflation rate. The discount rate representing the value of money. Discount rate net of inflation, that is R = r - i. R: Number of replenishment periods during the time horizon H. n: Total time elapsed up to and including the j-th replenishment period, j = 1 , . . . , n, where t/ t o = O a n d tn = H . Length of the j-th replenishment period, j -- 1 , . . . , n, where T/= tj - ty_ 1T]: I(t): Inventory level at time t, 0 ~
r:
3. A time-varying lot sizing model without shortages
3.1. Preliminaries The inventory level at any time t during the replenishment cycle is solution of the following differential equation with boundary condition I(tj) = O.
(dl(t)/dt)=-D(t),
tj_l<_t<_tj,
t0=0,
tn=H,
or
I(t) = fttJD(u) du,
ti_ 1 < t <_tj.
Next, we determine the expression of the present value of the various costs involved for a cycle. The present value of the inventory holding cost incurred during the j-th cycle is H C ( t j _ l , t j ) = h o f ti e ( i - r ) t l ( t ) d t = h o f t' e - R t ( f t ' D ( u ) d u ) a t . tj_ 1 tj_ 1 \ t Integrating by parts, we obtain HC(t/_l, tj)= -~
ftlJl(e-Rt~-l--e--Rt)D(t)._ dt.
(1)
Note that as R --+ O, it can be shown that (1) reduces to
hof tj ( t - t j _ l ) D ( t )
dt
~j- 1 which corresponds to the case of no inflation (Hariga [10]). The present value of the total variable cost incurred during the j-th cycle is
-1 + Coe-Rt-,Q(ti_l,
+ nc(tj_,
tj)
M.A. Hariga, M. Ben-Daya ~European Journal of Operational Research 89 (1996) 313-325
316
where Q(tj_ 1, tj) is the lot received at time tj_ 1 to satisfy the demand of the j-th cycle, that is,
The present value of the total cost over the time horizon is n
n
h0
K ( n ) = A 0 ~ e -R`J-a + C o Y'~ e-ms-'ftts D ( t ) dt + j=l
j=l
j-I
-
R j=l
/,
e -Rtj-' + C E e-RtJ-i tj D ( t ) dt -
=A 0 j=l
j=l
n
Y'. fts (e_ms_~_ e _ m ) D ( t ) dt
j-1
tj-1
- fo "e -RtD(t)
at
(2)
where C = C O+ ho/R. In the next lemma, we show that under a certain condition, it is optimal to place only one order during the planning horizon. L e m m a 1. I f the discount rate net of inflation is negative and less than the inventory carrying cost rate
(excluding the cost of capital invested in inventories), then the present value of the total cost is minimized by placing only one order during the planning horizon. Proof. Note that for negative discount rate net of inflation, C > 0 if RC o < - h o. Since h 0 = ioC o where i 0 is the inventory carrying cost rate excluding the cost of capital, the above inequality becomes R < - i 0. In this case it is easy to see that = A 0 e - m l + C(e -m' - 1)
K(2) -K(1)
(t) dt,
which is clearly a positive number. H e n c e K(2) > K(1).
[]
For fixed n, leaving out the constant term, the problem becomes:
tj-1
j=l
subject to
tj>tj_l,
t0=0,
t,=H.
In order to determine tj, j = 1 , . . . , n - 1, we proceed as follows. Taking the derivative of the objective function of the above optimization problem with respect to tj and equating to zero, we get
Re-l%(Ao+CftS+xD(t) t ""tj
dt} = C D ( t j ) ( e - R t i - l - e - R t J ) ,
j= 1,2 ..... n-1,
(3)
]
which can be written as
O(ti)[(eR
- 1)/R]
-Ao/C= f"+'D(t) dt. tj
Observe that as R --* 0, (4) becomes
D ( t i )-T i = l t~S + i D ( t ) d- t , ts
j=l
.... ,n-l,
(4)
M.A. Hariga, M. Ben-Daya/ European Journal of Operational Research 89 (1996) 313-325
317
which corresponds to the optimality conditions in the case of no inflation (Hariga [10]). Now since D ( t ) = a + bt, it follows that
ftJ+lD(t) d t =
1 2 T j + I D ( t i ) + ~bTj+ 1.
tj
Substituting in (4), we get 1 2 = Ty+lD(ti) + gbTj+l,
O ( t y ) [ ( e RTj - 1 ) / R ] - A o / C
j = 1,...,n-
1,
(5)
12)]
or
A o / C + ~bTy+ 1 + D(tj)
'
(6)
j = 1 . . . . . n - 1.
