International Journal of Mathematics and Mathematical Sciences
Volume 2009 (2009), Article ID 506376, 15 pages
doi:10.1155/2009/506376
Research Article

Optimal Transfer-Ordering Strategy for a Deteriorating Inventory in Declining Market

Nita H. Shah1 and Kunal T. Shukla2

1Department of Mathematics, Gujarat University, Ahmedabad 380 009, India
2JG College of Computer Application, Drive-in Road, Ahmedabad 380 054, India

Received 16 July 2009; Accepted 19 November 2009

Academic Editor: Heinrich Begehr

Copyright © 2009 Nita H. Shah and Kunal T. Shukla. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The retailer's optimal procurement quantity and the number of transfers from the warehouse to the display area are determined when demand is decreasing due to recession and items in inventory are subject to deterioration at a constant rate. The objective is to maximize the retailer's total profit per unit time. The algorithms are derived to find the optimal strategy by retailer. Numerical examples are given to illustrate the proposed model. It is observed that during recession when demand is decreasing, retailer should keep a check on transportation cost and ordering cost. The display units in the show room may attract the customer.

1. Introduction

The management of inventory is a critical concern of the managers, particularly, during recession when demand is decreasing with time. The second most worrying issue is of transfer batching, the integration of production and inventory model, as well as the purchase and shipment of items. Goyal [1], for the first time, formulated single supplier-single retailer-integrated inventory model. Banerjee [2] derived a joint economic lot size model under the assumption that the supplier follows lot-for-lot shipment policy for the retailer. Goyal [3] extended Banerjee’s [2] model. It is assumed that numbers of shipments are equally sized and the production of the batch had to be finished before the start of the shipment. Lu [4] allowed shipments to occur during the production period. Goyal [5] derived a shipment policy in which, during production, a shipment is made as soon as the buyer is about to face stock out and all the produced stock manufactured up to that point is shipped out. Hill [6] developed an optimal two-stage lot sizing and inventory batching policies. Yang and Wee [7] developed an integrated multilot-size production inventory model for deteriorating items. Law and Wee [8] derived an integrated production-inventory model for ameliorating and deteriorating items using DCE approach. Yao et al. [9] argued the importance of supply chain parameters when vendor-buyer adopts joint policy. The interesting papers in this areas are by Wee [10], Hill [11, 12], Vishwanathan [13], Goyal and Nebebe [14], Chiang [15], Kim and Ha [16], Nieuwenhuyse and Vandaele [17], Siajadi et al. [18], and their cited references. The aforesaid articles are dealing with integrated Vendor-buyer inventory model when demand is deterministic and known constant.

The aim of this paper is to determine the ordering and transfer policy which maximizes the retailer’s profit per unit time when demand is decreasing with time. It is assumed that on the receipt of the delivery of the items, retailer stocks some items in the showroom and rest of the items is kept in warehouse. The floor area of the showroom is limited and wellfurnished with the modern techniques. Hence, the inventory holding cost inside the showroom is higher as compared to that in warehouse. The problem is how often and how many items are to be transferred from the warehouse to the showroom which maximizes the retailer’s total profit per unit time. Here, demand is decreasing with time. This paper is organized as follows. Section 2 deals with the assumptions and notations for the proposed model. In Section 3, a mathematical model is formulated to determine the ordering-transfer policy which maximizes the retailer’s profit per unit time. Section 4 deals with the establishment of the necessary conditions for an optimal solution. Using these conditions, the algorithms are developed. In Section 5, numerical examples are given. The sensitivity analysis of the optimal solution with respect to system parameter is carried out. The research article ends with conclusion in Section 5.

2. Mathematical Model

2.1. The Total Cost per Cycle in the Warehouse

The retailer orders 𝑄 -units per order from a supplier and stocks these items in the warehouse. The 𝑞 -units are transferred from the warehouse to the showroom until the inventory level in the warehouse reaches to zero. Hence 𝑄 = 𝑛 𝑞 . The total cost per cycle during the cycle time 𝑇 in the warehouse is the sum of ( 1 ), the ordering cost 𝐴 , and ( 2 ) the inventory holding cost, ℎ 𝑤 [ ( 𝑛 ( 𝑛 − 1 ) / 2 ) 𝑞 ] 𝑡 1 .

2.2. The Total Cost per Unit Cycle in the Showroom

Initially, the inventory level is 𝐿 0 ≤ 𝐿 due to the unit’s transfer from the warehouse to the display area. The inventory level then depletes to 𝑅 due to time-dependent demand and deterioration of units at the end of the retailer’s cycle time, “ 𝑡 1 .” A graphical representation of the inventory system is exhibited in Figure 1.

506376.fig.001
Figure 1: Combined inventory status for items in the warehouse and showroom.

