An inventory control problem for deteriorating items with back-ordering and financial considerations

Applied Mathematical Modelling 38 (2014) 93–109

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Applied Mathematical Modelling journal homepage: www.elsevier.com/locate/apm

An inventory control problem for deteriorating items with back-ordering and financial considerations Ata Allah Taleizadeh a,⇑, Mohammadreza Nematollahi b a b

School of Industrial and Systems Engineering, College of Engineering, University of Tehran, Tehran, Iran Department of Industrial Engineering, Iran University of Science and Technology, Tehran, Iran

a r t i c l e

i n f o

Article history: Received 15 June 2012 Received in revised form 28 February 2013 Accepted 31 May 2013 Available online 26 June 2013 Keywords: Inventory control Delayed payment Deterioration Inflation Time value of money Shortage

a b s t r a c t This paper investigates the effects of time value of money and inflation on the optimal ordering policy in an inventory control system. We proposed an economic order quantity model to manage a perishable item over the finite horizon planning under which backordering and delayed payment are assumed. The demand and deterioration rates are constant. The present value of total cost during the planning horizon in this inventory system is modeled first, then a three phases solution procedure is proposed to derive the optimal order and shortage quantities, and the number of replenishment during the planning horizon. Finally, the proposed model is illustrated through numerical examples and the sensitivity analysis is reported to find some managerial insights. Ó 2013 Elsevier Inc. All rights reserved.

1. Introduction In real world problem, deterioration of many items such as chemicals, volatile liquids, blood banks, medicines and some other goods during storage period is non-negligible. In general, deterioration is defined as the decay, damage, spoilage, evaporation and obsolescence of stored items and it results in decreasing usefulness [1]. So, the management and holding of inventories of perishable items becomes an important problem for inventory managers. In the other hand, delayed payment is an important form of financing for businesses in a broad range of industries and economies that extremely well-developed and are being used in financial markets. Furthermore, both inflation and time value of money issues will have main effects in financial markets. All of the above mentioned issues (deteriorating items, delayed payment, inflation and time value of money) are separately regarded in some inventory models. But as is shown in Table 1 there are a few researches in which only number of all of the considered topics are mentioned together. Also there is no research in which all of them had been considered. For instance, Ghare and Schrader [22] were the pioneers to establish an inventory model for deteriorating items. Covert and Philip [18] extended Ghare and Schrader’s constant deterioration rate to a two-parameter Weibull distribution. In continue this topic was investigated with many researchers. Aggarwal and Jaggi [2] proposed an inventory control model for deteriorating items in which shortage was not permitted. Hariga and Ben-daya [24] developed lot-sizing problem with timedependent demand under inflationary conditions. Ray and Chaudhuri [51] developed a finite time-horizon EOQ model with backordering where the varying demand rate whereas the effects of inflation and the time value of money are taken into account. Chen [10] considered situation in which the demand rate is time-proportional and shortages are fully Backordered. He investigated the effects of inflation and time⇑ Corresponding author. Tel.: +98 9363447171. E-mail addresses: [email protected], [email protected] (A.A. Taleizadeh), [email protected] (M. Nematollahi). 0307-904X/$ - see front matter Ó 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.apm.2013.05.065

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Table 1 Brief literature review. No

Inventory Control System

[2] [3] [4] [5] [6] [7] [8] [1] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45] [46] [47] [48] [49] [50] [51] [52] [53] [54] [55] [56] [57] [58] [59] [60] [61] [62] [63]

EOQ EPQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EPQ EOQ EOQ EPQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EOQ EPQ EOQ EOQ EOQ EOQ EOQ EOQ EPQ EOQ EOQ EOQ EOQ EPQ EPQ EOQ EPQ EOQ EOQ EOQ

Shortage

[ x

Deterioration

Delayed Payment

[ [

[

x

x x

Inflation

[ x

Time Value of Money

[

x [

[ [ [

[ [ [ [ [

[

[ [

[

[ [

x [ [

[

[

[ [ [

[ [

[

[

[

[ [

[ [ [ [

[ [ [ [ [ [

[ [ [

[ [ [

[ [

[

[

[ [ [

[ [ [

[ [ x x

x

[ [ [ [ [ [ [ [

[ [

[ [ [ [

[

[

[ [

[ [

[

[ [ [

[ [ [ [ [ [

[

[ [ [

[ [

[ [

[

[

[ [ [ [

[ [ [ [ [

[

[ [

[ [ [ [

[ [

[ [

[ [

[

[ [ [ [ [

[ [ [

[ [ [

[

Solution Method Non-closed-form Closed-form Closed-form Closed-form Closed form Closed-form Closed-form Closed-form Closed-form Non-closed-form Non-closed-form Non-closed-form Closed-form Closed-form Non-closed-form Closed-form Closed-form Closed-form Non-closed-form Non-closed-form Non-closed-form Closed-form Non-closed-form Non-closed-form Non-closed-form Non-closed-form Closed-form Closed-form Closed-form Closed-form Non-closed-form Closed-form Non-closed-form Non-closed-form Closed-form Closed-form Non-closed-form Closed-form Non-closed-form Non-closed-form Closed-form Closed-form Non-closed-form Non-closed-form Non-closed-form Closed-form Non-closed-form Non-closed-form Closed-form Closed-form Non-closed-form Non-closed-form Non-closed-form Non-closed-form Closed-form Non-closed-form Non-closed-form Non-closed-form Closed-form Closed-form Non-closed-form Non-closed-form Closed-form

value of money too. Chung and Lin [15] developed an EOQ model for deteriorating items taking account of time value of money over a fixed planning horizon. Liao and Chen [41] developed an EOQ model under a situation in which the effects of the inflation, deterioration, initial stock-dependent demand rates and wholesaler’s permissible delay in payment are dis-

