Abstract: This research tries to deal with the retailer`s lot-sizing problem under two warehouses and two levels of delay permitted using algebraic method. In addition, we develop easy-to-use procedures to find the optimal lot-sizing policy for the retailer under minimizing annual total relevant cost. This study provides this algebraic approach that could be used easily to introduce the basic inventory theories to younger students who lack the knowledge of calculus. Furthermore, we develop three easy-to-use theorems to efficiently determine the optimal cycle time and optimal lot sizing for the retailer. Finally, a numerical example is given to illustrate these theorems obtained in this study. In addition, we obtain a lot of managerial insights from this numerical example.
INTRODUCTION
In most business transactions, the supplier will allow a specified credit period to the retailer for payment without penalty to stimulate the demand of his/her products. All previously published models discussed delay permitted assumed that the supplier would offer the retailer a delay period but the retailer would not offer the delay period to his/her customer. That is one level of delay permitted. Recently, Huang (2003) modified this assumption to assume that the retailer will adopt the delay permitted policy to stimulate his/her customer demand to develop the retailer’s replenishment model. That is two levels of delay permitted. This new viewpoint is more matched real-life situations in the supply chain model. Many papers have appeared in the literature that treat inventory problems with varying conditions under one level of delay permitted. Some of the prominent papers are discussed below.
Goyal (1985) established a single-item inventory model under permissible delay in payments. Teng (2002) assumed that the selling price not equal to the purchasing price to modify Goyal’s model (1985). Huang (2003) extended this issue under two levels of trade credit and developed an efficient solution procedure to determine the optimal lot-sizing policy of the retailer. Such delay permitted policy is one kind of encouragement of the retailer to order large quantities because a delay of payments indirectly reduces inventory cost. Hence, the retailer may purchase more goods than that can be stored in its own warehouse. Therefore, these excess quantities are stored in a rented warehouse. In general, the inventory costs for rented warehouse are higher than those for own warehouse.
Therefore, the main purpose of this paper tries to modify Huang’s model (2003) by considering two warehouses. In addition, we try to use the more easily algebraic method to find the optimal solution in this paper. In previous all published papers which have been derived using differential calculus to find the optimal solution and the need to prove optimality condition with second-order derivatives. The mathematical methodology is difficult to many younger students who lack the knowledge of calculus. In recent papers, Cárdenas-Barrón (2001) and Grubbström and Erdem (1999) showed that the formulae for the EOQ and EPQ with backlogging derived without differential calculus. They mentioned that this approach must be considered as a pedagogical advantage for explaining the basic inventory concepts to students that lack knowledge of derivatives, simultaneous equations and the procedure to construct and examine the Hessian matrix. This algebraic approach could be used easily to introduce the basic inventory theories to younger students who lack the knowledge of calculus.
So, this study tries to deal with the retailer’s lot-sizing problem under two warehouses and two levels of delay permitted using algebraic method. In addition, we develop easy-to-use procedures to find the optimal lot-sizing policy for the retailer under minimizing annual total relevant cost.
MODEL FORMULATION
Here, we want to develop the retailer’s inventory model under two warehouses and two levels of delay permitted. For convenience, most notation and assumptions similar to Huang (2003) will be used in this paper.
