IFSM 300 – Class 19

April 8, 2002

Preliminaries

  • Assignment 19: read chapter 14.
  • Homework 4 (covering chapters 11 and 12) on our Web site in assignment 17. Due today, Monday, April 8.
  • Class Web site reminder:

Inventory Management – Introduction

  • Complete notes below. Summary here.
  • Define inventory. Discuss its role.
  • Describe the special-purpose optimization model, of which inventory models are examples.
  • Discuss how an OR professional works with such special purpose models.
  • Note that inventory may be a component of a supply chain management model, to be covered in the next inventory lecture.

Economic Order Quantity

  • Complete notes below. Summary here.
  • State the questions answered in a simple inventory model.
  • Give an overview of the simplest inventory model: the EOQ model.
  • List the assumptions made to derive the EOQ model.
  • Review derivation of the EOQ equation.
  • Show the inventory-versus-time graph.
  • Find Q = Q* that minimizes total cost.
  • Check that the result is a minimum.
  • Give a numerical example of Q*.
  • Find the order point R*.
  • Give a numerical example of R*.
  • Discuss the summary table 14.7 on page 711 in the book.

Quantity Discounts

  • Complete notes below. Summary here.
  • Refer to Discount Buying, situation 14.2, page 692.
  • Derive the refined EOQ procedure.
  • Consider total cost curve, as in figure 14.2, page 688.
  • Review the concept of multiple cost curves, illustrated in figure 14.4, page 693.
  • Review the procedure summarized on page 695.
  • Numerical example: Data in table 14.2, page 692.Calculations in table 14.3, page 695.

Shortages

  • Handwritten notes below. Summary here.
  • Refer to Managing Shortages, situation 14.3, page 695.
  • Inventory versus time graph.
  • Find Q* and S* to minimize total cost. Find R*.
  • Numerical example.

Millionaire Q1

In a plain economic-order-quantity (EOQ) model, you are assuming:

  1. Periodic review.
  2. Known constant demand rate.
  3. Out-of-stock conditions are permitted.
  4. Costs may change over time.

Millionaire Q2

The total cost in an EOQ model does NOT include:

  1. Ordering cost.
  2. Holding cost.
  3. Shortage cost.
  4. Procurement cost.

Millionaire Q3

In an EOQ model with discounts, the best quantity Q* at each price level is the quantity in the available price range that is:

  1. Largest.
  2. Smallest.
  3. Farthest from the EOQ.
  4. Closest to the EOQ.

Millionaire Q4

The final step applying an EOQ model with discounts is to select the price level that has:

  1. The smallest EOQ.
  2. The smallest total cost.
  3. The largest EOQ.
  4. The largest total cost.

Millionaire Q5 (overhead)

In an EOQ model with backorders, in addition to computing Q* we compute S*, which represents:

  1. Shortage cost.
  2. Amount ordered in addition to Q* to cover backorders.
  3. Shortage cycle time.
  4. Amount to be taken out of Q* to cover backorders.

Millionaire Q6

The reorder point R in an EOQ model with backorders is:

  1. DL.
  2. DL + S.
  3. DL – S.
  4. (DL)(S).

Notes: Inventory Management

Introduction

Previously we reviewed methods that may be used to develop mathematical models in numerous different applications. Now we consider models for specific applications – in this case inventory management.

Inventory is a supply of goods held in storage for future sale or use. In many businesses, such as manufacturing and retailing, and in the military services, the investment in inventory is large in relation to total investment.

Inventory models illustrate a different type of optimization model -- optimization for which the application is specific and the optimization result has been worked out in advance in the form of solution equations.

You might call this a special-purpose optimizationmodel (my own term, not an official part of the OR vocabulary). We may contrast this with the general optimization we studied before, where we began a model from scratch.

Here the framework, described by particular standard assumptions, is the method. To use this framework in practice, the OR professional considers which of the standard model variations (which standard set of assumptions) best matches the application, balancing simplicity against fit. If the standard models being considered won’t match satisfactorily, the OR professional would then search for a better-matching variation, or try to develop a new and different model that matches better.

Ordinary inventory models have many uses. These days, however, the inventory decisions they address often are analyzed in conjunction with other related decisions, in a more comprehensive approach called supply chain management, which we will discuss in the next lecture.

Notes: Inventory Management

Economic Order Quantity

The simpler versions (some not so simple) of inventory models are optimization models that answer the questions: How many units should Ireorder? When should I reorder? The solution in this case comes not as a procedure, such as solving an LP, but rather as equations (formulas) that give the solution.

