An Inventory Model for Deteriorating Items under Conditionally Permissible Delay in Payments Depending on the Order Quantity

Applied Mathematics, 04, 5, 675-695 Published Online October 04 in SciRes. http://www.scirp.org/journal/am http://dx.doi.org/0.436/am.04.5756 An Inventory Model for Deteriorating Items under Conditionally Permissible Delay in Payments Depending on the Order Quantity Sumana Bera, Samarjit Kar, ripti Chakraborti 3, Bani Kumar Sinha 4 Department of Basic Science and Humanities, Future Institute of Engineering & Management, Kolkata, India Department of Mathematics, National Institute of echnology Durgapur, Durgapur, India 3 Department of Applied Mathematics, University of Calcutta, Kolkata, India 4 Department of Operations, Supply Chain and Retail Management, Calcutta Business School, Kolkata, India Email: kar_s_k@yahoo.com Received 5 July 04; revised 9 August 04; accepted 0 September 04 Copyright 04 by authors and Scientific Research Publishing Inc. his work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/ Abstract he purpose of this inventory model is to investigate the retailer s optimal replenishment policy under permissible delay in payments. In this paper, we assume that the supplier would offer the retailer partially permissible delay in payments when the order quantity is smaller than a predetermined quantity (W). he most inventory systems are usually formed without considering the effect of deterioration of items which deteriorate continuously like fresh fruits, vegetables etc. Here we consider the loss due to deterioration. In real world situation, the demand of some items varies with change of seasons and occasions. So it is more significant if the loss of deterioration is time dependent. Considering all these facts, this inventory model has been developed to make more realistic and flexible marketing policy to the retailer, also establish the result by ANOVA analysis by treating different model parameters as factors. Keywords Inventory, Economic Order Quantity (EOQ), Deterioration, Permissible Delay in Payment, Weibull Distribution. Introduction he general, economic order quantity (EOQ) model assumes that the retailer must be paid for the items as soon as the items are received. However, in practical situation, the supplier offers to the retailer many incentives such How to cite this paper: Bera, S., Kar, S., Chakraborti,. and Sinha, B.K. (04) An Inventory Model for Deteriorating Items under Conditionally Permissible Delay in Payments Depending on the Order Quantity. Applied Mathematics, 5, 675-695. http://dx.doi.org/0.436/am.04.5756

as a cash discount to motivate faster payment and stimulate sales, or a permissible delay in payments to attract new customers and increase sales. Before the end of the permissible delay period, the retailer can sell the goods and accumulate revenue and earn interest. On the other hand, a higher interest is charged if the payment is not settled by the end of the trade credit period. herefore, it makes economic sense for the retailer to delay the settlement of the replenishment account up to the last moment of the permissible period allowed by the supplier. Moreover, the most inventory systems are usually formed without considering the effect of deterioration. In real-life situations there are products such as volatile, liquids, some medicines, food materials, etc., in which the rate of deterioration is very large with time. herefore, the loss due to deterioration should not be neglected. So in this model, we are considering the items such as fresh fruits and vegetables which have the exponential distribution for the time to deterioration. Several authors discussing this topic have appeared in the literatures that investigate inventory problems under varying conditions. Some of the papers are discussed below. Goyal [] developed an EOQ model under the conditions of permissible delay in payments. Aggarwal and Jaggi [] extended Goyal s [] model to consider the deteriorating items. Chang, Ouyang and eng [3] then established an EOQ model for deteriorating items under supplier s trade credits linked to order quantity. Chung and Liao [4] studied a similar lot-sizing problem under supplier s trade credits depending on the retailer s order quantity. However, most of the papers dealing with EOQ in the presence of permissible delay in payments assume that the supplier only offers the retailer fully permissible delay in payments if the retailer orders a sufficient quantity. Otherwise, permissible delay in payments would not be permitted. We know that this policy of the supplier to stimulate the demands from the retailer is very practical. But this is just an extreme case. hat is, the retailer would obtain 00% permissible delay in payments if the retailer ordered a large enough quantity. Otherwise, 0% permissible delay in payments would happen. Huang [5] established an EOQ model in which the supplier offers a partially permissible delay in payments when the order quantity is smaller than the prefixed quantity W. In the above paper, a partially permissible delay in payments means the retailer must make a partial payment to the supplier when the order is received to enjoy some portion of the trade credit. hen, the retailer must pay off the remaining balances at the end of the permissible delay period. For example, the supplier provides 00% delay payment permitted if the retailer orderes a predetermined quantity, otherwise only λ% (0 λ 00) delay payment permitted. From the viewpoint of supplier s marketing policy, the supplier can use the fraction of the permissible delay in payments to attract and stimulate the demands from the retailer. Ouyang [6] studied the similar EOQ model with constant deterioration of the quantity. Das et al. [7] presented an EPQ model for deteriorating items under permissible delay in payment. eng et al. [8] developed an EOQ model for stock dependent demand to supplier s trade credit with a progressive payment scheme. Min et al. [9] developed an EPQ model with inventory-level dependent demand and permissible delay in payment. Recently, Ouyang and Chang [0] proposed an optimal production lot with imperfect production process under permission delay in payment and complete backlogging. he present EOQ model based on the fact that the suppliers would offer a partially permissible delay in payment if the retailer ordered more than or equal to a predetermined quantity W. If the ordering quantity is less than W, then the retailer has to pay off a certain amount (which is decided by the supplier) at the ordering time. In the real-world situation, we generalize the inventory model by relaxing some facts as ) the retailer s selling price per unit is higher than its purchase unit cost; ) the interest rate charged by the bank is not necessarily higher than the retailer s investment return rate; 3) many items like as fresh fruits and vegetables deteriorate exponentially with time. In this regard, we model a retailer s inventory system as a cost minimization problem to determine the retailer s optimal inventory cycle time and optimal order quantity. Several theorems are established to describe the optimal replenishment policy for the retailer under the more general framework and use an approach to solve this complex inventory problem. Finally, numerical example has been given to illustrate all these theorems and sensitivity analysis has been done. Also we have established the result by ANOVA analysis by treating different model parameters as factors.. Mathematical Notations and Assumptions In this section, the present study develops a retailer s inventory model under conditionally permissible delay in payments. he following notation and assumptions are used throughout the paper. 676