An alternative formula can be obtained by solving (5) as a second order polynomial for Ty+l, leading to Tj+I= ~
(tj.)Z+2b D ( t j )
-R-
-
-D(tj)
,
j= 1..... n-
1.
(7)
Note that it is enough to know Tn to determine all other T/s using (6). Alternatively, if T 1 is known, one can determine all other T/s using (7). It has been shown in [10] that when R = 0 and b > 0,
(8)
> Tn.
TI> T2> ""
This is not true in general here. The following lemma identifies a case where (8) still holds. 1.emma 2. I f b > 0, R < 0 and C > 0, then T 1 > Z 2 > . . . > T n. Proof. Using the mean value theorem,. ftt/+lD(t) dt = T~+ 1D(0) where tj < 0 < tj+ 1. Since b > 0, it follows • . J a . that D ( t ) < D(O). Multiplying both sldes by Tj+I, we obtain
rj+lO(tj) <
rj+lV(0
) .
(9)
Also, for R < 0 it can be shown that > (e
(10)
- O/R.
From (5) we have O ( t j ) [ ( e RTj - 1 ) / R ] = A o / C
+ Ti+ID(0 ).
Using (9) and (10), we obtain [ e RTj - -
D(tj)Tj>D(tj)[
1
-~
Hence (Tj - Tj+I)D(tj) > A o / C
Ao
Ao
=--~-+ Tj+ID(O)>~-+D(tj)Tj+
> 0 or Tj > Tj+p
[]
1.
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
318
3.2. Solution procedure Recall that the problem is to find the number of orders and their replenishment epochs tl, t2,... , t~_ 1 which solve the following problem. Minimize
j ~ l e -nt~-I A°+CftJtj_ 1D(t) dt
subject to
t1 > t j_l,
t0=0, t~=H.
We propose a solution procedure for finding the optimal n and In particular, we do not consider the usual [6] restriction that method rely on the development of the previous section. For enough to find T~ and then use (6) to determine the length determined as follows. Let
T. = T = f o ( T ),
1I(
t/s without any additional assumptions. the length of the cycles are equal. Our fixed n, it is clear, from (6), that it is of the other intervals. Now Tn can be
1 2
Ao/C + ~bfj_l(T )
T,,_j=fi(T ) = ~ln 1 +R fj_l(T) + D(H) -bE~-ifk(T)
)] ' j = 1.... ,n - 1.
Then T. is the solution of the equation n--1
F(T)=
E fj(T) -H= j=0
0.
(al)
It can be shown by induction that the function fj(T), j = 1. . . . , n - 1, is increasing in T. Observe that
f l ( T ) is increasing in T. Therefore standard search procedures can be used to solve (11) for Tn. Once Tn is found one can use (6) to find all T/s and consequently all t/s. So far we assumed that n is known. To find the optimal value of n, we repeat the scheme of finding the length of the replenishment periods for increasing values of n (n = 1, 2 . . . ). We stop when the total variable cost starts increasing. We can now outline the complete procedure for finding the optimal (T1, T 2 , . . . , T~).
Algorithm Step 1. Start with n = 1. Step 2. Find the optimal replenishment schedule (n, T 1, T 2 .... , Tn) by solving equation (11) for Tn and then using (6) to determine the remaining T/s.
