Chapter 11 Coordinated Replenishments at a Single Stocking Point 11.1 Advantages and Disadvantages of Coordination Advantages of Coordination 1. Savings on unit purchase costs.. Savings on unit transportation costs. 3. Savings on ordering costs. 4. Ease of scheduling. Disadvantages of Coordination 1. A possible increase in the average inventory level.. An increase in system control costs. 3. Reduced flexibility. 1
11. The Deterministic Case: Selection of Replenishment Quantities in a Family of Items 11..1 Assumptions (i The demand rate of each items is constant and deterministic. (ii The replenishment quantity of an item need not be an integer. (iii The unit variable cost of any of the items does not depend on the quantity. (iv The replenishment lead time is of zero duration. (v No shortages are allowed. (vi The entire order quantity is delivered at the same time. 11.. The Decision Rule Notation A= major setup cost for the family, in dollars = minor setup cost for ite, in dollars D i = demand rate of ite, in units/unit time v i = unit variable cost of ite, in $/unit n = number of items in the family = the integer number of T intervals that the replenishment quantity of ite will last (decision variable T = time interval between replenishments of the family (decision variable Derivation Q i = D i T 11.18 T RC(T, m 1,, m n = A + n T + D i T v i r 11.19
T RC T = A + n T + D i v i r = 0 T (m 1,, m n = (A + n r n D i v i 11.0 ( T RC (m 1,, m n = A + r D i v i 11.1 Minimize m1,,m n T RC (m 1,, m n ( Minimize m1,,m n F (m 1,, m n = A + r If we ignore the integrality of s, then F (m 1,, m n m j = a j m j D i v i + D j v j (A + = 0 D i v i 11. m j = a n j D i v i ( D j v j A + n j 11.3 m j m k = a j D j v j D k v k a k m j aj D k v k = m k D j v j a k j k a j D j v j < D kv k a k m j m k The ite having the smallest value of /D i v i should have the lowest value of, namely, 1. Without loss of generality, we number the items such that item 1 has the smallest value of /D i v i. m 1 = 1 11.5 aj n D i v i Equation 11.3 m j = ( D j v j A + n j =, 3, n 11.5 3
n D i v i ( A + n = C 11.6 aj Equation 11.5 m j = C j =, 3, n 11.7 D j v j ai D i v i = D 1 v 1 + C D i v i = D 1 v 1 + C D i v i 11.8 i= D i v i i= = a 1 + 1 C Substituting equations 11.8 & 11.9 into equation 11.6, we get i= D 1 v 1 + C n i= ai D i v i A + a 1 + 1 ni= = C ai D C i v i D i v i 11.9 Algorithm C = D1 v 1 A + a 1 aj D 1 v 1 m j = j =, 3, n 11.30 D j v j A + a 1 (Step 1 Number the items such that /D i v i is smallest for item 1. Set m 1 =1. (Step Evaluate rounded to the nearest integer. ai D 1 v 1 = 11.3 D i v i A + a 1 (Step 3 Evaluate T using the s of Step. T (m 1,, m n = (A + n r n 11. D i v i (Step 4 Determine (Numerical Illustration (p49 Q i = D i T i 11.4 4
11..3 A Bound on the Cost Penalty of the Heuristic Solution In order to find a lower bound on the minimum cost, we use m 1 = 1 and m j s (j =,, n in equation 11.30 which are not necessarily integer values. T RC LB = (A + a 1 D 1 v 1 r + j= a j D j v j r 11.5 The cost of heuristic is T RC H = A + n T + D i T v i r 11.6 (Example T RC LB =$,054.14/year vs. T RC H =$,067.65/year ratio=1.007 11.3 The Deterministic Case with Group Discounts Unit price or freight rate discounts may be offered on the total dollar value or the total volume of a replenishment made up of several items. Main Idea of the Algorithm Compare the Following Three Possible Solutions (1 coordinated solution, assuming a quantity discount, leads to total quantities that are always sufficient to achieve the discount the best to use if it is feasible ( breakpoint solution (3 coordinated solution without a quantity discount Algorithm (Step 1 Compute the s and T as in Section 11. using v i = v oi (1 d Let Q sm be summation of order quantities of all items having =1. If Q sm Q b, use the s, T, and Q i s just found. If not, proceed to Step. 5
(Step Scale up the family cycle time T until the smallest replenishment size equals the quantity breakpoint. Q b T b = 11.8 i {i =1} D i The associated cost of this breakpoint solution is as follows: T RC(T b, m 1,, m n = (1 d D i v oi + A + n T b + D i T b v oi (1 dr 11.9 ( T RC(T, m 1,, m n = D i v oi + A + r D i v oi 11.10 (Step 4 Compare the TRC values found in Steps and 3 and use the s, T, and Q i s associated with the lower of these. (Numerical Illustration (p43 11.4 The Case of Probabilistic Demand and No Quantity Discounts 11.4.1 (S, c, s or Can-Order System A special type of continuous review system for controlling coordinated items savings in setup costs are of primary concern Whenever ite s inventory position drops to or below s i (must-order point, it triggers a replenishment action that raises ite s level to its order-up-to level S i. At the same time any other item j (within the associated family with its inventory position at or below its can-order point c j is included in the replenishment. Figure 11. Behavior of an Item under (S, c, s Control 6
11.6 The Production Environment 11.6.1 The Case of Constant Demand and Capacity: The Economic Lot Scheduling Problem (ELSP Find a cycle length, a production sequence, production times, and idle times, so that the production sequence can be completed in the chosen cycle, the cycle can be repeated over time, demand can be fully met, and annual inventory and setup costs can be minimized. NP-hard problem the need to satisfy a production capacity constraint and the need to have only one product in production at a time (synchronization constraint Notation T = common order interval, or time supply, for each product, in units of time p i = production rate of ite, in units/unit time A i = setup cost for ite, in dollars K i = setup time for ite, in dollars D i = demand rate of ite, in units/unit time v i = unit variable cost of ite, in $/unit n = number of items in the family Minimize T RC i (T = [ Ai T + rv ] id i (p i D i T p i subject to ( K i + D it T 11.1 p i T = Max (Numerical Illustration (p446 n A i n rv i D i (p i D i, T p i n K i 1 n D i p i 7
Relevant Literature Hahm and Yano (199, 1995a, 1995b: ELDSP (ELSP+ delivery schedule Gallego and Moon (199: In the realm of changing the givens, they examine a multiple product factory that employs a cyclic schedule to minimize holding and setup costs. When setup times can be reduced, at the expense of setup costs, by externalizing setup operations, they show that dramatic savings are possible for highly utilized facilities. See also Moon, Gallego, and Simchi-Levi (1991, Hwang, Kim, and Kim (1993, Gallego (1993, Gallego and Moon (1995, and Moon (1994. 11.7 Shipping Consolidation Shipment Consolidation Decisions (Higginson and Bookbinder (1994 (i Which orders will be consolidated and which will be shipped individually? (ii When will orders be released for possible shipping? Immediately, or after some time or quantity-trigger? (iii Where will the consolidation take place? At the factory or at an off-site warehouse or terminal? (iv Who will consolidate? The manufacturer, customer, or a third party? Three Possible Policies (Higginson and Bookbinder (1994 (i a time policy that ships at a prespecified time (ii a quantity policy that ships when a given quantity is achieved (iii a time/quantity policy that ships at the earliest of the time and quantity values The shipper must trade off cost per unit with customer service in deciding on which policy to use. 8