.. Notation D: the annual demand A: the ordering cost per order W: the quantity at which the fully delay payment permitted per order P: the purchasing cost per unit H: the unit holding cost per year excluding interest charge S: the selling price per unit I e : the interest earned per dollar per year I k : the interest charged per dollar in stock per year M: the period of permissible delay in settling accounts λ: the fraction of delay payment permitted by the supplier per order, 0 λ Z(t) = t : is two parameter Weibull distribution function representing time to deterioration, where = scale parameter, (0 < ), = shape parameter ( > ) : the replenishment cycle time in years Q: the order quantity CR(): the annual total relevant cost, which is a function of * : the optimal replenishment cycle time of CR() Q * : the optimal order quantity = D *.. Assumptions ) Replenishments are instantaneous. ) Demand rate, D, is known and constant. 3) Shortages are not allowed. 4) Inventory system involves only one type of inventory. 5) ime horizon is infinite. 3. Mathematical Formulation he inventory level decreases due to demand as well as deterioration. hus, the change of inventory level can be represented by the following differential equation: di t dt + Z t I t = D, 0< t < () Z t where = t, with boundary condition I(t) = 0. he solution of () with t e = + t (as ) is t + + I( t) = De ( t) + ( t ) ; 0 t + Hence, the order quantity for each cycle is () Q I( 0) D + = = + + From (3), we can obtain the time interval w that W units are depleted to zero due to both demand and deterioration. We put Q = W in (3) and get w from (3) W D w + = + w + If Q W (i.e., w ), the fully delayed payment is permitted, otherwise, partial delayed payment is permitted if Q < W (i.e., < w ), the retailer must have to pay supplier, the partial payment of ( λ)pq at time 0. From the constant sales revenue sd, the retailer will be able to pay off the loan ( λ)pq at time (3) (4) 677

+ ( )( p s) + λ + ( = G( say) ) At time 0, the pay of time G of the partial payment is shorter or equal to the trade credit period M. i.e., G M. herefore we get 0 from the following relation p s 0 + 0 M + + ( λ ) It is obvious that always M < 0 and if 0 then G M and vice-versa. Based on the values of M, w, 0, we have three possible cases: ) w M 0, ) M < w 0, 3) M < 0 < w A a) Annual ordering cost =, b) Annual stock holding cost excluding interest charge t + + H= h I( t) dt= h De ( t) + ( t ) dt 0 0 + t Simplifying with e = t (as ) H = Dh + ( + )( + ) c) Annual deteriorating cost [ ] 3.. Case : w M < 0 H = Dh + ( + )( + ) + (Neglecting term, since ) = pq pd = Dp + ( + ) here are three sub-cases in terms of annual opportunity cost of the capital which are depicted in Figure. 3... Sub-Case.: M he retailer starts paying the interest for the items in stock after time M with rate I k and during time 0 to M, from the sale revenue the retailer earns the interest with rate I e, therefore in this sub-case, the annual opportunity cost of capital is Inventory level [I(t)] Inventory level [I(t)] Inventory level [I(t)] Q Q Q O G w M 0 Sub-case. M time O G w M 0 Sub-case. M w time O G w M 0 Sub-case. 0 w time Figure. Graphical representation of three different situations of case. 678

M + + siedm M M M + + + M pikd ( + )( + ) + 3... Sub-Case.: w M In this case, there is no interest paid for financing inventory, therefore in this sub-case, the annual opportunity cost of capital is sied M 3..3. Sub-Case.3: 0 < < w If < w, then the retailer must have to pay the partial payment ( λ)pq at time 0 to the supplier, and retailer pays off the loan to the bank from sales revenue at time G ( λ )( p s) + = + + Consequently, the interest is charged on the partial payment from time 0 to G. Hence the annual interest payable is Ik G ID k ( ) {( ) } d ( ) ( + λ pq sdt t λ p s) = 0 + (5) + Similarly, the interest earned starts from time G to M, and the annual earned interest is Ie M { ( λ) } + { ( λ) } G sdt pq dt sd pq dt ( + ) sdie ( + ) sdi e = ( λ)( p s) + + ( M ) ( λ)( p s) + + + herefore in this sub-case, the annual opportunity cost of capital is (6) IkD ( + ) sdi e ( + ) ( λ) ( p s) + ( λ)( p s) + + + sdi e ( M ) ( λ )( p s) + + ( + ) herefore in case, the annual relevant cost can be expressed as RC, M RC = RC, w M RC3, 0< < w where, A pik D M RC = + Dh + + Dp + + M ( + )( + ) + + + siedm + M + M M + + + A RC = + Dh + + Dp sied M ( + )( + ) + (7) (8) (9) (0) 679

A RC3 = + Dh + + Dp ( + )( + ) + ID k p ( + ) + ( λ ) + s + sied p ( + ) ( λ ) + s + sied p ( + ) ( M ) ( λ ) + s + () 3.. Case : M < w 0 (cf. Figure ) Similar to case, three different sub-cases are as follows: 3... Sub-Case.: w his case is similar to the sub-case. (where M ). Since M < w, therefore the total relevant cost is same as RC (). 3... Sub-Case.: M < w Since < w, the retailer must have to pay the partial payment ( λ)pq at time 0 to the supplier, and retailer pays off the loan to the bank from sales revenue at time G. he annual interest payable from 0 to time G is given by + Ik G ID k p ( + ) ( λ) pq sdt dt ( λ) 0 = + s Again since M, the retailer has to pay interest from time M to time. herefore, the annual payable interest is ( + ) ( + ) ( ) pik pikd M I ( t) dt M M M M = + + + M ( + )( + ) + Similarly, the interest earned starts from time G to M, and thus the annual interest earned is M e + sdt ( λ) pqdt = M ( λ) + G Ie si D p s ( + ) In this sub-case, the annual relevant cost is Inventory level [I(t)] Inventory level [I(t)] Inventory level [I(t)] Q Q Q O G M w 0 Sub-case. w time O G M w 0 Sub-case. M w time O G M w 0 Sub-case. M time Figure. Graphical representation of three different situations of Case. 680