The differential equation representing inventory status at any instant of time 𝑡 is given by

𝑑 𝐼 ( 𝑡 ) 𝑑 𝑡 = − 𝐷 ( 𝑡 ) − 𝜃 𝐼 ( 𝑡 ) , 0 ≤ 𝑡 ≤ 𝑡 1 ( 2 . 1 ) with boundary condition 𝐼 ( 𝑡 1 ) = 𝑅 . The solution of (2.1) is

𝐼 ( 𝑡 ) = 𝑅 𝑒 𝜃 ( 𝑡 1 − 𝑡 ) e + 𝑎 𝜃 ( 𝑡 1 − 𝑡 ) − 1 ( 𝜃 + 𝑏 ) 𝜃 2 − 𝑏 𝑡 1 𝑒 𝜃 ( 𝑡 1 − 𝑡 ) − 𝑡 𝜃 ; 0 ≤ 𝑡 ≤ 𝑡 1 . ( 2 . 2 ) The total cost incurred during the cycle time 𝑡 1 is the sum of the ordering cost, G and the inventory holding cost, where

i n v e n t o r y h o l d i n g c o s t = ℎ 𝑑 𝑡 1 0 𝐼 ( 𝑡 ) 𝑑 𝑡 = ℎ 𝑑 − 𝑅 𝜃 + 𝑎 𝑏 𝜃 2 𝑡 2 1 − 2 𝜃 − 2 𝑏 − 2 𝜃 2 𝑡 1 2 𝜃 3 − ℎ 𝑑 𝑒 𝜃 𝑡 1 𝑎 𝜃 𝑏 𝑡 1 − 𝜃 − 𝑏 𝜃 3 − 𝑅 𝜃 ( 2 . 3 ) Using (2.2) and 𝐼 ( 0 ) = 𝑞 + 𝑅 , we get

𝑞 = 𝑅 𝑒 𝜃 𝑡 1 𝜃 2 + 𝑎 𝑒 𝜃 𝑡 1 𝜃 + 𝑎 𝑒 𝜃 𝑡 1 𝑏 − 𝑎 𝜃 − 𝑎 𝑏 − 𝑎 𝑏 𝑡 1 𝑒 𝜃 𝑡 1 𝜃 − 𝑅 𝜃 2 𝜃 2 . ( 2 . 4 ) The revenue per cycle is

( 𝑃 − 𝐶 ) 𝑞 = ( 𝑃 − 𝐶 ) 𝑅 𝑒 𝜃 𝑡 1 𝜃 2 + 𝑎 𝑒 𝜃 𝑡 1 𝜃 + 𝑎 𝑒 𝜃 𝑡 1 𝑏 − 𝑎 𝜃 − 𝑎 𝑏 − 𝑎 𝑏 𝑡 1 𝑒 𝜃 𝑡 1 𝜃 − 𝑅 𝜃 2 𝜃 2 . ( 2 . 5 ) Then inventory holding cost in the warehouse is ℎ 𝑤 𝑛 ( 𝑛 − 1 ) 𝑡 1 𝑅 𝑒 𝜃 𝑡 1 𝜃 2 + 𝑎 𝑒 𝜃 𝑡 1 𝜃 + 𝑎 𝑒 𝜃 𝑡 1 𝑏 − 𝑎 𝜃 − 𝑎 𝑏 − 𝑎 𝑏 𝑡 1 𝑒 𝜃 𝑡 1 𝜃 − 𝑅 𝜃 2 2 𝜃 2 . ( 2 . 6 ) Hence, the total profit, 𝑍 𝑃 per cycle during the period [0, 𝑇 ] is

[ ] − [ ] = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 𝑛 𝑍 𝑃 = R e v e n u e − t o t a l c o s t i n t h e w a r e h o u s e t o t a l c o s t i n t h e s h o w r o o m ( 𝑃 − 𝐶 ) 𝑅 𝑒 𝜃 𝑡 1 𝜃 2 + 𝑎 𝑒 𝜃 𝑡 1 𝜃 + 𝑎 𝑒 𝜃 𝑡 1 𝑏 − 𝑎 𝜃 − 𝑎 𝑏 − 𝑎 𝑏 𝑡 1 𝑒 𝜃 𝑡 1 𝜃 − 𝑅 𝜃 2 𝜃 2 − ℎ − 𝐴 𝑤 𝑛 ( 𝑛 − 1 ) 𝑡 1 𝑅 𝑒 𝜃 𝑡 1 𝜃 2 + 𝑎 𝑒 𝜃 𝑡 1 𝜃 + 𝑎 𝑒 𝜃 𝑡 1 𝑏 − 𝑎 𝜃 − 𝑎 𝑏 − 𝑎 𝑏 𝑡 1 𝑒 𝜃 𝑡 1 𝜃 − 𝑅 𝜃 2 2 𝜃 2 − 𝑛 𝐺 − 𝑛 ℎ 𝑑 − 𝑅 𝜃 + 𝑎 𝑏 𝜃 2 𝑡 2 1 − 2 𝜃 − 2 𝑏 − 2 𝜃 2 𝑡 1 2 𝜃 3 + 𝑛 ℎ 𝑑 𝑒 𝜃 𝑡 1 𝑎 𝜃 𝑏 𝑡 1 − 𝜃 − 𝑏 𝜃 3 − 𝑅 𝜃 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . ( 2 . 7 ) During period [0, 𝑇 ], there are 𝑛 -transfers at every 𝑡 1 -time units. Hence, 𝑇 = 𝑛 𝑡 1 . Therefore, the total profit per time unit is 𝑍 𝑛 , 𝑅 , 𝑡 1 = 𝑍 𝑃 𝑇 = 𝑛 ( 𝑃 − 𝐶 ) 𝑅 𝑒 1 𝜃 𝑡 𝜃 2 + 𝑎 𝑒 1 𝜃 𝑡 𝜃 + 𝑎 𝑒 1 𝜃 𝑡 𝑏 − 𝑎 𝜃 − 𝑎 𝑏 − 𝑎 𝑏 𝑡 1 𝑒 1 𝜃 𝑡 𝜃 − 𝑅 𝜃 2 / 𝜃 2 − 𝐴 − 𝑛 𝐺 + ℎ 𝑤 𝑛 ( 𝑛 − 1 ) 𝑡 1 𝑅 𝑒 1 𝜃 𝑡 𝜃 2 + 𝑎 𝑒 1 𝜃 𝑡 𝜃 + 𝑎 𝑒 1 𝜃 𝑡 𝑏 − 𝑎 𝜃 − 𝑎 𝑏 − 𝑎 𝑏 𝑡 1 𝑒 1 𝜃 𝑡 𝜃 − 𝑅 𝜃 2 / 2 𝜃 2 − 𝑛 ℎ 𝑑 − 𝑅 / 𝜃 + 𝑎 𝑏 𝜃 2 𝑡 2 1 − 2 𝜃 − 2 𝑏 − 2 𝜃 2 𝑡 1 / 2 𝜃 3 + 𝑛 ℎ 𝑑 𝑒 1 𝜃 𝑡 𝑎 𝜃 𝑏 𝑡 1 − 𝜃 − 𝑏 / 𝜃 3 − 𝑅 / 𝜃 𝑛 𝑡 1 . ( 2 . 8 )