We begin with the very simplest model, called the economic order quantity(EOQ) model. In the book, refer to Stocking Office Supplies, situation 14.1, page 685.

The basic idea is to list the costs that are affected by the inventory decisions and minimize them in total. Any cost unaffected by an inventory decision (such as allocated overhead) doesn’t belong in this analysis.

The relevant costs we consider are:

  • Procurement cost – the cost of the material itself.
  • Ordering cost – cost of each replenishment action.
  • Holding cost – cost of having material in inventory (also called carrying cost).

As we shall see, only the ordering cost and holding cost are truly relevant. We could have omitted the procurement cost.

EOQ assumptions. The assumptions of a simple, deterministic EOQ model include:

  1. Demand for the inventory is at a known constant rate.
  2. Replenishment quantity is not limited, and need not have an integer value.
  3. No quantity discounts are offered by supplier.
  4. Costs do not change over the time considered.
  5. There is no benefit in pooling different items held in inventory for this analysis (the items are independent).
  6. The replenishment lead time (time from placing an order to receiving and stocking the material) is known.
  7. The full order arrives all at the same time.
  8. No inventory out-of-stock is permitted.

Note that this basic model will be deterministic. That is, we will not recognize uncertainty explicitly, using probability, in the model. Advanced inventory models are probabilistic.

Notation. To put together an EOQ model, we need notation. We’ll use symbols introduced in the book:

c – cost per item (that is, per unit) purchased ($ per unit)

D – demand that causes units to be withdrawn from inventory (units per

designated time period)

c0 – cost to prepare and place one replenishment order ($ per order)

Q – quantity ordered in a replenishment (units)

ch – cost of holding one item in inventory over the designated time period time

($ per designated time period)

T – cycle time, meaning the time between replenishments (time)

I – inventory level, meaning the total number of units in inventory (units)

t – time elapsed since the beginning of the designated time period (time)

TC – total cost related to inventory decisions ($ per designated time period).


Inventory-Versus-Time Graph. To set the stage for an EOQ model, we graph the inventory level I on the y axis versus elapsed time t on the x axis:

The idea is that the quantity ordered, Q, arrives just as the inventory level reaches zero. Because we assume that the items withdrawn from inventory, due to demand, are withdrawn continuously rather than in single discrete units (one among many simplifying assumptions), the curve for inventory level is a smooth straight line, moving down from Q to zero. That’s between two consecutive replenishments. Then the process repeats, giving us a saw-tooth-shaped graph.

The time between replenishments, or cycle time, is T = Q / D. During a cycle, the inventory level is I(t) = Q – D t where t is the time since replenishment. That’s a straight line with slope –D.

Total Related Cost. The approach we take is to write an equation for total related cost TC and then minimize it. In words, we have

TC = (procurement cost) + (ordering cost) + (holding cost).

To assess these costs, we need to specify what period of time is under consideration. That designated time period usually is taken to be one year. Then all costs are annual costs.

With the designated time period of one year in mind, we can see that the procurement cost in a year is the cost per item times the number of items demanded (in a year) = c D.

To write expressions for ordering cost and holding cost, we make one more simplifyingassumption -- namely, that the number of cycles in a year is an integer. For example, we can have 4 cycles, or 5 cycles, but not 4.6 cycles. Then the number of cycles in the designated time period of a year is:

(number of units in a year) / (number of units in a cycle) = D / Q.

The ordering cost in the designated time period (one year) is cost per order times the number of orders in a year. Since the number of orders is the number of cycles, the annual ordering cost is c0 (D / Q).

It sounds reasonable that the annual cost of holding inventory would be the annual cost of holding one unit times the average number of units held. The average number of units held over a cycle is Q / 2. Since this repeats in each cycle, and we have an integer number of cycles (no partial cycles), the average over the year would be Q / 2. That reasoning gives us an annual holding cost of ch (Q / 2).

Confirming the Holding Cost. We can confirm the expression for holding cost by using calculus 101. Although the inventory level varies, in a small time interval  t the level stays fixed (in the limit). So when the level is I(t), the holding cost over the interval from t to t +  t is equal to (annual holding cost per unit) (number of units) (fraction of a year over which these units are held) = ch I(t)  t. To obtain the cost over one cycle, we add up these  t costs:

Q/D Q/D Q/D

 ch I(t) dt =  ch (Q – D t) dt = ch [Q t – D t2 / 2  = ch [Q2 / 2 D].