Notation:
| A | = | Ordering cost per order |
| c | = | Unit purchasing price per item |
| D | = | Demand rate per year |
| h | = | Unit stock holding cost per item per year excluding interest charges |
| Ie | = | Interest earned per $ per year |
| Ip | = | Interest charged per $ in stocks per year by the supplier |
| k | = | Unit stock holding cost of rented warehouse per item per year, (k>h) |
| M | = | The retailer’s trade credit period offered by supplier in years |
| N | = | The customer’s trade credit period offered by retailer in years |
| tw | = | The rented warehouse time in years, |
| T | = | The cycle time in years |
| W | = | Retailer’s storage capacity |
| TRC (T) | = | The annual total relevant cost, which is a function of T |
| T* | = | The optimal cycle time of TRC (T) |
| Q* | = | The optimal order quantity= DT*. |
Assumptions:
| • | Demand rate is known and constant. |
| • | Shortages are not allowed. |
| • | Time horizon is infinite. |
| • | Replenishments are instantaneous with a known and constant lead time. |
| • | k>h and Ip>Ie. |
| • | If the order quantity is larger than retailer’s storage capacity W, the retailer will rent the warehouse to storage these exceeding items. And the rented warehouse has unlimited capacity. When the demand occurs, it first is replenished from the warehouse which storages those exceeding items. |
| • | When T > M, the account is settled at T = M and the retailer starts paying for the interest charges on the items in stock with rate Ip. When T<M, the account is settled at T = M and the retailer does not need to pay any interest charge. |
| • | The retailer can accumulate revenue and earn interest after his/her customer pays for the amount of purchasing cost to the retailer until the end of the trade credit period offered by the supplier. That is, the retailer can accumulate revenue and earn interest during the period N to M with rate Ie under the condition of trade credit. |
The model: The total annual relevant cost consists of the following elements. Three situations may arise. (I) M>W/D>N, (II) M>N>W/D and (III) W/D>M>N.
Case I: Suppose that M>W/D > N.
| • | Annual ordering cost = A/T. |
| • | According to assumption (6), annual stock holding cost (excluding interest charges) can be obtained as follows. |
| • | W/D < T. |
| In this case, the order quantity is larger than retailer’s storage capacity. So the retailer needs to rent the warehouse to storage the exceeding items. Hence | |
| Annual stock holding cost = annual stock holding cost of rented warehouse+annual stock holding cost of the storage capacity W |
| • | T<W/D. |
| In this case, the order quantity is not larger than retailer’s storage capacity. So the retailer will not necessary to rent warehouse to storage items. Hence | |
| Annual stock holding cost = DTh/2. | |
| • | According to assumption (7), cost of interest charges for the items kept in stock per year can be obtained as follows. |
| • | M<T. |
| Cost of interest charges for the items kept in stock per year = |
|
| • | T<M. |
| In this case, no interest charges are paid for the items kept in stock. | |
| • | According to assumption (8), interest earned per year can be obtained as follows. |
| • | M<T. |
| • | N<T<M. |
| • | 0<T<N. |
| Annual interest earned = cIeDT(M-N)/T |
From the above arguments, the annual total relevant cost for the retailer can be expressed as TRC (T) = ordering cost+stock-holding cost+interest payable-interest earned.
We show that the annual total relevant cost, TRC(T), is given by
| (1) |
where
| (2) |
| (3) |
| (4) |
and
| (5) |
Since TRC1(M) = TRC2(M), TRC2(W/D) = TRC3(W/D) and TRC3(N) = TRC4(N), TRC(T) is continuous and well defined on T > 0. All TRC1(T), TRC2(T), TRC3(T), TRC4(T) and TRC(T) are defined on T > 0.
Case II: Suppose that M > N > W/D.
If M > N > W/D, equations 1(a, b, c, d) will be modified as
| (6) |
When W/D < T<N, the annual total relevant cost, TRC5(T), consists of the following elements.
| • | Annual ordering cost = A/T |
| In this case, the order quantity is larger than retailer’s storage capacity. So the retailer needs to rent the warehouse to storage the exceeding items. Hence |
| • | In this case, no interest charges are paid for the items kept in stock. |
| • | Annual interest earned = cIeDT(M-N)/T. |
Combined above elements, we get
| (7) |
Since TRC1(M)=TRC2(M), TRC2(N)=TRC5(N) and TRC5(W/D)=TRC4(W/D), TRC(T) is continuous and well defined on T > 0. All TRC1(T), TRC2(T), TRC5(T) , TRC4(T) and TRC(T) are defined on T > 0.
Case III: Suppose that W/D > M > N.
If W/D > M > N, equations 1(a, b, c, d) will be modified as
| (8) |
When M<T<W/D, the annual total relevant cost, TRC6(T), consists of the following elements.
| • | Annual ordering cost = A/T. |
| • | In this case, the order quantity is not larger than retailer’s storage capacity. So the retailer will not necessary to rent warehouse to storage items. Hence Annual stock holding cost = DTh/2. |
| • | Cost of interest charges for the items kept in stock per year |
| • | Annual interest earned = cIeD(M2-N2)/2T. |
Combined above elements, we get
| (9) |
Since TRC1(W/D) = TRC6(W/D), TRC6(M) = TRC3(M) and TRC3(N) = TRC4(N), TRC(T) is continuous and well defined on T > 0. All TRC1(T), TRC6(T), TRC3(T) , TRC4(T) and TRC(T) are defined on T > 0.
