Supply chain coordination with controllable lead time under imperfect production process

3.1 Notations

The following notations are used throughout this article.

Buyer’s annual demand rate in units per unit time

Decision variables

Buyer’s order quantity in units

3.2 Assumptions

The inventory model deals with single-buyer and single-vendor with single item. The buyer orders a lot size of Q units and the vendor produces nQ units with a finite production rate P, (P > D) in units per unit time in one setup but ships in quantity Q units to the buyer over n times. The buyer’s shortages are completely backordered.

OPEN IN VIEWER

Fig.1

Behavior of buyer’s inspection policy.

4.1 Buyer’s expected total cost

The buyer places an order of Q units, therefore for expected cycle time of Q D , the expected ordering and lead time crashing cost per unit time can be given by kD Q and DR ( L ) Q . The expected net inventory level just before the arrival of a procurement quantity Q is only the safety stock τ = r - DL. Each shipment takes time Q D to be consumed by the buyer, thus T = Q D and during this time the buyer’s average inventory is Q/2. Also, the expected inventory over the cycle is Q 2 + τ σ L . Therefore, the holding cost for the cycle stock per unit time is c b h b Q 2 . On the other hand, the safety stock τ σ L is held throughout the cycle. Therefore, according to credit policy, the interest is charged at the rate of I _p to this portion of stock by the vendor and the buyer must pay the holding cost for this portion of stock as well. Hence, the total cost for safety stock per unit time as sum of holding cost and interest charged, i.e. ( h b + I p ) c b τ σ L .

According to our assumption, the credit period cannot be greater than the ordering time. Therefore, when the buyer’s permissible delay period expires on or before all inventories are depleted completely, the buyer can sell the items and earn interest with the rate of I _e until the end of the credit period M. Hence, the buyer’s interest earned per unit time is D 2 M 2 pI e 2 Q . In addition, the expected shortage B(r) is completely backordered in the previous cycle, which is cleared in the beginning of the current cycle, therefore, during the trade credit period the buyer earns an interest of DMpI e B ( r ) Q per unit time. On the other hand, the buyer still has (Q - DM) units unsold at the end of the permissible delay. Hence, the buyer has a loan for the unpaid purchase cost of unsold units with the interest charge of I _p . Therefore, the opportunity interest cost per cycle time for the unsold items is obtained by ( Q - DM ) 2 c b I p 2 Q . The transportation cost is given by FD Q . The lead time crashing cost per order R(L), is assumed to be an exponential function of L.

Hence the total expected annual cost for the buyer consists of ordering cost, holding cost, safety stock cost, opportunity interest cost, lead time crashing cost, interest earned and the transportation cost is expressed as

EAC b ( Q , L ) = kD Q + Qc b h b 2 + ( h b + I p ) c b τ σ L + ( Q - DM ) 2 c b I p 2 Q + DR ( L ) Q - D 2 M 2 pI e 2 Q - DMpI e B ( r ) Q + FD Q(1)

Once the buyer orders a lot size of Q units, the vendor produces the items in a lot size of nQ units in each production cycle of length nQ/D with constant production rate P units per unit time, and the buyer will receive the supply in n lots each of size Q units. The first lot size of Q units is ready for shipment after time Q/P′ just after the start of the production. During the production period, nQ/P′, the vendor’s inventory is building up at a constant rate, and simultaneously supplies a lot of size Q units to the buyer on every Q/D units of time, thus T = Q/D. By assumption, the production processes are imperfect and the probability of an item to be defective is ɛ, and hence, the vendor’s effective annual production rate P′ = P (1 - ɛ). Since ɛ < 1 - D/P, (i, e . , P (1 - ɛ) > D) ,  Q/P′, the time taken to produce the first shipment of Q good quality items, is less than the cycle time T. Thus, a new production run will start at time T - Q/P′ after the last shipment of each production run. Hence, the vendors holding cost per year for non-defective items are obtained using the same approach as Uthayakumar and Priyan [35]. { ( nQ [ Q P ( 1 - ɛ ) + ( n - 1 ) T ] - nQ ( nQ ) 2 P ( 1 - ɛ ) ) - T ( Q + 2 Q + ⋯ + ( n - 1 ) Q ) } = { ( nQ [ Q P ( 1 - ɛ ) + ( n - 1 ) T ] - n 2 Q 2 2 P ( 1 - ɛ ) ) - Q D ( n ( n - 1 ) Q 2 ) } = { nQ 2 P ( 1 - ɛ ) [ 1 - n 2 ] + n ( n - 1 ) Q 2 2 D }