A ID k p + RC4 = + Dht + + Dp + ( λ ) + ( + )( + ) ( + ) s ( + ) + + + + pik D M M M M M sdi e p + M ( λ ) + s ( + ) ( + + ) ( ) + + + () 3..3. Sub-Case.3: M his sub-case is similar to the Sub-Case.3 (where < w < M). Since M < w, the annual total relevant cost is same as CR 3 (). herefore in Case the annual relevant cost can be expressed as 3.3. Case 3: M < 0 < w (cf. Figure 3) here are four sub-cases as following: RC, w RC = RC4, M RC3, < M 3.3.. Sub-Case 3. w his case is similar to the Sub-Case. (where M ). Since M < 0 < w, therefore the total relevant cost is RC (). w (3) 3.3.. Sub-Case 3. 0 < w p + If 0, then M < G, i.e. M < ( λ ). s + + Inventory level [I(t)] Inventory level [I(t)] Q Q O M G 0 w Sub-case 3. w time O M G 0 w Sub-case 3. 0 < w time Inventory level [I(t)] Inventory level [I(t)] Q Q time O M G 0 w O M G 0 w Sub-case 3.3 M 0 Sub-case 3.4 M Figure 3. Graphical representation of four different situations of Case 3. time 68

In this sub-case, at the beginning i.e. at time 0, the retailer must take a loan to pay the supplier the partial payment of ( λ)pq. Since M < G, the retailer have to take another loan to pay the rest of λpq at time M. p + Again since the first loan will be paid from the sale revenue until t = λ. s + + Hence, the retailer gets the second loan at time M but can start paying off from the sales revenue after time t = G (>M). As a result, there is no interest earned, but have to pay the interest. he st payable interest is G ID k p + Ik ( λ) pq sdt dt ( λ). 0 = + s + For the nd loan λpq, retailer has to pay interest at I k rate per year from M to G. herefore, the nd payable interest is G + p + Ik λpqdt = IkλpD ( G M ) IkλpD ( λ) M M + = + + s + Again, since the retailer has started to pay off the loan λpq from the sales revenue after time G the loan will λ pq be completely paid off at time. sd herefore, the 3 rd λ pq IQ k p payable interest is sd + Ik ( λpq sdt) dt = λ 0 + s + herefore the annual total payable interest is IkD p + Ikλ pd p + ( λ) + + ( λ) + M s + s + IQ k p + + λ + s + (4) 3.3.3. Sub-Case 3.3 M 0 his case is similar to the Sub-Case. (where M < w ). Since M 0 < w, the annual total relevant cost is RC 4 (). 3.3.4. Sub-Case 3.4 M his case is similar to the Sub-Case.3 (where 0 < < w ). Since M < 0 < w, therefore the annual total relevant cost is RC 3 (). In Case 3, the annual relevant cost can be expressed as where, RC + M RC, w RC5, 0 = RC, M < RC3, M 4 0 A ID k p + RC5 = + Dh + + Dp + ( λ+ λ ) + ( + )( + ) + s + Ikλ pq p + ( λ ) + s + w (5) (6) 68

4. heoretical Results Now we try to determine the optimal replenishment cycle time () that minimizes the annual relevant cost. 4.. Case w M < 0 o minimize RC () in (9) for M we get RC = 0 + Dh + + Dp + + + M + pikd + + M M ( + ) ( + )( + ) ( + ) + + + siedm + A = 0 Now there exists a value in [M, ) at which we get minimum value of RC (). Let + + M M M RC siedm Dh + + Dp + A + + = M (7) (8) hen we have the following lemma. Lemma : a) If Δ 0, then the annual total relevant cost RC () has the unique minimum value at the point =, where [M, ) and satisfies (7). b) If Δ > 0, the annual total relevant cost RC () has a minimum value at the boundary point = M. Proof: Let F = Dh + + Dp + pi D ( M ) + + + + [ ) M, + + + k + + + M M + siedm A + + + aking the derivative of F () with respect to (M, ), we get df + k + + = Dh + + Dp + pi D + M k = Dh + + Dp + pi D + M > 0 herefore, F () is strictly increasing function of in [M, ). From (A), we get herefore, if 0, and lim F M = F = + F M = 0, then by applying the Darboux s theorem, a unique [M, ) such that F ( ) = 0. Again taking the second order derivative of RC () with respect to at, we have Hence, d d d d =, = = + 3 RC F RC F F F (A) (A) 683

d RC = { + } k { } = Dh + + Dp + pi D + M > 0 (A4) herefore [M, ) is the unique minimum solution to RC (). On the other hand, if F( M ) = > 0, we have F ( ) > 0, [ M, ). Consequently, we know that RC F = > 0, ( M, ). hat is, RC () is a strictly increasing function of in [M, ). herefore, RC () has a minimum value at the boundary point = M. his completes the proof. Darboux s heorem: If a function f is derivable on a closed in terval [a, b] and f'(a), f'(b) are of opposite signs then there exist at least one point c [a, b] such that f'(c) = 0. RC Again for w M, RC () in (0) is minimum when = 0 si D Dh Dp + + A e + + + = + + sied Dh + + Dp A 0 + = + + o prove that there exist a value of in the interval [ w, M] at which minimizes RC (), we let RC = W W + + sied Dh + W + Dp W W A + + + 0 (9) (0) It is obvious that Δ Δ if M w, then we have the following lemma: Lemma : a) If Δ 0 Δ, then the annual total relevant cost RC () has the unique minimum value at the point =, where [ w, M] and (9) is satisfied by. b) If Δ > 0, the annual total relevant cost RC () has a minimum value at the lower boundary point = w. c) If Δ < 0, the annual total relevant cost RC () has a minimum value at the upper boundary point = M. Proof: he proof is similar to that in Lemma so we omit it. RC3 Similarly, for 0 < < w, RC 3 () in () is minimum when = 0. hat is, M A 0 ( ) ( + ) + + ID k p Dh + + Dp + ( λ) + + + + s + sdi e p p + + ( λ) + + ( λ) + s + s + p + ( )( λ ) = s + () Again there exist a value of in the interval (0, w ) which minimizes RC 3 (), we let 684