DECISION RULE OF THE OPTIMAL CYCLE TIME T*
In this section, we shall determine optimal cycle time for above three cases under minimizing annual total relevant cost using algebraic method.
Case I: Suppose that M>W/D>N.
Then, we can rewrite
| (10) |
Equation (10) represents that the minimum of TRC1(T) is obtained when the quadratic non-negative term, depending on T, is made equal to zero. Therefore, the optimum value T1* is
| (11) |
Therefore, Eq. (10) has a minimum value for the optimal value of T1* reducing TRC1(T) to
| (12) |
Similarly, we can derive TRC2(T) without derivatives as follows.
| (13) |
Equation (13) represents that the minimum of TRC2(T) is obtained when the quadratic non-negative term, depending on T, is made equal to zero. Therefore, the optimum value T2* is
| (14) |
Therefore, Eq. (13) has a minimum value for the optimal value of T2* reducing TRC2(T) to
| (15) |
Likewise, we can derive TRC3(T) algebraically as follows.
| (16) |
Equation (16) represents that the minimum of TRC3(T) is obtained when the quadratic non-negative term, depending on T, is made equal to zero. Therefore, the optimum value T3* is
| (17) |
Therefore, Eq. (16) has a minimum value for the optimal value of T3* reducing TRC3(T) to
| (18) |
At last, we can derive TRC4(T) algebraically as follows.
| (19) |
Equation (19) represents that the minimum of TRC4(T) is obtained when the quadratic non-negative term, depending on T, is made equal to zero. Therefore, the optimum value T4* is
| (20) |
Therefore, Eq. (19) has a minimum value for the optimal value of T4* reducing TRC4(T) to
| (21) |
Equation (11) implies that the optimal value of T for the case of T > M, that is T1*> M. We substitute equation (11) into T1*> M, then we can obtain that
if and only if
Similarly, Eq. (14) implies that the optimal value of T for the case of W/D < T<M, that is W/D < T2*<M. We substitute Eq. (14) into W/D < T2*<M, then we can obtain that
if and only if
and
if and only if
Likewise, Eq. (17) implies that the optimal value of T for the case of N<T<W/D, that is N<T3*<W/D. We substitute Eq (17) into N<T3*<W/D, then we can obtain that
if and only if
and
if and only if -2A+DN2h≤0
Finally, Eq. (20) implies that the optimal value of T for
the case of T<N, that is T4*< N. We substitute
Eq. (20) into T4*< N, then we can obtain
that
if and only if -2A+DN2h≥0
Furthermore, we let
| (22) |
| (23) |
and
| (24) |
Equations (22), (23) and (24) imply that Δ1>Δ2>Δ3. From above arguments, we can summarize following results.
Theorem 1: Suppose that M > W/D > N, then
| • | If Δ3 > 0, then TRC(T*) = TRC(T4*) and T* = T4*. |
| • | If Δ2 > 0 and Δ3 < 0, then TRC(T*) = TRC(T3*) and T* = T3*. |
| • | If Δ1 > 0 and Δ2 < 0, then TRC(T*) = TRC(T2*) and T* = T2*. |
| • | If Δ1<0, then TRC(T*) = TRC(T1*) and T* = T1*. |
Case II: Suppose that M > N > W/D.
If M > N > W/D, we know TRC(T) as follows from Eq. 6 (a, b, c, d).
From Eq. (7), we can derive TRC5(T) without derivatives as follows.
| (25) |
Equation (25) represents that the minimum of TRC5 (T) is obtained when the quadratic non-negative term, depending on T, is made equal to zero. Therefore, the optimum value T5* is
| (26) |
Therefore, Eq. (25) has a minimum value for the optimal value of T5* reducing TRC5(T) to
| (27) |
Similar as above procedure in Case I. We substitute Eq. (11) into T1*> M, then we can obtain that
if and only if
Substitute Eq. (14) into N<T2*<M, then we can obtain that
and
Substitute Eq. (26) into W/D < T5*< N, then we can obtain that
and
Substitute Eq. (20) into T4*< W/D, then we can obtain that
Furthermore, we let
| (28) |
and
| (29) |
Equations (22), (28) and (29) imply that Δ1>Δ4>Δ5. From above arguments, we can summarize following results.
