Inventory models with ramp type demand rate, partial backlogging and general deterioration rate

Applied Mathematics and Computation 219 (2013) 4288–4307

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Inventory models with ramp type demand rate, partial backlogging and general deterioration rate q M.A. Ahmed, T.A. Al-Khamis, L. Benkherouf ⇑ Department of Statistics and Operations Research, College of Science, Kuwait University, P.O. Box 5969, Safat 13060, Kuwait

a r t i c l e

i n f o

a b s t r a c t This paper is concerned with finding the economic order quantity, EOQ, for an inventory model with ramp type demand-rate, partial backlogging and general deterioration rate. A new method for finding the EOQ policy is proposed. Existence of a unique EOQ is shown for two different replenishment policies ((a) periods starting with shortages and (b) periods starting with no shortages). Two different cost functions are considered. Ó 2012 Elsevier Inc. All rights reserved.

This paper is dedicated to the memory of Dr. M.A. Ahmed who passed away shortly before the completion of the paper Keywords: Inventory model Optimal schedule EOQ Ramp type demand Deteriorating item

1. Introduction This paper is an attempt to formulate single-item inventory models in the most general form, where the planning horizon is assumed to be infinite and the objective is to minimize the total inventory costs per unit time. These costs will be made precise below. In the absence of intervention, the level of stock IðtÞ, at time t, depletes according to some rule which is a function of time and the level of stock. This may be generally represented by the differential equation

I0 ðtÞ ¼ Fðt; IðtÞÞ;

ð1:1Þ

where F is some smooth function in some subspace of X, where



X ¼ ðx; yÞ 2 R2 ;

 with x P 0 ;

and I0 is the derivative of the function I. The second argument in the function F is allowed to take negative values to model shortages. In the present paper we shall allow for the possibility of partial backlogging periodic review models with the first period being of length T, say, and examine two different scenarios. In the first scenario we assume that the period starts with a positive inventory level and ends with shortage whereas the second scenario starts with shortage and ends with zero inventory level as depicted in Figs. 2 and 1 respectively. In scenario 1, at time t ¼ 0, the inventory level Ið0Þ is assumed positive and depletes according to (1.1) up to level zero and this occurs at a time t 1 , say. The dynamics of the inventory is then governed by the differential equation:

I0 ðtÞ ¼ Fðt; IðtÞÞ;

q

IðtÞ ! 0;

as t " t 1 :

The authors would like to thank the referees for useful comments on an earlier version of the paper.

⇑ Corresponding author.

E-mail addresses: [email protected] (T.A. Al-Khamis), [email protected] (L. Benkherouf). 0096-3003/$ - see front matter Ó 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2012.09.068

ð1:2Þ

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level of stock

time t1

T

Fig. 1. A typical behavior of the level of stock for model 2.

On the interval ½t1 ; TÞ, the system experiences shortages in which case the dynamics of the level of inventory is given by

I0S ðtÞ ¼ F S ðt; IðtÞÞ;

IS ðt 1 Þ ! 0;

as t # t1 :

ð1:3Þ

The notation in (1.3) contains the subscript S to emphasize shortage and to allow for the possibility for (1.3) to have a different representation from that in scenario (1.2). In scenario 2, at time t ¼ 0, the inventory level starts at level 0, then depletes up to time t 1 according to the dynamics

I0S ðtÞ ¼ F S ðt; IðtÞÞ;

IS ð0Þ ¼ 0:

ð1:4Þ

A replenishment is made at time t 1 , and on the interval ½t 1 ; TÞ, the inventory level depletes according to the dynamics

I0 ðtÞ ¼ Fðt; IðtÞÞ;

IðtÞ ! 0;

as t " T:

ð1:5Þ

It is worth noting at this stage that the forms (1.2) and (1.3) encompass all EOQ inventory models existing in the literature. Possible forms of F include (i)

Fðt; IðtÞÞ ¼ D;

ð1:6Þ

where D > 0, and known, which is the classical constant demand rate model: see Silver et al. [16] (ii)

Fðt; IðtÞÞ ¼ DðtÞ  hðtÞIðtÞ;

ð1:7Þ

where D and h are some positive function on Rþ . The function D models the demand rate and h some deterioration rate. The function D as presented in (1.7) is a function of time and general. It can be linear, exponential, increasing, decreasing or have other forms. This allows the possibility of modeling fluctuation of sales if required. Earlier work on time-varying demand included Donaldson [7]. The general form D contains the ramp type demand function. This type of function, although continuous in its argument, may not be differentiable at a point. Ramp type demand models a situation where demand stabilizes after a certain time. This has been observed for newly launched products: for more details see: Hill [9], Deng et al. [6], and Skouri et al. [17]. In fact it has been noticed that the demand rate for a new product in the market increases initially up to a certain time, after which it stabilizes and become constant. Moreover, the deterioration rate h plays an important role in modeling items that experience spoilage, damage, vaporization etc. during their lifetime. Food stuff, electronic components, blood are examples of items that deteriorate over time: for more details see Raafat [13], Goyal and Giri [8], and Bakker et al. [2]. (iii)

Fðt; IðtÞÞ ¼ DðtÞFðeHðtÞ IðtÞÞ  hðtÞIðtÞ; with

HðtÞ ¼

Z

ð1:8Þ

t

hðuÞdu;

ð1:9Þ

0

where the term DðtÞFðeHðtÞ IðtÞÞ models a demand rate which is stock dependent. It is a well known phenomena in the marketing literature that large piles of goods displayed attract customers and hence the influence of the level of stock on demand: see Benkherouf and Gilding [3], Balkhi and Benkherouf [4], Corstjens and Doyle [5], and Urban [21]. One last word concerns F S in that it can take any of the forms (1.6–1.8) with alternative interpretation as no deterioration will be present during shortages: see [3] for the discussion of this issue.

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level of stock

time t1

T

Fig. 2. A typical behavior of the level of stock for model 1.

In the present paper we shall be interested in the form

F S ðtÞ ¼ DðtÞbðg  tÞ;

ð1:10Þ

where D is a strictly positive function defined in Rþ , and b is some positive function of time with image the interval ½0; 1, with g ¼ T in scenario 1, and g ¼ t 1 in scenario 2. The case b  0 represents lost sales, while the case b  1, represents complete backlogging and finally the case 0 < bðuÞ < 1 represents partial backlogging, and in the no shortage period we assume that the changes in the inventory level evolves according to (1.7). We further assume that 1. D is continuous having left and right derivatives. The function D may take the form

 DðtÞ ¼

f ðtÞ;

t < l;

f ðlÞ; t P l;

ð1:11Þ

where f is a continuous strictly positive function in Rþ , and l is a known strictly positive constant: see [17]. In [17] the length of the period was fixed and the objective was to determine the changeover time t 1 which minimizes the total inventory costs: see Figs. 1 and 2. 2. The deterioration rate h is a continuously differentiable function in Rþ . 3. The function b is twice differentiable. 4. Deteriorated items are neither repaired nor replaced. 5. Replenishment occurs instantaneously at an infinite rate. 6. The cost structure is: (a) a fixed set-up cost, K > 0, (b) a purchasing cost per unit of item c1 > 0, (c) a holding cost per unit in stock per unit time, c2 > 0, (d) a shortage cost per unit in stock per unit time, c3 > 0, (e) a lost sales cost per unit of time c4 > 0, with inventory holding cost charged only for good units. In this paper we shall be interested in finding an optimal inventory policy which minimizes the total costs per unit time of two separate inventory models. The costs will be made precise in the next section. The presence of the parameter l in (1.11) leads to the definition of the cost functions on two separate branches of some subspace of the domain of definition of the functions based on whether some variable is less than or equal to l. Existing approaches in the inventory control literature for finding the optimal inventory policy for such problems are based on examining the costs for the branches separately: see [17] and Skouri and Konstantaras [18], Skouri, Konstantaras, Manna and Chaudhuri [19], Skouri et al. [20] and Mandal [12]. It is generally assumed that the cost function on the branches is smooth and consequently differential calculus is applied. The approach, although successful in some cases, cannot handle situations where cost functions are not twice differentiable on the branches. This paper proposes a new approach for tackling such problems which is applicable to demand rate functions which are simply required to be continuous. Furthermore, the approach avoids the examination of branches of cost functions separately. This allows the possibility to model demand rates which are for example increasing initially, then are fixed for a certain period time period, and finally are decreasing as considered in [18]. The method may also be a basis for examining inventory control optimization problems with piecewise objective functions as in inventory models with permissible delay in payments: see [20] and the references therein and for recent work Khanra et al. [11]. In the next section we examine scenario 1 and propose a method for finding the EOQ. Section 3 treats scenario 2. The last section contains numerical examples with some sensitivity analysis along with concluding remarks and suggestions for possible future work.