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95

cussed while shortage is not allowed. Yang [61] developed two-warehouse inventory management problem for perishable products with constant demand rate and backordering shortages under inflation. In this paper in contrast to the traditional deterministic two-warehouse inventory model with shortages at the end of each replenishment cycle, an alternative model in which each cycle begins with shortages and ends without shortages is proposed here [61]. Chang [7] extended an EOQ model under a situation in which the retailer considers permissible delay of payments for purchaser if he/she orders a large quantity. In this work shortages are not permitted and the effects of the deterioration and inflation rates, and delayed payment policy are as well discussed. Hou [25] developed an EOQ model over finite planning horizon for perishable products with stock-dependent demand rate and full backordering under inflation and time value money. Ghosh and Chaudhuri [23] developed an EOQ model with a quadratic demand, shortages and time-proportional deterioration in all cycles while Mukhopadhyay et al. [48] developed and EOQ model for perishable items in which demand is price dependent. Dye [21] extended joint pricing and ordering policy for perishable products whereas shortage is partially being backlogged. Mohan and Gopalakrishnan [46] proposed a multi products EOQ model with delayed payment and budget limitation. Chang et al. [9] investigated an inventory model for deterioration items in which both delayed payment and inflation are considered. Chung and Huang [17] developed an EOQ model with delayed payment and full backordering shortage. In continue, Hu and Liu [26] developed the EPQ model with delayed payment and full backordering shortage. Jaggi et al. [36] developed EOQ model for perishable products whereas partial backordering, linear trend in demand and inflation are assumed. Sarkar and Moon [58] investigated the effects of inflation in an imperfect production system. The effects of time-value of money and inflation on order quantity of the product with random life cycle are investigated by Moon and Lee [64]. Sarkar et al. [65] extended an EOQ model for perishable items under inflation, permissible delay of payment and backordering. Huang [66] developed an integrated single vendor single buyer model under conditions of order processing cost reduction and permissible delay in payments when shortage is not permitted. Chen and Kang [67] developed and integrated inventory control model with permissible delay in payments for determining the optimal order quantity time and replenishment frequency. Chung [68] studied a supply chain problem in which optimal order quantity with partially permissible delay in payments derived analytically. Cheng et al. [69] developed an inventory control model for deteriorating items in which trapezoidal type demand and partial backlogging are assumed. Chung and Lin [70] studied trade credit issue in a supply chain model proposed to determine the order quantity of a deteriorating item. Urban [71] developed the work of Min et al. [19] with relaxing the boundary condition imposed in their model that ensures the entire stock is depleted at the end of each order cycle and resolving the potential unbounded solution resulting from a linear demand function by constraining the maximum inventory level. More over some related articles are Cárdenas-Barrón [72–74], Widyadana et al. [75], Chung et al. [76] and Wee et al. [77,78]. Above literature review of the mentioned topic and Table 1 show EOQ model is not developed under situation in which shortage, inflation, time value of money and delayed payment over the fixed planning horizon are investigated. So in this paper we want to investigate these issues together and derive a comprehensive model to determine the economic order quantity.

2. Assumptions and notations The mathematical model in this paper is developed with the following assumptions: Planning horizon is finite. Single item inventory control is mentioned. Demand and deterioration rates are constant. Deterioration occurs as soon as the items are received into inventory. There is no replacement or repair of deteriorating items during the period under consideration. Shortage is allowed and will totally be backordered. Delayed payment strategy is offered by the seller. Inflation rate is constant and time value of money is considered. The lead time is zero. The inventory level at the end of the planning horizon will be zero. The cost factors are deterministic, but will be inflated under constant inflation rate. The order and shortage quantities are continues decision variables. The number of replenishments are restricted to integer one. The total relevant cost consists of fixed ordering, purchasing, holding, shortage, interest payable, and interest earned from sales revenue during the permissible period.  The last order is only being placed to satisfy the shortage of last period.  The goal of the model is to determine the decision variables under which the net present value of total cost during the horizon planning is minimized.              

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Under above assumptions following notations are used to develop the model where ðj ¼ 0; 1; 2; . . . ; NÞ. 2.1. Parameters: A : The fixed ordering cost per replenishment, $/order, C : The unit purchasing price at time zero, $/item, CðtÞ : The unit purchasing cost at time t, ðCðtÞ ¼ CeRt Þ, D: The constant demand rate per unit time, H: The length of the finite planning horizon, i: The constant inflation rate, IðtÞ: The inventory level at time t, Ic : The interest charged per $ per year by the supplier, Ie : The interest earned per $ per year by the purchaser, Ih : The holding cost rate per unit time excluding interest charges, M: The permissible delay in settling account, r: The discount rate, representing the time value of money, R: r  i, representing the net constant discount rate of inflation, T: The length of each replenishment cycle, T j : The total time that is elapsed up to and including the jth replenishment cycle where T 0 ¼ 0, T 1 ¼ T, and T N ¼ H, tj = The time at which the inventory level in the jth replenishment cycle drops to zero, V : The unit selling price at time zero, $/item, VðtÞ: The selling price per unit at time t, VðtÞ ¼ VeRt , h: The constant deterioration rate, units/unit time, p: The shortage cost per unit time, $/unit/unit time, pðtÞ :: The shortage cost per unit per unit time at time t, ðpðtÞ ¼ peRt Þ. 2.2. Decision variables F: The fraction of replenishment cycle where the net stock is positive, N: The number of replenishments during the planning horizon, N ¼ HT. 2.3. Dependent variables C s : The present value of shortage cost during the first replenishment cycle, C P : The present value of holding cost during the first replenishment cycle, Ib : The maximum shortage quantity, Im : The maximum inventory level, IC: The interest payable during the first replenishment cycle, IE: The interest earned during the first replenishment cycle, Q : The order quantity in each replenishment, TC A : The total fixed ordering cost during, ð0; HÞ TC h : The total holding cost during, ð0; HÞ TC P : The total purchasing cost during, ð0; HÞ TC S : The total shortage cost during, ð0; HÞ TIC: The total interest cost during, ð0; HÞ TIE: The total interest earned during, ð0; HÞ TC: The total relevant cost during, ð0; HÞ. 3. Model formulation As it shown in Fig. 1, we assume the length of planning horizon H ¼ NT; where N is an integer decision variable representing the number of replenishments to be made during H; and T is time between two replenishments. The inventory level at time t decreases because of demand and deterioration. So according to this description the changes of inventory respect to time can be shown using the following differential Equations;

