Figure 1. Inventory Level vs. Time - EOQ Problem

IEOR 254 Spring, 2009 Prof. Leachman

Notes on Economic Lot Scheduling and Economic Rotation Cycles

1. The Economic Order Quantity (EOQ). Consider an inventory item in isolation with demand rate D, holding cost h per unit per unit time, and replenishment cost A

(independent of quantity replenished). What is the optimal replenishment lot size Q? Figure 1 plots inventory level vs. time for this problem for the case D = 1,000 and Q is taken to be 200. Note the “sawtooth” pattern, whereby inventory is replenished by the order quantity (200 in this case), and then steadily falls to 0, whereupon the cycle repeats itself.

Figure 1. Inventory Level vs. Time -

EOQ Problem

0 50 100 150 200 250 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Time In vent o ry Level

We formulate a function expressing the total cost rate as a function of Q. The relevant costs are the inventory holding cost and replenishment cost. Note that replenishment orders will be placed once every Q/D units of time. The inventory level has a maximum of Q and falls linearly to 0; thus the average inventory level is Q/2. The total cost rate is therefore expressed as . 2 / / ) (Q AD Q hQ TRC = +

. / 2 2 / / / Q AD Q2 h Q* AD h TRC ∂ =− + ⇒ = ∂

The optimal time between replenishments is given by . / 2 / * * hD A D Q T = =

Figure 2 plots TRC (Q) vs. Q for the case of h = 0.5, D = 1,000 and A = 10. The best value of Q is where the ordering cost rate and the holding cost rate are equalized; this occurs at Q = 200 in this case. Note that the total cost rate curve is very flat around the best order size.

Figure 2. Total Cost Rate vs. Order Quantity

0 50 100 150 200 250 50 70 90 ₁₁0 ₁₃0 ₁₅0 ₁₇0 ₁₉0 ₂₁0 ₂₃0 ₂₅0 ₂₇0 ₂₉0 ₃₁0 ₃₃0 ₃₅0 Order Quantity, Q Tot al C o st R a te AD/Q hQ/2 TRC(Q)

2. The Economic Manufacturing Quantity (EMQ). Now suppose that replenishment is not instantaneous but occurs at a finite production rate P > D. Now what is the optimal replenishment quantity Q and optimal time between replenishments T?

While replenishment is underway, we add to inventory at rate P but inventory is consumed by demands at rate D. The net rate of change in inventory during

replenishment is P – D. Figure 3 plots inventory versus time for this problem in the case D = 1,000, P = 4,000 and Q is taken to be 200. We still have a sawtooth pattern for the

inventory cycle, but now replenishment occurs gradually rather than all at once, so the triangles are no longer right triangles. During replenishment, inventory accumulates at rate P – D. After a quantity Q has been produced, inventory declines at rate D until the next replenishment batch is started.

Figure 3. Inventory Level vs. Time -

EMQ Problem

0 20 40 60 80 100 120 140 160 180 200 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Time In ven tor y Level

If we replenish a quantity Q, the duration of the replenishment is Q/P and the height of the inventory at the completion of the replenishment is therefore [Q/P](P – D) = Q(1 – D/P). The average inventory level is half of this height (why?). The total cost rate is therefore expressed as . 1 2 1 ) ( ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + = P D hQ Q D A Q TRC

Taking a derivative and solving for Q, we find

. 1 2 * ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = P D h AD Q

Figure 4 plots TRC (Q) vs. Q for the case h = 0.5, A = 10, D = 1,000, P = 4,000. This is the same data as in Figure 1 except we have added a finite replenishment rate P = 4,000. For this case, the optimal batch size is 230, i.e., about 15% bigger than the optimal order

size in the case of an infinite replenishment rate. As before, the total cost rate curve is very flat in the vicinity of the best point.

Figure 4. Total Cost Rate vs. Batch Size

0 50 100 150 200 250 50 70 90 ₁₁0 ₁₃0 _{150 17}0 ₁₉0 ₂₁0 _{230 250 27}0 ₂₉0 ₃₁0 _{330 35}0 Batch Size, Q T o ta l Cos t Ra te AD/Q h(1-D/P)Q/2 TRC(Q)

The optimal time between replenishments in this case is expressed as

. 1 2 / * ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = = P D hD A D Q T_opt

Note that we could have formulated this problem with the time between replenishments as the decision variable:

, 1 2 1 ) ( T P D hD T A T TCR ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + =

and taking a derivative of this expression with respect to T and solving for T results in

. 1 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = P D hD A T_opt

When we consider replenishment of multiple items, it will be more convenient to work with this variable rather than with the order quantity variable. This time between replenishments variable also is commonly referred to as the cycle length.