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2. The EOQ for models starting with positive inventory Note that from now onwards we shall use the words model and scenario interchangeably. This section is concerned with model 1 as depicted in Fig. 1. For these models, the dynamics of the inventory level I is assumed to be given

I0 ðtÞ ¼ DðtÞ  hðtÞIðtÞ;

with IðtÞ ! 0 as t " t 1 ;

0 6 t < t1 ;

ð2:1Þ

and the shortage level IS evolves according to

I0S ðtÞ ¼ DðtÞbðT  tÞ;

with IðtÞ ! 0 as t # t1 ;

t 1 6 t < T;

ð2:2Þ

where bð0Þ ¼ 1. Let

HðtÞ ¼

Z

t

ð2:3Þ

hðuÞdu; Z t1 jðt; t1 Þ ¼ eHðtÞ eHðuÞ DðuÞdu; t Z t jS ðt1 ; tÞ ¼ bðT  uÞDðuÞdu; 0

ð2:4Þ ð2:5Þ

t1 t

rS ðt1 ; tÞ ¼

Z

f1  bðT  uÞgDðuÞdu:

ð2:6Þ

t1

Direct computations show that j and jS are the solutions of (2.1) and (2.2) respectively. It follows from (2.4) and (2.5) that the total purchasing cost is given by

c1 fjð0; t1 Þ þ jS ðt 1 ; TÞg;

ð2:7Þ

the holding cost by

c2

Z

t1

jðt; t1 Þdt;

ð2:8Þ

0

the shortage cost by

c3

Z

T

jS ðt1 ; tÞdt;

ð2:9Þ

t1

the deterioration cost by

c1

Z

t1

hðtÞjðt; t1 Þdt;

ð2:10Þ

0

and the lost sales cost by

c4 rS ðt 1 ; TÞ:

ð2:11Þ

The total cost, TCðt1 ; TÞ in the period can be given by the sum of ordering cost, holding cost, shortage cost, and purchasing cost. In which case (2.7)–(2.9) give

TCðt1 ; TÞ ¼ K þ c1 fjð0; t1 Þ þ jS ðt 1 ; TÞg þ c2

Z

t1

jðt; t1 Þdt þ c3

Z

T

jS ðt1 ; tÞdt:

ð2:12Þ

t1

0

We call this cost the OHP cost. One can also adopt the cost given by the sum of ordering cost, holding cost, shortage cost, deterioration cost, and lost sales cost. In which case (2.7)–(2.11) give

TCðt1 ; TÞ ¼ K þ c4 rS ðt 1 ; TÞ þ

Z 0

t1

ðc2 þ hðtÞc1 Þjðt; t 1 Þdt þ c3

Z

T

jS ðt1 ; tÞdt:

ð2:13Þ

t1

This cost will be called the OHDL cost. This cost is the same cost adopted in [17]. The objective is to solve the optimization problem

min TCUðt1 ; TÞ ¼

TCðt 1 ; TÞ ; T

ð2:14Þ

subject to the constraint that T P t1 P 0, and where TCðt 1 ; TÞ is either given by (2.12) or (2.13). We shall be concerned in the present paper with the optimization problem (2.14) for the OHP model and the OHDL model. However, before we do that we present two examples which show that multiple optima or optimal boundary points cannot be ruled out for the optimization problem (2.14). This paper proposes a methodology for handling these cases.

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T

TCU

t1 Fig. 3. The plot of the total cost per unit time for Example 1.

Example 1. Consider the OHDL starting with shortage and assume that c4 ¼ 0, and that hðtÞ ¼ 0; K ¼ 8; c2 ¼ 1; c3 ¼ 2; b  1, and

DðtÞ ¼



2 þ sinð10tÞ þ cosð10tÞ; for 0 6 t 6 5 2 þ sinð50Þ þ cosð50Þ

for t > 5

:

This demand function is a ramp demand function, but which is not increasing. Fig. 3 shows a plot of TCU. It is clear that this function has multiple local optima. Example 2. Consider the OHP starting with positive inventory c2 ¼ 1; b  1; c3 ¼ 999999 (that is shortage cost is very large), and

DðtÞ ¼



2 þ 3t; for 0 6 t < 5 17

for t > 5

and

assume

that

c1 ¼ 0,

hðtÞ ¼ 0; K ¼ 8;

:

This demand function is a non-decreasing ramp demand function. Fig. 4 shows a plot of TCU. This function has a minimum which occurs at the boundary leading to a policy of prohibiting shortages. This example, although extreme, serves to illustrate the possibility of existence of optimal boundary solutions. We begin in the next subsection with the treatment of the OHD model. 2.1. Derivation of the EOQ for the OHP model We shall be concerned with solving (2.14), where TCðt 1 ; TÞ is given by (2.12). We begin with some notations: @ x represents the partial derivative of a bivariate function with respect to the first variable, @ y represents the partial derivative of a bivariate function with respect to the second variable. Note that since Hð0Þ ¼ 0, by (1.9), it follows from (2.4) that

jð0; t1 Þ ¼

Z

t1

eHðtÞ DðtÞdt:

ð2:15Þ

0

Also, it can be shown using (2.5) and integrating by parts that

Z

T

t1

jS ðt1 ; tÞdt ¼

Z

T

t1

ðT  tÞbðT  tÞDðtÞdt:

ð2:16Þ

M.A. Ahmed et al. / Applied Mathematics and Computation 219 (2013) 4288–4307

4293

t1

TCU

T

Fig. 4. The plot of the total cost per unit time for Example 2.

Therefore, (using (2.15) and (2.16)), TCðt1 ; TÞ given by (2.12) may be rewritten as

TCðt1 ; TÞ ¼ K þ

Z

t1

fc1 eHðtÞ DðtÞ þ c2 jðt; t 1 Þgdt þ

Z

T

tðT  tÞDðtÞdt;

ð2:17Þ

t1

0

where

tðxÞ ¼ ðc1 þ c3 xÞbðxÞ:

ð2:18Þ

The first order condition of optimality, assuming that the solution of (2.14) is in the interior of the feasible region implies, by noting that jðt 1 ; t 1 Þ ¼ 0, and setting ð@TCUÞx ðt 1 ; TÞ ¼ ð@TCUÞy ðt1 ; TÞ ¼ 0, that

Z t1 c1 eHðt1 Þ Dðt 1 Þ þ c2 ð@ jÞy ðt; t 1 Þdt  tðT  t 1 ÞDðt 1 Þ ¼ 0; 0   Z T t0 ðT  tÞDðtÞdt  TCðt1 ; TÞ ¼ 0; T tð0ÞDðTÞ þ

ð2:19Þ

t1

where t is the derivative of t. In order to complete the computations, we need the following 0

ð@ jÞy ðt; t 1 Þ ¼ eHðtÞþHðt1 Þ Dðt 1 Þ;

ð2:20Þ

which follows from (2.4). Using (2.20) with the knowledge that tð0Þ ¼ c1 (since bð0Þ ¼ 1, by assumption), the system of nonlinear Eqs. (2.20) reduces to

Z

eHðt1 Þ fc1 þ c2 Tfc1 DðTÞ þ

Z

t1

eHðuÞ dug  tðT  t1 Þ ¼ 0;

0 T

ð2:21Þ

t0 ðT  uÞDðuÞdug  TCðt1 ; TÞ ¼ 0:

t1

Our objective next is to examine under what conditions on the model’s components the system of equations described by (2.21) has a unique solution. Write the left hand side of the first equation in (2.21) as a function of x and y to get

Rðx; yÞ :¼ eHðxÞ fc1 þ c2

Z

x

eHðuÞ dug  tðy  xÞ:

0

We make the following assumption about the function t. A1. The function t is non-decreasing.

