Supply Chains: EOQ and Extensions - University of...

Department of Industrial Engineering

Supply Chains:EOQ and Extensions

Jayant Rajgopal, Ph.D., P.E.


University of Pittsburgh

Pittsburgh, PA 15261



Recall that there are three main decisionswith an inventory control system:

1. The review interval (I) (how often…)

2. The reorder point (R) (when…)

3. The order quantity (Q) (how much…)

We will focus on continuous review systems (I=0),and initially, on demand that is static (i.e., steadyover time).

Let us first look at decision 3 – the order quantity (Q)

© Jayant Rajgopal, 2016


Order Quantity - Reorder Point Systemaka (Q,R) System

“When your pills get down to four,

Go ahead, and order some more!”

These models are applicable to independentdemand items that have static demand (i.e.,demand that is “reasonably” constant over time).



Assumptions

1. Continuous Review: The system inventory ismonitored continuously and when it reaches avalue R an order is immediately placed for anamount Q

2. The entire order arrives after a time L (thereplenishment lead-time) and in general, this time isstochastic

3. Demand arrivals occur stochastically, one at a time

4. There is a backorder or shortage penalty for unfilleddemand



Sample Inventory ProfileON-HAND INVENTORY

TIME

R

AverageLead-TimeDemand

Q

Q

Q

Lead-TimeDemand

Lead-TimeDemand

Lead-TimeDemand

Lead-Time Lead-Time Lead-Time

ORDER CYCLE 1 ORDER CYCLE 2 ORDER CYCLE 3

Lead-Time Demand (DLT):

• In Cycle 1: DLT < R

• In Cycle 2: DLT < R

• In Cycle 3: DLT > R ⇒STOCKOUT!!

R: Reorder Point

Q: Order Quantity

0


NotationD = Stationary demand per unit time (assumed to be “steady” with some

constant mean, but varying stochastically around the mean)

µD = Expected value of demand-per-unit-time

σD = Standard Deviation of demand-per-unit-time

L = Replenishment lead-time

µL = Expected value of lead-time

σL = Standard Deviation of lead-time

DLT= Lead-time demand (random variable…)

µDLT = Expected value of lead-time demand

σDLT = Standard Deviation of lead-time demand

NOTE: All time units above MUST be consistent!© Jayant Rajgopal, 2016


The Economic Order Quantity (EOQ)Model (F. W. Harris, 1914)

Let us start with a very simple (simplistic?) model assumingthat everything is deterministic so that we don’t need anysafety stock as long as we plan correctly. In particular, wealso assume that

• The demand rate is constant at D units per unit time (i.e.,σD = 0 so that we can simply use D instead of mD).

• Lead time = 0

• Entire order is available at one time



Inventory Profile with the EOQassumptions

Ordercycle 1

Ordercycle 2

Ordercycle 3

Q

T TTime

Slope = -D

Inventory



Economic Order Quantity (EOQ) Model(F. W. Harris, 1914)

Let Q = order quantity (to be determined)

Clearly, there are

• n = D/Q orders per year, and

• each order cycle covers T=1/n=Q/D time periods.

So, average annual order costs= n(A+cQ)=�

�

Suppose X = area under each triangle in the inventory profile

⇒ Average Inventory =�

�

�

��

�

�

�

So, average annual inventory costs =�

�=

�

�


Q

T T

X


Total Annual Cost (TAC) per unit time (order costs + holding costs) is

The quantity��

�

�

�that depends on Q is also called the Total

Variable Cost (TVC) per unit time, i.e.,

To find optimal Q we differentiate TAC(Q) w.r.t. Q and set it to zero:

�


Derivation of the EOQ Formula


The value��

�is called the Economic Order Quantity:

∗

In particular, the minimum total annual cost using the EOQ is equal to

∗∗

∗



0

2000

4000

6000

8000

10000

12000

14000

16000

18000

0.00 20.00 40.00 60.00 80.00 100.00 120.00

EOQ


1Note that 2

2 2

*

*

AD Q hADh

Q= =

EOQ=Q* Q

AD/QhQ/2

TVC(Q)

Cost


The EOQ is very robust w.r.t. the estimatesof the various parameters in the formula

• Very little increase in the TVC for even ±20% variation in the valuecomputed for the EOQ

To see this, suppose we use some other Q instead of Q*= EOQ = ,

so that TVC(Q) = AD/Q+hQ/2; while TVC(Q*)= AD/Q* +hQ*/2= .

Then 1 £ ∗

��

��

��

��

∗

∗

• If Q/Q* (or Q*/Q) is 1.20 then the above ratio = 1.0167; if Q/Q* is 1.5 it is1.083 and if Q/Q*=2 it is 1.25. So, a 20% error in Q increases costs byless than 2%; even a 100% error in Q increases costs by only 25%!