0 0 0

Then to obtain the annual cost, we multiply this cost-over-a-cycle times the number of cycles in a year. Annual cost = ch (Q2 / 2 D)(D / Q) = ch (Q / 2).

Minimizing the Total Related Cost. When we reorder, we’ll order a number of units Q. The question is: what is the cost-minimizing Q?

To derive the best Q – call it Q* -- we will work with the equation for total related cost:

TC = c D + c0 (D / Q) + Ch (Q / 2).

Drawing once more on calculus 101, we know that a minimum (or a maximum) will occur when the slope of the curve “total-cost-versus-Q” becomes zero. To find that value of Q, we differentiate:

TC = c D + (c0 D) Q-1 + (ch / 2) Q

d/dQ TC = 0 + (c0 D)(-1)(Q-2) + (ch / 2)(1) = -c0 D Q-2 + ch / 2.

Set this equal to zero and solve for Q:

ch / 2 = c0 D Q-2 ,

Q2 = 2 c0 D/ ch ,

Q* =  2 c0 D / ch .

This is the famous economic-order-quantity formula. While its conciseness is appealing, keep in mind that the assumptions leading to it are ultra simple, and quite often not sufficiently realistic for practical application.

Checking That This Is a Minimum (Not a Maximum). To confirm that Q* corresponds to the minimum total cost, not the maximum, we find the second derivative and check that it is positive. We are working with TC expressed as a function of Q. The first derivative is the slope of the curve TC-versus-Q. That slope is zero at a minimum or maximum. The second derivative is the rate at which the slope is changing as Q increases. A positive second derivative shows that the slopes are increasing as Q increases, which indicates a minimum.


This may seem more abstract that it really is. If we were instead working with distance traveled as a function of time elapsed, the first derivative (slope) is speed, as shown on a speedometer. The second derivative is the rate at which speed is changing as time passes, called acceleration. When the speed increases as time passes, we are accelerating -- that is, acceleration is positive.

To check the EOQ, we have:

d/dQ TC = -c0 D Q-2 + ch / 2

d2/dQ2 TC = -c0 D (-2 Q-3) + 0

= 2 c0 D / Q3 ,

which is positive for any positive Q.

EOQ Numerical Example. Suppose our data are:

c – $0.50 per package

c0 – $5.00 per order

D – 81,000 packages per year

ch – $0.09 per package in inventory per year.

Then:

Q* =  2 c0 D / ch =  2(5.00)(81,000)/(0.09) =  810,000/0.09

=  9,000,000 = 3,000 packages in an order.

TC = c D + c0 (D/Q) + ch (Q/2)

= (0.50)(81,000) + (5.00)(81,000)/(3,000) + (0.09)(3,000)/(2)

= 40,500 + 135 + 135 = $40,770 per year.

Reorder Point. Q* is the best quantity to reorder. That leaves the question: when to reorder? The answer is reorder when the inventory level falls to the reorder point R*. We assume that the inventory level is always being monitored, called “continuous review.”

Thus we will know precisely when the inventory level reaches R*.

Let the designated time period for analysis be one year. Define the new notation:

L – lead time to receive an order (in years)

R – reorder point: inventory level at which we reorder (units).

We must reorder when the time remaining before running out of inventory is the lead time it takes to place and receive an order. The inventory still available, therefore, must

cover the demand over the lead time. In other words,

(demand over lead time) = (demand in a year)(fraction of a year in a lead time).


In symbols, R* = D L.

We could change the units of measurement without changing the result. So if l is lead time in days, and d is demand in units per day, then

R* = d l.

Reorder Point Numerical Example. Suppose the data are:

l – 5 days (out of 250 working days of operation)

D – 81,000 packages per year.

Let’s use years. We convert l to L as

L = 5 days/250 working days = 0.02 years.

Then, R* = D L = (81,000)(0.02) = 1,620 packages in inventory.

Or if we work in days, convert D to d as

d = 81,000 packages per year/250 days in a year = 324 packages per day.

Now, R* = d l = (324)(5) = 1,620 packages in inventory – the same answer.

Summary Table in the Book. The equations, or “formulas,” just derived are summarized in the book in table 14.7, page 711. Note the differences between notation there and here. In the book’s table:

R is our R*,

L is our l (lead time in days),

T is cycle time in days, where N means the number of working days in the year.

Notes: Inventory Management

Quantity Discounts

Concept. Now we begin refining the EOQ model to accommodate possible realities that differ from the original assumptions. First, we consider the availability of quantity discounts.