RC3 3 = W or ( + ) ( + ) + + ID k p 3 Dh W + W + Dp W + W ( λ) + ( W ) + W + + s + sdi e p p + W ( λ) + W + ( λ) + W s + s + p + ( M W )( λ ) W A = 0 s + () hen we have the following lemma: Lemma 3: a) If Δ 3 0, then the annual total relevant cost RC 3 () has the unique minimum value at the point = 3, where 3 (0, w ) and satisfies (). b) If Δ 3 < 0, then the value of (0, w ) which minimizes RC 3 () does not exist. Proof: Let + ( + ) ( + ) + + ID k p F3 = Dh + + Dp + λ + + + + s + sdi e p p + ( λ) + + ( λ) + s + s + p ( M )( λ ) s + A = 0 (B) Differentiating F 3 () with respect to (0, w ), we have df Since ( ) herefore we have have ( k e)( ) = Dh + + Dp + D I I λ p s 3 + ( + ) ( + ) + ( + ) + + sdie { + ( λ)( p s) M } { ( λ) } ( λ) ( ) { } ( λ) ( ) ( k λ ) ( λ) ( + ) ( + ) h + h + p + ( I I )( λ) ( p s) + ( + ) + k e + si + p s M h + I I p s + si e k e e = h+ p s I + s p s I > 0 s p s > 0 for s p and 0 λ df 3 ( ) 3 ( ) e (B) > 0 F is a strictly increasing function of in (0, w ). Now from (B), we lim F = A < 0 and lim F =. (B3) 3 3 3 0 W 685

herefore, if F ( ) lim = 0, then applying Darboux s theorem, a unique 3 (0, W ) such that W 3 3 F 3 ( 3 ) = 0. Again, taking the second order derivative of RC 3 () with respect to at 3, we have d RC3 = + + + λ = 3 3 3 3 ( k e)( ) D h h p I I p s ( + ) + ( + ) 3 + 3 + sie { + ( λ)( p s) M3 } > 0 + herefore, 3 (0, W ) is the unique minimum solution to RC 3 (). Again if F ( ) F3 < 0 0, W. We have df3 F3 = < 0 ( 0, W ). 3 3 (B4) lim = < 0, then herefore, RC 3 () is a strictly decreasing function of in (0, W ), but we cannot find the value of in the open interval (0, W ) which minimizes RC 3 (). his completes the proof. For 0 λ, it can be written that Δ Δ 3. Again w M, we that Δ Δ. Now combining Lemmas - 3 and including the fact that RC (M) = RC (M), for Case- we can obtain a theoretical result to determine the optimal cycle time * as: heorem : For w M < 0, the optimal replenishment cycle time *, that minimizes the annual total relevant cost is given as: W Condition RC( * ) * Δ 0 and Δ 3 < 0 RC ( ) Δ 0 and Δ 3 0 Min {RC ( ), RC 3( 3)} or 3 Δ > 0, Δ 3 0 and Δ < 0 Min {RC ( ), RC 3( 3)} or 3 Δ 0 Min {RC ( w), RC 3( 3)} w or 3 Δ > 0 and Δ 3 < 0 RC ( ) 4.. Case M < w 0 For w, similar approach used in Case, the st order condition for RC () of (9) is the same as (7), so there exist a unique value of in [ w, ) at which RC () is minimized. RC Let 4, then = W 4 Dh + Dp pi D ( M ) + + + + + + + + + w w w k w w sied + + + M Mw + M A ( + )( + ) ( + ) We have the following lemma: Lemma 4: a) If Δ 4 0, then the annual total relevant cost RC () has the unique minimum value at the point =, where [ w, ) and satisfies (7). b) If Δ 4 > 0, the annual total relevant cost RC () has a minimum value at the boundary point = w. (3) 686

Proof: he proof is similar to that in Lemma so we omit it. Again for M < w, the total relevant cost RC 4 () in (3) is minimum when RC4 = 0 + + p ( + ) Dh + Dp IkD ( λ) + + + + + + s ( + ) + pikd M + + M M + ( + )( + ) + + + + p + p + + + sied M ( λ) + M + ( λ) + A = 0 s + s + (4) o prove that there exist a unique value of in [M, w ) at where RC 4 () is minimum, we let RC4 5, then = M 5 Dh M + M + Dp M + + + + ( + ) ( + ) p + Ik DM ( λ) + ( M ) + M s ( ) p p + + siedm ( λ) + M + ( λ) + M A s + s + (5) and let RC4 6 = W, then + + + k W W W ( + ) + + p 6 Dh W + W Dp W IkDW ( λ) ( W ) W + + + + + + s + + pi D M + + M M + ( + )( + ) + p + p + + + sied M ( λ ) W + W M + ( λ ) W + W A s + s + (6) hen we have the following lemma: Lemma 5: a) If Δ 5 0 Δ 6, then the annual total relevant cost RC 4 () has the unique minimum value at the point = 4, where 4 [M, w ) and (4) is satisfied by 4. b) If Δ 5 > 0, the annual total relevant cost RC 4 () has a minimum value at the lower boundary point = M. c) If Δ 6 < 0, then [M, w ) which minimizes RC 4 () does not exist. Proof: he proof of (a) and (b) is similar to that in Lemma and that of (c) is similar to that in Lemma 3 b. RC3 Again in (0, M], the total relevant cost RC 3 () in () is minimum when = 0, which is same as in (), since at = M, then Δ 3 = Δ 5, now we have the following lemma: Lemma 6: 687