A warehouse regularly orders four products from four different distributors. Thecharacteristics of the items are as follows; the last column gives the order quantities thatare currently in use.

It is estimated by the company's accountants that the holding costs are about 25 cents onthe dollar annually. However, there is no clear agreement on this point - one section of theproduction team is of the view that holding costs are a lot higher (perhaps as high as 40%of an item’s value annually), while another section feels they are a lot lower (perhaps aslow as 15%). Similarly, there is no clear consensus on the order placement costs. Thecurrent is estimate is $7.50 per order; however estimates run anywhere between $5 and$20 per order. The current order quantities in use have basically "come down through theyears" and are consensus values.

As a new industrial engineer at the company, demonstrate how you could use the EOQformula to find order quantities. Also show how this would reduce costs under differentpossibilities for the order placement and holding costs.

Suppose that no more than 20 orders can be handled annually. How would you modifyyour plans?

Item Annual Usage (D) Price (c) Dollar Volume (D$) Order Quantity Q

A 1,000 $30 $30,000 500

B 200 $5 $1,000 20

C 3,000 $1 $3,000 3,000

D 500 $50 $25,000 1,000



We can readily adapt the EOQ toaccount for uncertainty or constraints

Item Demand Value Dollar Volume Order Quantities

j D C D$ Current Q Q*=EOQ Q+

A 1000 $30 30,000 500 45 122

B 200 $5 1,000 20 49 133

C 3000 $1 3,000 3000 424 1154

D 500 $50 25,000 100 24 65


Suppose we compute the EOQ for each item j byusing the consensus estimates for i and A, that is,

=


Item (Qj/2)×cj nj=Dj/Qj ( �∗/2)×cj nj

*=Dj/ �∗ ( �

�/2)×c ��=Dj/ �

�

A 7500 2 675 22.2 1830 8.2

B 50 10 122.50 4.1 332.5 1.5

C 1500 1 212 7.1 577 2.6

D 2500 5 600 20.8 1625 7.7

(Σ) 11,500 18 1609.50 54.2 4364.5 20

= (0.25*11,500) + (7.50*18)= 2887.50 + 135 =$3022.50

0.25*1609.50 + 7.50*54.2= 402.38+406.50 =$808.88

0.25*4364.5 + 7.50*20= 1091.13+150 =$1241.13


Note: n+=(20/54.2)×n*, or Q+=(54.2/20)×Q*

Note that total cost given any Qj and nj =

� � �


Performance/Cost Analysis

From our analysis: With Q, we get 18 orders & average inventory = $11,550

With Q* we get 54.2 orders & average inventory =$1,609.5

Base Case: A = $7.50, i= $0.25/$/year

Q ≡(18*7.50)+(11,550*0.25) = 135+2,888 =$3,023

Q*≡(54.2*7.50)+(1,609.5*0.25) = 407+403= $810 (savings=$2,213)

Worst Possible Case: A = $20, i = $0.15/$/year

Q ≡(18*20)+(11,550*0.15) = 360+1,733 = $2,092

Q*≡(54.2*20)+(1,609.5*0.15) = 1,084+241= $1,325 (savings=$767)

Best Possible Case: K= $5, i= $0.40/$/year

Q ≡(18*5)+(11,550*0.4) = 90+4,620 = $4,710

Q*≡(54.2*5)+(1,609.5*0.4) = 271+644= $915 (savings=$3,795)



Quantity DiscountsCase 1: “All-units” discount schedule

c1 is the discounted price and Qb is called the “break-point.” Consider the valueof TAC(Q) as a function of Q:

Qb

c0 c1

TAC(Q) TAC(Q)

TAC(Q)

c0

c0

c0

c1c1

c1

Qb Qb

Qb

Q* =

Q* =

Q* =


If 0<Q<Qb then unit price=c0

TAC(Q) = AD/Q + (ic0)Q/2 + Dc0

If Q≥Qb then unit price=c1 (<c0)TAC(Q) = AD/Q + (ic1)Q/2 + Dc1


MULTIPLE PRICE BREAKS:

0 <Q<Q0 ⇒ c=c0

Q0≤Q<Q1 ⇒ c=c1

: : :Qj-1≤Q<Qj ⇒ c=cj etc., where c0>c1>c2>...

The TAC(Q) curves may look as shown below

DEFINITION: An EOQ computed via EOQ = � is said to be

realizable if its value lies within the range for which cj holds

TAC(Q)

Q



FACT: The optimal lot size Q* is equal to the largest realizableEOQ, or a break-point above the largest realizable EOQ.