dIðtÞ þ hIðtÞ ¼ D; dt dIðtÞ ¼ D; dt

0 6 t 6 t1 ;

t1 6 t 6 T:

Considering the boundary conditions, Iðt1 Þ ¼ 0, for the first differential equation gives;

ð1Þ

ð2Þ

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I (t )

Im

Q t1 =

FH

t2 =

N

( F + 1) H

tN =

N

Ib T0 = 0

T1 =

H

T2 =

N

( F + N − 1) H

2H N

T N −1 =

( N − 1) H

N

t

TN = H

N

Fig. 1. Inventory control diagram.

IðtÞ ¼

 D  hðt1 tÞ e 1 ; h

0 6 t 6 t1

ð3Þ

And for the second differential equation we have;

IðtÞ ¼ Dðt  t 1 Þ;

t 1 6 t 6 T:

ð4Þ

According to Eqs. (3) and (4) the maximum inventory quantity at the beginning of each period and maximum shortage quantity at the end of each period are:

i  D h h FH D  hðt1 Þ e 1 ¼ e N 1 ; h h   H FH : Ib ¼ DðT  t 1 Þ ¼ D  N N Im ¼ Ið0Þ ¼

ð5Þ

ð6Þ

Since we have mentioned the delayed payment policy, the total inventory cost of buyer depends on the delayed payment time, M, which may be before t 1 ðM 6 t 1 Þ or after this time ðM > t1 Þ. So firstly we determine the same costs which exist in both cases and in continue the specific costs of each case will be determined. 3.1. Fixed ordering cost Since the numbers of replenishment or period is N, the fixed ordering coat over the planning horizon under net present value and inflation considerations is;

TC A ¼

 ðNþ1ÞRT  N N X X e 1 : AðjTÞ ¼ AeðjRTÞ ¼ A eRT  1 j¼0 j¼0

ð7Þ

Since the independent decision variables are N and F, using T ¼ HN, Eq. (7) can be changed to:

TC A ðN; FÞ ¼ A

" Nþ1 # eð N ÞRH  1 RH

e N  1

:

ð8Þ

3.2. Holding cost excluding interest cost In order to determine the holding cost, firstly the average inventory quantity should be determined. Using Eq. (3) the average inventory is;

I¼

Z 0

t1

IðtÞ dt ¼

Z 0

t1

    D  hðt1 tÞ D 1 eht1 D ¼ 2 eht1  ht 1  1 : e  1 dt ¼   t1 þ h h h h h

ð9Þ

Then, using Eq. (9) the holding cost over the planning horizon under net present value and inflation considerations is;

TC h ¼

  N1 N1 X X  eNRT  1 Ih CD  ; Ih C ðjÞ I ¼ Ih CeðjRTÞ I ¼ 2 eht1  ht 1  1 eRT  1 h j¼0 j¼0

ð10Þ

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T ¼ HN yields to;

Ih CD

TC h ðN; FÞ ¼

h2

  RH  hFH e 1 hFH : eN  1 RH N e N  1

ð11Þ

3.3. Shortage cost Similar to the holding cost, firstly the average shortage should be determined which is shown in Eq. (12).

¼ B

Z

H

T

N

IðtÞ dt ¼

t1

  2  FH  D HN  FH D H FH N N ¼  : 2 2 N N

ð12Þ

Using Eq. (12), the shortage cost over the planning horizon under net present value and inflation considerations is;

TC S ¼

N1 X

N1 X

j¼0

j¼0

pðjÞB ¼

peðjRTÞ B ¼ p



 eNRT  1 B; RT e 1

ð13Þ

T ¼ HN yields to;



TC S ðN; FÞ ¼

pD H FH 2

N



2  RH  e 1 RH

N

e N  1

ð14Þ

:

3.4. Purchasing cost According to Fig. 1, the purchasing cost of jth cycle is being calculated by Eq. (15).

C p ðjÞ ¼ C ðjÞ Im þ C ðjþ1ÞT Ib ¼

  i C ðjÞ D h hFH H FH ; e N  1 þ C ðjþ1ÞT D  N N h

j ¼ 0; 1; 2; . . . ; N:

ð15Þ

So, the total purchasing cost over the planning horizon with considering T ¼ HN will be;

TC p ðN; FÞ ¼

   N 1 ieRH  1 X CD h hFH H FH eRH  1 RH N þ CDe : C p ðjÞ ¼ e N 1  RH RH h N N e N  1 e N  1 j¼0