3. Necessary and Sufficient Condition for an Optimal Solution

The total profit per unit time of a retailer is a function of three variables, namely, 𝑛 , 𝑅 and 𝑡 1 :

𝜕 2 𝑍 𝑛 , 𝑅 , 𝑡 1 𝜕 𝑛 2 = − 2 𝐴 𝑛 3 𝑡 1 < 0 . ( 3 . 1 ) Thus, the retailer’s total profit per unit time is a concave function of 𝑛 for fixed 𝑅 and 𝑡 1 .

Next, to determine the optimum cycle time for showroom, for given 𝑛 , we first differentiate 𝑍 ( 𝑛 , 𝑅 , 𝑡 1 ) with respect to 𝑅 . We get

𝜕 𝑍 𝑛 , 𝑅 , 𝑡 1 = 𝜕 𝑅 1 − 𝑒 𝜃 𝑡 1 𝑡 1 ℎ − ( 𝑃 − 𝐶 ) + 𝑤 ( 𝑛 − 1 ) 𝑡 1 2 + ℎ 𝑑 𝜃 . ( 3 . 2 ) Depending on the sign of ( 𝑃 − 𝐶 ) 𝜃 − ℎ 𝑑 three cases arise: Define Δ = ( 𝑃 − 𝐶 ) 𝜃 − ℎ 𝑑 .