3. Replenishment Lead Time and Reorder Point. Suppose there is a delay time L between the time of request for replenishment and the time replenishment starts to arrive (finite production rate case) or the time the replenishment arrives (EOQ) case. Then the replenishment order must be placed before the inventory is exhausted.

If the demand occurs exactly at the continuous rate D, then the replenishment order may be made when the inventory falls to the level s = DL. This level is termed the reorder point.

Now suppose the demand is uncertain with mean rate D and suppose the standard deviation of demand during the lead time is σL. Suppose we wish the probability of a stock-out to be very low. It is customary to set the reorder point to be s = DL + kσL where k is called the safety factor. For example, if k = 3 and if demand during the lead time has a normal distribution, then the probability of stock-out during the lead time is about one in one thousand.

4. Changeover time. Suppose there is a fixed delay c between the time a production-replenishment is requested and the time the production-replenishment of inventory begins. (This is an example of an inventory replenishment lead time as defined above.) This delay is

sometimes termed the changeover time or the setup time. The existence of a setup time enforces a constraint on the time between replenishments: there must be time to make the setup and complete production before the replenishment cycle expires, i.e.,

, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ≥ P TD c T or, equivalently, . 1 P D c T − ≥

That is, the minimum cycle length that is feasible is

. 1 min P D c T − =

{

min,

}

. * opt T T Max T =

5. Simple Common Rotation Cycle for Replenishment of Multiple Items. Now suppose N different inventory items must be replenished using the same machine. Only one item may be replenished at a time. There is a changeover time ci when the machine is switched to produce item i after producing some other item. The demand rate for item i is Di, the production rate is Pi, the changeover cost is Ai, and the holding cost is hi.

One way to handle this situation is establish a common rotation cycle of length T for all items. That is, the items are set up in sequence and a quantity lasting T time units is produced of each item. After a time T has elapsed, the cycle starts over again. Obviously, T must be long enough to make all the changeovers and produce a cycle quantity of each item, i.e., , 1 1

∑

∑

= = + ≥ N i i i N i i P TD c T or . 1 min 1 1 T P D c T _N i i i N i i ≡ − ≥

∑

∑

= =

Considering the relevant costs, the total cost rate is

T P D D h T A T TCR N i i i i i N i i

∑

∑

= = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + = 1 1 1 2 1 ) (

and the unconstrained optimal cycle length is given by

. 1 2 1 1

∑

∑

= = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = N i i i i i N i i opt P D D h A T

The best feasible cycle length is therefore given by

{

_min,

}

. * opt T T Max T =

6. Common Rotation Cycle with Item Multipliers. Consider the same situation as above but suppose some of the items have very low demand rates relative to the others. Forcing all items onto the same replenishment cycle may seem wasteful. Instead, it might be more reasonable that a low demand item i should not be produced every cycle but instead be produced once every ki cycles. Here, the parameter ki is called the “multiplier”. At least one item will have ki = 1, i.e., it will be produced every cycle, and its cycle length is termed the fundamental cycle length T. All other items are replenished in an integer multiple of T. The total cost rate in this case is expressed as

. 1 2 1 ) , ( 1 1 T k P D D h T k A T k TRC _i N i i i i i N i i i i

∑

∑

= = ⎟⎟⎠ ⎞ ⎜⎜ ⎝ ⎛ − + =

As simple as this generalization of the common rotation cycle problem is to state, there is no polynomial algorithm for determining optimal values of the ki and T. Fortunately, several practical heuristic algorithms have been developed that generate near-optimal solutions for this problem. One such algorithm, known as the Doll and Whybark Procedure (named after its authors), works as follows:

Step 1. For each item i, compute its optimal cycle length in isolation, i.e.,

. 1 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = i i i i i i P D D h A T

Step 2. Set the fundamental cycle length T to be

{ }

i .

i T

Min T =

Step 3. For each i, set ki to the integer round-up or integer round-down of Ti/T, whichever minimizes . 1 2 1 T k P D D h T k A TRC i i i i i i i i ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + =

Step 4. Considering all possible combinations of items that would be produced in the same cycle, compute the minimum feasible cycle length, i.e.,

. 1 min

∑

∑

∈ ∈ − = j i i i i j i i j cycles P D k c Max T

Step 5. Re-compute the fundamental cycle length using the ki’s from Step 3 and the Tmin from Step 4 as . , 1 2 min 1 1 ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − =

∑

∑

= = T P D D k h k A Max T N i i i i i i N i i i

Step 6. If the value of T is unchanged, terminate the algorithm. Otherwise, return to Step 3 to re-compute the multipliers.