ð16Þ

3.5. Interest charged and earned Regarding interests charged and earned, we have the following two possible cases based on the values of M 6 t 1 and M 6 t1 which are described as follow. 3.5.1. Case 1: M 6 t 1 3.5.1.1. Interest earned. As items are sold and before the replenishment account is settled, the sales revenue is used to earn interest. At the beginning of the time interval, the backordered quantity which is Ib , should be replenished first and the maxRM imum accumulated sold until M is equal to 0 Dtdt. For the first case which is shown in Fig. 2, the interest earned for the first cycle is;

 Z IE1 ¼ Ie Vð0Þ Ib M þ 0

M

" #    H FH DM2 þ Dtdt ¼ Ie Vð0Þ MD  N N 2

ð17Þ

And total interest earned using T ¼ HN will be;

TIE1 ðN; FÞ ¼

" #    N1 N1 X X H FH DM 2 eRH  1 þ : IE1 ðjÞ ¼ IE1 eðjRTÞ ¼ VIe MD  RH N N 2 e N  1 j¼0 j¼0

ð18Þ

3.5.1.2. Interest charged. When the replenishment account is settled, the situation is reversed and effectively the items still in Rt stock, which is equal to M1 IðtÞ dt, have to be financed at interest rate Ic . So for the first period the interest payable is;

IC 1 ¼ Ic Cð0Þ

Z

t1

M

 Z IðtÞ dt ¼ Ic Cð0Þ

t1

M

    D  hðt1 tÞ Ic Cð0ÞD 1 ehðt1 MÞ e  1 dt ¼   t1 þ þM h h h h

And total interest payable over the planning horizon using T ¼ HN will be;

ð19Þ

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99

I (t )

D

M

t1

Time

Ib Fig. 2. Calculating interested charged and earned when M 6 t1.

" #  FH N1 N1 X X CIc D 1 FH ehð N MÞ eRH  1 ðjRTÞ : TIC 1 ðN; FÞ ¼ IC 1 ðjÞ ¼ IC 1 e ¼   þ þM RH h h N h e N  1 j¼0 j¼0

ð20Þ

So the total cost function of the first case, M 6 t1 , is;

1 RH  þ CDe N HN  FH N i h i C B hFH I CD C hFH CD hFH " Nþ1 # B B þ h e N  1 þ hh2 e N  N  1 C RH  ð N ÞRH C e B e 1 1   C B FHM þ TC 1 ðN; FÞ ¼ A h RH C eRHN  1 : B CIc D e ð N Þ e N  1 þ M  1h  FH C Bþ h N h C B A @ h H FH DM2 i VIe MD N  N þ 2 0

 pD H  FH 2 2

N

h

N

ð21Þ

3.5.2. Case 2: M 6 t1 3.5.2.1. Interest earned. Similar to the first case, at the beginning of the time interval, the backordered quantity which is Ib , should be replenished first and its interest earned for the first cycle will be Ie Vð0ÞMIb . Then the maximum accumulated sold Rt until M is equal to 01 Dtdt while the interest earned will be

"

Ie Vð0Þ ðM  t 1 ÞDt1 þ

# Dt21 : 2

So, for the first case which is shown in Fig. 3, the interest earned for the first cycle is;

3 3 zfflfflfflfflfflffl ffl }|fflfflfflfflfflffl ffl { " #     7 6z}|{ zfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflffl{ Z M H FH FH FH DF 2 H2 7 6 þ M D IE2 ¼ Ie Vð0Þ6 Ib M þ ðM  t1 ÞDt1 þ Dt1 dt 7 ¼ Ie Vð0Þ MD  þ 5 4 N N N N 2N2 0 2

2

1

ð22Þ

And total interest earned over the horizon planning using T ¼ HN will be;

" #      N1 N1 X X H FH FH FH DF 2 H2 eRH  1 ðjRTÞ þ M D : TIE2 ðN; FÞ ¼ IE2 ðjÞ ¼ IE2 e ¼ Ie Vð0Þ MD  þ RH N N N N 2N2 e N  1 j¼0 j¼0

ð23Þ

3.5.2.2. Interest charged. In this case when the replenishment account is settled, the number of items which are in stock is zero, so the interest payable will be zero. So the total cost function of the first case, M > t 1 , is;

TC 2 ðN; FÞ ¼ A

" Nþ1 # eð N ÞRH  1 RH

e N  1

i h i1 hFH FH e N  hFH  1 þ CD eh N  1  N h C eRH  1 C B þ B þCDeRHN H ð1  FÞ þ pD H  FH 2 C RH : A e N 1 @ N 2 N N    FH  VIeNDH Mð1  FÞ þ M  2N F 0

Ih CD B h2

h

ð24Þ

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A.A. Taleizadeh, M. Nematollahi / Applied Mathematical Modelling 38 (2014) 93–109

I (t )

D

Dt 1

(2)

(3)

(1) T

t1

Ib

M

t

Fig. 3. Calculating interested charged and earned when M > t1.