a) If Δ 5 0, then the annual total relevant cost RC 3 () has the unique minimum value at the point = 3, where 3 (0, M] and satisfies (). b) If Δ 5 < 0, the annual total relevant cost RC 3 () has a minimum value at the boundary point = M. Proof: he proof is similar to that in Lemma so we omit it. From (3) and (6), we get Δ 6 Δ 4 for 0 λ. Again since M < w, we get Δ 6 Δ 5. Now combining Lemmas 4-6 and the fact that RC (M) = RC (M), we can obtain a theoretical result to determine the optimal cycle time * for Case. heorem : For M < w 0, the optimal replenishment cycle time *, that minimizes the annual total relevant cost is given as follows: Situation Condition RC( * ) * Δ 6 < 0 Δ 4 < 0, Δ 5 < 0 Min{RC ( ), RC 3(M)} or M........... Δ 4 < 0, Δ 5 < 0 Min{RC ( ), RC 4( 4)} or 4 Δ 6 0 Δ 4 < 0, Δ 5 0 Min{RC ( ), RC 3( 3)} or 3 Δ 4 0, Δ 5 < 0 Min{RC ( w), RC 4( 4)} w or 4 Δ 4 0, Δ 5 0 Min{RC ( w), RC 3( 3)} w or 3 4.3. Case 3 M < 0 < w For [ w, ), the annual total relevant cost is similar as in (9) i.e. RC (). From Lemma 4 of the Case, if Δ 4 0, RC () has the unique minimum value at =, where [ w, ) and satisfies (7) and if Δ 4 > 0, then RC () has minimum value at the boundary point = w. RC5 Again in [ 0, w ), the annual total relevant cost RC 5 () in (6) is minimum when = 0. ( + ) + + p Dh + Dp IkD ( λ λ ) + + + + + + + s + p ( ) ( ) + + + + + λpikd ( λ) + + M A 0 = s ( ) + + o prove that there exist a value of in [ 0, w ) at which minimizes RC 5 (), we let RC5 7 = 0 (7) = Dh + + Dp + + + + 7 0 0 0 ( + ) ( + ) p ( + + ) ID k ( λ + + λ ) 0 + ( 0 ) + 0 s p ( ) ( ) + + + ( + ) + λpikd ( λ) 0 + ( 0 ) + 0 M 0 A s ( ) + + (8) and RC5 8 = W 688

8 = Dh + + Dp + + + + W W W ( + ) ( + ) p ( + + ) ID k ( λ + + λ ) W + ( W ) + W s (9) p ( ) ( ) + + + ( + ) + λpikd ( λ) W + ( W ) + W M W A s ( ) + + Consequently, we have the following lemma: Lemma 7: a) If Δ 7 0 Δ 8, then the annual total relevant cost RC 5 () has the unique minimum value at the point = 5, where 5 [ 0, w ) and (7) is satisfied by 5. b) If Δ 7 > 0, the annual total relevant cost RC 5 () has a minimum value at the lower boundary point = 0. c) If Δ 8 < 0, then [ 0, w ) which minimizes RC 5 () does not exist. Proof: he proof of a) and b) is similar to that in Lemma and that of (c) is similar to that in Lemma 3 b. Again in [M, 0 ], the annual relevant cost is similar to RC 4 () in (3). RC 4 () is minimum when RC4 = 0, which is same as in (4). o prove that there exists a unique value of in [M, 0 ] at which minimizes RC 4 (), we let RC4 9 = = 0 + + + 0 0 0 ( + ) + + p 9 = Dh 0 + 0 Dp 0 IkD0 ( λ) ( 0 ) 0 + + + + + + s + + Ik D M + + M M + + + + p + p + + + sdie M ( λ ) 0 0 s ( + ) M + ( λ ) 0 0 A s ( + ) (30) hen we have the following lemma: Lemma 8: a) If Δ 5 0 Δ 9, then the annual total relevant cost RC 4 () has the unique minimum value at the point = 4, where 4 [M, 0 ] and (4) is satisfied by 4. b) If Δ 5 > 0, the annual total relevant cost RC 4 () has a minimum value at the lower boundary point = M. c) If Δ 9 < 0, the annual total relevant cost RC 4 () has a minimum value at the upper boundary point = 0. Proof: he proof is similar to that in Lemma so we omit it. Again in (0, M], the annual total relevant cost is similar to RC 3 () in (5). We know that at = M, Δ 3 = Δ 5, so from Lemma 6, if Δ 5 0, RC 3 () has unique minimum value at = 3, where 3 (0, M) and satisfies (). On the other hand if Δ 5 < 0, then RC 3 () has a minimum value at boundary point = M. Since M < 0 < w, from (8) and (30) we can get Δ 7 Δ 9. Again we know that Δ 5 Δ 9 and Δ 5 Δ 7 Δ 8 for 0 λ. Consequently, combining Lemmas 4, 6, 7 and 8, and the fact that RC 3 (M) = RC 4 (M), we can obtain the theoretical result to get the optimal cycle time * for Case 3 as: heorem 3: For M < 0 < w, the optimal replenishment cycle time *, that minimizes the annual total relevant cost is given as follows: 689

