1

Inventory Theory – Part IIQuantity Discounts

Constrained Inventory ManagementSingle Facility EOQ

A quantity of commodity held for some time to satisfy some future demand.

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Check it out!

The Engineering Manager’s and Management Scientist’s

Handy-Dandy Cheat Sheet for

Deterministic Continuous Review Inventory Models

This assignment is easy with the cheat sheet

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Quantity Discounts

• All Units Discounts: the discount is applied to ALL of the units in the order.

• Incremental Discounts: the discount is applied only to the number of units above the breakpoint.

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All Units Discount

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A Really nice property of the all unit discount solution

• For all units discounts, the optimal will occur at the bottom of one of the cost curves or at a breakpoint. (It is generally at a breakpoint).

• Compare the cost at the largest (Q) realizable EOQ and all of the breakpoints succeeding it.

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An Example

.30 0 500

( ) .29 500 1,000

.28 1,000

Q for Q

C Q Q for Q

Q for Q

Assume D = 600 per year, K = $8, and I = .20

(0)

0

(1)

1

(2)

2

2 (2)(8)(600)400

(0.2)(0.30)

2 (2)(8)(600)406

(0.2)(0.29)

2 (2)(8)(600)414

(0.2)(0.28)

KDQ

Ic

KDQ

Ic

KDQ

Ic

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ALL Units Discount

Q

TC(Q)

500 1000

TC0(Q)

TC1(Q)

TC2(Q)

400

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The Example Continued

0

1 0

2 0

(600)(8)( ) 600 .2

2 2

(400) $204 .30

( ) (500) $198.10 .29

(1,000) $200.80 .28

j j j j

o

DK Q QTC Q Dc Ic c c

Q Q

TC for c

TC Q TC for c

TC for c

I choose Q = 500.

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Incremental Discount

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Not so nice properties of the incremental discounting solution

• Minimum cost point will never occur at a breakpoint

• If the EOQ for an interval is in the interval, it still may not be optimal

• Must compute the EOQ for each price break– if it falls within the interval, compute the

average cost– pick the best

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Incremental Discounting

( ) ( )( )

2

C Q kD C Q QG Q D I

Q Q Q

Average annual inventory cost

C(Q) is the cumulative cost to purchase Q units

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Incremental Discounting Example

( ) ( )( )

2

C Q kD C Q QG Q D I

Q Q Q

Minimize average annual cost

0

8 600.30 .30( ) 600 .20

2

8 600600 .30 (.20)(.30) / 2

Q Q QG Q

Q Q Q

QQ

(0)

0

2 (2)(8)(600)400

(0.2)(0.30)

KDQ

Ic

.30 for 0 500

( ) 150 .29( 500) 5 .29 for 500 100

295 .28( 1000) 15 .28 for 1000

Q Q

C Q Q Q Q

Q Q Q

C(Q) / Q is averageunit cost

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Incremental Discounting Example

( ) ( )( )

2

C Q kD C Q QG Q D I

Q Q Q

1

8 600( ) 600 .29 5 / .20 .29 5 /

2

QG Q Q Q

Q

(1)

1

2 (2)(13)(600)519

(0.2)(0.29)

KDQ

Ic

.30 for 0 500

( ) 150 .29( 500) 5 .29 for 500 100

295 .28( 1000) 15 .28 for 1000

Q Q

C Q Q Q Q

Q Q Q

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Incremental Discounting Example.30 for 0 500

( ) 150 .29( 500) 5 .29 for 500 100

295 .28( 1000) 15 .28 for 1000

Q Q

C Q Q Q Q

Q Q Q

( ) ( )( )

2

C Q kD C Q QG Q D I

Q Q Q

2

8 600( ) 600 .28 15 / .20 .28 15 /

2

QG Q Q Q

Q

(2)

2

2 (2)(23)(600)702

(0.2)(0.28)

KDQ

Ic Not realizable

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Incremental Discounting( ) ( )

( )2

C Q kD C Q QG Q D I

Q Q Q

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Incremental Discounting Example

(0)

0

(1)

1

(2)

2

2 (2)(8)(600)400

(0.2)(0.30)

2 (2)(13)(600)519

(0.2)(0.29)

2 (2)(23)(600)702

(0.2)(0.28)

KDQ

Ic

KDQ

Ic

KDQ

Ic

Not realizable

( ) ( )( )

2

C Q kD C Q QATC Q D I

Q Q Q

Minimize average annual cost

G0(400) = $204.00G1(519) = $204.58

.30 for 0 500

( ) 150 .29( 500) 5 .29 for 500 100

295 .28( 1000) 15 .28 for 1000

Q Q

C Q Q Q Q

Q Q Q

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Resource ConstrainedMulti-Item Inventories

Consider an inventory system of n items in which the total amount available to spend is C and items cost respectively c1, c2, . . ., cn. Then this imposes the following constraint on the system:

1 1 2 2 ... n nc Q c Q c Q C

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The Model

1

1

( )2

:

nj j j

jj j

n

j jj

K D QT Q Ic

Q

subject to c Q C

Obviously this can be solved using the

generalized Lagrangian approach!

An EMS graduate

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Form the Lagrangian

1 1

2

1

( , )2

( , )0 1,2,...,

2

( , )0

j

n nj j j

j j jj jj

j j jj

n

j jj

K D QL Q Ic c Q C

Q

K D IcL Qc for j n

Q Q

L Qc Q C

Q

is called the Lagrangian multiplier

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Solve the Lagrangian

2

2

2

2 2;

2 2

j

j

j j jj

j j j jj

j j j j

Ic K Dc

Q

K D K DQ Q

Ic c Ic c

The approach:1. Pick a value for

2. Compute Qj and

3. Repeat 1. and 2. until1

n

j jj

c Q C

1

n

j jj

c Q

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What about other constraints?