At the beginning of each production period, there are zero defective items. After the period of production (which has duration of nQ P ( 1 - ɛ ) ), the number of defective items produced is nQ ɛ ( 1 - ɛ ) . Hence the holding cost for defective items and the good quality are h v { ( nQ 2 [ ɛ 1 - ɛ ] ) ( nQ P [ 1 1 - ɛ ] ) } and h v { nQ 2 p ( 1 - ɛ ) [ 1 - n 2 ] + n ( n - 1 ) Q 2 2 D } respectively. Since the production quantity for a vendor in a lot is nQ units, the set-up cost for the vendor is given by k v D nQ . In addition, the buyer has a loan from a bank for the unsold products cost. Therefore, during the credit period of delayed payment, the vendor has an opportunity interest loss with annual rate I _v , by assumption I _v  = I _e . Thus, the vendor’s opportunity interest loss per unit time is I v c b MnQ nQ / D = I v c b MD .

Hence, the vendor’s total expected annual cost comprising, the setup cost, inspection cost, the cost of defective items, holding cost and opportunity interest loss, can be expressed as

EAC v ( Q , n ) = k v D nQ + c i DE [ 1 1 - ɛ ] + c d DE [ ɛ 1 - ɛ ] + h v { QD P [ 1 - n 2 ] E [ 1 1 - ɛ ] + ( n - 1 ) Q 2 + nQD 2 P E [ ɛ ( 1 - ɛ ) 2 ] } + I v c b MD(2)

The expected total annual cost for the vendor-buyer integrated inventory system under permissible delay in payment is given by

EAC ( Q , L , n ) = EAC b ( Q , L ) + EAC v ( Q , n ) EAC ( Q , L , n ) = kD Q + Qc b h b 2 + ( h b + I p ) c b τ σ L + ( Q - DM ) 2 c b I p 2 Q + DR ( L ) Q - D 2 M 2 pI e 2 Q - DMpI e B ( r ) Q + FD Q + k v D nQ + c i DE [ 1 1 - ɛ ] + c d DE [ ɛ 1 - ɛ ] + h v { QD P [ 1 - n 2 ] E [ 1 1 - ɛ ] + ( n - 1 ) Q 2 + nQD 2 P E [ ɛ ( 1 - ɛ ) 2 ] } + I v c b MD(3) where E [x] denotes the expected value of x.

5.1 Normal distribution model

The lead time demand X follows a normal probability distribution function with mean DL and standard deviation σ L and the reorder point, r = DL + τ σ L , where τ is the safety factor and σ, the standard deviation of the lead time demand. Therefore, the buyer’s expected demand shortages at the end of the cycle is given by E ( X - r ) + = B ( r ) = ∫ r ∞ ( x - r ) f ( x ) dx = σ L ψ ( τ ) where ψ (τ) = φ (τ) - τ [1 - φ (τ)] > 0 and φ, φ are the standard normal p.d.f and cumulative distribution function respectively. In developing, the joint total expected cost of the system, the stockout cost term for the buyer is not considered. Normally, it is complicated to calculate the penalty costs associated with a shortage, as a stock-out event may include uncertain manipulates. As a result, authors like Jha and Shankar [27], Moon et. al. [36], Annadurai and Uthayakumar [37] assumes that buyer has set a target service level corresponding to proportion of demands to be satisfied directly from available stock. Hence, a service level constraint puts a limit on the proportion of demands not met from stock. From assumption (5), the actual proportion of demands not met from stock should not exceed the desired value of α. Hence, the Service level constraint can be expressed as

Expected demand shortages at the end of the cycle for a given safety factor Quantity available for satisfying the demand per cycle ≤ α i.e., B ( r ) Q ≤ α

When the lead time demand follows normal distribution, service level constraint is given by σ L ψ ( τ ) Q ≤ α

The total expected cost per unit time for the vendor-buyer integrated inventory system under permissible delay in payment is given by

EAC N ( Q , L , n ) = D Q { k + k v n + R ( L ) - MpI e σ L ψ ( τ ) + F + DM 2 2 ( c b I p - pI e ) } + Q 2 [ ( h b + I p ) c b + h v { D P [ 2 - n ] E [ 1 1 - ɛ ] + ( n - 1 ) + nD P E [ ɛ ( 1 - ɛ ) 2 ] } ] + ( h b + I p ) c b τ σ L + c i DE [ 1 1 - ɛ ] + c d DE [ ɛ 1 - ɛ ] + DMc b ( I v - I p )(4)

Now, the problem is to find the optimal lot size, lead time, setup cost and the total number of deliveries in a production cycle that minimize the joint total expected cost min EAC N ( Q , L , n ) = D Q { k + k v n + R ( L ) - MpI e σ L ψ ( τ ) + F + DM 2 2 ( c b I p - pI e ) } + Q 2 [ ( h b + I p ) c b + h v { D P [ 2 - n ] E [ 1 1 - ɛ ] + ( n - 1 ) + nD P E [ ɛ ( 1 - ɛ ) 2 ] } ] + ( h b + I p ) c b τ σ L + c i DE [ 1 1 - ɛ ] + c d DE [ ɛ 1 - ɛ ] + DMc b ( I v - I p ) subject to σ L ψ ( τ ) Q ≤ α .