Situation Condition Sub-condition RC( * ) * Min{RC ( ), RC 4( 0)} or 0 Δ 4 < 0 Δ 8 < 0, Δ 9 < 0 Δ 8 < 0, Δ 9 0 Δ 8 0, Δ 9 < 0 Δ 8 0, Δ 9 0 Min{RC ( ), RC 4( 4)} or 4 Min{RC ( ), RC 5( 5), RC 4( 0)} or 5 or 0 Min{RC ( ), RC 5( 5), RC 4( 4)} or 5 or 4 Min{RC ( ), RC 5( 0), RC 4( 4)} or 0 or 4 Min{RC ( ), RC 5( 0), RC 3( 3)} or 0 or 3 Min{RC ( w), RC 4( 0)} w or 0 Δ 4 0 Δ 8 < 0, Δ 9 < 0 Δ 8 < 0, Δ 9 0 Δ 8 0, Δ 9 < 0 Δ 8 0, Δ 9 0 Min{RC ( w), RC 4( 4)} w or 4 Min{RC ( w), RC 5( 5), RC 4( 0)} w or 5 or 0 Min{RC ( w), RC 5( 5), RC 4( 4)} w or 5 or 4 Min{RC ( w), RC 5( 0), RC 4( 4)} w or 0 or 4 Min{RC ( w), RC 5( 0), RC 3( 3)} w or 0 or 3 5. Solution Procedures Here we develop the following algorithm to solve this complex inventory problem by using the characteristics of heorems -3 above. Algorithm: Step Step Step 3 Step 4 Step 4. Step 4. Step 5 Compare the values of 0, M and w, if w M < 0, then go to step. If M < w 0, then go to Step 3. Otherwise, if M < 0 < w, then go to Step 4. Calculate Δ, Δ and Δ 3 which are shown as in Equation (8), (0) and (), respectively. ) If Δ < 0 and Δ 3 < 0, then RC( * ) = RC ( ) and * =. Go to Step 5. ) If Δ < 0 and Δ 3 0, then RC( * ) = min {RC ( ), RC 3( 3)} and * = or 3. Go to Step 5. 3) If Δ 0, Δ < 0 and Δ 3 0, then RC( * ) = min{rc ( ), RC 3( 3)} and * = or 3. Go to Step 5. 4) If Δ 0, then RC( * ) = min{rc ( w), RC 3( 3)} and * = w or 3. Go to Step 5. 5) If Δ 0 and Δ 3 < 0, then RC( * ) = RC ( ) and * =. Go to Step 5. Calculate Δ 4, Δ 5 and Δ 6 which are shown as in Equations (3), (5) and (6), respectively. ) If Δ 6 < 0, then RC( * ) = min {RC ( ), RC 3(M)} and * = or M. Go to Step 5. ) If Δ 4 < 0, Δ 5 <0 and Δ 6 0, then RC( * ) = min {RC ( ), RC 4( 4)} and * = or 4. Go to Step 5. 3) If Δ 4 < 0 and Δ 5 0, then RC( * ) = min{rc ( ), RC 3( 3)} and * = or 3. Go to Step 5. 4) If Δ 4 0 and Δ 5 < 0, then RC( * ) = min{rc ( w), RC 4( 4)} and * = w or 4. Go to Step 5. 5) If Δ 4 0 and Δ 5 0, then RC( * ) = min{rc ( w), RC 3( 3)} and * = w or 3. Go to Step 5. Calculate Δ 4, Δ 5, Δ 7, Δ 8 and Δ 9 which are shown as in Equation (3), (5), (8), (9) and (30) respectively. If Δ 4< 0, then go to Step 4.. Otherwise, go to Step 4.. ) If Δ 8 < 0 and Δ 9 < 0, then RC( * ) = min{rc ( ), RC 4( 0)} and * = or 0. Go to Step 5. ) If Δ 8 < 0 and Δ 9 0, then RC( * ) = min{rc ( ), RC 4( 4)} and * = or 4. Go to Step 5. 3) If Δ 8 0 and Δ 9 < 0, then RC( * ) = min{rc ( ), RC 4( 0), RC 5( 5)} and * = or 0 or 5. Go to Step 5. 4) IfΔ 7 < 0, Δ 8 0 and Δ 9 0, then RC( * ) = min{rc ( ), RC 4( 4), RC 5( 5)} and * = or 4 or 5. Go to Step 5. 5) If Δ 5 < 0 and Δ 7 0, then RC( * ) = min{rc ( ), RC 4( 4), RC 5( 0)} and * = or 4 or 0. Go to Step 5. 6) If Δ 5 0, then RC( * ) = min{rc ( ), RC 3( 3), RC 5( 0)} and * = or 3 or 0. Go to Step 5. 7) If Δ 8 < 0 and Δ 9 < 0, then RC( * ) = min{rc ( ), RC 4( 0)} and * = or 0. Go to Step 5. 8) If Δ 8 < 0 and Δ 9 0, then RC( * ) = min{rc ( ), RC 4( 4)} and * = or 4. Go to Step 5. 9) If Δ 8 0 and Δ 9 < 0, then RC( * ) = min{rc ( ), RC 4( 0), RC 5( 5)} and * = or 0 or 5. Go to Step 5. 0) IfΔ 7 < 0, Δ 8 0 and Δ 9 0, then RC( * ) = min{rc ( ), RC 4( 4), RC 5( 5)} and * = or 4 or 5. Go to Step 5. ) If Δ 5 < 0 and Δ 7 0, then RC( * ) = min{rc ( ), RC 4( 4), RC 5( 0)} and * = or 4 or 0. Go to Step 5. ) If Δ 5 0, then RC( * ) = min{rc ( ), RC 3( 3), RC 5( 0)} and * = or 3 or 0. Go to Step 5. ) If Δ 8 < 0 and Δ 9 < 0, then RC( * ) = min{rc ( w), RC 4( 0)} and * = w or 0. Go to Step 5. ) If Δ 8 < 0 and Δ 9 0, then RC( * ) = min{rc ( w), RC 4( 4)} and * = w or 4. Go to Step 5. 3) If Δ 8 0 and Δ 9 < 0, then RC( * ) = min{rc ( w), RC 4( 0), RC 5( 5)} and * = w or 0 or 5. Go to Step 5. 4) If Δ 7 < 0, Δ 8 0 and Δ 9 0, then RC( * ) = min{rc ( w), RC 4( 4), RC 5( 5)} and * = w or 4 or 5. Go to Step 5. 5) If Δ 5 < 0 and Δ 7 0, then RC( * ) = min {RC ( w), RC 4( 4), RC 5( 0)} and * = w or 4 or 0. Go to Step 5. 6) If Δ 5 0, then RC( * ) = min {RC ( w), RC 3( 3), RC 5( 0)} and * = w or 3 or 0. Go to Step 5. Stop 690

