dtp\ankur.doc

Lecture No.15

Held on: Wednesday, 13 September 2000.

Prepared By: Ankur Jain, 1997232

The motivation behind keeping inventory has already been explained in the previous lecture. In this lecture, we look at some mathematical formulations to model inventory decisions. The following models would be explained:

(1) Economic Order Quantity (EOQ) model.

(2) Economic Manufacturing Quantity (EMQ) model.

(3) EOQ with multiple products & resource constraints.

(4) EOQ model with shortage permitted.

(5) EOQ model with existence of quantity discounts.

1. Economic Order Quantity (EOQ) model

The basic objective in all inventory models to be considered is to minimize cost per unit time.

The present EOQ model is valid for only one product and assumes

(a) Linearly varying holding cost, h being the cost of holding per unit per unit time.

(b) Demand is known with certainty, demand rate given by units per unit time.

(d) Ordering cost independent of order quantity, given by k rupees per order.

(e) Price per unit being independent of order quantity, given by rupees c per unit.

(f) No backordering permitted.

Fig.1 shows the way inventory varies with time under the stated assumptions. One may order different quantities in different cycles, but it can be shown that ordering the same amount every cycle is economical. Moreover it makes sense to order only when current inventory is exhausted since instantaneous delivery is assumed.

In general, holding cost = . In this case, since inventory decreases linearly, the holding cost is simply given by .

Now, cost per cycle = ordering cost + holding cost

C(Q), cost per unit time = ... (1)

where is the cycle period.

\ C(Q) = ... (2)

Thus, there are two contributors to C(Q), one of which increases with Q, the other decreases with increase in Q. C(Q) is clearly convex in Q. The optimal Q at which C(Q) is minimized can be found by calculus to be

... (3)

and optimal cost per unit time to be,

C(Q*) = ... (4)

It may also be seen that

... (5)

Thus, Q* is fairly robust with respect to l and h because of the square root. Also equation (5) shows that even large deviations from Q* result in relatively small deviation of cost from the minimum possible, e.g., if Q = 2Q* then C(Q) = 1.25·C(Q*), i.e., ordering double the economic quantity results in increased cost by only 25%.

2. Economic Manufacturing Quantity MODEL

Now instead of assuming instantaneous production, it is assumed that stock is produced at a finite rate y (y > l) and not supplied instantaneously as in EOQ.

Highest inventory level attained =

In this case,

... (6)

where as before.

The economic quantity is found to be

... (7)

which is higher than in EOQ. It may be seen that y = reduces this model to EOQ.

3. EOQ with multiple products and resource constraints

The basic EOQ model considered only one product and no constraints on Q. We may generalize it to analyze N (>1) products, with constraints on the possible order quantities.

As an example, consider a department that stores three items. Suppose the warehouse has limited space L.

Now, the cost per unit time function is given by

C(Q_i) = ... (8)

and the constraints are ,

The optimal Q_i’s can be found as follows:

define ... (9)

The optimal point must satisfy

for all i;

and

where ;

If then it means optimal solution set of the unconstrained problem also satisfies the constraint and hence is the solution set of constrained problem as well (i.e., use l = 0 in (11)).

If not, then one must obtain l from

... (10)

and use it in

... (11)

to obtain the respective economic orders.

4. EOQ model with shortages permitted

In this model, back ordering is permitted at a cost of p per unit demand unfulfilled per unit time. By backordering it would be possible to save on ordering cost and hence possibly come up with a scheme more economical than without backordering.

In this case,

holding cost =

backordering cost =

ordering cost = k

We may also include the production cost c·Q although it would not affect the value of Q* as it only contributes a constant to C(Q, S). The cost per unit function which now depends on Q as well as S is given by

C(Q, S) = ...(12)

The optimal values of Q and S can be found by equating the first order partial derivatives to zero and solving the two equations simultaneously (the usual method of calculus). The values are found to be

... (13)

and ... (14)

is the cycle time.

It may be noted that p = (i.e., specifying a very very high penalty on backordering) reduces this model to the basic EOQ model.

5. EOQ with quantity discounts

The supplier may offer discount in per unit cost if bulk purchases are made i.e., c may vary with Q.

In this case,

C(Q) = ... (15)

where c_i(Q) is the price per unit time.

For example, we may have

c_i(Q) = Rs. 2 if Q < 1000.

Rs. 1.90 if 1000 Q 2000.

Rs. 1.86 if Q > 2000.

This model can be solved by the following steps:

(a) Calculating EOQ for each price range.

(b) Eliminating the EOQ’s falling outside their respective range.

(d) Selecting the optimal quantity (i.e., one with lowest total cost).

For example, if ordering cost=Rs. 20/order, inventory cost 16% of average inventory per unit per year, the EOQ’s at Rs. 2, 1.90 and 1.86 are found to be 500, 573 and 518.4 units respectively. The total cost may be plotted against order quantity (Fig .4). Possible values for Q* could be 500, 1000 and 2000 (not 573 and 518.4 since these fall outside the range of the cost function). It is found that Q* =1000 when TC=3992 is the minimum.

CONCLUSION

Thus, this lecture looked at the various inventory models, their assumptions and nature of solutions. It may be noted that beginning with the very simple EOQ model, it is possible to keep increasing model complexity in order to capture more and more features found in the realistic scenario.