Step 3. Compute the total variable cost using (2). Step 4. Increase n by 1 and repeat Step 2 and 3. Step 5. Repeat Steps 2 - 4 until the cost increases for the first time. The optimal value of n is the one corresponding to the minimal cost. This value of n and the associated T/s form the optimal replenishment schedule. Note that a similar procedure can be used for decreasing market demands (b < 0) using the recursive equation (7) instead of (6). A more efficient way of searching for the optimal value of n may be to start close to the optimal n using some approximation. Then we find the optimal value of n above or below this value. A simple formula for an approximate value of n is
n=[ l °rnl
M,4. Hariga, M. Ben-Daya ~European Journal of Operational Research 89 (1996) 313-325
319
Table 1 Computational results for R = 0.2, with A 0 = 2, C O= 1, h 0 = 0.32, H = 10
D(t)= 10+ t
D(t)= 1 0 - t
(n * = 14, K * = 82.12)
(n * = 7, K * = 44.75)
TI = 0.831 Tz = 0.804 T3 = 0.780 T4 0.758 T5 = 0.740 T6 = 0.723 T7 = 0.708
T1 = 0.910 T2 = 0.964 T3 = 1.035 T4 = 1.131 T5 = 1.278 T6 = 1.547 T7 = 3.135
=
Ts = 0.695 T9 = 0.683 T10 = 0.672 Tll = 0.663 Tt2 = 0.655 7"i3= 0.647 T14 = 0.641
where T * is the economic order interval using an equivalent constant demand equal to the total demand over (0, H ) divided by H and a unit holding cost equal to (h 0 + RC0); [x] is the smallest integer greater t h a n or e q u a l to x. B e i n g t h e s u m o f a h o l d i n g c o s t w h i c h is d e c r e a s i n g i n n a n d a n o r d e r i n g c o s t w h i c h is i n c r e a s i n g i n n , t h e t o t a l c o s t f u n c t i o n is c o n v e x i n n . T h e r e f o r e t h e m i n i m u m o f t h e c o s t f u n c t i o n is g l o b a l a n d c a n b e determined efficiently.
3.3. N u m e r i c a l examples
We illustrate the above procedure using the following examples for both growing and declining markets. The value of the parameters for the first example are:
D ( t ) = 10 + t , H =
10, A 0 = 2, h 0 = 0 . 3 2 , C O = 1.
F o r t h e s e c o n d e x a m p l e t h e d e m a n d is g i v e n b y D ( t ) = 10 - t a n d t h e r e m a i n i n g p a r a m e t e r s as t h e s a m e a s f o r t h e f i r s t e x a m p l e . B o t h p r o b l e m s a r e s o l v e d f o r R = - 0 . 2 a n d R = 0.2. T a b l e s 1 a n d 2 summarize the computational results using the algorithm discussed above. For the case of increasing demand
w i t h n o e f f e c t o f i n f l a t i o n ( R = 0), t h e o p t i m a l r e p l e n i s h m e n t
s c h e d u l e h a s 11 o r d e r s a n d a t o t a l c o s t o f 193.84. T h e o p t i m a l i n v e n t o r y p o l i c y f o r t h e s e c o n d p r o b l e m w i t h o u t i n f l a t i o n is t o p l a c e 6 o r d e r s d u r i n g t h e p l a n n i n g h o r i z o n w i t h a t o t a l c o s t o f 72.92. We run some experiments to study the effect of varying the discount rate net of inflation on the o p t i m a l n u m b e r of o r d e r s a n d total cost for t h e two e x a m p l e s . T h e r e s u l t s i n d i c a t e t h a t as R gets s m a l l e r t h e s o l u t i o n g e n e r a t e d b y o u r a l g o r i t h m c o n v e r g e s t o t h e n o i n f l a t i o n s o l u t i o n ( R = 0) a s e x p e c t e d . W e also observed that, for both examples, the optimal number of orders increases with R and the total cost decreases.
Table 2 Computational results for R = -0.2, with A 0 = 2, C o = 1, h 0 = 0.32, H = 10 D ( t ) = 10+ t (n * = 6, K * = 611.438)
D(t)= 10- t (n * = 3, K ~"= 137.21)
T~ = 2.069 T2 = 1.831 T3 = 1.673 T4 = 1.557 T5 = 1.470 Z 6 = 1.400
T1 = 2.129 T2 = 2.580 T3 = 5.291
M.A. Hariga, M. Ben-Daya/European Journal of OperationalResearch 89 (1996)313-325
320
4. A time-varying lot sizing model with shortages In this section, we develop an optimal time-varying lot-sizing model with the relaxed assumption that shortages are allowed and are completely backordered. We also assume that an extra order is placed at the end of the time horizon (t = H ) just to cover up the shortages during the last replenishment cycle. Moreover, each unsatisfied unit of demand results in a cost of ~'0 per unit of time. The lot-size received at the beginning of each replenishment cycle is partly used to satisfy the accumulated backorders in the previous cycle. The amount of inventory left after clearing the backorder position is depleted at time sy (sy > ty_l). The length of time until the inventory level reduces to zero is Iy that is, Iy = s y - ty_l. The demand from time sy to ty is satisfied when the lot size of ( j + 1)st replenishment is received. The length of time over which shortages are incurred is Sy, that is, Sy = ty - sy, These times are illustrated in Fig. 1 in case of increasing demand.