ð2:22Þ

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Note that in the case of complete backlogging, that is b  1, assumption A1 is trivially satisfied. Moreover, Assumption A1 implies that

c3 bðxÞ þ ðc1 þ c3 xÞb0 ðxÞ P 0; or

b0 ðxÞ c3 P : bðxÞ c1 þ c3 x

ð2:23Þ

In other words the rate of change of the backlog to itself is bounded below by some strictly negative increasing function. The class of functions satisfying (2.23) contains the class of functions for which b is non-decreasing. The following result is immediate from assumption A1 and direct differentiation. Lemma 1. Under assumption A1, the function R satisfies for y > x > 0, (i) ð@RÞx ðx; yÞ > 0, (ii) ð@RÞy ðx; yÞ 6 0, (iii) ð@RÞx ðx; yÞ > ð@RÞy ðx; yÞ. Now, assume that y > 0 and known and consider solving the equation

Rðx; yÞ ¼ 0;

ð2:24Þ

as a function of y. Then we have the following result. Lemma 2. There is a unique x which is a function of y, with x :¼ sðyÞ which solves (2.24). Furthermore, 0 < s0 ðyÞ < 1. Proof. Note from the definitions of R in (2.22) and

Rðy; yÞ ¼ c2 eHðyÞ

Z

s is differentiable and

t in (2.18) that

y

eHðtÞ dt þ c1 feHðyÞ  1g > 0;

0

and Rð0; yÞ ¼ c1  tðyÞ. But, t is non-decreasing by assumption A1. Thus, c1  tðyÞ 6 c1  tð0Þ ¼ 0. Now, part (i) of Lemma 1 implies that there is a unique solution x ¼ sðyÞ to (2.24). Therefore RðsðyÞ; yÞ ¼ 0. Applying the implicit function theorem to (2.24) to get

s0 ðyÞð@RÞx ðsðyÞ; yÞ þ ð@RÞy ðsðyÞ; yÞ ¼ 0: The rest of the theorem is then immediate from parts (ii) and (iii) of Lemma 1. This completes the proof. h Before we proceed further, note that direct computations on (2.17) give

  Z x ð@TCÞx ðx; yÞ ¼ eHðxÞ fc1 þ c2 eHðuÞ dug  tðy  xÞ DðxÞ; 0 Z y 0 t ðy  uÞDðuÞdu: ð@TCÞy ðx; yÞ ¼ c1 DðyÞ þ x

For fixed y > 0, let x ¼ sðyÞ be the unique solution of the equation

RðsðyÞ; yÞ ¼ 0;

ð2:25Þ

and substitute this in the left-hand side of the second equation of the system given by (2.21), with the notation T ¼ y and t1 ¼ sðyÞ to get

HðyÞ :¼ yfc1 DðyÞ þ

Z

y

t0 ðy  uÞDðuÞdug  TCðsðyÞ; yÞ:

ð2:26Þ

sðyÞ

The following two assumptions are needed before we proceed further A2.The function D is non-decreasing. A3. The function t is twice differentiable and convex. (recall that a twice differentiable function t is convex if t00 ðxÞ P 0 for all x in its domain of definition) Note that assumption A2 is found in [17]. Assumption A3 is satisfied if 0 6 b 6 1 and is constant, which corresponds to the classical partial backlogging. Lemma 3. Under assumptions A1-A3, the function H defined in (2.26) is continuous and increasing in y.

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Proof. The continuity is immediate since TC is continuous in both its argument and s is continuous. It follows from (2.26), by noting that (2.24) implies that ð@TCÞx ðsðyÞ; yÞ ¼ 0, and after some tedious but direct algebra that

H0 ðy Þ ¼ yft0 ðy  sðyÞÞDðsðyÞÞs0 ðyÞ þ c1 D0 ðy Þ þ t0 ð0ÞDðyÞ þ

Z

y

t00 ðy  uÞDðuÞdug;

ð2:27Þ

sðyÞ

where H0 ðy Þ stands for the right and left derivatives of H since H is not necessarily differentiable. The same remark can be made about D. But assumptions A2 and A3 imply that

Z

y

t00 ðy  uÞDðuÞdu P DðsðyÞÞ

Z

sðyÞ

y

t00 ðy  uÞdu ¼ DðsðyÞÞft0 ð0Þ þ t0 ðy  sðyÞÞg:

sðyÞ

Also,

t0 ðy  sðyÞÞDðsðyÞÞs0 ðyÞ 6 t0 ðy  sðyÞÞDðsðyÞÞ; by Lemma 2. Consequently,

H0 ðy Þ > yfc1 D0 ðy Þ þ t0 ð0ÞðDðyÞ  DðsðyÞÞg: This is greater than zero by Assumption A2. This leads to the required result. h Theorem 1. Assume further to assumptions A1–A3 that A4. there exists c > 0 such that bðxÞ > c; (that is b is bounded below by some strictly positive constant), then there exists a unique solution y which solves HðyÞ ¼ 0. Proof. Note that if y ¼ 0, then by (2.22) sðyÞ ¼ 0, and consequently (2.26) gives Hð0Þ ¼ K, which is strictly negative. Now, we set to show that

lim HðyÞ ¼ 1:

ð2:28Þ

y!1

First, note that as y ! 1; sðyÞ ! 1. To see that assume that sðyÞ is bounded and write RðsðyÞ; yÞ in the form

eHðsðyÞÞ fc1 þ c2

Z sðyÞ

eHðuÞ dug  tðy  sðyÞÞ:

0

Here tðy  sðyÞÞ ! 1 as y ! 1, and therefore the above expression is strictly negative, contradicting the assertion that RðsðyÞ; yÞ ¼ 0. Direct computations, using (2.26) and (2.17), show that

( HðyÞ ¼ K þ y c1 DðyÞ þ

Z

)

y

t0 ðy  tÞDðtÞdt 

sðyÞ

Z sðyÞ Z  HðtÞ  c1 e DðtÞ þ c2 jðt; sðyÞÞ dt 

y

tðy  tÞDðtÞdt:

ð2:29Þ

sðyÞ

0

Assumptions A1 and A2 imply that

Z

y

t0 ðy  tÞDðtÞdt P DðsðyÞÞ

sðyÞ

Z

y

t0 ðy  tÞdt ¼ DðsðyÞÞftðy  sðyÞÞ  tð0Þg;

sðyÞ

and

Z

y

tðy  tÞDðtÞdt 6 ðy  sðyÞÞtðy  sðyÞÞDðyÞ:

sðyÞ

Also, Assumption A2 with (2.4) lead to

  Z sðyÞ Z sðyÞ  HðtÞ  c1 e DðtÞ þ c2 jðt; sðyÞÞ dt 6 eHðsðyÞÞ c1 sðyÞ þ c2 eHðtÞ ðsðyÞ  tÞdt DðsðyÞÞ 0

0

  Z sðyÞ Z sðyÞ eHðtÞ dt DðsðyÞÞ  c2 eHðsðyÞÞ DðsðyÞÞ teHðtÞ dt: 6 sðyÞeHðsðyÞÞ c1 þ c2 0

Recalling that tð0Þ ¼ c1 , and that as y ! 1, sðyÞ ! 1 and that D stabilizes after large, we get that

HðyÞ P K þ c2 eHðsðyÞÞ DðsðyÞÞ

Z sðyÞ

teHðtÞ dt  sðyÞDðsðyÞÞRðsðyÞ; yÞ:

0

The result is then immediate since RðsðyÞ; yÞ ¼ 0. This completes the proof. h The following theorem is a consequence of Theorem 1.

0

l: that is DðyÞ ¼ DðsðyÞÞ ¼ f ðlÞ for y

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Theorem 2. Under assumptions A1–A4, the system of equations by (2.21) has a unique solution ðt 1 ; T  Þ. Further, this solution satisfy 0 < t1 < T  < 1. Remark 1. Note that the maximum of TCUðt 1 ; TÞ can be obtained by making t1 and T arbitrarily small, in which case TCUðt1 ; TÞ is unbounded above and consequently ðt1 ; T  Þ represents the minimal solution in the interior of the feasible region. To complete the analysis we need to check optimality at the boundary of the feasible region. Namely, the cases t 1 ¼ 0, and t1 ¼ T. The case t 1 ¼ 0 corresponds to a period with complete shortage and the case t 1 ¼ T corresponds to a period of no shortage. Let us first examine the case t 1 ¼ T. That is, the case where there is no shortage in the period. Then, it follows from (2.12) that the total cost per unit time TCUðTÞ is given by

TCUðTÞ ¼

K þ c1 jð0; TÞ þ c2 T

RT 0

jðt; TÞdt

ð2:30Þ

:

0

Then, TCU ðTÞ ¼ 0 implies that

 Z T c1 þ c2

T

  Z eHðuÞ du eHðTÞ DðTÞ  K þ c1

0

T

eHðuÞ DðuÞdu þ c2

0

Z

T

eHðtÞ

Z

0

T

  eHðuÞ DðuÞdu dt ¼ 0:

ð2:31Þ

t

Let HðTÞ be the left-hand side of (2.31). Note that Hð0Þ ¼ K < 0. We claim that HðTÞ ! 1, as ! 1. Indeed, it can be shown that

 Z HðTÞ P T c1 þ c2

   Z Z T eHðuÞ du eHðTÞ DðTÞ  eHðTÞ DðTÞ c1 T þ c2 ðT  tÞeHðtÞ dt  K ¼ c2 eHðTÞ DðTÞ

T

0

0

T

teHðtÞ dt

0

 K ! 1; as T ! 1: It follows that there exists a solution T > 0, to the equation HðTÞ ¼ 0, since H is continuous. Next, we examine the variation of H with respect to T. This can be shown to be equal to:

H0 ðT  Þ ¼

 Z c1 þ c2

   eHðuÞ du hðtÞDðTÞ þ D0 ðT  Þ TeHðTÞ þ c2 TDðTÞ;

T

0

e. which is positive. Therefore H is strictly increasing in T. Thus, there exits a unique solution to the equation HðTÞ ¼ 0, call it T It remains to check that this solution gives the minimum of TCUðTÞ which is given by (2.31). To do that we just need to eliminate 1 and 0 as possible solutions. It is clear that TCUðTÞ ! 1, as T ! 0. So, 0 is excluded. Further, TCUðTÞ ! 1, as T ! 1. To see that note that

TCUðTÞ P

K þ c2

RT 0

jðt; TÞdt

¼ T 2 1 K þ 2 c2 T Dð0Þ P ! 1; T

K þ c2

RT 0

eHðtÞ

nR T t

eHðuÞ DðuÞdu

o

P

T

K þ c2

RT 0

ðT  tÞDðtÞdt T

as T ! 1:

e is the minimum of TCUðTÞ. Consequently, the point T We shall next examine the case t1 ¼ 0 and assume that the expression

 Z y    ðc1 þ c3 ðy  tÞÞb0 ðy  tÞDðtÞdt  < 1   0

which is clearly satisfied if b is fixed. The cost per unit time reduces to:

TCUðTÞ ¼

Kþ

RT 0

tðT  tÞDðtÞdt T

ð2:32Þ

:

0

Setting TCU ðTÞ ¼ 0 gives

 LðTÞ :¼ T

tð0ÞDðTÞ þ

Z



T



t0 ðT  tÞDðtÞdt  K þ

0

Z

T



tðT  tÞDðtÞdt ¼ 0:

0

It is not difficult to see that Lð0Þ ¼ K < 0, and that

t0 ðxÞ ¼ c3 bðxÞ þ ðc1 þ c3 xÞb0 ðxÞ: We have from direct substitution that

 Z LðTÞ ¼ T c1 DðTÞ þ c3 þ c3

Z 0

T

bðT  tÞDðtÞdt þ

0

Z

T 0

 Z ðc1 þ c3 ðT  tÞÞb0 ðT  tÞDðtÞdt  fK þ c1

T

bðT  tÞDðtÞdt

0

T

ðT  tÞbðT  tÞDðtÞdtg:

ð2:33Þ

M.A. Ahmed et al. / Applied Mathematics and Computation 219 (2013) 4288–4307

4297

Therefore, LðTÞ asymptotically (as T goes to 1)

LðTÞ ’ c1 TDðTÞ  c1

Z

T

bðT  tÞDðtÞdt þ c3

0

Z

T

tbðT  tÞDðtÞdt;

0

which ! 1, as T ! 1, since the integral with coefficient c3 dominates the integral with coefficient c1 . We conclude that there is a solution to the equation LðTÞ ¼ 0. It remains to show that this solution is unique. Differentiating L with respect to T gives after some algebra

 Z L0 ðT  Þ ¼ T c1 D0 ðT  Þ þ c



T

t00 ðT  tÞDðtÞdt > 0

0

^ This is positive by assumptions A2 and A3. Thus, there exists a unique solution to LðTÞ ¼ 0. Call this solution T. The same argument used above leads to TCUðTÞ given by (2.32) to have limit 1 as T ! 0, and as T ! 1. Therefore, T^ is the point where TCUðTÞ achieves its minimum. e , and T^ we are in a position to find the optimal inventory policy for model 1. For With the existence of the points ðt 1 ; T  Þ; T this, we pick the point that achieves the minimum among the three objective functions. 2.2. Derivation of the EOQ for the OHDL model In this subsection we shall consider inventory models where the total cost per period is given by (2.13), that is

TCðt1 ; TÞ ¼ K þ c4 rS ðt 1 ; TÞ þ

Z

t1

ðc2 þ hðtÞc1 Þjðt; t 1 Þdt þ c3

0

Z

T

jS ðt1 ; tÞdt;

ð2:34Þ

t1

where rS ðt 1 ; TÞ; jðt; t 1 Þ, and jS ðt 1 ; tÞ are given by (2.6), (2.4), and (2.5) respectively. The objective is to find a pair ðt1 ; T  Þ with 0 6 t1 6 T which minimizes TCUðtT 1 ;TÞ. In order to achieve our objective we shall mimic the step of subSection 2.1. Let us rewrite (2.34) in the compact form

TCðt1 ; TÞ ¼ K þ

Z

t1

wðtÞjðt; t 1 Þdt þ

Z

0

T

tðT  tÞDðtÞdt;

ð2:35Þ

t1

where

wðtÞ :¼ c2 þ hðtÞc1 ;

ð2:36Þ

tðxÞ :¼ c4 ð1  bðxÞÞ þ c3 xbðxÞ:

ð2:37Þ

Note that the function t in this model differs from that of the previous model: see (2.18). The first order optimality condition leads by (2.20) and after some algebra to

eHðt1 Þ T

Z

Z

t1

wðtÞeHðtÞ dt  tðT  t 1 Þ ¼ 0;

ð2:38Þ

0 T

t0 ðT  tÞDðtÞdt  TCðt1 ; TÞ ¼ 0:

ð2:39Þ

t1

Define

Rðx; yÞ ¼ eHðxÞ

Z

x

wðtÞeHðtÞ dt  tðy  xÞ:

ð2:40Þ

0

We make the following assumptions before we proceed further. L1. The function t is non-decreasing in Rþ . Then we have the following result which is similar to Lemma 1 for this model. Lemma 4. For x–y and under assumption L1, the function R satisfies (i) ð@RÞx ðx; yÞ > 0, (ii) ð@RÞy ðx; yÞ < 0, (iii) ð@RÞx ðx; yÞ > ð@RÞy ðx; yÞ. Using Lemma 4 and noting that for y > 0; Rð0; yÞ ¼ tðyÞ < 0, and Rðy; yÞ > 0, then there exist a unique x, which is a function of y, x :¼ sðyÞ, such that RðsðyÞ; yÞ ¼ 0. Moreover, it can be shown along the lines of Lemma 2 that s is differentiable and that

0 < s0 ðyÞ < 1:

ð2:41Þ

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M.A. Ahmed et al. / Applied Mathematics and Computation 219 (2013) 4288–4307

Eq. (2.39) suggests that we consider the equation

HðyÞ :¼ y

Z

y

t0 ðy  tÞDðtÞdt  TCðsðyÞ; yÞ:

ð2:42Þ

sðyÞ

Examining the variation of H with respect to y, and using the fact that RðsðyÞ; yÞ ¼ 0, we get after some algebra that

( H0 ðyÞ ¼ y

t0 ð0ÞDðyÞ þ

Z

)

y

t00 ðy  tÞDðtÞdt  t0 ðy  sðyÞÞDðsðyÞÞs0 ðyÞ :

ð2:43Þ

sðyÞ

We assume that L2. The function D is non-decreasing. L3. The function t is convex. Under assumptions L2 and L3

Z

y

t00 ðy  tÞDðtÞdt P DðsðyÞÞ

Z

sðyÞ

y

t00 ðy  tÞdt ¼ DðsðyÞÞft0 ðy  sðyÞÞ  t0 ð0Þg:

sðyÞ

This implies, using assumption L1 and (2.41) that

H0 ðyÞ P y½t0 ð0ÞyfDðyÞ  DðsðyÞÞg þ t0 ðy  sðyÞÞsðyÞð1  s0 ðyÞÞ P 0: This implies that H is increasing in y. Note from (2.40) that sð0Þ ¼ 0. Therefore, Hð0Þ ¼ K < 0. Furthermore, HðyÞ ! 1, as y ! 1. This can be obtained by noting that sðyÞ also goes to 1 as y ! 1, otherwise we get a contradiction with RðsðyÞ; yÞ ¼ 0. Indeed, write