Case 1 ( Δ < 0 ). If Δ < 0 , then 𝑍 ( 𝑛 , 𝑅 , 𝑡 1 ) is a decreasing function of 𝑅 for fixed 𝑅 . It suggests that no transfer of units should be made from the warehouse to the showroom; so put 𝑅 = 0 in 𝑍 ( 𝑛 , 𝑅 , 𝑡 1 ) and differentiate resultant expression with respect to 𝑡 1 . We have 𝜕 𝑍 𝜕 𝑡 1 | | | | 𝑅 = 0 = 0 𝑎 ( 𝑃 − 𝐶 ) 1 − 𝑏 𝑡 1 𝑒 1 𝜃 𝑡 − ( 1 / 2 ) ℎ 𝑤 ( 𝑛 − 1 ) 𝑎 𝜃 2 𝑡 1 1 − 𝑏 𝑡 1 𝑒 1 𝜃 𝑡 ℎ + ( 1 / 2 ) 𝑤 ( 𝑛 − 1 ) 𝑎 1 − 𝑒 1 𝜃 𝑡 ( 𝜃 + 𝑏 ) + 𝑏 𝜃 𝑡 1 𝑒 1 𝜃 𝑡 / 𝜃 2 − ℎ 𝑑 𝑎 / 𝜃 2 𝑏 𝑡 1 − 1 1 − 𝑒 1 𝜃 𝑡 𝑡 1 − 𝑎 ( 𝑃 − 𝐶 ) 1 − 𝑒 1 𝜃 𝑡 ( 𝜃 + 𝑏 ) + 𝑏 𝑡 1 𝑒 1 𝜃 𝑡 𝜃 / 𝜃 2 + ℎ 𝑤 ( 𝑛 − 1 ) 𝑎 1 − 𝑒 1 𝜃 𝑡 ( 𝜃 + 𝑏 ) + 𝑏 𝑡 1 𝑒 1 𝜃 𝑡 𝜃 𝑡 1 / 2 𝜃 2 ℎ − 𝐴 / 𝑛 − 𝐺 − 𝑑 𝑎 𝑏 𝑡 1 2 + 𝜃 𝑡 1 / 2 𝜃 2 − ( 𝜃 + 𝑏 ) 1 + 𝜃 𝑡 1 / 𝜃 3 − ℎ 𝑑 𝑎 𝑏 𝜃 𝑡 1 𝑒 − 𝜃 − 𝑏 1 𝜃 𝑡 / 𝜃 3 𝑡 2 1 = 0 . ( 3 . 3 ) The sufficiency condition is 𝜕 2 𝑍 ( 𝑛 , 𝑅 , 𝑡 1 ) / 𝜕 𝑡 2 1 < 0 , that is, 1 2 𝜃 3 𝑛 𝑡 3 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ − 4 𝑛 𝑎 𝜃 3 𝑡 1 𝑃 𝑒 𝜃 𝑡 1 + 4 𝑛 𝑎 𝜃 3 𝑡 1 𝐶 𝑒 𝜃 𝑡 1 + 4 𝑛 𝜃 2 𝑃 𝑎 𝑒 𝜃 𝑡 1 − 4 𝑛 𝜃 2 𝐶 𝑎 𝑒 𝜃 𝑡 1 + 4 𝑛 𝜃 𝑃 𝑎 𝑏 𝑒 𝜃 𝑡 1 − 4 𝑛 𝜃 2 𝑃 𝑎 + 4 𝑛 𝜃 2 𝐶 𝑎 − 4 𝑛 𝐺 𝜃 3 − 4 𝐴 𝜃 3 − 4 𝑛 𝜃 𝑃 𝑎 𝑏 + 4 𝑛 𝜃 𝐶 𝑎 𝑏 + 4 𝑛 ℎ 𝑑 𝑎 𝜃 + 4 𝑛 ℎ 𝑑 𝑎 𝑏 − 4 𝑛 𝜃 2 𝑃 𝑎 𝑏 𝑡 1 𝑒 𝜃 𝑡 1 − 4 𝑛 𝜃 𝐶 𝑎 𝑏 𝑒 𝜃 𝑡 1 + 4 𝑛 𝜃 2 𝐶 𝑎 𝑏 𝑡 1 𝑒 𝜃 𝑡 1 − 4 𝑛 ℎ 𝑑 𝑎 𝜃 𝑒 𝜃 𝑡 1 − 4 𝑛 ℎ 𝑑 𝑎 𝑏 𝑒 𝜃 𝑡 1 + 4 𝑛 ℎ 𝑑 𝑎 𝑏 𝑡 1 𝜃 𝑒 𝜃 𝑡 1 + 2 𝑛 𝑎 𝜃 4 𝑡 2 1 𝑃 𝑒 𝜃 𝑡 1 + 2 𝑛 𝑎 𝜃 3 𝑡 2 1 𝑃 𝑏 𝑒 𝜃 𝑡 1 − 2 𝑛 𝑎 𝜃 4 𝑡 3 1 𝑃 𝑏 𝑒 𝜃 𝑡 1 − 2 𝑛 𝑎 𝜃 4 𝑡 2 1 𝐶 𝑒 𝜃 𝑡 1 − 2 𝑛 𝑎 𝜃 3 𝑡 2 1 𝐶 𝑏 𝑒 𝜃 𝑡 1 + 2 𝑛 𝑎 𝜃 4 𝑡 3 1 𝐶 𝑏 𝑒 𝜃 𝑡 1 − 𝑛 2 𝑎 𝜃 4 𝑡 3 1 ℎ 𝑤 𝑒 𝜃 𝑡 1 + 𝑛 2 𝑎 𝜃 3 𝑡 3 1 ℎ 𝑤 𝑏 𝑒 𝜃 𝑡 1 + 𝑛 2 𝑎 𝜃 4 𝑡 4 1 ℎ 𝑤 𝑏 𝑒 𝜃 𝑡 1 + 𝑛 𝑎 𝜃 4 𝑡 3 1 ℎ 𝑤 𝑒 𝜃 𝑡 1 − 𝑛 𝑎 𝜃 3 𝑡 3 1 ℎ 𝑤 𝑏 𝑒 𝜃 𝑡 1 − 𝑛 𝑎 𝜃 4 𝑡 4 1 ℎ 𝑤 𝑏 𝑒 𝜃 𝑡 1 − 2 𝑛 𝑎 𝜃 3 𝑡 2 1 ℎ 𝑑 𝑒 𝜃 𝑡 1 − 2 𝑛 𝑎 𝜃 2 𝑡 2 1 ℎ 𝑑 𝑏 𝑒 𝜃 𝑡 1 + 2 𝑛 𝑎 𝜃 3 𝑡 3 1 ℎ 𝑑 𝑏 𝑒 𝜃 𝑡 1 + 4 𝑛 𝑎 𝜃 2 𝑡 1 ℎ 𝑑 𝑒 𝜃 𝑡 1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ < 0 . ( 3 . 4 ) Thus, 𝑍 ( 𝑛 , 𝑡 1 ) , the total profit per unit time, is a concave function of 𝑡 1 for fixed 𝑛 . There exists a unique 𝑡 1 , denoted by 𝑡 1 ∗ 1 such that 𝑍 ( 𝑛 , 𝑡 1 ∗ 1 ) is maximum. Substituting 𝑡 1 ∗ 1 and 𝑅 ∗ = 0 into (2.5) are obtain number of units to be transferred (say) 𝑞 ∗ 1 for fixed 𝑛 .