4.1. Preliminaries Consider the j-th replenishment cycle ( j = 1, 2 . . . . . n). It is clear from Fig. 1 that the inventory level, and the shortage level S(t), are given by
I(t),
I ( t ) = ftSJD(t) dt,
tj_x <-t ~ s j,
S(t)
si
sy
dt,
The present value of the inventory holding cost as to time zero is HC(ty_ 1, sy) = h ° f sy (e - R g - ' - e - R t ) D ( t ) dt,
R Jty_l
j = 1, 2 , . . . n.
Inventorylevel
,o= o
=
T2 v
Fig. 1. Variation of the inventorylevel as function of time for increasing demand.
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
321
The present value of the shortage cost is ftt - R t
SC(sj, t j ) = .,oj~t e
S(t) dt
or, after an integration by parts, ~T 0
sc(sj, tj)=
ftJ(e-gt-e-gtOD(t )dt, st
j = 1, 2 . . . . . n.
The present value of the purchase cost as to time zero is
=Cofst e-gtt-~D(t)
P(sj_ 1, sj)
dt,
j=l,
2,...,n,
sj-1
where s o -- 0. Therefore, the present value of the total cost of the system during the j-th cycle is st
h.
s~
gJ=A°e-Rtt-l+C°e-Rtj-lfst - lD(t) dt + R ~t-1 (e -R%' - - e - R t ) D ( t ) dt
Jt•
f't(e-"t-e-"'OD(t) dt. S]
The present value of the total cost during the entire planning horizon is n
K(n) = E Kj+KH,
(13)
y=l where K H = A 0 e -RH +Coe-RHfs~D(t ) d t
(14)
represents the present value of the replenishment and purchasing costs of the extra order placed at time H. Substituting K i and rearranging terms, the present value of the total cost can be written as
g ( n ) = A o e - R H + ~__ e -mj-1 A o + -RJtt_D(t) dt j=l
~e-roD(t)
-~re-Rt~[t'D(t)R Jsi dt + ~r--P-°% R .e
D ( t ) dt
(15)
where h = h o + CoR and ~r = ~'o - CoR. For fixed n, the partial derivative of K(n) with respect to each sj gives
- - eh h+~"
-Rtt-~ +
'rr e_Rt j = e-RSt, h+~"
j = 1, 2 . . . . . n.
In the next lemma, we derive the optimality conditions of the s/s in case of no effect of inflation. Lemma 3. In case of no effect of inflation, (16) becomes
ho( sj - tj_ 1) - 7%( tj - si) = O.
(16)
MM. Hariga, M. Ben-Daya ~European Journal of Operational Research 89 (1996) 313-325
322
Proof. Recall that h = h o + CoR and 7r = ~'0 - CoR. T h e n (16) can be rewritten as
ho(e-Rtt-1 _ e-R*O C ° ( e - R t j - I -- e - R t o +
+ 7r°
R
(e-Rt~ _ e-Rs) R = O.
Clearly, as R goes to zero, the last e q u a t i o n b e c o m e s ho(s i - t j _ a) -7ro(s j - t ) = N o t e that (16) can also be rewritten as h
7r
h+~"
h+~"
--eRT:+--=eRSt,
j=l,
O.
2 . . . . . n,
(17)
or 'W
1+ ~-(1-
e - R S 0 = eRlt,
j=1,2
. . . . . n.
[]
(lS)
L e m m a 4. A feasible replenishment schedule that minimizes K(n) exists if
- h o / C o < R < zro/C o.
(19)
Proof. T h e p r o o f is trivial by noting that a r e p l e n i s h m e n t schedule g e n e r a t e d by solving (17) is feasible if 1 < eRSt < e RT~ for positive R , and e RT~ < e Rst < 1 for negative R. Observe that condition (19) simply states that h and ~- have to be positive for any value of R.
[]
Next, setting the partial derivation of K(n) with respect to each tj equal to zero yields
-RAo-hf*t+lD(t) dt +rrf"D(t) d t = 0 , tj
j= 1,2 ..... n-
1.