HðyÞ ¼ y

Z

(

y 0

t ðy  tÞDðtÞdt  K þ

Z sðyÞ

sðyÞ

wðtÞjðt; sðyÞÞdt þ

Z

)

y

tðy  tÞDðtÞdt :

sðyÞ

0

We claim that for large y we have

HðyÞ P K þ eHðsðyÞÞ DðsðyÞÞ

Z sðyÞ

twðtÞeHðtÞ dt:

ð2:44Þ

0

Indeed, we have by Assumption L2

Z sðyÞ

wðtÞjðt; sðyÞÞdt ¼

0

Z sðyÞ

wðtÞeHðtÞ

0

Z sðyÞ

Z sðyÞ eHðuÞ DðuÞdu dt 6 wðtÞeHðtÞþHðsðyÞÞ DðsðyÞÞðsðyÞ  tÞdt:

t

0

Also, assumptions L1 and L2 lead to

Z

y

tðy  tÞDðtÞdt 6 ðy  sðyÞÞDðyÞtðy  sðyÞÞ:

sðyÞ

Further, by assumptions L1 and L2 and the fact that tð0Þ ¼ 0 (see (2.37))

y

Z

y

t0 ðy  tÞDðtÞdt P yDðsðyÞÞftðy  sðyÞÞ  tð0Þg ¼ yDðsðyÞÞtðy  sðyÞ:

sðyÞ

Relation (2.44) is now immediate from the fact that RðsðyÞ; yÞ ¼ 0. Therefore, HðyÞ ! 1, as y ! 1. The following result is then immediate. Theorem 3. Under assumption L1–L3, the system of equation given by (2.38) and (2.39) has a unique solution ðt 1 ; T  Þ. Furthermore, this solution satisfies 0 < t1 < T  < 1. Next, following Remark 1 we consider the cases t 1 ¼ 0 and t 1 ¼ T separately. First, let t1 ¼ T. This case corresponds the situation of zero shortages, and therefore, using (2.14), the total cost per unit time is given by

TCUðTÞ :¼

Kþ

RT 0

wðtÞjðt; TÞdt : T

ð2:45Þ

Then, TCU 0 ðTÞ ¼ 0 gives by (2.20)

TeHðTÞ DðTÞ

Z 0

T

 Z wðtÞeHðtÞ dt  K þ 0

T

wðtÞjðt; TÞdt



¼ 0:

ð2:46Þ

M.A. Ahmed et al. / Applied Mathematics and Computation 219 (2013) 4288–4307

4299

Let HðTÞ be the left-hand side of (2.46), then Hð0Þ ¼ K < 0, and by assumption L2 we get

Z

T

wðtÞjðt; TÞdt 6 eHðTÞ DðTÞ

Z

0

T

ðT  tÞwðtÞeHðtÞ dt:

0

This leads, after some computations to

HðTÞ P K þ eHðTÞ DðTÞ

Z

T

twðtÞeHðtÞ dt;

0

which goes to 1, as T goes to 1. It is now immediate that there is a solution to TCU 0 ðTÞ ¼ 0, with T > 0. It remains to show the uniqueness of this by checking the H0 ðTÞ. Indeed, H0 ðTÞ can be shown to be equal to

 Z   T eHðTÞ hðTÞDðTÞ þ D0 ðT  Þ

T

 wðtÞeHðtÞ dt þ wðTÞDðTÞ > 0:

0

e. Therefore, the root of TCU 0 ðTÞ ¼ 0 is unique which is the minimum of the function TCU, call this T Now, we move to the case where t 1 ¼ 0. That is where complete shortage occurs in the period. In this case the total cost per unit time is given by

TCUðTÞ ¼

Kþ

RT 0

tðT  tÞDðtÞdt T

:

This case is equivalent to that treated in (2.32) with different representations for the t function. Here t is given by (2.37). Following the same approach to that adopted for (2.32) (this is left as an exercise), it can be shown that TCU has a unique ^ critical point which correspond to the minimum solution of TCU, call this T. e , and T. ^ Finally, the optimal policy is the one gives smallest value function for the points ðt ; T  Þ; T 1

3. The EOQ for models starting with shortages In this section, we shall mimic the steps taken in Section 2 to show the existence of the optimal EOQ for models starting with shortages. However, due to complications in obtaining simple conditions for the analysis to go through we shall assume that the deterioration rate hðtÞ is fixed to h. Moreover, we shall keep the set-up as general as possible to allow for possible generalizations in the future. Some details are skipped since they contain essentially the same ideas to those encountered previously. The changes in the inventory level take the form shown in Fig. 2 and is given by

I0 ðtÞ ¼ DðtÞbðt 1  tÞ;

with Ið0Þ ! 0;

as t # 0; 0 < t < t1 ;

ð3:1Þ

and

I0 ðtÞ ¼ DðtÞ  hðtÞIðtÞ;

with IðTÞ ! 0;

as t " T;

t 1 6 t < T:

ð3:2Þ

We shall keep the basic assumptions about b, namely bð0Þ ¼ 1, and 0 6 bðtÞ 6 1 for all t P 0. Write

jS ð0; tÞ ¼

Z

t

bðt 1  uÞDðuÞdu;

0

jðt; TÞ ¼ eHðtÞ

rS ð0; t1 Þ ¼

Z

ð3:3Þ

T

eHðuÞ DðuÞdu;

ð3:4Þ

t

Z

t1

f1  bðt 1  uÞgDðuÞdu:

ð3:5Þ

0

It can be shown that jS and j solve (3.1) and (3.2) respectively. A similar approach to that of Section 2 for the OHP model implies that the total costs on the interval ½0; T is given by

TCðt1 ; TÞ ¼ K þ c1 fjS ð0; t 1 Þ þ jðt 1 ; TÞg þ c3

Z

t1

jS ð0; tÞdt þ c2

Z

T

jðt; TÞdt;

ð3:6Þ

t1

0

and for the OHDL model

TCðt1 ; TÞ ¼ K þ c4 rS ð0; t 1 Þ þ c3

Z 0

t1

jS ð0; tÞdt þ

Z

T

ðc2 þ hðtÞc1 Þjðt; TÞdt:

ð3:7Þ

t1

The total cost per unit time for both models is given by:

TCUðt 1 ; TÞ ¼

TCðt1 ; TÞ : T

ð3:8Þ

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M.A. Ahmed et al. / Applied Mathematics and Computation 219 (2013) 4288–4307

The objective is to find a pair ðt 1 ; TÞ which minimizes TCUðt 1 ; TÞ, subject to the constraint T P t1 P 0, where TC is either given by (3.6) or (3.7). We next consider the OHP model. 3.1. Derivation of the EOQ for the OHP model Here, we will be interested in minimizing TCU where TC is given by (3.6). Recall the definition of t in (2.18) to get from (3.6) that

TCðt1 ; TÞ ¼ K þ

Z

t1

Z

tðt1  tÞDðtÞdt þ

T



 c1 eHðtÞHðt1 Þ DðtÞ þ c2 jðt; TÞ dt:

ð3:9Þ

t1

0

It follows from the first order necessary conditions for the minimum of (3.8) with TC given by (3.9), and noting that tð0Þ ¼ c1 , that

Z

t1

t0 ðt1  tÞDðtÞdt  wðt1 Þjðt1 ; TÞ ¼ 0;

0

 Z TeHðTÞ DðTÞ c1 eHðt1 Þ þ c2

ð3:10Þ



T

eHðtÞ dt  TCðt 1 ; TÞ ¼ 0;

ð3:11Þ

t1

where w is given by (2.36). We shall retain assumptions A1–A3 of Section 2 and assume that h is constant. This is in fact not essential. However, it shall make the computations simple. Moreover, we make the following additional assumption A4. ðt0 ð0Þ þ wðxÞÞDðxÞ þ ðhwðxÞ  w0 ðxÞÞjðx; yÞ P wðxÞehðyxÞ DðyÞ, for all y P x P 0. Assumptions A4 may look complicated but a close look at the expression reveals that it is trivially satisfied for constant demand functions. Also, letting b  1, and noting that t0 ð0Þ ¼ c3 , the left-hand side of the above inequality, by assumption A2, is greater than or equal to

c3 DðxÞ þ wðxÞDðxÞ þ hwðxÞehy DðxÞ

Z

y

  eht dt ¼ c3 þ ðc1 þ hc2 ÞehðyxÞ Þ DðxÞ:

ð3:12Þ

x

It is now an easy exercise to deduce from (3.12) and (1.11) that assumption A4 is satisfied if

c3 DðlÞ  Dð0Þ hl P e : Dð0Þ c1 þ hc2

ð3:13Þ

This is clearly satisfied if c3 is large. Define

Rðx; yÞ :¼

Z

x

t0 ðx  tÞDðtÞdt  wðxÞjðx; yÞ:

ð3:14Þ

0

Then, under assumptions A1-A4, it can be shown that for y > x > 0, ð@RÞx ðx; yÞ > 0; ð@RÞy ðx; yÞ 6 0, and ð@RÞx ðx; yÞ > ð@RÞy ðx; yÞ. Also, Rðy; yÞ P 0, and Rð0; yÞ < 0. Therefore, if y is known, there exists a unique x ¼ sðyÞ which solves

RðsðyÞ; yÞ ¼ 0;

ð3:15Þ

with sð0Þ ¼ 0, and

0 < s0 ðyÞ < 1:

ð3:16Þ

The proofs of the above statements are similar to the proof of Lemma 2 and are left as exercises. For a given y, let sðyÞ be such that (3.15) is satisfied. Guided by (3.11) set

( HðyÞ

HðyÞ :¼ ye

HðsðyÞÞ

DðyÞ c1 e

þ c2

Z

)

y

e

HðtÞ

dt

 TCðsðyÞ; yÞ:

ð3:17Þ

sðyÞ

Direct computations, using (3.15), show that

H0 ðy Þ ¼ yeHðyÞ

" ( Z   hðyÞDðyÞ þ D0 ðy Þ c1 eHðsðyÞÞ þ c2

)

y

eHðtÞ dt

# þ DðyÞfc2 eHðyÞ  ðc2 þ hðsðyÞÞc1 ÞeHðsðyÞÞ s0 ðyÞg :

sðyÞ

ð3:18Þ This is, by (3.16), P

yeHðyÞ c1 eHðsðyÞÞ

" Z     hðyÞDðyÞ þ D0 ðy Þ  hðsðyÞÞDðyÞ þ c2 hðyÞDðyÞ þ D0 ðy Þ

y

sðyÞ

#! eHðtÞ dt  DðyÞðeHðsðyÞÞ  eHðyÞ Þ

:

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M.A. Ahmed et al. / Applied Mathematics and Computation 219 (2013) 4288–4307

It can be shown by recalling that hðtÞ ¼ h (constant), and some direct algebra that the above is strictly positive. Therefore, H is increasing. Note that since sð0Þ ¼ 0; Hð0Þ ¼ K < 0. We claim that

HðyÞ ! 1;

as y ! 1:

ð3:19Þ

The key idea in proving (3.19) is to add the expression

sðyÞwðsðyÞÞjðsðyÞ; yÞ þ sðyÞ

Z sðyÞ

t0 ðsðyÞ  tÞDðtÞdt;

0

which is equal to zero by (3.15) and the definition of R, to HðyÞ. Then consider the terms involving each cost coefficient separately. To be precise, we write

( HðyÞ ¼ yeHðyÞ DðyÞ c1 eHðsðyÞÞ þ c2

Z

)

y

eHðtÞ dt

 TCðsðyÞ; yÞ  sðyÞwðsðyÞÞjðsðyÞ; yÞ þ sðyÞ

Z sðyÞ

sðyÞ

t0 ðsðyÞ  tÞDðtÞdt:

0

ð3:20Þ This is equal, by recalling the definitions of w; t, and computing

HðyÞ ¼ K þ c1 fyeHðyÞHðsðyÞÞ DðyÞ  " Z

Z sðyÞ Z

eHðtÞ dt 

sðyÞ

þ c3 fsðyÞ þ

Z sðyÞ

bðsðyÞ  tÞDðtÞdt 

0

y

þ c2 yeHðyÞ DðyÞ

y

eHðtÞ

bðsðyÞ  tÞDðtÞdt 

Z

y

ð1 þ sðyÞhðsðyÞÞÞeHðtÞHðsðyÞÞ DðtÞdtg #  Z

sðyÞ

Z

sðyÞ

y

Z sðyÞ

y

eHðuÞ DðuÞdu dt  sðyÞeHðsðyÞÞ

eHðtÞ DðtÞdt

sðyÞ

t

0

Z sðyÞ

t , to 0

ðsðyÞ  tÞbðsðyÞ  tÞDðtÞdtg

0

sðyÞðc1 þ c3 ðsðyÞ  tÞÞb0 ðsðyÞ  tÞDðtÞdt:

ð3:21Þ

0

We remark that sðyÞ ! 1, as y ! 1. Otherwise, sðyÞ cannot be the solution of (3.15). First, consider the term, of interest to us, involving the coefficient c1 . This equal to

yeHðyÞHðsðyÞÞ DðyÞ 

Z sðyÞ

bðsðyÞ  tÞDðtÞdt 

Z

y

eHðtÞHðsðyÞÞ DðtÞdt  sðyÞhðsðyÞÞ

sðyÞ

0

Z

y

eHðtÞHðsðyÞÞ DðtÞdt:

ð3:22Þ

sðyÞ

Now, note by integration by parts that

Z

y

eHðtÞ dt ¼ yeHðyÞ  sðyÞeHðsðyÞÞ 

Z

sðyÞ

y

thðtÞeHðtÞ dt:

sðyÞ

This leads by assumption A2 that the term involving c1 , that is (3.21), is greater than or equal to

DðyÞeHðsðyÞÞ

Z

y

ðthðtÞ  sðyÞhðsðyÞÞÞeHðtÞ dt;

sðyÞ

which is greater than zero. For the term involving c2 , we have

yeHðyÞ DðyÞ

Z

y

eHðtÞ dt 

sðyÞ

P yeHðyÞ DðyÞ

Z

Z

y

eHðtÞ

Z

sðyÞ

HðyÞ

¼ DðyÞ e

Z

t

y

eHðtÞ dt  eHðyÞ DðyÞ

sðyÞ

"

y

 Z eHðuÞ DðuÞdu dt  sðyÞeHðsðyÞÞ

Z

HðtÞ

te

dt  sðyÞe

sðyÞ

y

ðy  tÞeHðtÞ dt  sðyÞeHðsðyÞÞ

HðsðyÞÞ

Z

#

y

HðtÞ

e

dt :

sðyÞ

It can be shown that the term in square brackets reduces to

ehðysðyÞÞ  1  hðy  sðyÞÞ h2

:

This is >0, since the exponential function if expanded gives

ex ¼ 1 þ

1 X xn n¼1

n!

:

eHðtÞ DðtÞdt

sðyÞ

sðyÞ

y

y

Z

y

eHðtÞ DðtÞdt

sðyÞ

ð3:23Þ

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M.A. Ahmed et al. / Applied Mathematics and Computation 219 (2013) 4288–4307

The term involving c3 in the two lines before the end of expression (3.22), reduces to

Z sðyÞ

tbðsðyÞ  tÞDðtÞdt:

0

Finally, we assume that the expression

Z   

x

0

  xðc1 þ c3 ðx  tÞÞb0 ðx  tÞDðtÞdt < 1:

ð3:24Þ

This is satisfied when b is fixed. Grouping all the terms we have

HðyÞ P K þ

Z sðyÞ

tbðsðyÞ  tÞDðtÞdt:

0

It is then immediate to conclude that HðyÞ ! 1, as y ! 1. We have then the following theorem Theorem 4. Under the assumptions that (i) A1–A4 hold, (ii) the deterioration rate is fixed and (3.24) holds, the system of equations given by (3.10) and (3.11) has a unique solution ðt1 ; T  Þ. Moreover, this solution satisfies 0 < t1 < T  . To exclude the possibility that ðt 1 ; T  Þ is the maximum of the function TCU, note that it is easy to check that TCU is unbounded above by making T arbitrarily small. To complete the analysis one should examine the boundaries cases where t1 ¼ T, and t 1 ¼ 0. This has already been done in the previous section. The optimal policy is the one that gives the minimal value function from among the point ðt 1 ; T  Þ and the optimal boundary points. 3.2. Derivation of the EOQ for the OHDL model Here, we shall be interested in finding a pair ðt1 ; T  Þ which minimizes TCUðt1 ; TÞ given by (3.8) where TCðt 1 ; TÞ is given by (3.7). The following is a representation of TCðt 1 ; TÞ that we shall find useful in the sequel

