Inventory Policy for a Deteriorating Item with Time-Varying Demand Under Trade Credit and Inflation

Luqi Wang; Zhijian Chen; Mingyao Chen; Ruijie Zhang

doi:10.21078/JSSI-2019-115-19

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Inventory Policy for a Deteriorating Item with Time-Varying Demand Under Trade Credit and Inflation

Luqi Wang , Zhijian Chen , Mingyao Chen und Ruijie Zhang

Veröffentlicht/Copyright: 31. Mai 2019

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Aus der Zeitschrift Journal of Systems Science and Information Band 7 Heft 2

Abstract

It’s often the case that the supplier will provide the retailer with a permissible delay period in payments, during which the supplier charges the retailer no interest and the retailer accumulates interest earned from investment return. As a type of price reduction and an alternative to price discount, trade credit helps the supplier encourage the retailer’s ordering. This paper develops an inventory replenishment model for a deteriorating item with time-varying demand and shortages, taking account of trade credit and time value of money under inflation over a finite time horizon. This model is an extension and development of the existing studies related to the inventory system considering trade credit and time value of money and offers a more general model with more flexibility and resilience to handle the situation where demand of the end market is non-decreasing with regard to time.

Keywords: inventory; deteriorating items; time-varying demand; trade credit; inflation

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Appendix

Proof of Theorem 1

For the given n and i = 1, 2, ⋯, n − 1,

G1′(K)=−γHn(p+sγ+h+cicγ+θ)e−γHKn−θHn[ce−γM+cice−(θ+γ)M+hγ+θ]eθHKn<0,

therefore, G₁(K) is monotonically decreasing with regard to K.

For

G1(0)=(p+sγ−ce−γM)(1−e−γHn)+cic[1−e−(θ+γ)M]γ+θ−pieMe−γ(M+Hn),G1(1)=−(ce−γM+hγ+θ)(eθHn−e−γHn)−cice−γHn[e(Hn−M)(θ+γ)−1]γ+θ−pieMe−γ(M+2Hn)≤0,

we have the following lemmas.

Lemma 1

IfG₁(0) ≤ 0, thenTP₁(n, K) yields the maximum value atK_i = 0 (i = 1, 2, ⋯, n − 1).

Lemma 2

IfG₁(0) ≥ 0, then there exists a unique solution to, K^*, toG₁(K) = 0, andTP₁(n, K) yields the maximum value atK_i = 0 (i = 1, 2, ⋯, n − 1).

Proof of Lemma 1

If G₁(0) ≤ 0, then G₁(K) ≤ 0 and ∂TP1(n,K)∂K≤0, the latter of which means that TP₁(n, K) is decreasing with regard to K, and TP₁(n, K) will yield the maximum value at K_i = 0 (i = 1, 2, ⋯, n − 1).

Proof of Lemma 2

If G₁(0) ≥ 0, then the solution, K^*, to G₁(K) = 0 will be unique. For i, j = 1, 2, ⋯, n − 1, and i ≠ j, we can easily derive

∂2TP1∂Ki2Ki=Ki∗=Hne−γ(i−1)Hnf(Si∗)G1′(K∗),∂2TP1∂Kj∂Ki=0.

Thus, the k-order Hessian matrix of TP₁(n, K_i) with regard to K_i will be

D1k=∂2TP1∂2K1∂2TP1∂K1∂K2⋯∂2TP1∂K1∂Kk∂2TP1∂K2∂K1∂2TP1∂2K2⋯∂2TP1∂K2∂Kk⋮⋮⋱⋮∂2TP1∂Kk∂K1∂2TP1∂Kk∂K2⋯∂2TP1∂2Kk=Hnf(S1∗)G1′(K∗)0⋯00Hnf(S2∗)G1′(K∗)⋯0⋮⋮⋱⋮00⋯Hnf(Sk∗)G1′(K∗)=∏i=1kHnf(Si∗)G2′(Ki∗),k=1,2,⋯,n−1,

where f(S_i^*) > 0 and G₁'(K^*) < 0.

Therefore, D¹_k, the k-order Hessian matrix of TP₁(n, K) with regard to K, is a negative definite matrix, which means that TP₁(n, K) will yield the maximum value at K_i^* = K^*(i = 1, 2, ⋯, n − 1).

With the above lemmas, we complete the proof of Theorem 1.