So, the final objective function of our model becomes;

 Min : TCðN; FÞ ¼

TC 1 ðN; FÞ; M 6 t 1 TC 2 ðN; FÞ; M > t1

S:t : 06F61

ð25Þ

N P 0; Integer: In the following section we will discuss about the solution method. 4. Solution method In real world problem, the values of deterioration and inflation rates are usually very small. Utilizing a truncated Taylor series expansion for the exponential term, we have;

1 2 ehk ¼ 1 þ hk þ ðhkÞ : 2

ð26Þ

Using the above approximation, the total relevant cost of the first case, when M 6 t1 , (shown in Eq. (21)) can be rewritten as;

h  i 8 9  2 1 hFH 2 > > A þ IhhCD þ p2D HN ð1  FÞ2 > > 2 2 N > > > > < =eRH  1 h i   RH N H CD hFH 1 hFH 2 FH þ AeRH : TC 1 ðN; FÞ ¼ þ h N þ 2 N  þ CDe RH N N > > e N  1 > > h i h i > > > : þ CIc D 1 h2 FH  M 2  VI MDH  FH þ DM2 > ; e 2 N N N 2 h2

ð27Þ

The first derivative of TC 1 ðN; FÞ is;

@TC 1 ¼ @F

(

2 Ih CDH2 F þ pDH ðF  1Þ þ CDH N N2 N2   þ CIcNDH F HN  M þ VIe MDH N



 CDH RH ) RH  1 þ F hH  N e N e 1 N RH

e N  1

ð28Þ

And the second order derivative is;

@ 2 TC 1 @F 2

¼

( Ih CDH2 N2

þ

pDH2 N2

þ

    ) RH  CDH hH CIc DH H e 1 þ þ P 0: RH N N N N e N  1

ð29Þ

Which is strongly positive and shows the first objective function for any given integer N, is convex over F. Setting Eq. (28) equal to zero gives; RH

F1 ¼

pH  CN þ CNe N þ ðCIc  VIe ÞMN : ðp þ Cðh þ Ic þ Ih ÞÞH

ð30Þ

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101

Since we have assumed that inflation rate is ever more than discount rate, so Eq. (30) is always non-negative. In continue, using Eq. (26), the total relevant cost of the second case, when M > t1 , (shown in Eq. (24)) can be rewritten as;

h  i 8 9  2 Ih CD 1 hFH 2 > > A þ þ p2D HN ð1  FÞ2 > > 2 2 N h > > > > < =eRH  1 h i  RH  2  hFH 1 hFH N H  FH þ AeRH : TC 2 ðN; FÞ ¼ þ CD þ þ CDe h N 2 N N N RH > > N  1 e > > h i> > > : VI MDH  FH þ M  FH D FH þ DF 2 H2 > ; e N N N N 2N2

ð31Þ

The first derivative of TC 2 ðN; FÞ is;

8   9 Ih CDH2  pDH2 CDH hH @TC 2 < N2 F þ N2 ðF  1Þ þ N 1 þ F N = eRH  1

¼ :  CDH eRHN þ VI DH2 F ; eRH @F N  1 e N2 N

ð32Þ

And the second order derivative is;

@ 2 TC 2 @F 2

¼

( Ih CDH2 N2

þ

pDH2 N2

þ

!)    CDH hH DH2 eRH  1 þ VIe P 0: RH N N N2 e N  1

ð33Þ

Which is strongly positive and shows the second objective function for any given integer N, is convex over F. Setting Eq. (32) equal to zero gives; RH

F2 ¼

pH  CN þ CNe N : ðIh C þ p þ Ch þ VIe ÞH

ð34Þ

Since we have assumed that inflation rate is ever more than discount rate, so Eq. (34) is always non-negative. In order to determine the optimal policy, following solution procedure is proposed. 4.1.1. The first phase A. Deriving the optimal value of N 1 and F 1 . A.1. Firstly consider n1 ¼ 1 and using Eq. (30), calculate F 1 ðn1 Þ. A.2. Using Eq. (27), calculate the objective function value (TC 1 ðn1 ; F 1 ðn1 ÞÞ. A.3. Consider n1 ¼ n1 þ 1 and using Eq. (30) calculate F 1 ðn1 Þ. A.4. Using Eq. (27), calculate the objective function value (TC 1 ðn1 ; F 1 ðn1 ÞÞ. A.5. If TC 1 ðn1 ; F 1 ðn1 Þ > TC 1 ðn1  1; F 1 ðn1  1ÞÞ, consider N 1 ¼ n1  1, F 1 ¼ F 1 ðn1  1Þ and go to step B, else go to step A.3 and continue this procedure until the stopping criteria is met. B. Deriving the optimal value of N 2 and F 2 . B.1. Firstly consider n2 ¼ 1 and using Eq. (34) calculate F 2 ðn2 Þ. B.2. Using Eq. (31), calculate the objective function value (TC 2 ðn2 ; F 2 ðn2 ÞÞ. B.3. Consider n2 ¼ n2 þ 1 and using Eq. (34) calculate F 2 ðn2 Þ. B.4. Using Eq. (31), calculate the objective function value (TC 2 ðn2 ; F 2 ðn2 ÞÞ. B.5. If TC 2 ðn2 ; F 2 ðn2 ÞÞ > TC 2 ðn2  1; F 2 ðn2  1ÞÞ, consider N 2 ¼ n2  1, F 2 ¼ F 2 ðn2  1Þ and go to the second phase, else go to step B.3 and continue this procedure until the stopping criteria is met. 4.1.2. The second phase According to the results of the first phase, following four possible cases may occur.  The First Possibility: M 6 F 1 ðnn111ÞH & M > F 2 ðnn221ÞH : 1 1 In this case, do the following steps: C. Compare the both objective functions of TC 1 ðn1  1; F 1 ðn1  1ÞÞ and TC 2 ðn2  1; F 2 ðn2  1ÞÞ. C.1. If TC 1 ðn1  1; F 1 ðn1  1ÞÞ 6 TC 2 ðn2  1; F 2 ðn2  1ÞÞ then N  ¼ n1  1 and F  ¼ F 1 ðn1  1Þ. C.2. If TC 1 ðn1  1; F 1 ðn1  1ÞÞ > TC 2 ðn2  1; F 2 ðn2  1ÞÞ then N  ¼ n2  1 and F  ¼ F 2 ðn2  1Þ. C.3. Now go to the third phase. & M > F 2 ðnn221ÞH :  The second possibility: M > F 1 ðnn111ÞH 1 1 In this case, since the first criterion is not met following steps should be performed to obtain the new values of ðn1 ; F 1 ðn1 ÞÞ. C. Deriving the optimal value of new N 1 and F 1 .