(20)
sj
Clearly as R goes to zero, (20) b e c o m e s
hf*J+lD(t) d t = r r f t J D ( t ) dt, .t.
d~
ti
j=1,2 .... ,n-
1.
sj
L e m m a 5. I f shortages are not allowed, then (20) reduces to (4). Proof. If shortages are not permitted, t h e n ~-o --+ ~ implies, using (16), that sj ~ tj. Next, observe that (20) can be rewritten as
-Rao-hfS~+'D(t)
dt q- (h + r : ) f " D ( t )
st
dt=0.
(21)
st
Moreover, using (17), we have h + 7r = h ( e R ~ -
1)/(e Rs, - 1).
(22)
Substituting (22) into (21), we obtain
-*t4o - h fS'+'D(t) dt + h(e~, "- 1) ft'D(t) dt//(e R(tt-sj) st
-
-
1) = 0,
sj
which, as sj ~ tj and si+ 1 ~ tj+ 1, reduces to egTJ -- 1
-RAo-hlt'+'D(t)-a. " dt + h - - D ( t j ) = 0 . ti
R
(23)
M.A. Hariga, M. Ben-Daya /European Journal of Operational Research 89 (1996) 313-325 F i n a l l y , (4) c a n b e o b t a i n e d Now substituting
after dividing (23) by h.
323
[].
D ( t ) = a + bt i n t o (20), w e g e t
rr . l_~bSy2 rrD(t,)Sy + hZj+l[ D(ty) + ½bZj+, ] + R A 0 = 0 ' _
j = 1, 2 . . . . , n .
(24)
S o l v i n g ( 2 4 ) as a s e c o n d p o l y n o m i a l f o r Sy, w e o b t a i n
Sy=(rrD(tj)+_x/A)/(rrb), depending
on the signs of b and
j=1,2
(25)
..... n-l,
(rrD(tj) - v ~ ) , w h e r e
A = T r 2 D 2 ( t j ) - 2 r r b { h l j + l [ D ( t j ) + ½bIj+l] + R A 0 } . 4.2. Solution procedure N o t e t h a t if T n is k n o w n , t h e n S n c a n b e e a s i l y o b t a i n e d f r o m (17). T h e n , I n c a n b e d e t e r m i n e d b y s u b s t i t u t i n g t h e v a l u e o f S n i n t o (18). T h i s i n t u r n l e a d s t o S n _ 1 b y u s i n g (25). T h e r e f o r e , b y r e p e a t i n g t h i s p r o c e s s all t h e T / s , I/s, a n d S / s c a n b e d e t e r m i n e d . H o w e v e r , t h e v a l u e o f T n s h o u l d l e a d t o a f e a s i b l e s c h e d u l e ; t h a t is all T / s m u s t s u m t o H . A o n e - d i m e n s i o n a l s e a r c h t e c h n i q u e b a s e d o n t h e bisection method can be used for this purpose. The optimal number of replenishments n can be found u s i n g a s c h e m e s i m i l a r t o t h a t d i s c u s s e d i n S e c t i o n 3.2.
4.3. Numerical examples T h e s o l u t i o n p r o c e d u r e o u t l i n e d a b o v e is n o w i l l u s t r a t e d u s i n g t h e s a m e f o u r e x a m p l e s s o l v e d i n S e c t i o n 3. W e f u r t h e r a s s u m e a u n i t s h o r t a g e o f 0.76 p e r u n i t o f t i m e ; t h a t is, ~ 0 = 0.76. T h e r e f o r e , t h e f e a s i b l e v a l u e s f o r R a r e i n t h e i n t e r v a l ( - 0 . 3 2 , 0.76). T a b l e s 3 a n d 4 s u m m a r i z e t h e r e s u l t s o b t a i n e d b y solving the four example problems. I n c a s e o f n o - i n f l a t i o n ( R = 0), t h e s e t w o e x a m p l e p r o b l e m s ( D ( t ) = 10 + t, a n d D ( t ) = 10 - t ) w e r e s o l v e d u s i n g H a r i g a ' s o p t i m a l s o l u t i o n p r o c e d u r e [10]. F o r i n c r e a s i n g d e m a n d t h e o p t i m a l s o l u t i o n is t o p l a c e 9 o r d e r s a t a c o s t o f 188.69. F o r d e c r e a s i n g d e m a n d t h e o p t i m a l r e p l e n i s h m e n t s c h e d u l e h a s 5 o r d e r s w i t h a t o t a l c o s t o f 71.82.