TCðt1 ; TÞ ¼ K þ

Z

t1

tðt1  tÞDðtÞdt þ

0

Z

T

wðtÞjðt; TÞdt;

ð3:25Þ

t1

where

tðxÞ ¼ c4 ð1  bðxÞÞ þ c3 xbðxÞ;

ð3:26Þ

and wðtÞ is given by (2.36). Note that the function t differs from the one used for the OHP model. The first order optimality condition leads to the system of non-linear equations:

Z

t1

t0 ðt1  tÞDðtÞdt  wðt1 Þjðt1 ; TÞ ¼ 0;

0

Z

TeHðTÞ DðTÞ

ð3:27Þ

T

wðtÞeHðtÞ dt  TCðt1 ; TÞ ¼ 0:

ð3:28Þ

t1

We shall retain assumptions L1–L3 mentioned in the subsection of the ODHL model. We also modify assumption A4 above to get the following assumption: L4. ðt0 ð0Þ þ wðxÞÞDðxÞ þ ðhwðxÞ  w0 ðxÞÞjðx; yÞ P wðxÞehðyxÞ DðyÞ, for all y P x P 0. Define

Rðx; yÞ ¼

Z

x

t0 ðx  tÞDðtÞdt  wðxÞjðx; yÞ:

ð3:29Þ

0

Then, under assumptions L1-L4, it can be shown that if y is known, then there exists a unique x ¼ sðyÞ which solves

RðsðyÞ; yÞ ¼ 0;

ð3:30Þ

with sð0Þ ¼ 0, and

0 < s0 ðyÞ < 1:

ð3:31Þ

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Eq. (3.28) suggests that we consider the function

Z

HðyÞ :¼ yeHðyÞ DðyÞ

y

wðtÞeHðtÞ dt  TCðsðyÞ; yÞ;

ð3:32Þ

sðyÞ

where sðyÞ satisfies RðsðyÞ; yÞ ¼ 0. It follows that

H0 ðy Þ ¼ yFðsðyÞ; yÞ; where

  FðsðyÞ; yÞ ¼ eHðyÞ hðyÞDðyÞ þ D0 ðy Þ

Z

y

  wðtÞeHðtÞ dt þ wðyÞ  wðsðyÞÞeHðyÞHðsðyÞÞ s0 ðyÞ DðyÞ:

sðyÞ

This is, by (3.31), greater than or equal to

  eHðyÞ hðyÞDðyÞ þ D0 ðy Þ

Z

y

  wðtÞeHðtÞ dt þ wðyÞ  wðsðyÞÞeHðyÞHðsðyÞÞ DðyÞ:

ð3:33Þ

sðyÞ

The objective at this stage is to show that FðsðyÞ; yÞ > 0. Note that (since h is fixed) we have wðtÞ ¼ wðyÞ ¼ wðsðyÞÞ. Moreover, (3.33) leads to

( wðyÞ e

HðyÞ

DðyÞ

Z

)

y

he

ht

dt þ DðyÞ  e

hðysðyÞÞ

DðyÞ

þ wðyÞeHðyÞ D0 ðy Þ

sðyÞ

Z

y

eHðtÞ dt ¼ wðyÞeHðyÞ D0 ðy Þ

sðyÞ

Z

y

eHðtÞ dt P 0;

sðyÞ

which gives the required result. Recall that sð0Þ ¼ 0, so Hð0Þ ¼ K < 0. We next show that

HðyÞ ! 1;

as y ! 1:

We shall follow the same approach of the previous subsection. Write

HðyÞ ¼ yeHðyÞ DðyÞ

Z

y

wðtÞeHðtÞ dt  TCðsðyÞ; yÞ  sðyÞwðsðyÞÞjðsðy; yÞ þ sðyÞ

sðyÞ

Z sðyÞ

t0 ðsðyÞ  tÞDðtÞdt:

0

Grouping some terms with coefficients c1 ; c2 ; c3 , and c4 separately we get for the c1 terms

yeHðyÞ DðyÞ

Z

y

hðtÞeHðtÞ dt 

sðyÞ

Z

y

hðtÞeHðtÞ

Z

sðyÞ

y

 Z eHðuÞ DðuÞdu dt  sðyÞhðsðyÞÞeHðsðyÞÞ

t

y

eHðtÞ DðtÞdt:

t

It can be shown, since D is non-decreasing, that the above expression is greater than or equal to

( HðyÞ

hDðyÞ e

Z

y HðtÞ

te

dt  sðyÞe

HðsðyÞÞ

sðyÞ

Z

)

y

e

HðtÞ

dt ;

sðyÞ

which was shown to be P 0 in (3.23). The term with c2 coefficient is given by exactly the same expression as that for the term for c1 but without the factor h. Therefore it is also P 0. The remaining terms are



sðyÞ

c3

Z sðyÞ 0

þ

Z sðyÞ

bðsðyÞ  tÞDðtÞdt 

Z sðyÞ

 Z sðyÞ ðsðyÞ  tÞbðsðyÞ  tÞDðtÞdt  c4 ð1  bðsðyÞ  tÞÞDðtÞdt

0 0

sðyÞðc1 þ c3 ðsðyÞ  tÞÞb ðsðyÞ  tÞDðtÞdt ¼ c3

0

þ

Z sðyÞ

Z sðyÞ

0

tbðsðyÞ  tÞDðtÞdt  c4

0

sðyÞðc1 þ c3 ðsðyÞ  tÞÞb0 ðsðyÞ  tÞDðtÞdt:

Z sðyÞ

ð1  bðsðyÞ  tÞÞDðtÞdt

0

ð3:34Þ

0

Assume that

Z   

x

0

  xðc1 þ c3 ðx  tÞÞb0 ðx  tÞDðtÞdt  < 1:

ð3:35Þ

Also, assume that bðxÞ is bounded below by some constant c > 0. It follows that expression (3.34) behaves asymptotically (at 1) as

’

1 c3 cðsðyÞÞ2 Dð0Þ  c4 sðyÞDðsðyÞÞ ! 1; 2

as

sðyÞ ! 1;

which leads, since D stabilizes for large values of its argument, to the required result.

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The following theorem is then immediate. Theorem 5. Under the assumptions that (i) L1–L4 hold, (ii) the deterioration rate is fixed, (3.35) holds, and b is bounded below by some strictly positive constant, the system of equations given by (3.27) and (3.28) has a unique solution ðt 1 ; T  Þ. Moreover, this solution satisfies 0 < t 1 < T  . Remark 2. Note that it was assumed that bð0Þ ¼ 1, for the analysis to go through. However, a close re-examination of TCU in (2.12) and (2.13) reveals that this assumption is not essential. What is required is that 0 6 bðxÞ 6 1 , with bð0Þ ¼ max fbðxÞ; x 2 Rþ g; and bð0Þ > 0. We close this section by noting that hidden in the analysis of the previous section and this section a simple methodology for finding the optimal EOQ. As a matter of fact this methodology is very general and may be presented as a method for finding the root of a system of non-linear equation with two variables. Consider the problem of finding y > x > 0, which solves the system

Rðx; yÞ ¼ 0;

ð3:36Þ

Gðx; yÞ ¼ 0:

ð3:37Þ

Then, under some technical conditions (these are assumptions related to each particular model) the following algorithm gives the solution of the system of equations given by (3.36) and (3.37). Step 0. Let lb < ub be the limit of an initial estimate of an interval which contains the solution y (the second variable in the system of non-linear equations), and let  be a tolerance value. Goto 1. Step 1. Let y ¼ ðlb þ ubÞ=2, goto 2. Step 2. Find sðyÞ, which solves RðsðyÞ; yÞ ¼ 0, and compute GðsðyÞ; yÞ. Step 3. If jGðsðyÞ; yÞj < , stop, the solution has been found, elseif GðsðyÞ; yÞ > , let ub ¼ y, goto 1, else GðsðyÞ; yÞ < , let lb ¼ y, goto 1. 4. Numerical computations and conclusions This section presents two illustrative examples for two different demand rate functions. Namely, a linear rate function and an exponential rate function. To make things simple we take b  1; hðtÞ ¼ 0:0001, with K ¼ 45; c1 ¼ 10; c2 ¼ 3; c3 ¼ 24; c4 ¼ 30. With these all the assumptions required for the results of the previous sections to go through are satisfied. The objective functions for the 4 possible models were minimized and the results are summarized below. The notation OHP 1 refers to OHP models starting with positive inventory and OHP 2 refers to OHP models starting with shortages. The same interpretation applies to OHDL 1 and OHDL 2 models. These two examples are the base example for a sensitivity analysis which follows after the presentation of the examples: Example 3. Let