Note. Since 𝑞 ∗ 1 ≤ 𝐿 for all 𝑞 , 𝑞 ∗ 1 = 𝐿 . If 𝑞 ∗ 1 > 𝐿 , then obtain 𝑡 1 ∗ 1 using 𝑡 1 ∗ 1 = 1 𝜃 l n 1 + 𝐿 𝜃 2 𝑎 ( 𝜃 + 𝑏 ) . ( 3 . 5 )

Case 2 ( Δ = 0 ). In this case, we made (2.8) as 𝑍 𝑛 , 𝑅 , 𝑡 1 = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ℎ 𝑤 𝑅 𝑒 𝜃 𝑡 1 2 + ℎ 𝑤 𝑎 𝑒 𝜃 𝑡 1 + ℎ 2 𝜃 𝑤 𝑎 𝑏 𝑒 𝜃 𝑡 1 2 𝜃 2 − ℎ 𝑤 𝑎 − ℎ 2 𝜃 𝑤 𝑎 𝑏 2 𝜃 2 − 𝑡 1 ℎ 𝑤 𝑎 𝑏 𝑒 𝜃 𝑡 1 − ℎ 2 𝜃 𝑤 𝑅 2 − 𝐺 𝑡 1 − 𝐴 𝑛 𝑡 1 − 𝑛 ℎ 𝑤 𝑅 𝑒 𝜃 𝑡 1 2 − 𝑛 ℎ 𝑤 𝑎 𝑒 𝜃 𝑡 1 − 2 𝜃 𝑛 ℎ 𝑤 𝑎 𝑏 𝑒 𝜃 𝑡 1 2 𝜃 2 + 𝑛 ℎ 𝑤 𝑎 + 2 𝜃 𝑛 ℎ 𝑤 𝑎 𝑏 2 𝜃 2 + 𝑛 𝑡 1 ℎ 𝑤 𝑎 𝑏 𝑒 𝜃 𝑡 1 + 2 𝜃 𝑛 ℎ 𝑤 𝑅 2 + ℎ 𝑑 𝑎 𝜃 − 𝑡 1 ℎ 𝑑 𝑎 𝑏 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 2 𝜃 . ( 3 . 6 ) Here, 𝜕 𝑍 𝑛 , 𝑅 , 𝑡 1 ℎ 𝜕 𝑅 = − 𝑤 2 𝑒 ( 𝑛 − 1 ) 𝜃 𝑡 1 − 1 < 0 . ( 3 . 7 ) that is, 𝑍 ( 𝑛 , 𝑅 , 𝑡 1 ) is decreasing function of 𝑅 for given 𝑛 . So no transfer should be made from the warehouse to the showroom, that is, 𝑅 = 0. So (3.6) becomes 𝑍 𝑛 , 𝑡 1 = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ℎ 𝑤 𝑎 𝑒 𝜃 𝑡 1 + ℎ 2 𝜃 𝑤 𝑎 𝑏 𝑒 𝜃 𝑡 1 2 𝜃 2 − ℎ 𝑤 𝑎 − ℎ 2 𝜃 𝑤 𝑎 𝑏 2 𝜃 2 − 𝑡 1 ℎ 𝑤 𝑎 𝑏 𝑒 𝜃 𝑡 1 − 𝐺 2 𝜃 𝑡 1 − 𝐴 𝑛 𝑡 1 − 𝑛 ℎ 𝑤 𝑎 𝑒 𝜃 𝑡 1 − 2 𝜃 𝑛 ℎ 𝑤 𝑎 𝑏 𝑒 𝜃 𝑡 1 2 𝜃 2 + 𝑛 ℎ 𝑤 𝑎 + 2 𝜃 𝑛 ℎ 𝑤 𝑎 𝑏 2 𝜃 2 + 𝑛 𝑡 1 ℎ 𝑤 𝑎 𝑏 𝑒 𝜃 𝑡 1 + ℎ 2 𝜃 𝑑 𝑎 𝜃 − 𝑡 1 ℎ 𝑑 𝑎 𝑏 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 2 𝜃 . ( 3 . 8 ) The optimal value of 𝑡 1 ∗ 2 can be obtained by solving 𝜕 𝑍 𝑛 , 𝑡 1 𝜕 𝑡 1 = ⎛ ⎜ ⎜ ⎜ ⎝ ℎ 𝑤 𝑎 𝑒 𝜃 𝑡 1 2 − 𝑡 1 ℎ 𝑤 𝑎 𝑏 𝑒 𝜃 𝑡 1 2 + 𝐺 𝑡 2 1 + 𝐴 𝑛 𝑡 2 1 − 𝑛 ℎ 𝑤 𝑎 𝑒 𝜃 𝑡 1 2 + ℎ 𝑤 𝑡 1 𝑛 𝑎 𝑏 𝑒 𝜃 𝑡 1 2 − ℎ 𝑑 𝑎 𝑏 ⎞ ⎟ ⎟ ⎟ ⎠ 2 𝜃 = 0 . ( 3 . 9 ) The sufficiency condition is 𝜕 2 𝑍 𝑛 , 𝑡 1 𝜕 𝑡 2 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ = − 𝑛 ℎ 𝑤 𝑎 𝜃 𝑒 𝜃 𝑡 1 2 − 𝑛 𝑎 𝑏 ℎ 𝑤 𝑒 𝜃 𝑡 1 2 − 𝑛 𝑎 𝑏 𝑡 1 𝜃 ℎ 𝑤 𝑒 𝜃 𝑡 1 2 − 𝑎 𝜃 ℎ 𝑤 𝑒 𝜃 𝑡 1 2 + 𝑎 𝑏 ℎ 𝑤 𝑒 𝜃 𝑡 1 2 + 𝑡 1 ℎ 𝑤 𝑎 𝑏 𝜃 𝑒 𝜃 𝑡 1 2 + 2 𝐺 𝑡 3 1 + 2 𝐴 𝑛 𝑡 3 1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ < 0 , f o r 𝑡 1 = 𝑡 1 ∗ 2 . ( 3 . 1 0 ) Then, 𝑍 ( 𝑛 , 𝑡 1 ∗ 2 ) is a concave function of 𝑡 1 ∗ 2 and hence 𝑍 ( 𝑛 , 𝑡 1 ∗ 2 ) is the maximum profit of the retailer. 𝑞 ∗ 2 can be obtained by substituting value of 𝑡 1 ∗ 2 in (2.5).