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D.1. Firstly consider n01 ¼ 1 and F 01 ðn01 Þ ¼ 1H . D.2. Using Eq. (27), calculate the objective function value (TC 1 ðn01 ; F 01 ðn01 ÞÞ. D.3. Consider n01 ¼ n01 þ 1 and calculate F 01 ðn01 Þ ¼

n01 M . H

D.4. Using Eq. (27), calculate the objective function value (TC 1 ðn01 ; F 01 ðn01 ÞÞ). D.5. If TC 1 ðn01 ; F 01 ðn01 ÞÞ > TC 1 ðn01  1; F 01 ðn01  1ÞÞ, consider N 1 ¼ n01  1, F 1 ¼ F 01 ðn01  1Þ and go to step E, else go to step D.3 and continue this procedure until the stopping criterion is met. D. Compare the both objective functions of new TC 1 ðn01  1; F 01 ðn01  1ÞÞ and TC 2 ðn2  1; F 2 ðn2  1ÞÞ. E.1. If TC 1 ðn01  1; F 01 ðn01  1ÞÞ 6 TC 2 ðn2  1; F 2 ðn2  1ÞÞ then N  ¼ n01  1 and F  ¼ F 01 ðn01  1Þ. E.2. If TC 1 ðn01  1; F 01 ðn01  1ÞÞ > TC 2 ðn2  1; F 2 ðn2  1ÞÞ then N  ¼ n2  1 and F  ¼ F 2 ðn2  1Þ. E.3. Go to the third phase.  The third possibility: M 6 F 1 ðnn111ÞH & M 6 F 2 ðnn221ÞH : 1 1 In this case, since the second criterion is not met, following steps should be performed to obtain the new values of ðn2 ; F 2 ðn2 ÞÞ. G. Deriving the optimal value of new N 2 and F 2 . G.1. Firstly considern02 ¼ 1 and calculateF 02 ðn02 Þ ¼

n02 M . H

G.2. Using Eq. (31), calculate the objective function value (TC 2 ðn02 ; F 02 ðn02 ÞÞ. G.3. Consider n02 ¼ n02 þ 1 and calculate F 02 ðn02 Þ ¼

n02 M . H

G.4. Using Eq. (31), calculate the objective function value (TC 2 ðn02 ; F 02 ðn02 ÞÞ. G.5. If TC 2 ðn02 ; F 02 ðn02 ÞÞ > TC 2 ðn02  1; F 02 ðn02  1ÞÞ, consider N 2 ¼ n02  1, F 2 ¼ F 02 ðn02  1Þ and go to step J, else go to step G.3 and continue this procedure until the stopping criteria is met. J. Compare the both objective functions of new TC 1 ðn1  1; F 1 ðn1  1ÞÞ and TC 2 ðn02  1; F 02 ðn02  1ÞÞ. E.1. If TC 1 ðn1  1; F 1 ðn1  1ÞÞ 6 TC 2 ðn02  1; F 02 ðn02  1ÞÞ then N  ¼ n1  1 and F  ¼ F 1 ðn1  1Þ. E.2. If TC 1 ðn1  1; F 1 ðn1  1ÞÞ > TC 2 ðn02  1; F 02 ðn02  1ÞÞ then N  ¼ n02  1 and F  ¼ F 02 ðn02  1Þ. E.3. Go to the third phase.  The fourth possibility: M > F 1 ðnn111ÞH & M 6 F 2 ðnn221ÞH : 1 1 In this case, since the both criteria are not met, following steps should be performed to obtain the new values of ðn1 ; F 1 ðn1 ÞÞ and ðn2 ; F 2 ðn2 ÞÞ. J. Deriving the new optimal value of ðN 1 ; F 1 Þ and ðN 2 ; F 2 Þ. K.1. Go to step D of the second possibility and do D.1–D.5 until getting new value of ðN 1 ; F 1 Þ. K.2. Go to step G of the third possibility and do G.1–G.5 until getting new value of ðN 2 ; F 2 Þ. K. Compare the both objective functions of new TC 1 ðn1  1; F 1 ðn1  1ÞÞ and TC 2 ðn02  1; F 02 ðn02  1ÞÞ. L.1. If TC 1 ðn01  1; F 01 ðn01  1ÞÞ 6 TC 2 ðn02  1; F 02 ðn02  1ÞÞ then N  ¼ n01  1 and F  ¼ F 01 ðn01  1Þ. L.2. If TC 1 ðn01  1; F 01 ðn01  1ÞÞ > TC 2 ðn02  1; F 02 ðn02  1ÞÞ then N  ¼ n02  1 and F  ¼ F 02 ðn02  1Þ. L.3. Go to the third phase. 4.1.3. The third phase According to the results of the second phase, following steps should be done to calculate the dependent variables. P. Calculate dependent variables. h F H i P.1. Calculate maximum inventory level using Im ¼ Dh eh N  1 .    P.2. Calculate maximum shortage level using Ib ¼ D NH  FNH . P.3. P.4.

Calculate order quantity using Q ¼ Im þ Ib . Calculate relevant total cost factor using N  and F  obtained in the second phase.