DðtÞ ¼



1 þ t; 0 6 t 6 2 3;

t P 2;

The optimal replenishment schedule are given in Table 1 where Type column takes values 0 or 1 or 2, where 0 means that none of the variables t 1 and T are below l; 1 means that t 1 is the only variable below l and 2 indicates that both variables are below l. Example 4. Let

 DðtÞ ¼

expðbtÞ; 0 6 t 6 2; 3; t P 2;

Table 1 Optimal solutions for the linear ramp demand case. Model

t1

T

TCU

Type

OHP 1 OHP 2 OHDL 1 OHDL 2

2.0361 0.48626 2.84229 0.686257

2.29073 2.47767 3.19881 3.42847

48.3328 47.9259 25.6167 24.6941

0 1 0 1

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where b ¼ 12 lnð3Þ. Note that b was selected so that the demand rates in Examples 3 and 4 coincides at t ¼ 0 and t ¼ 2. The optimal replenishment schedule are given in Table 2. Note that the examples illustrates the fact that a slight difference in the cost function may lead to different optimal replenishment policies. However, OHP models compared with each other, although may have different type of policies as in the linear demand rate case, the optimal costs have the same magnitude. This also applies to OHDL models. We shall next present a small scale sensitivity analysis to examine the effect of changes of the parameters K and h (while keeping all other parameters unchanged) on the value of the objective function as well as on the position of the optimal solutions of t1 and T with respect to l. The base are Examples 3 and 4. Tables 3 and 4 contains the results of the sensitivity analysis. The rate of change is computed using the formula



TCUðModelÞ  TCUðBaseÞ %: 100 TCUðBaseÞ It is clear from Tables 3 and 5 that as K increases the optimal values of t1 and T increase for all models. This may lead to the change in the value of Type. Also, a decrease of 44% in the value of K leads to a drop of approximately of 21% to 29% in the costs value depending on the model. A decrease of 22% leads to decrease in the costs between around 10% to 13%, while an increase of 22% leads to an increase between 8% and 12%. Finally, an increase of 44% increases the costs in the range of 14% to 23%. However, for small value of h changes for up to more 400% (while keeping h small) has very small effect on the optimal costs: see Tables 4 and 6. A more thorough numerical study is may be warranted to examine this case. This paper was concerned with finding the optimal order quantity (EOQ) for inventory models with ramp type demand rate, partial backlogging and general deterioration rate. The EOQ for four inventory models were investigated. Each of these models called for its own function of the cost per unit time to be optimized. The fact that the ramp demand rate function is not differentiable, made the use of the second order condition for optimality not possible. To circumvent this problem, this paper proposed a methodology for finding the optimal EOQ. The EOQ was shown to be unique. Also, numerical examples were presented along with a sensitivity analysis. The method proposed in this paper seems to be able to handle more general models with dynamics given by (1.9), or models with inflation or even two-warehouse inventory models as discussed in Agrawal et al. [1]. However, the computations may become messier. But, the essential ideas for the theory to go through are already apparent and present in the paper. Before closing, we would like to go back to [17] and make a number of comments. In the paper cited the time horizon was fixed and the first order condition in their models reduced to solving Rðx; yÞ ¼ 0, for y fixed. This was considered in the pres-

Table 2 Optimal solutions for the exponential ramp demand case. Model

t1

T

TCU

Type

OHP 1 OHP 2 OHDL 1 OHDL 2

1.84616 0.419848 2.80596 0.718932

2.07702 2.20242 3.15687 3.42995

46.1673 46.0500 25.2656 24.4106

1 1 0 1

Table 3 The results for sensitivity analysis with respect to K for Example 3. K

Model

t1

T

Type

TCU

% Change

25

OHP1 OHP2 OHDL1 OHDL2

1:472 0:297167 2:0364 0:485624

1:65608 1:67325 2:29083 2:47654

2 2 0 1

37:8883 37:7637 18:3328 17:9259

21.610 21.204 28.434 27.408

35

OHP1 OHP2 OHDL1 OHDL2

1:70166 0:360131 2:47319 0:59889

1:91445 1:9386 2:78163 2:99604

2 2 0 1

43:4855 43:2988 22:2745 21:5805

10.029 9.655 13.047 12.609

55

OHP1 OHP2 OHDL1 OHDL2

2:47395 0:598035 3:17344 0:759453

2:78333 2:99466 3:57036 3:80879

0 1 0 1

52:2744 51:5805 28:5707 27:4572

8:155 7:626 11:532 11:189

65

OHP1 OHP2 OHDL1 OHDL2

2:84554 0:686022 3:46868 0:823204

3:20163 3:42952 3:90373 4:1516

0 1 0 1

55:6167 54:694 31:2468 29:9697

14:070 14:122 21:978 21:364

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Table 4 The results for sensitivity analysis with respect to h for Example 3. h

Model

t1

T

Type

TCU

% Change

0.001

OHP1 OHP2 OHDL1 OHDL2

2:03075 0:485997 2:84235 0:685977

2:28569 2:47181 3:19827 3:42599

0 1 0 1

48:3562 47:9497 25:626 24:7022

0 0:050 0:036 0:032

0.002

OHP1 OHP2 OHDL1 OHDL2

2:02483 0:48641 2:83944 0:686186

2:28014 2:46656 3:1955 3:42343

0 1 0 1

48:3821 47:9758 25:6362 24:7111

0 0:10 0:076 0:069

0.003

OHP1 OHP2 OHDL1 OHDL2

2:01893 0:48682 2:83654 0:686408

2:2746 2:46135 3:19274 3:42089

0 1 0 1

48:4078 48:0018 25:6463 24:72

0 0:16 0:12 0:10

0.004

OHP1 OHP2 OHDL1 OHDL2

2:01309 0:487228 2:83365 0:686614

2:26911 2:45617 3:18999 3:41837

0 1 1 1

48:4334 48:0277 25:6565 24:7289

0 0:21 0:16 0:14

Table 5 The results for sensitivity analysis with respect to K for Example 4. K

Model

t1

T

Type

TCU

% Change

25

OHP1 OHP2 OHDL1 OHDL2

1:4765 0:280303 1:98194 0:480393

1:66114 1:66632 2:22979 2:43492

2 2 1 1

35:3995 35:3651 17:8388 17:5983

23.32 23.20 29.39 27.91

35

OHP1 OHP2 OHDL1 OHDL2

1:66086 0:33241 2:42886 0:614757

1:86855 1:87588 2:73261 2:97933

2 2 0 1

41:0581 41:0038 21:8697 21:2908

11.066 10.84 13.44 12.78

55

OHP1 OHP2 OHDL1 OHDL2

2:27829 0:570979 3:13805 0:805008

2:5632 2:79752 3:5305 3:82222

0 1 0 1

50:5138 50:0478 28:2563 27:1681

9:415 8:67 11:83 11:30

65

OHP1 OHP2 OHDL1 OHDL2

2:67669 0:684045 3:43821 0:878835

3:01143 3:27616 3:8682 4:17376

0 1 0 1

54:1014 53:3398 30:9595 29:6691

17:18 15:83 22:53 21:54

Table 6 The results for sensitivity analysis with respect to h for Example 4. h

Model

t1

T

Type

TCU

% Change

0.001

OHP1 OHP2 OHDL1 OHDL2

1:84309 0:419941 2:80338 0:719139

2:07445 2:19785 3:15442 3:42771

1 1 0 1

46:1852 46:0688 25:2743 24:4183

0:04 0:04 0:03 0:03

0.002

OHP1 OHP2 OHDL1 OHDL2

1:83971 0:42004 2:80053 0:719367

2:07164 2:1928 3:1517 3:42522

1 1 0 1

46:205 46:0897 25:2839 24:4269

0:08 0:09 0:07 0:07

0.003

OHP1 OHP2 OHDL1 OHDL2

1:83637 0:420135 2:79768 0:719595

2:06886 2:18777 3:14898 3:42275

1 1 0 1

46:2248 46:1105 25:2936 24:4354

0:12 0:13 0:11 0:10

0.004

OHP1 OHP2 OHDL1 OHDL2

1:83307 0:420226 2:79485 0:719823

2:06611 2:18277 3:14628 3:42028

1 1 0 1

46:2445 46:1312 25:3032 24:444

0:17 0:18 0:15 0:14

M.A. Ahmed et al. / Applied Mathematics and Computation 219 (2013) 4288–4307

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