Note. Since 𝑞 ∗ 2 ≤ 𝐿 for all 𝑞 , then 𝑞 ∗ 2 = 𝐿 . If 𝑞 ∗ 2 > 𝐿 , then obtain 𝑡 1 ∗ 2 using, 𝑡 1 ∗ 2 = 1 𝜃 l n 1 + 𝐿 𝜃 2 𝑎 ( 𝜃 + 𝑏 ) . ( 3 . 1 1 )

Case 3 ( Δ > 0 ). There are three subcases.Subcase 3.1. ( ( 𝑃 − 𝐶 ) 𝜃 − ℎ 𝑑 ) / 𝜃 𝑡 1 < ℎ 𝑤 ( 𝑛 − 1 ) / 2 and then 𝜕 𝑍 ( 𝑛 , 𝑅 , 𝑡 1 ) / 𝜕 𝑅 < 0 . It is same as Case 1.
The optimal transfer level of units in showroom is zero and there exists a unique 𝑡 1 (say) 𝑡 1 ∗ 3 . 1 such that 𝑍 ( 𝑛 , 𝑡 1 ∗ 3 . 1 ) is maximum.Note. ( 1 ) 𝑡 1 ∗ 3 . 1 ≤ 2 ( ( 𝑃 − 𝐶 ) 𝜃 − ℎ 𝑑 ) / 𝜃 𝑡 1 ℎ 𝑤 ( 𝑛 − 1 ) and then 𝑡 1 ∗ 3 . 1 is infeasible. ( 2 ) Because 𝑞 ≤ 𝐿 for all 𝑞 , 𝑞 ∗ 3 . 1 = 𝐿 . If 𝑞 > 𝐿 , then obtain 𝑡 1 ∗ 3 . 1 using (2.5). ( 3 ) The number of transfers from the warehouse to the showroom must be at least 2.Subcase 3.2 3 . 2 . ( ( 𝑃 − 𝐶 ) 𝜃 − ℎ 𝑑 ) / 𝜃 𝑡 1 > ℎ 𝑤 ( 𝑛 − 1 ) / 2 . Here, 𝜕 𝑍 ( 𝑛 , 𝑅 , 𝑡 1 ) / 𝜕 𝑅 > 0 . Therefore, raise the inventory level to the maximum allowable quantity. So from 𝐿 = 𝑞 + 𝑅 and (2.5), we get 𝑅 = 𝐿 𝜃 2 − 𝑎 𝜃 𝑒 𝜃 𝑡 1 − 𝑎 𝑏 𝑒 𝜃 𝑡 1 + 𝑎 𝜃 + 𝑎 𝑏 + 𝑎 𝑏 𝑡 1 𝜃 𝑒 𝜃 𝑡 1 𝜃 2 𝑒 𝜃 𝑡 1 . ( 3 . 1 2 ) Then 𝑅 is a function of 𝑡 1 . Substitute (3.12) into (2.8). The resultant expression for the total profit per unit time is function of 𝑛 and 𝑡 1 . The necessary condition for finding the optimal time 𝑡 1 ∗ 3 . 2 in showroom is 𝜕 𝑍 𝑛 , 𝑡 1 𝜕 𝑡 1 = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 𝑃 𝑎 𝑏 𝜃 𝑡 1 𝑒 𝜃 𝑡 1 − ℎ 𝑑 𝑎 𝑏 + 𝐺 2 𝜃 𝑡 2 1 + 𝐴 𝑛 𝑡 2 1 − ( 𝑃 − 𝐶 ) 𝐿 𝑡 2 1 − ( 𝑃 − 𝐶 ) 𝑎 𝜃 𝑡 2 1 + ℎ 𝑑 𝑎 𝜃 2 𝑡 2 1 + ℎ 𝑑 𝐿 𝜃 𝑡 2 1 + 𝑛 ℎ 𝑤 𝑎 𝑏 − ( 2 𝜃 𝑃 − 𝐶 ) 𝑎 𝑏 𝜃 2 𝑡 2 1 + ℎ 𝑑 𝑎 𝑏 𝜃 3 𝑡 2 1 − ℎ 𝑤 𝑎 𝑏 − 2 𝜃 𝑛 ℎ 𝑤 𝐿 𝜃 2 𝑒 𝜃 𝑡 1 − 𝐶 𝐿 𝑡 2 1 𝑒 𝜃 𝑡 1 − 𝐶 𝐿 𝜃 𝑡 1 𝑒 𝜃 𝑡 1 + ℎ 𝑤 𝑎 2 𝑒 𝜃 𝑡 1 − ℎ 𝑑 𝐿 𝜃 𝑡 2 1 𝑒 𝜃 𝑡 1 − ℎ 𝑑 𝐿 𝑡 1 𝑒 𝜃 𝑡 1 + 𝑃 𝐿 𝑡 2 1 𝑒 𝜃 𝑡 1 + 𝑃 𝐿 𝜃 𝑡 1 𝑒 𝜃 𝑡 1 + 𝑃 𝑎 𝜃 𝑡 2 1 𝑒 𝜃 𝑡 1 + 𝑃 𝑎 𝑡 1 𝑒 𝜃 𝑡 1 + 𝑃 𝑎 𝑏 𝜃 2 𝑡 2 1 𝑒 𝜃 𝑡 1 − 