Table 3 Computational results for R = 0.2, with A 0 = 2, C o = 1, h 0 = 0.32, ~'0 = 0.76, H = 10
D(t) = 10 + t
D(t) = 1 0 - t
( N * = 10, K * = 75.63 (74.18))
( N * = 6, K * = 40.76 (40.33))
T 1 = 1.183 T2 = 1.124 T 3 = 1.075 T4 = 1.033 Ts = 0.998 T6 = 0.967 T7 = 0.939 Ts = 0.915 T9 = 0.893 Ta0 = 0,873
TI = T2 = T3 = T4 = T5 = T6 =
11 = 0.578 I 2 = 0.551 I 3 = 0.528 /4 = 0.509 I s = 0.493 16 = 0.478 /7 = 0.465 I 8 = 0.454 19 = 0.443 110 = 0.434
1.220 1.300 1.406 1.560 1.830 2.684
11 = 12 = 13 = 14 = 15 = 16 =
0.595 0.632 0.679 0.748 0.865 1.213
M.A. Hariga, M. Ben-Daya ~European Journal of Operational Research 89 (1996) 313-325
324
Table 4 Computational results for R = -0.2, with D(t) = 10 + t (N * = 6, K * = 621.25 (585.64)) T1 = T2 = T3 = T4 = T5 = T6 =
1.976 1.804 1.681 1.585 1.509 1.445
Ii = I2 = 13 = I4 = Is = 16 =
A 2 = 2 , C O = 1, h 0 =
0.32, ~'0 = 0.76, H = 10 D(t) = 1 0 - t (N * = 3, K * = 151.28 (135.99))
1.791 1.633 1.519 1.432 1.362 1.304
7"i = 2.181 T2 = 2.660 T3 = 5.159
11 = 1.980 I 2 = 2.426 13 = 4.788
T h e c o m p a r i s o n o f T a b l e 1 with T a b l e 3 r e v e a l s t h a t t h e fact o f allowing s h o r t a g e s d e c r e a s e d t h e n u m b e r o f r e p l e n i s h m e n t s a n d t h e t o t a l cost for positive d i s c o u n t r a t e n e t o f inflation. By c o m p a r i n g T a b l e s 2 a n d 4, n o t i c e t h e u n e x p e c t e d i n c r e a s e in t h e t o t a l cost. This is d u e to t h e cost o f t h e e x t r a o r d e r p l a c e d at t i m e H . A s a m a t t e r o f fact, t h e t o t a l cost excluding this e x t r a cost, i n d i c a t e d b e t w e e n p a r e n t h e s i s in T a b l e s 3 a n d 4, is l o w e r t h a n t h e cost w i t h o u t s h o r t a g e s . It c a n b e s e e n f r o m (13) a n d (14) that, for R < 0 a n d l a r g e H , t h e cost o f t h e e x t r a o r d e r , K H, h a s a significant effect o n t h e p r e s e n t value o f t o t a l cost. U s i n g t h e s e e x a m p l e , we s t u d i e d t h e effect o f c h a n g i n g t h e d i s c o u n t r a t e n e t o f inflation on t h e o p t i m a l n u m b e r o f o r d e r s a n d t o t a l costs. T h e results o b t a i n e d i n d i c a t e t h a t m o r e o r d e r s a r e p l a c e d for s m a l l e r v a l u e s o f R. M o r e o v e r , as it is e x p e c t e d t h e t o t a l cost d e c r e a s e s as t h e d i s c o u n t r a t e n e t o f inflation decreases.