Procedure:

1. Set c = the cheapest per-unit cost available

2. Compute EOQ =

3. If the value is realizable, go to Step 5

4. Else, set c = the next cheapest per-unit cost and go back to Step 2

5. Compute TAC(Q) with Q equal to the EOQ just found as well asfor all Qb values that are greater than this EOQ - be careful to usethe correct values for c in all cases.

6. Set Q* to the value of Q in Step 5 that has the least value forTAC(Q)



All-units Quantity Discounts: An Example

D=4000/yr A=$40 i=20%

Unit price c uses the following discount schedule

Lot Size (Q) Unit Price (c)

0<Q<500 $ 2.55

500≤Q<2250 $ 2.50

2250≤Q<3200 $ 2.45

3200≤Q<5250 $ 2.40

Q≥5250 $ 2.35

© Jayant Rajgopal,2016


Case 2: Incremental Quantity Discounts

If Q<Qb we spend C(Q) = c0Q to buy the batch of items in the order

On the other hand, if Q>Qb we spend C(Q) = c0Qb + c1(Q-Qb) to buy the batch ofitems in the order. Thus the average price per item is c*= C(Q)/Q

Since there are D/Q orders per year TAC =Annual average setup cost = AD/Q

+ Annual average purchase cost = Dc*

+ Annual average holding cost = (ic*)(Q/2)

Qb

c0c1

Qb Q

TAC


If 0<Q<Qb thenunit price=c0

f Q≥Qb then unit price=c0 forthe first Qb units and c1 (<c0) forthe remaining items


The same idea can be extended to multiple break points...

The minimum always occurs at the minimum of one of the totalcost curves representing the different c* =C(Q)/Q.

So the procedure would be

1. Find c*= C(Q)/Q as an algebraic expression for each priceinterval,

2. Substitute c* from Step 1 into TAC = {AD/Q + (ic*)(Q/2) +Dc*}, and find the value of Q for which this function isminimized. Do this separately for each price interval.

3. Check if the corresponding Q values are realizable or not.

4. For all realizable values, check the TAC and pick the valuethat yields the smallest value for the TAC.



Note that

• each C(Q)/Q reduces to an expression of the form α1+(α2/Q) where α1 and α2

are constants

• There is thus a simple closed form expression for expressing the minimizingvalue of Q in Step 2

An Example

D=600/yr.; A=$8; i=20%

Unit price c* uses the following discount schedule

Lot Size (Q) C(Q) c* = C(Q)/Q

0<Q<500 $ 0.30Q 0.30

500≤Q<1,000 $150+0.29(Q-500) (0.29 + 5/Q)

Q≥1,000 $295+0.28(Q-1000) (0.28 + 15/Q)



Ordering with Announced Price ChangeSuppose we KNOW that unit price is going up at some timet* in the future. What should be done to take advantage oftoday’s lower price?

• Order too little ⇒ not taking sufficient advantage of thelower price

• Order too much ⇒ the extra holding cost might negatesavings from lower price/unit

On the last regularly scheduled order before c increases,we order a larger than normal quantity of Q′ units

This order will last for t′=Q′/D time units

After this order runs out we use the new EOQ found using(c+Dc)



So the holding cost incurred for the last unit used from this batch = ht′ = hQ′/D. Thesavings on this last item from the lower price paid = $Dc.

Question: Should we have bought this last unit or used a smaller Q′ ?

The marginal costs and benefits of the last unit (cost from holding it in inventoryand benefit from the lower unit price) will offset each other if

Dc = hQ′ /D

Thus we should use Q′ = DDc/h (unless of course, the current EOQ is alreadylarger than this…).


$c per unit $(c+Dc) per unit

t*

current EOQ New (smaller) EOQ

t′

Q′


Extension to Finite Production Rates

T1T2

Production(uptime)

No Production(downtime)

Time

Inventory

P-D -D

Q

IMAX


Suppose P = production rate (in units produced/unit time)• For a stable system, P must be greater than D.


Finite Production Rates

T=T1+T2 = cycle time =

n = no. of cycles per unit time

Uptime T1= (time taken to produce a batch of Q units at the rate of P units/unit

time)

Downtime T2=T-T1 = =

IMAX = (P-D)T1 = (P-D) = Q =Q∆ (where∆=1- )

� ��

� �

��


T1T2

P-D -D

Q

IMAX


Notice that this is identical to the TAC for the EOQ derivation exceptthat the holding cost is now h∆ (rather than h)!

Thus, if we define h′=h∆ = h(1-D/P), then the optimum batch size(also referred to as the Economic Production Quantity, or EPQ) isgiven by

∗

and


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