𝐶 𝑎 𝜃 𝑡 2 1 𝑒 𝜃 𝑡 1 − 𝐶 𝑎 𝑡 1 𝑒 𝜃 𝑡 1 − 𝐶 𝑎 𝑏 𝜃 2 𝑡 2 1 𝑒 𝜃 𝑡 1 − 𝐶 𝑎 𝑏 𝜃 𝑡 1 𝑒 𝜃 𝑡 1 − 𝑛 ℎ 𝑤 𝑎 2 𝑒 𝜃 𝑡 1 − 𝑛 ℎ 𝑤 𝑎 𝑏 2 𝜃 𝑒 𝜃 𝑡 1 + ℎ 𝑤 𝐿 𝜃 2 𝑒 𝜃 𝑡 1 + ℎ 𝑤 𝑎 𝑏 2 𝜃 𝑒 𝜃 𝑡 1 − ℎ 𝑑 𝑎 𝜃 2 𝑡 2 1 𝑒 𝜃 𝑡 1 − ℎ 𝑑 𝑎 𝜃 𝑡 1 𝑒 𝜃 𝑡 1 − ℎ 𝑑 𝑎 𝑏 𝜃 3 𝑡 2 1 𝑒 𝜃 𝑡 1 − ℎ 𝑑 𝑎 𝑏 𝜃 2 𝑡 1 𝑒 𝜃 𝑡 1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ . ( 3 . 1 3 ) The obtained 𝑡 1 = 𝑡 1 ∗ 3 . 2 maximizes the total profit, 𝑍 ( 𝑛 , 𝑡 1 ∗ 3 . 2 ) , per unit time because 𝜕 2 𝑍 𝑛 , 𝑡 1 𝜕 𝑡 2 1 = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ − 2 𝐶 𝐿 𝑡 3 1 + ℎ 𝑑 𝐿 𝜃 𝑡 1 𝑒 𝜃 𝑡 1 − 2 𝑃 𝐿 𝑡 3 1 𝑒 𝜃 𝑡 1 − 2 𝑃 𝐿 𝜃 𝑡 2 1 𝑒 𝜃 𝑡 1 − 𝑃 𝐿 𝜃 2 𝑡 1 𝑒 𝜃 𝑡 1 − 2 𝑃 𝑎 𝜃 𝑡 3 1 𝑒 𝜃 𝑡 1 − 2 𝑃 𝑎 𝑡 2 1 𝑒 𝜃 𝑡 1 − 𝑃 𝑎 𝜃 𝑡 1 𝑒 𝜃 𝑡 1 − 2 𝑃 𝑎 𝑏 𝜃 2 𝑡 3 1 𝑒 𝜃 𝑡 1 + 2 𝐶 𝑎 𝜃 𝑡 3 1 𝑒 𝜃 𝑡 1 + 2 𝐶 𝑎 𝑡 2 1 𝑒 𝜃 𝑡 1 + 𝐶 𝑎 𝜃 𝑡 1 𝑒 𝜃 𝑡 1 − 2 𝐺 𝑡 3 1 + 2 ℎ 𝑑 𝑎 𝜃 𝑡 2 1 𝑒 𝜃 𝑡 1 + ℎ 𝑑 𝑎 𝑏 𝜃 𝑡 1 𝑒 𝜃 𝑡 1 + 2 𝐶 𝑎 𝑏 𝜃 2 𝑡 3 1 𝑒 𝜃 𝑡 1 + 2 𝐶 𝑎 𝑏 𝜃 𝑡 2 1 𝑒 𝜃 𝑡 1 + 𝐶 𝑎 𝑏 𝑡 1 𝑒 𝜃 𝑡 1 + 𝑛 ℎ 𝑤 𝑎 𝜃 2 𝑒 𝜃 𝑡 1 + 𝑛 ℎ 𝑤 𝑎 𝑏 2 𝑒 𝜃 𝑡 1 − ℎ 𝑤 𝐿 𝜃 2 2 𝑒 𝜃 𝑡 1 − 2 𝐶 𝑎 𝑏 𝜃 2 𝑡 3 1 − 2 𝐴 𝑛 𝑡 3 1 − ℎ 𝑤 𝑎 𝑏 2 𝑒 𝜃 𝑡 1 + 2 ℎ 𝑑 𝑎 𝜃 2 𝑡 3 1 𝑒 𝜃 𝑡 1 + ℎ 𝑑 𝑎 𝑡 1 𝑒 𝜃 𝑡 1 + 2 ℎ 𝑑 𝑎 𝑏 𝜃 3 𝑡 3 1 𝑒 𝜃 𝑡 1 + 2 ℎ 𝑑 𝑎 𝑏 𝜃 2 𝑡 3 1 𝑒 𝜃 𝑡 1 − 2 𝐶 𝑎 𝜃 𝑡 3 1 + 2 𝑃 𝑎 𝜃 𝑡 3 1 − 2 ℎ 𝑑 𝑎 𝜃 2 𝑡 3 1 − 2 ℎ 𝑑 𝐿 𝜃 𝑡 3 1 + 2 ℎ 𝑑 𝐿 𝑡 2 1 𝑒 𝜃 𝑡 1 − 𝑃 𝑎 𝑏 𝑡 1 𝑒 𝜃 𝑡 1 − 2 ℎ 𝑑 𝑎 𝑏 𝜃 3 𝑡 3 1 + 𝑛 ℎ 𝑤 𝐿 𝜃 2 2 𝑒 𝜃 𝑡 1 + 2 𝐶 𝐿 𝑡 3 1 𝑒 𝜃 𝑡 1 + 2 𝐶 𝐿 𝜃 𝑡 2 1 𝑒 𝜃 𝑡 1 + 𝐶 𝐿 𝜃 2 𝑡 1 𝑒 𝜃 𝑡 1 − ℎ 𝑤 𝑎 𝜃 2 𝑒 𝜃 𝑡 1 + 2 ℎ 𝑑 𝐿 𝜃 𝑡 3 1 𝑒 𝜃 𝑡 1 − 2 𝑃 𝑎 𝑏 𝜃 𝑡 2 1 𝑒 𝜃 𝑡 1 + 2 𝑃 𝑎 𝑏 𝜃 2 𝑡 3 1 + 2 𝑃 𝐿 𝑡 3 1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ < 0 . ( 3 . 1 4 ) Subcase 3.3. ( ( 𝑃 − 𝐶 ) 𝜃 − ℎ 𝑑 ) / 𝜃 𝑡 1 = ℎ 𝑤 ( 𝑛 − 1 ) / 2 and then 𝑡 1 ∗ 3 . 3 = 2 ( 𝑃 − 𝐶 ) 𝜃 − ℎ 𝑑 𝜃 ℎ 𝑤 ( 𝑛 − 1 ) . ( 3 . 1 5 ) Hence, one can obtain retransfer level of items in the showroom 𝑅 ∗ 3 . 3 and optimal units 𝑞 ∗ 3 . 3 transferred.