5. Conclusion I n this p a p e r w e d e v e l o p e d two o p t i m a l s o l u t i o n p r o c e d u r e s for a d e t e r m i n i s t i c i n v e n t o r y p r o b l e m with a l i n e a r t r e n d in d e m a n d a n d t a k i n g into a c c o u n t i n f l a t i o n a n d t i m e v a l u e o f m o n e y . T h e first s o l u t i o n p r o c e d u r e g e n e r a t e s o p t i m a l r e p l e n i s h m e n t s c h e d u l e s w h e n s h o r t a g e s a r e n o t allowed. T h e second solution method provides optimal ordering policies when shortages are permitted and are c o m p l e t e l y b a c k o r d e r e d . B o t h s o l u t i o n p r o c e d u r e s d o n o t p u t any r e s t r i c t i o n o n t h e l e n g t h o f t h e r e p l e n i s h m e n t cycles. T o t h e b e s t o f o u r k n o w l e d g e t h e s e a r e t h e first o p t i m a l p r o c e d u r e s for this t y p e o f inventory problems. T h e d e v e l o p m e n t o f a p p r o x i m a t e s o l u t i o n p r o c e d u r e s for t h e i n v e n t o r y r e p l e n i s h m e n t p r o b l e m with l i n e a r t r e n d in d e m a n d u n d e r i n f l a t i o n a r y c o n d i t i o n s is a p o s s i b l e t o p i c for f u t u r e r e s e a r c h . T h e s i m p l e s t e x t e n s i o n a l o n g this line o f r e s e a r c h is to use t h e E O Q m o d e l with t h e d e m a n d p e r u n i t o f t i m e e q u a l to t h e t o t a l d e m a n d over t h e interval (0, H ) d i v i d e d by H , h = h o + CoR, a n d ¢r = ~'0 - CoR. I n this case, t h e n u m b e r o f o r d e r s p l a c e d over t h e p l a n n i n g h o r i z o n is given by n = v/H 2h'n'(a + ½ b H ) / ( 2 A o ( h
+ ~r)),
a n d t h e l e n g t h o f t h e r e p l e n i s h m e n t cycles a r e Tj=T~
j=
1, 2 , . . . , n -
1,
T,,=H-(n-1)T,
where T= ~2A0(h + rr)/(h'n'(a + ½bH)) .
M.A. Hariga, M. Ben-Daya / European Journal of Operational Research 89 (1996) 313-325
325
Another approximate solution procedure can be obtained by extending Silver's Heuristic [17] to accommodate for inflation and time value money.
References [1] Aggrawal, S.C., "Purchase inventory decision models for inflationary conditions", Interfaces 11 (1981) 18-23. [2] Banks, J., Mangiameli, P.M., and Schwarzbach, H., "Static inventory models and inflationary increases", The Engineering Economist 26/2 (1982) 91-112. [3] Bierman, H., Thomas, J., "Inventory decision under inflationary conditions", Decision Sciences 8 (1977) 151-155. [4] Buzacott, J.A., "Economic order quantity with inflation", Operational Research Quarterly 26 (1975) 553-558. [5] Chandra, M.J., and Bahner, M.L., "The effects of inflation and the time value of money on some inventory systems", Intematioual Journal of Production Research 23/4 (1985) 723-730. [6] Datta, T.K., and Pal, A.K., "Effects of inflation and time-value of money on an inventory model with linear time-dependent demand rate and shortages", European Journal of Operational Research 52 (1991) 326-333. [7] Donaldson, W.A., "Inventory replenishment policy for a linear trend in demand: An analytical approach", Operational Research Quarterly 28 (1970) 663-670. [8] Goyal, S.K., "A heuristic for replenishment of trended inventories considering shortages", Journal of the OperationalResearch Society 39 (1988) 885-887. [9] Guruani, C., "Economic analysis of inventory systems", International Journal of Production Research 21/2 (1983) 261-277. [10] Hariga, M., "The inventory replenishment problem with a linear trend in demand", Computers and Industrial Engineering 24/2 (1993) 143-150. [11] Misra, R.B., " A study of inflationary effects on inventory systems", Logistics Spectrum 9/3 (1975). [12] Misra, R.B., " A note on optimal inventory management under inflation", Naval Research Logistics Quarterly 26 (1979) 161-165. [13] Mitra, A., Cox, J.F., and Jesse, Jr., R.R., " A note on determining order quantities with a linear trend in demand", Journal of the Operational Research Society 35 (1984) 141-144. [14] Murdeshwar, T.M., "Inventory replenishment policy for linearly increasing demand considering shortages-an optimal solution", Journal of the Operational Research Society 39 (1988) 687-692. [15] Phelps, R.I., "Optimal inventory rule for a linear trend in demand with constant replenishment period", Journal of the Operational Research Society 31 (1980) 439-442. [16] Ritchie, E., "The EOQ for linear increasing demand; a simple optimal solution", Journal of the Operational Research Society 35 (1984) 949-952. [17] Silver, E.A., " A simple inventory replenishment decision rule for a linear trend in demand", Journal of the Operational Research Society 30 (1979) 71-75.