5. Numerical examples In order to demonstrate the solution method, in this section several examples are prepared. 5.1. The first example 2 For the first example suppose, A ¼ 150, C ¼ 3, D ¼ 1000, H ¼ 2, Ic ¼ 0:10, Ie ¼ 0:06, Ih ¼ 0:3, M ¼ 12 , p ¼ 2, R ¼ 0:05, V ¼ 5, h ¼ 0:1. From the proposed solution method we have N  ¼ 3, F  ¼ 0:529 and finally TC  ðN  ; F  Þ ¼ 6747:14. In continue we have;

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T ¼

H 2 ¼ ¼ 0:666; N 3

t1 ¼ F  T  ¼ 0:352; Im ¼

i D h hFH e N  1 ¼ 359:15; h

  H FH ¼ 313:81; Ib ¼ D  N N Q  ¼ Im þ Ib ¼ 672:96: In this case, some other numerical examples, prepared based on changing h, R, M values and taking one parameter at a time, keeping the remaining parameters at their original levels. The results of this numerical analysis are shown in Table 2. Also some numerical examples, prepared based on changing two of h, R, M values and at a time, keeping the remaining parameters at their original levels. The results of only objective are shown in Table 3. Also Fig. 4 shows that the objective function for any given N is convex over F and the optimal value is derived at N  ¼ 3; F  ¼ 0:529; TC  ¼ 6747:14. 5.2. The second example 3 For the second example suppose, A ¼ 500, C ¼ 5, D ¼ 1000, H ¼ 5, Ic ¼ 0:12, Ie ¼ 0:08, Ih ¼ 0:5, M ¼ 12 , p ¼ 3, R ¼ 0:2,      V ¼ 7, h ¼ 0:1. From the proposed solution method we have N ¼ 7, F ¼ 0:315 and finally TC ðN ; F Þ ¼ 19628:60. In continue we have;

T ¼

H 5 ¼ ¼ 0:714; N 7

t1 ¼ F  T  ¼ 0:225; Im ¼

i D h hFH e N  1 ¼ 227:74; h

  H FH ¼ 489:10; Ib ¼ D  N N Q  ¼ Im þ Ib ¼ 716:84 Similar to the first case, In this case, some other numerical examples, prepared based on changing h, R, M values and taking one parameter at a time, keeping the remaining parameters at their original levels. The results of this numerical analysis are shown in Table 4. Also some numerical examples, prepared based on changing two of h, R, M values and at a time, keeping the remaining parameters at their original levels. The results of only objective are shown in Table 5. Also Fig. 5 shows that the objective function for any given N is convex over F and the optimal value is derived at N  ¼ 7; F  ¼ 0:315; TC  ¼ 19628:60.

Table 2 Some numerical examples on the data set of the first example. Parameters

Values

N⁄

F⁄

T⁄

Im

Ib

Q⁄

TC  ðN  ; F  Þ

h

0.05 0.10 0.20 0.50 0.05 0.10 0.15 0.20 0.08 0.25 0.50 1.00

3 3 4 4 3 4 4 4 3 3 3 3

0.553 0.529 0.487 0.394 0.529 0.488 0.448 0.408 0.529 0.529 0.529 0.529

0.667 0.667 0.500 0.500 0.667 0.500 0.500 0.500 0.667 0.667 0.667 0.667

372.07 359.15 249.70 207.04 359.16 246.91 226.30 206.24 359.15 359.15 359.15 359.15

297.02 313.81 256.33 302.99 313.81 256.09 276.22 296.85 313.81 313.81 313.81 313.81

670.08 672.97 506.03 510.03 672.97 502.99 502.52 502.10 672.97 672.97 672.97 672.97

6718 6747 6785 6868 6747 6433 6133 5849 6795 6698 6553 6263

R

M

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Table 3 Other analysis on the first example. Total cost

h

M

0.05 0.10 0.20 0.50

0.08

0.25

0.5

1

6767 6795 6833 6916

6670 6699 6737 6820

6525 6554 6592 6675

6235 6263 6303 6386

0.10 6414 6433 6463 6532

0.15 6119 6134 6158 6214

0.20 5838 5849 5869 5914

0.25 6699 6386 6089 5806

0.5 6554 6247 5954 5676

1 6263 5968 5685 5416

R

h 0.05 0.10 0.20 0.50 R

0.05 6719 6747 6785 6868 M

0.05 0.10 0.15 0.20

0.08 6795 6479 6178 5892

Fig. 4. Showing the best result of the first example.

6. Sensitivity analysis To study the effects of the parameter together, changes on the optimal results derived by the proposed method, a sensitivity analysis is performed by changing the values of some parameters and taking one parameters at a time, keeping the remaining parameters at their original levels. This analysis is performed on the both first and second numerical examples. The results are shown in Tables 6 and 7 respectively.

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Values

N⁄

F⁄

T⁄

Im

Ib

Q⁄

TC  ðN  ; F  Þ

h

0.05 0.10 0.20 0.50 0.05 0.10 0.15 0.20 0.08 0.25 0.50 1.00

6 7 7 7 6 6 7 7 7 7 6 6

0.329 0.315 0.293 0.24 0.419 0.384 0.349 0.315 0.314 0.315 0.316 0.317

0.833 0.714 0.714 0.714 0.833 0.833 0.714 0.714 0.714 0.714 0.833 0.833

276.29 227.74 213.68 180.24 356.56 324.90 252.35 227.74 226.86 227.74 267.60 267.60

558.94 489.10 505.04 541.71 483.95 513.60 465.07 489.09 489.95 489.10 569.25 569.25

835.22 716.84 718.73 721.95 839.51 838.50 717.42 716.84 716.82 716.84 836.85 836.85