• Max warehouse space

• Holding costs

• Average Stock on-hand

1

n

j jj

w Q W

1 2

nj

jj

QIc H

1

Stock2

nj

j

Q

I bet you can’t work an example of this

so called generalized Lagrangian approach.

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Warehouse space constraint

1 1

2

1

( , )2

( , )0 1,2,...,

2

( , )0

j

n nj j j

j j jj jj

j j jj

n

j jj

K D QL Q Ic w Q W

Q

K D IcL Qw for j n

Q Q

L Qw Q W

Q

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Solving…

2

2

2

2 2;

2 2

j

j

j j jj

j j j jj

j j j j

Ic K Dw

Q

K D K DQ Q

Ic w Ic w

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Example of the so-called generalized Lagrangian Approach

Unit Cost

Order Cost

Demand rate

Holding cost

Storage space

$/unit ($/order) (Units per day)

($ per unit per day)

(Sq. ft. / unit)

1 2.5 10 20 0.5 4.152 6 23 34 1.2 5.63 10 15 54 2 2.84 11.7 18 67 2.34 3.95 8.9 36 28 1.78 4.76 13.35 45 17 2.67 2.857 15.5 32 47 3.1 5.48 4.75 26 38 0.95 4.19 3.15 9 60 0.63 3.25

Item

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Find EOQ for each item independentlyEOQ unconstrn

inv order cost

inv hold cost

Total inv cost

storage

28.28 7.07 7.07107 14.14 117.436.10 21.66 21.661 43.32 202.228.46 28.46 28.4605 56.92 79.732.11 37.56 37.5635 75.13 125.233.65 29.95 29.952 59.90 158.223.94 31.96 31.9574 63.91 68.231.15 48.28 48.2825 96.57 168.245.61 21.66 21.6633 43.33 187.041.40 13.04 13.0422 26.08 134.6

$239.65 $239.65 $479.31 1240.6

Q x sq. ft., / unit

1,000 sq. ft. limitation

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Optimal Constrained SolutionEOQ EOQ cost storage

18.51 15.43 76.827.28 45.03 152.825.71 57.22 72.028.51 75.66 111.228.19 60.85 132.522.11 64.12 63.027.53 97.30 148.735.03 44.84 143.630.60 27.29 99.5

$487.73 1000.0

EOQ unconstrn

28.2836.1028.4632.1133.6523.9431.1545.6141.40

$479.31let Excel show the way…

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Single Facility Production Model

1. n items to produce2. Dj = constant demand rate for product j3. Pj = constant production rate for product j4. hj = holding cost per unit per unit time for product j5. Kj = fixed setup cost for product j6. sj = setup time for product j (not sequence dependent)7. no stock outs permitted

Determine production lot sizes (Qj) that will minimize relevant costs.

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Single Facility Production Model

Let T = cycle time in which one lot of each productis produced.

Then Qj = DjT in order to meet demands

And G(Qj) = Kj Dj / Qj + (1 - Dj/Pj) hj Qj /2

Note that 1

1n

j

j j

D

P

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1 1

1

(1 / )( )

2

(1 / )

2

n nj j j j j j

jj j j

nj j j j j

j

K D D P h QG Q

Q

K D P h D T

T

Total relevant costs:

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1

1

(1 / ){}0

2

2

*(1 / )

nj j j j j

j

n

jj

n

j j j jj

K D P h Dd

dT T

K

Th D D P

P-D -D

Q/P

(P-D) Q/P = (1-D/P) Q

Qj = DjT

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T* must include production time + setup time

1 1 1 1

1 1

1min

1

1

1 /

n n n nj j j

j j jj j j jj j j

n nj

jj jj

n

jjn

j jj

Q D T DT S S S T

P P P

DT S

P

S

T TD P

SolutionSet T = Maximum{T*,Tmin}

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The Inevitable Example

Product 1 2 3 4 5 6 7 sum

Demand rate Dj (units/yr) 200 300 150 400 1000 250 2000 4300Prod Rate Pj (units/yr) 4000 4000 1500 2000 8000 2500 20000 42000Holding cost hj ($/item/yr) 2 3 4 2.5 3 1 5 20.5

Adj holding cost h'j ($/item/yr) 1.9 2.775 3.6 2 2.625 0.9 4.5 18.3

Setup time sj (yr) 0.004 0.008 0.002 0.008 0.004 0.002 0.008 0.036Set-up Cost (Kj) 20 40 10 40 20 10 40 180

Dj / Pj 0.05 0.075 0.1 0.2 0.125 0.1 0.1 0.75

(1-D/P) h1

1

2

*(1 / )

n

jj

n

j j j jj

K

Th D D P

to Excel and beyond…

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Single Facility Production ModelSome final thoughts

1

1n

j

j j

D

P

is the fraction of time the

facility is not producing1.

2. If products are made on different facilitiesThen use independent lot models.

3. If setup times or setup costs depend upon therotation order, then have a more difficultsequencing problem.

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EOQ Homework

I always work extra problems – not only the ones that are assigned!

Text Chapter 4: 4,8,10,12,13,14,17,20,21,2224,26,27,28,29,30,33,35,40 + Handout