The idea of quantity discounts, common in purchasing, is shown in the following table:

Category / Quantity Ordered
(number of units) / Discount
(% of category 1 price) / Unit Price
($ per unit)
1 / Less than 3,100 / 0.0 / $0.50
2 / 3,100 to 3,999 / 10.0 / $0.45
3 / 4,000 to 9,999 / 24.0 / $0.38
4 / 10,000 and over / 24.2 / $0.379

To find the best order quantity in this situation, follow these steps:

Step 1: Each price offered (c in our original notation) will have its own associated total cost TC that varies with quantity Q, pictured in this graph:


At first we ignore the fact that not all quantities are offered at each price. On each TC-versus-Q curve (that is, at each price c), find the EOQ quantity Q in the normal way.

Step 2: Now we consider what quantities actually are offered. On each TC-versus-Q curve (that is, at each price c), we want to find the Q* that gives us the lowest feasible TC – in other words, gives us the lowest TC that really is available. If the EOQ is an offered quantity, then Q* = EOQ. If the EOQ is not offered, however, then Q* is the feasible Q that is closest to the EOQ. This closest feasible Q will be at a boundary of the offered quantities:


Step 3: At each price c,using the Q* from step 2 calculate TC. Then, finally, select the Q* that has produced the lowest of all TC’s.

To summarize, for each price we find the quantity that’s the best we can do (gives the lowest total cost) at that price. Then we compare the best total costs to select the best of the best quantities (the quantity that provides the overall lowest total cost).

Numerical Example. With the data already introduced, in each price category we can (1) calculate the EOQ, then (2) determine Q* (either the EOQ if available, or, if not available, the closest available Q), and then (3) calculate TC using Q*. The computations are as before with the basic EOQ model. See the book, pages 692 to 695. The results are:

Category / Quantity Ordered
(units) / Unit Price
($ per unit) / EOQ
(units) / Q*
(units) / TC(Q*)
($ per year)
1 / Less than 3,100 / $0.50 / 3,000 / 3,000 / $40,770.00
2 / 3,100 to 3,999 / $0.45 / 3,162.28 / 3,162,28 / $36,706.14
3 / 4,000 to 9,999 / $0.38 / 3,441.24 / 4,000 / $31,018.05
4 / 10,000 and over / $0.379 / 3,445.77 / 10,000 / $31,080.60

Thus in this example we select the quantity 4,000 units. That quantity has the overall lowest total cost: $31,018.05.

Summary Table in the Book. The foregoing procedure is summarized, very briefly, in the book in table 14.7, page 711.

Notes: Inventory Management

Shortages

Concept. The second refinement we consider is to permit going out of stock, creating what is often termed a shortage. We assume that demand remains the same during a period of shortage, but it accumulates in the form of customer backorders to be filled when the next replenishment arrives.

Notation. The symbols we’ll use this time are:

c – cost per unit purchased ($ per unit)

c0 – cost to prepare and place one replenishment order ($ per order)

ch – annual cost of holding one item in inventory ($ per year)

cb – annual cost of holding one item in backorder ($ per year)

D – annual rate of demand for units (units per year)

Q – quantity of units in a replenishment order (units)

S – maximum number of units that will be backordered (units)

L – lead time to receive an order (years)

R – reorder point: inventory level at which we reorder (units)

t1 – cycle time when in stock (years)

t2 – cycle time when out of stock (years)

T – total cycle time: the time between replenishments (years)

I – inventory level: the total number of units in inventory (units)

t – time elapsed since the beginning of the analysis period (years)

TC – total cost related to inventory decisions ($ per year).

Inventory-Versus-Time Graph. To develop the appropriate model for the case of backorders, again we start with a graph showing inventory level I on the y axis and elapsed time t on the x axis.

Now we allow inventory to be depleted. During time t1 inventory is in stock, and is drawn down by demand as before. During time t2 we are out of stock, accumulating backorders. When the reorder point is reached, we’ll order a quantity Q as before (but Q is a different amount now, because of taking the backorders into account). When the order amount Q arrives, we fill the accumulated backorders of amount S and place the remaining quantity Q - S into inventory. Thus the maximum inventory level is Q - S.

It’s still true that the total cycle time is T = Q / D. We can divide this time into two parts: the in-stock time t1 = (Q – S) / D and the backorder time t2 = S / D. The inventory level during a cycle is I(t) = (Q – S) – D t.