6. Numerical Example In this section, the present study provides the following numerical example as shown in Huang [5] to illustrate all the theoretical results. he values of the parameters are taken randomly. We assume that selling price per unit s = $50, ordering cost A = $50/order, demand D = 000 units/year, purchasing cost p = $0, holding cost h = $5/unit/year, period of permissible delay M = 0. year, interest earned I e = $0.07/$/ year, interest charged I k = $0./$/ year, scale parameter = 0.0, shape parameter =.5. We obtain the optimal cycle time and optimal order quantity for different parameters of the fraction of the delay payment λ = {0., 0.5, 0.8} and the prefix quantity W = {50, 50, 50} as shown in able. 7. Sensitivity Analysis he purpose of the sensitivity analysis is to identify parameters to the changes of which the solution of the model is sensitive. he following inferences can be made based on above solution table. able. Optimal solutions of deterministic model under different parametric values. λ W p * Q * RC( * ) 0 = 0.079 07.977 504.8680 50 0 = 0.074 07.4866 507.6956 0. 0.5 0.8 50 50 50 50 50 50 50 50 30 = 0.069 07.0048 50.5040 0 w = 0.499 50.0000 548.074 0 w = 0.499 50.0000 555.6495 30 w = 0.499 50.0000 563.87 0 3 = 0.077 07.733 574.584 0 3 = 0.065 06.543 650.3540 30 3 = 0.049 04.9506 730.4759 0 = 0.079 07.977 504.8680 0 = 0.074 07.4866 507.6956 30 = 0.069 07.0048 50.5040 0 3 = 0.078 07.8809 547.6896 0 w = 0.499 50.00 555.6495 30 w = 0.499 50.00 563.87 0 3 = 0.078 07.8809 547.6896 0 3 = 0.070 07.3 594.939 30 3 = 0.06 06.860 643.736 0 = 0.079 07.977 504.8680 0 = 0.074 07.4866 507.6956 30 = 0.069 07.0048 50.5040 0 3 = 0.079 07.96 5.803 0 3 = 0.073 07.456 54.80 30 3 = 0.068 06.87 56.0734 0 3 = 0.079 07.96 5.803 0 3 = 0.073 07.456 54.80 30 3 = 0.068 06.87 56.0734 69

) For fixed W and p, increasing the value of λ will result in a significant increase in the value of the optimal order quantity and a significant decrease in the value of the annual total relevant costs as the retailer s order quantity is smaller and only the partially delayed payment is permitted. For example when W = 50, p = 30 and λ increases from 0. to 0.5, the optimal order quantity will increase.7%((06.860 04.9506)/04.9506) and the annual total relevant costs will decrease.87%((730.4759 643.736)/730.4759). However, if the fully delayed payment is permitted, the optimal order quantity and the annual total relevant cost are independent of the value of λ. It implies that the retailer will order a larger quantity since the retailer can enjoy greater benefits when the fraction of the delay payments permitted is increasing. So the supplier can use the policy of increasing λ to stimulate the demands from the retailer. Consequently, the supplier s marketing policy under partially permissible delay in payments will be more attractive than fully permissible delay in payments. ) For fixed λ and p, increasing the value of W will result in a significant decrease in the value of the optimal order quantity and a significant increase in the value of the annual total relevant costs. For example, when λ = 0., p = 30 and W increases from 50 to 50, the optimal order quantity will decrease 30.03%((50.00 04.9506)/50.00) and the annual total relevant costs will increase 9.68%((730.4759 563.87)/563.87). It implies that the retailer will not order a quantity as large as the minimum order quantity as required to obtain fully permissible delay in payments. Hence, the effect of stimulating the demands from the retailer turns negative when the supplier adopts a policy to increase the value of W. 3) Last, for fixed λ and W, increasing the value of p will result in a significant decrease in the value of the optimal order quantity and a significant increase in the value of the annual total relevant cost. However, for the case with λ = 0. and W = 50 in the numerical example, the optimal replenishment cycle and optimal order quantity are fixed and are not affected by the increase of the unit purchase price. he reason is that in this situation, the retailer trades off the benefits of full delay in payment against the partial delay in payment and enjoys the full delay in payment. 8. ANOVA Analysis If the values of λ and W are taken randomly (able ), the wo-way ANOVA analysis on otal Relevant Cost (RC) shown in able 3: 8.. Does W Value Affect the Result? Since from able 3, the calculated values are F cal = 8.7, df n =, df d = 4, = 0.05, F table = 6.9443 and F cal > F table, we conclude that effect of prefixed quantity (W value) on the result (otal relevant cost) is considered extremely significant. able. Cost table: (when p = 0). W = 50 W- = 50 W = 50 λ = 0. 504.8680 548.074 574.584 λ = 0.5 504.8680 547.6896 547.6896 λ = 0.8 504.8680 5.803 5.803 able 3. wo way ANOVA table. Source of Variation df Sum-of-squares Mean square Calculated F cal value abular F table value W value 303 55 8.70 6.9443 Fraction of permissible delay 064 53.8.800 6.9443 Residual (error) 4 759.6 89.9 otal 8 496 69