19595 19629 19685 19813 27599 24527 21892 19629 19947 19629 19152 18191

R

M

Table 5 Sensitivity analysis on the second example. Total Cost

h

M

0.05 0.10 0.20 0.50

0.08

0.25

0.5

1

19916 19947 20003 20130

19595 19629 19685 19813

19115 19152 19210 19338

18154 19191 18258 18389

0.10 24462 24527 24627 24856

0.15 21845 21892 21967 22139

0.20 19595 19629 19685 19813

0.25 27599 24527 21892 19629

0.5 26965 23952 21371 19152

1 25701 22804 20323 18191

R h

0.05 0.10 0.20 0.50

0.05 27513 27599 27735 28038 M

R

0.05 0.10 0.15 0.20

0.08 28027 24912 22241 19947

According to the Table 6, F is slightly sensitive respect to the changes in the values of h, R, V, Ic , and Ie while is moderately sensitive respect to the changes in the values of unit purchasing cost. Period length and order quantity are moderately sensitive respect to the all changes and both Im and Ib are moderately sensitive respect to the changes in the values of h, R, V, Ic , Ie and highly sensitive respect to unit purchasing cost. Also total cost is highly sensitive respect to the changes of unit purchasing cost while is slightly sensitive respect to the changes of other parameters. According to the Table 7, F is slightly sensitive respect to the changes in the values of h, V, Ic , and Ie while is moderately and highly sensitive respect to the changes in the values of R and C. Period length and order quantity are moderately sensitive respect to the all changes and both Im and Ib are moderately sensitive respect to the changes in the values of h, R, V, Ic , Ie and highly sensitive respect to unit purchasing cost. Also total cost is highly sensitive respect to the changes of unit purchasing cost while is slightly sensitive respect to the changes of other parameters. It should be noted that in the second example, since the second case has occurred (there is no interest charged), so the changes of all decision variables respect to Ic are zero.

7. Conclusion In our proposed model some realistic features are incorporated. The effects of inflation and time value of money in an inventory control model are investigated. We developed this inventory control model to manage the deteriorating items when shortage is permitted and supplier offers delayed payment strategy. We assumed the demand and deteriorating rates are constant and planning horizon is finite. The present value of total cost during the planning horizon in this inventory system is developed first, then a three-phase solution procedure is proposed to obtain the economic order and shortage quantities, and number of replenishments during the planning horizon. Finally, the proposed model is illustrated through several numerical examples and the sensitivity analysis is performed. For the further researches, our proposed model can be

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Fig. 5. Showing the best result of the second example.

Table 6 Sensitivity analysis of the first example. Changes

F⁄

T⁄

Im

Ib

Q⁄

TC⁄

h

+50 +20 20 50

4.137 1.719 1.738 4.478

25.004 25.004 0 0

28.006 26.439 1.409 3.597

21.429 23.551 1.961 5.035

24.976 25.093 0.163 0.428

0.296 0.118 0.163 0.429

R

+50 +20 20 50

3.967 1.606 1.568 3.949

25.004 25.004 0 0

28.324 26.549 1.599 4.021

21.663 23.643 1.767 4.439

25.218 25.194 0.029 0.076

2.357 2.431 0.948 2.371

C

+50 +20 20 50

18.742 8.407 10.221 29.756

25.004 25.004 0 0

39.479 31.683 10.432 30.452

9.193 17.913 11.504 33.467

25.356 25.262 0.203 0.646

43.293 17.385 18.260 44.731

V

+50 +20 20 50

2.739 1.115 0.812 Infeasible

25.004 25.004 0 Infeasible

27.398 26.173 0.824 Infeasible

22.696 24.062 0.910 Infeasible

25.205 25.189 0.015 Infeasible

0.459 0.177 0.163 Infeasible

Ic

+50 +20 20 50

1.549 0.661 0.907 2.362

25.004 25.004 0 0

26.507 25.828 0.937 2.406

23.537 24.446 1.035 2.656

25.193 25.184 0.017 0.045

0.044 0.015 0.044 0.118

Ie

+50 +20 20 50

2.739 1.115 0.812 2.021

25.004 25.004 0 0

27.398 26.173 0.824 2.061

22.696 24.061 0.910 2.276

25.205 25.189 0.015 0.038

0.459 0.178 0.163 0.415

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F⁄

T⁄

Im

Ib

Q⁄

TC⁄

h

+50 +20 20 50

3.679 1.522 1.522 4.440

0 0 0 16.659

3.185 1.299 1.333 21.314

1.690 0.691 0.712 14.279

0.141 0.059 0.062 16.517

0.148 0.056 0.066 0.173

R

+50 +20 20 50

20.235 8.278 8.499 21.694

0 0 0 16.659

20.409 8.350 8.608 42.658

9.313 3.805 3.915 5.010

0.129 0.057 0.063 16.971

18.544 8.069 9.083 24.952

C

+50 +20 20 50

Infeasible 16.651 20.774 62.638

Infeasible 0 16.659 16.659

Infeasible 16.793 41.572 91.704

Infeasible 7.659 5.500 16.989

Infeasible 0.109 16.961 17543

Infeasible 15.951 16.246 41.541

V

+50 +20 20 50

4.091 1.681 1.713 Infeasible

0 0 0 Infeasible

4.138 1.697 1.757 Infeasible

1.885 0.773 0.799 Infeasible

0.029 0.012 0.012 Infeasible

1.044 0.417 0.413 Infeasible

Ic

+50 +20 20 50

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

Ie

+50 +20 20 50

4.091 1.681 1.713 4.503

0 0 0 16.659

4.138 1.697 1.757 22.247

1.885 0.773 0.799 14.236

0.028 0.011 0.012 16.781

1.044 0.418 0.413 1.029

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