Proof of Theorem 2

(i) For the given n and i = 1, 2, ⋯, n − 1,

G2′(Ki)=−γHn(p+sγ+hγ+θ)e−γHKin−θHn(ce−γM+hγ+θ)eθHKin−Hnpiee−γM−Hnpie[γe−γHKin∫Ti−1Sif(t)dtf(Si)+f′(Si)(e−γHKin−e−γM)∫Ti−1Sif(t)dtf2(Si)+e−γM−e−γHKin]=−Hne−γHKin[(p+sγ+hγ+θ)γ−pie]−θHn(ce−γM+hγ+θ)eθHKin−Hnpiee−γM−Hnpie[γe−γHKin∫Ti−1Sif(t)dtf(Si)+f′(Si)(e−γHKin−e−γM)∫Ti−1Sif(t)dtf2(Si)+e−γM]≤0.

therefore, G₂(K_i)is monotonically decreasing with regard to K_i. For

G2(0)=(p+pieMe−γM+sγ−ce−γM)(1−e−γHn)≥0,G2(1)=−(ce−γM+hγ+θ)(eθHn−e−γHn)−pie(e−γM−e−γHn)∫Ti−1Tif(t)dtf(Ti)−piee−γM(Hn−M+Me−γHn)≤0,

then the solution, K_i^*, to G₂(K_i) = 0 will be unique. (ii) For i, j = 1, 2, ⋯, n − 1, and i ≠ j, we can easily derive

∂2TP1∂Ki2Ki=Ki∗=Hne−γ(i−1)Hnf(Si∗)G1′(Ki∗),∂2TP2∂Kj∂Ki=0.

Thus, the k-order Hessian matrix of TP₂(n, K_i) with regard to K_i will be

D2k=∂2TP2∂2K1∂2TP2∂K1∂K2⋯∂2TP2∂K1∂Kk∂2TP2∂K2∂K1∂2TP2∂2K2⋯∂2TP2∂K2∂Kk⋮⋮⋱⋮∂2TP2∂Kk∂K1∂2TP2∂Kk∂K2⋯∂2TP2∂2Kk=Hnf(S1∗)G2′(K1∗)0⋯00Hnf(S2∗)G2′(K2∗)⋯0⋮⋮⋱⋮00⋯Hnf(Sk∗)G2′(Kk∗)=∏i=1kHnf(Si∗)G2′(Ki∗),k=1,2,⋯,n−1,

where f(S_i^*) > 0 and G₂^′(K^*) ≤ 0.

Therefore, D²_k, the k-order Hessian matrix of TP₂(n, K_i) with regard to K_i, is a negative definite matrix, which means that TP₂(n, K_i) will yield the maximum value at (K1∗,K2∗,⋯,Kn−1∗).

Thus, we complete the proof of Theorem 2.

Proof of Theorem 3

(i) for the given n and i = 1, 2, ⋯, n − 1,

GNTC′(Ki)=−γHn(p+sγ+hγ+θ)e−γHKin−θHn(c+hγ+θ)eθHKin<0,

therefore, G^NTC(K_i) is monotonically decreasing with regard to K_i. For

GNTC(1)=(c+hγ+θ)(e−γHn−eθHn)e−γ(i−1)Hn<0,GNTC(0)=(p+sγ−c)(1−e−γHn)>0,

then the solution, Ki∗, to G^NTC(K_i) = 0 will be unique. (ii) for i, j = 1, 2, ⋯, n − 1, and i ≠ j, we can easily derive

∂2TPNTC∂Ki2Ki=Ki∗=Hnf(Si∗)GNTC′(Ki∗),∂2TP∂Kj∂Ki=0.

Thus, the k-order Hessian matrix of TP^NTC(n, K_i) with regard to K_i will be

DkNTC=∂2TPNTC∂2K1∂2TPNTC∂K1∂K2⋯∂2TPNTC∂K1∂Kk∂2TPNTC∂K2∂K1∂2TPNTC∂2K2⋯∂2TPNTC∂K2∂Kk⋮⋮⋱⋮∂2TPNTC∂Kk∂K1∂2TPNTC∂Kk∂K2⋯∂2TPNTC∂2Kk=Hnf(S1∗)GNTC′(K1∗)0⋯00Hnf(S2∗)GNTC′(K2∗)⋯0⋮⋮⋱⋮00⋯Hnf(Sk∗)GNTC′(Kk∗)=∏i=1kHnf(Si∗)GNTC′(Ki∗),k=1,2,⋯,n−1,

where f(S_i^*) > 0 and G^{NTC^′}(K_i^*) < 0 . Therefore, D_k^NTC, the k-order Hessian matrix of TP^NTC(n, K_i) with regard to K_i, is a negative definite matrix, which means that TP^NTC(n, K_i) will yield optimal value at (K1∗,K2∗,⋯,Kn−1∗).

Thus, we complete the proof of Theorem 3.

Received: 2017-04-24

Accepted: 2018-03-19

Published Online: 2019-05-31

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Artikel in diesem Heft

https://doi.org/10.21078/JSSI-2019-115-19

Schlagwörter für diesen Artikel

inventory; deteriorating items; time-varying demand; trade credit; inflation