In order to find the minimum total cost for this nonlinear programming problem, we temporarily relax the integer requirement on n, and then try to find the optimal solution of EAC ^N  (Q, L, n) with classical optimization technique by taking the first order partial derivatives of EAC ^N  (Q, L, n) with respect to Q.

For fixed Q and L ∈ [L _b , L₀] ,  EAC ^N  (Q, L, n) can be proved to be a convex function of n.

Proposition 1. For fixed Q and L ∈ [L _b , L₀] , EAC ^N  (Q, L, n) is convex in n.

Proof. Taking the first and second partial derivatives of EAC ^N  (Q, L, n) with respect to n, we have ∂ EAC N ( Q , L , n ) ∂ n = - Dk v Qn 2 + Qh v 2 { - D P E [ 1 1 - ɛ ] + 1 + D P E [ ɛ ( 1 - ɛ ) 2 ] } ∂ 2 EAC N ( Q , L , n ) ∂ n 2 = 2 Dk v Qn 3 > 0 .

Therefore, for fixed Q and L ∈ [L _b , L₀] , EAC ^N  (Q, L, n) is convex in n.

Now, for fixed n, we take the first partial derivatives of EAC ^N  (Q, L, n) with respect to Q. ∂ EAC N ( Q , L , n ) ∂ Q = - D Q 2 [ k + k v n + R ( L ) - MpI e σ L ψ ( τ ) + F + DM 2 2 ( c b I p - pI e ) ] + 1 2 [ ( h b + I p ) c b + h v { D P [ 2 - n ] E [ 1 1 - ɛ ] + ( n - 1 ) + nD P E [ ɛ ( 1 - ɛ ) 2 ] } ]

By setting the above equation equal to zero, for a given value of L ∈ [L _b , L₀], we obtain

Q = 2 D [ k + k v n + R ( L ) - MpI e σ L ψ ( τ ) + F + DM 2 2 ( c b I p - pI e ) ] ( h b + I p ) c b + h v { D P [ 2 - n ] E [ 1 1 - ɛ ] + ( n - 1 ) + nD P E [ ɛ ( 1 - ɛ ) 2 ] }(5)

Theoretically, for fixed n and L ∈ [L _b , L₀], we can obtain the optimal value of Q. Moreover, for fixed n, the Hessian matrix of EAC ^N  (Q, L, n) is positive definite. (See Appendix A)

Computational Algorithm 1

Step 1. Set n = 1.

Step 2. For each L ∈ [L _b , L₀], perform steps (2.1) to (2.3).

Step (2.1). Determine the value of Q using Equation (5).

Step (2.2). Let Q ¯ = max { Q , ( σ / α ) L ψ ( τ ) } .

Step (2.3). Compute the corresponding EAC N ( Q ¯ , L , n ) using Equation (4).

Step 3. Find min L ∈ [ L b , L 0 ] EAC N ( Q ¯ , L , n ) . Let EAC N ( Q ( n ) * , L ( n ) * , n ) = min L ∈ [ L b , L 0 ] EAC N ( Q ¯ , L , n ) , then ( Q ( n ) * , L ( n ) * , n ) is the optimal solution for fixed n.

Step 4. Set n = n + 1, repeat steps 2 and 3 and find EAC N ( Q ( n ) * , L ( n ) * , n ) .

Step 5. If EAC N ( Q ( n ) * , L ( n ) * , n ) ≤ EAC N ( Q ( n - 1 ) * , L ( n - 1 ) * , n - 1 ) , then go to step 4, otherwise go to step 6.

Step 6. Set EAC N ( Q * , L * , n * ) = EAC N ( Q ( n - 1 ) * , L ( n - 1 ) * , n - 1 ) , then (Q^*, L^*, n^*) is the set of optimal solution and EAC ^N  (Q^*, L^*, n^*) is the minimum joint total expected cost.