Algorithm
Step 1. Assign parametric values to 𝐴 , 𝐺 , ℎ 𝑑 , ℎ 𝑤 , 𝑃 , 𝐶 , 𝑎 , 𝑏 , 𝜃 , 𝐿 .Step 2. If Δ < 0 , then go to Algorithm 3.1.Step 3. If Δ = 0 , then go to Algorithm 3.2.Step 4. If Δ > 0 , then go to Algorithm 3.3.

Algorithm 3.1. Step 1. Set 𝑅 = 0 and 𝑛 = 1.Step 2. Obtain 𝑡 1 ∗ 1 by solving (3.3) with Maple 11 (mathematical software) and 𝑞 ∗ 1 from (2.5).Step 3. If 𝑞 ∗ 1 < 𝐿 , then 𝑡 1 ∗ 1 obtained in Step 2 is optimal; otherwise, 𝑡 1 ∗ 1 = 1 𝜃 l n 1 + 𝐿 𝜃 2 𝑎 ( 𝜃 + 𝑏 ) . ( 3 . 1 6 ) Step 4. Compute 𝑍 ( 𝑛 , 𝑡 1 ∗ 1 ) .Step 5. Increment 𝑛 by 1.Step 6. Continue Steps 2 to 5 until 𝑍 ( 𝑛 , 𝑡 1 ∗ 1 ) < 𝑍 ( ( 𝑛 − 1 ) , 𝑡 1 ∗ 1 ) .

Algorithm 3.2. Step 1. Set 𝑅 = 0 and 𝑛 = 2.Step 2. Obtain 𝑡 1 ∗ 2 from (3.8) and 𝑞 ∗ 2 from (2.5).Step 3. If 𝑞 ∗ 2 < 𝐿 , then 𝑡 1 ∗ 2 obtained in Step 2 is optimal; otherwise, 𝑡 1 ∗ 2 = 1 𝜃 l n 1 + 𝐿 𝜃 2 𝑎 ( 𝜃 + 𝑏 ) . ( 3 . 1 7 ) Step 4. Compute 𝑍 ( 𝑛 , 𝑡 1 ∗ 2 ) .Step 5. Increment 𝑛 by 1.Step 6. Continue Steps 2 to 5 until 𝑍 ( 𝑛 , 𝑡 1 ∗ 2 ) < 𝑍 ( ( 𝑛 − 1 ) , 𝑡 1 ∗ 2 ) .

Algorithm 3.3. Step 1. Set 𝑛 = 2.Step 2. Solve (3.3) to compute 𝑡 1 ∗ 3 . 1 and determine 𝑞 ∗ 3 . 1 from (2.5) and 𝑅 = 0.Step 3. If 𝑞 ∗ 3 . 1 ≤ 𝐿 , then 𝑡 1 ∗ 3 . 1 obtained in Step 2 is optimal; otherwise, 𝑡 1 ∗ 3 . 1 = 1 𝜃 l n 1 + 𝐿 𝜃 2 𝑎 ( 𝜃 + 𝑏 ) ( 3 . 1 8 ) is optimal.Step 4. If ( ( 𝑃 − 𝐶 ) 𝜃 − ℎ 𝑑 ) / 𝜃 𝑡 1 < ℎ 𝑤 ( 𝑛 − 1 ) / 2 then Compute 𝑍 ( 𝑛 , 𝑡 1 ∗ 3 . 1 ) , otherwise set 𝑍 ( 𝑛 , 𝑡 1 ∗ 3 . 1 ) = 0 .Step 5. Solve (3.13) to compute