Theorem 2: Suppose that M > N > W/D, then
| • | If Δ5 > 0, then TRC(T*)= TRC(T4*) and T* = T4*. |
| • | If Δ4 > 0 and Δ5 < 0, then TRC(T*)= TRC(T5*) and T* = T5*. |
| • | If Δ1 > 0 and Δ4 < 0, then TRC(T*)= TRC(T2*) and T* = T2*. |
| • | If Δ1<0, then TRC(T*)= TRC(T1*) and T* = T1*. |
Case III: Suppose that W/D > M > N.
If W/D>M > N, we know TRC(T) as follows from Eq. 8(a, b, c, d).
From Eq. (9), we can derive TRC6(T) without derivatives as follows.
| (30) |
Equation(30) represents that the minimum of TRC6(T) is obtained when the quadratic non-negative term, depending on T, is made equal to zero. Therefore, the optimum value T6* is
| (31) |
Therefore, Eq. (30) has a minimum value for the optimal value of T6* reducing TRC6(T) to
| (32) |
Similar as above procedures in Case I and Case II. We substitute Eq. (11) into T1*> W/D, then we can obtain that
if and only if
Substitute Eq. (31) into M<T6*<W/D, then we can obtain that
if and only if
and
if and only if
Substitute Eq. (17) into N<T3*<M, then we can obtain that
if and only if
and
if and only if -2A+DN2h≤0
Substitute Eq. (20) into T4*< N, then
we can obtain that
if and only if -2A+DN2h≥0
Furthermore, we let
| (33) |
and
| (34) |
Equations (24), (33) and (34) imply that Δ6>Δ7>Δ3. From above arguments, we can summarize following results.
Theorem 3: Suppose that W/D > M > N, then
| • | If Δ3 > 0, then TRC(T*)= TRC(T4*) and T* = T4*. |
| • | If Δ7 > 0 and Δ3 < 0, then TRC(T*)= TRC(T3*) and T* = T3*. |
| • | If Δ6 > 0 and Δ7 < 0, then TRC(T*)= TRC(T6*) and T* = T6*. |
| • | If Δ6 < 0, then TRC(T*)= TRC(T1*) and T* = T1*. |
Numerical example: To illustrate the results obtained in this study, let us apply the proposed method to efficiently solve the following numerical example.
| Table 1: | The optimal cycle time with various values of W, k and c |
For convenience, the numbers of the parameters are selected randomly.
From Table 1, we can observe the optimal cycle time with various parameters of W, k and c, respectively. The following inferences can be made based on Table 1.
| • | The optimal cycle time for the retailer will increase when retailer’s storage capacity W is increasing. The retailer will order more quantity since the retailer owns larger storage space to storage more items. |
| • | When the unit stock holding cost of rented warehouse k is increasing, the optimal cycle time for the retailer will not increase. The retailer will order less quantity to avoid renting expensive warehouse to storage these exceeding items. |
| • | And last, we can find the optimal cycle time for the retailer will decrease when the unit purchasing cost c is increasing. This result implies that the retailer will order less quantity to take the benefits of the delay permitted more frequently. |
CONCLUSIONS
This research adopts the algebraic method to modify Huang’s model (2003) by considering two warehouses. Using this approach presented in this paper, we can find the optimal cycle time without using differential calculus. This should also mean that this algebraic approach is a more accessible approach to ease the learning of basic inventory theories for younger students who lack the knowledge of differential calculus. Furthermore, we develop three easy-to-use theorems to help the retailer in accurately and quickly determining the optimal lot-sizing policy. Finally, a numerical example is given to illustrate all theorems developed in this paper and we obtain a lot of managerial insights from this numerical example.
ACKNOWLEDGMENTS
This research is supported by NSC Taiwan, no. NSC 95-2416-H-324-006 and CYUT.