In the previous section, a model is developed under the assumption that the lead time demand follows a normal distribution. Practically, it is seen that the information about the probability distribution of the lead time is limited. Hence in this section, we assume that the density function F of lead time demand X belongs to the class of density functions with a known finite mean DL and standard deviation σ L and we

try to use a mini-max distribution free procedure to solve this problem. Min Q , L , n Max F ˇ ∈ Ω EAC U ( Q , L , n )(6)

To solve our problem, we apply the following proposition which was asserted by Gallego and Moon [38], Annadurai and Uthayakumar [39].

Proposition 2. For any F ˇ ∈ Ω ,

E ( X - r ) + ≤ 1 2 { σ 2 L + ( r - DL ) 2 - ( r - DL ) }(7)

Substituting r = DL + τ σ L into (7), we obtain the following inequality E ( X - r ) + ≤ 1 2 { σ L ( 1 + τ 2 - τ ) }(8)

Now using (3) and inequality (8), (6) reduces to minimize EAC U ( Q , L , n ) = D Q { k + k v n + R ( L ) - MpI e σ L ( 1 + τ 2 - τ ) + F + DM 2 2 ( c b I p - pI e ) } + Q 2 [ ( h b + I p ) c b + h v { D P [ 2 - n ] E [ 1 1 - ɛ ] + ( n - 1 ) + nD P E [ ɛ ( 1 - ɛ ) 2 ] } ] + ( h b + I p ) c b τ σ L + c i DE [ 1 1 - ɛ ] + c d DE [ ɛ 1 - ɛ ] + DMc b ( I v - I p ) subject to σ L ψ ( τ ) Q ≤ α(9) where EAC ^U  (Q, L, n) is the least upper bound of EAC (Q, L, n).

As on before, for fixed n, it can be verified that the values of Q, L and n satisfy the second order sufficient conditions and we can show that EAC ^U  (Q, L, n) is a convex in Q. Hence, to find the optimal value, we take the partial derivative of EAC ^U  (Q, L, n) with respect to Q. ∂ EAC U ( Q , L , n ) ∂ Q = - D Q 2 [ k + k v n + R ( L ) - MpI e σ L ( 1 + τ 2 - τ ) + F + DM 2 2 ( c b I p - pI e ) ] + 1 2 [ ( h b + I p ) c b + h v { D P [ 2 - n ] E [ 1 1 - ɛ ] + ( n - 1 ) + nD P E [ ɛ ( 1 - ɛ ) 2 ] } ]

By setting the above equation equal to zero, for a given value of L ∈ (L _b , L₀), we obtain

Q = 2 D [ k + k v n + R ( L ) - MpI e σ L ( 1 + τ 2 - τ ) + F + DM 2 2 ( c b I p - pI e ) ] ( h b + I p ) c b + h v { D P [ 2 - n ] E [ 1 1 - ɛ ] + ( n - 1 ) + nD P E [ ɛ ( 1 - ɛ ) 2 ] }(10)

Theoretically, for fixed n and L ∈ [L _b , L₀], we can obtain the optimal value of Q. Moreover, for fixed n, the Hessian matrix of EAC ^U  (Q, L, n) is positive definite. (See Appendix B)

Computational Algorithm 2

Step 1. Set n = 1.

Step 2. For each L ∈ [L _b , L₀], perform steps (2.1) to (2.3).

Step (2.1). Determine the value of Q using Equation (10).

Step (2.2). Let Q ¯ = max { Q , ( σ / α ) L ψ ( τ ) } .

Step (2.3). Compute the corresponding EAC U ( Q ¯ , L , n ) using Equation (9).

Step 3. Find min L ∈ [ L b , L 0 ] EAC U ( Q ¯ , L , n ) .

Let EAC U ( Q ( n ) * , L ( n ) * , n ) = min L ∈ [ L b , L 0 ] EAC U ( Q ¯ , L , n ) , then ( Q ( n ) * , L ( n ) * , n ) is the optimal solution for fixed n.

Step 4. Set n = n + 1, repeat steps 2 and 3 and find EAC U ( Q ( n ) * , L ( n ) * , n ) .

Step 5. If EAC U ( Q ( n ) * , L ( n ) * , n ) ≤ EAC U ( Q ( n - 1 ) * , L ( n - 1 ) * , n - 1 ) , then go to step 4, otherwise go to step 6.

Step 6. Set EAC U ( Q * , L * , n * ) = EAC U ( Q ( n - 1 ) * , L ( n - 1 ) * , n - 1 ) then (Q^*, L^*, n^*) is the set of optimal solution and EAC ^U  (Q^*, L^*, n^*) is the minimum joint total expected cost.