8.. Does Fraction of Permissible Delay Affect the Result? From the calculated values of able 4, F cal =.80, df n =, df d = 4, = 0.05, F table = 6.9443 and since F cal < F table, we can conclude that effect of fraction of permissible delay (λ value) on the result (otal relevant cost) is considered not quite significant. From the above analysis we can conclude that the fraction of permissible delay has no effect overall i.e., the effect is considered not significant. 8.3. Does W Value Affect the Result? Since from able 5 the calculated values of F cal = 6.6, df n =, df d = 4, = 0.05, F table = 6.9443 and F cal < F table, we conclude that effect of prefixed quantity (W value) on the result (otal relevant cost) is considered not quite significant. 8.4. Does Fraction of Permissible Delay Affect the Result? From the calculated values of able 6, F cal =.44, df n =, df d = 4, = 0.05, F table = 6.9443 and F cal < F table, we can conclude that effect of fraction of permissible delay (λ value) on the result (otal relevant cost) is considered not quite significant. From the above analysis we can conclude that after increasing the price rate the effect of W value and the fraction of permissible delay on the result is considered not significant overall i.e., the effect is considered not significant. 8.5. Does W Value Affect the Result? From the values of able 7, F cal = 5.93, df n =, df d = 4, = 0.05, F table = 6.9443 and F cal < F table, we can conclude that effect of prefixed quantity (W value) on the result (otal relevant cost) is considered not quite significant. able 4. Cost table: (when p = 0). W = 50 W- = 50 W = 50 λ = 0. 507.6956 555.6495 650.3540 λ = 0.5 507.6956 555.6495 594.939 λ = 0.8 507.6956 54.80 54.80 able 5. wo way ANOVA table. Source of Variation df Sum-of-squares Mean square Calculated F cal value abular F table value W value 60 580 6.6 6.9443 Fraction of permissible delay 503 5.44 6.9443 Residual (error) 4 355 878.8 otal 8 7640 able 6. Cost table: (when p = 30). W = 50 W- = 50 W = 50 λ = 0. 50.5040 563.87 730.4759 λ = 0.5 50.5040 563.87 643.736 λ = 0.8 50.5040 56.0734 56.0734 693

able 7. wo way ANOVA table. Source of Variation df Sum-of-squares Mean square Calculated F cal value abular F table value W value 7760 3880 5.93 6.9443 Fraction of permissible delay 4795 398.0 6.9443 Residual (error) 4 9390 347 otal 8 4950 8.6. Does Fraction of Permissible Delay Affect the Result? Since F cal =.0, df n =, df d = 4, = 0.05 and F table = 6.9443 and F cal < F table, we also conclude that effect of fraction of permissible delay (λ value) on the result (otal relevant cost) is considered not quite significant. From the above analysis we can conclude that after increasing the price rate the effect of W value and the fraction of permissible delay on the result is considered not significant overall i.e., the effect is considered not significant. 800 600 400 Data W=50 W-=50 W=50 00 0 λ.0. λ=0.5 λ=0.8 800 600 400 Data W=50 W-=50 W=50 00 0 λ=0. λ=0.5 λ=0.8 800 600 400 Data 3 W=50 W-=50 W=50 00 0 λ=0. λ=0.5 λ=0.8 694

9. Conclusion In this paper, we develop a deterministic inventory model under the conditions of permissible delay in payments by considering the following situations simultaneously: ) the retailer s selling price per unit is higher than the purchase price; ) the interest charged by a bank is not necessarily higher than the retailer s investment return rate; 3) many selling items deteriorate continuously such as fresh fruits and vegetables and 4) the supplier may offer a partial permissible delay in payments even if the order quantity is less than W. Considering all these facts, this inventory model has been developed to make more realistic and flexible marketing policy to the retailer. References [] Goyal, S.K. (985) Economic Order Quantity under Conditions of Permissible Delay in Payments. Journal of the Operational Research Society (JORS), 36, 335-338. http://dx.doi.org/0.057/jors.985.56 [] Aggarwal, S.P. and Jaggy, C.K. (995) Ordering Policies of Deterioration Items under Permissible Delay in Payments. Journal of the Operational Research Society (JORS), 46, 658-66. http://dx.doi.org/0.057/jors.995.90 [3] Chang, C.., Ouyang, L.Y. and eng, J.. (003) An EOQ Model for Deteriorating Items under Supplier Credits Linked to Ordering Quantity. Applied Mathematical Modelling (AMM), 7, 983-996. http://dx.doi.org/0.06/s0307-904x(03)003-8 [4] Chang, K.J. and Liao, J.J. (004) Lot-Sizing Decisions under rade Credit Depending on the Ordering Quantity. Computers & Operations Research (COR), 3, 909-98. http://dx.doi.org/0.06/s0305-0548(03)00043- [5] Huang, Y.F. (007) Economic Order Quantity under Conditionally Permissible Delay in Payments. European Journal of Operational Research (EJOR), 76, 9-94. http://dx.doi.org/0.06/j.ejor.005.08.07 [6] Ouyang, L.Y., eng, J.., Goyal, S.K. and Yang, C.. (009) An Economic Order Quantity Model for Deteriorating Items with Partially Permissible Delay in Payments Linked to Order Quantity. European Journal of Operational Research, 94, 48-43. http://dx.doi.org/0.06/j.ejor.007..08 [7] Das, D., Roy, A. and Kar, S. (00) Improving Production Policy for a Deteriorating Item under Permissible Delay in Payments with Stock-Dependent Demand Rate. Computers and Mathematics with Applications, 60, 973-985. http://dx.doi.org/0.06/j.camwa.00.07.03 [8] eng, J.., Krommyda, I.P., Skouri, K. and Lou, K.R. (0) A Comprehensive Extension of Optimal Ordering Policy for Stock-Dependent Demand under Progressive Payment Scheme. European Journal of Operational Research, 5, 97-04. http://dx.doi.org/0.06/j.ejor.0.05.056 [9] Min, J., Zhou, Y.W., Liu, G.Q. and Wang, S.D. (0) An EPQ Model for Deteriorating Items with Inventory-Level- Dependent Demand and Permissible Delay in Payments. International Journal of Systems Science, 43, 039-053. http://dx.doi.org/0.080/00077.0.659685 [0] Ouyang, L.Y. and Chang, C.. (03) Optimal Production Lot with Imperfect Production Process under Permissible Delay in Payments and Complete Backlogging. International Journal of Production Economics, 44, 60-67. http://dx.doi.org/0.06/j.ijpe.03.04.07 695