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Inventory

Lecturer: Stanley B. Gershwi

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Storage

• Storage is fundamental in nature, manageengineering.

⋆ In nature, energy is stored. Life can only exist if

acquisition of energy can occur at a different time

the expenditure of energy.

⋆ In management, products are stored.

⋆ In engineering, energy is stored (springs, batteri

capacitors, inductors, etc.); information is stored

hard disks); water is stored (dams and reservoir

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Storage

• The purpose of storage is to allow systemssurvive even when important events are

unsynchronized. For example,⋆ the separation in time from the acquisition or pro

something and its consumption; and

⋆ the occurrence of an event at one location (such machine failure or a power surge) which can prev

performance or do damage at another.

•

Storage improves system performance bydecoupling parts of the system from one a

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Storage

• Storage decouples creation/acquisition an

emission.

• It allows production systems (of energy or

be built with capacity less than the peak de

•

It reduces the propagation of disturbancesreduces instability and the fragility of comp

expensive systems.

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Manufacturing InveStorage

Motives/bene

• Reduces the propagation of disturbances

machine failures).

•

Allows economies of scale:⋆ volume purchasing

⋆ set-ups

• Helps manage seasonality and limited cap

• Helps manage uncertainty:

⋆ Short term: random arrivals of customers⋆ Long term: Total demand for a product n

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Manufacturing InveStorage

Costs

•

Financial: raw materials paid for, but no re

comes in until item is sold.

•

Demand risk: item may not be sold due to example)

⋆ time value (newspaper)

⋆ obsolescence

⋆ fashion

•

Holding cost (warehouse space)•Damage/theft/spoilage/loss

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NewsvendorProblem

... formerly called the “Newsboy Problem”.

Motive: demand risk.

Newsguy buys x newspapers at c dollars per paper (Demand for newspapers, at price r > c is w. Unsold

newspapers are redeemed at salvage price s < c.

Demand W is a continuous random variable with disfunction F (w) = prob(W ≤ w); f (w) = dF (w)/d

all w.

Note that w ≥ 0 so F (w) = 0 and f (w) = 0 for w

Problem: Choose x to maximize expected

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NewsvendorProblem

rx if x ≤ wRevenue = R =

rw + s(x − w)if

x > w

(r − c)x if x ≤ wProfit = P = rw + s(x − w) − cx

= (r − s)w + (s − c)x

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NewsvendorProblem

700 RevenueProfit

-100

0

100

200

300

400

500

600

-2000 200 400 600 800 1000 1200 1400

stock

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NewsvendorProblem

Revenue=Revenue=rw+s(x−w)

x < w x > w

Profit=(r−Profit=rw+s(x−w)−cx

f(w)

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

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xNewsvendorProblem

Expected Profit = EP (x) =

[(r − s)w + (s − c)x]f (w)dw−∞ ∞+ (r − c)xf (w)dw

x

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NewsvendorProblem

or EP (x)

x x

= (r − s) wf (w)dw + (s − c)x f−∞ −∞ ∞

+(r − c)x f (w)dwx

x

= (r − s) wf (w)dw−∞( ) F ( ) ( ) (1 F (

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NewsvendorProblem

EP is 0 when x = 0.

When x → ∞, the first term goes to a finite(r − s)Ew. The last term goes to 0.

The remaining term,

x(s − c)F (x) → −∞.

Therefore when x → ∞, EP → −∞.

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NewsvendorProblem

EP(x)

x

?

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NewsvendorProblem

EP(x)

x

?

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NewsvendorProblem

dEP= (r − s)xf (x)+ (s − c)F (x)+(s

dx +(r − c)(1 − F (x)) − (r − c)xf (x)

= xf (x)(r − s + s − c − r + c)

+r − c + (s − c − r + c)F (x)

= r − c + (s − r)F (x)

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NewsvendorProblem

Note that

• d2EP/dx2 = (s − r)f (x) ≤ 0. Therefor

concave and has a maximum.

• dEP/dx > 0 when x = 0. Therefore EP

maximum which is greater than 0 for some∗x = x > 0.dEP

• x∗ satisfies (x∗) = 0.dx r − c

Therefore F (x∗)

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NewsvendorProblem

EP(x)

x

x*

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NewsvendorProblem

x*

f(w)

r−c

r−s

F(x*)=

0

0.05

0.1

0.15

0.2

µ−4σ µ−2σ µ µ+2σ µ+4σ

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NewsvendorProblem

This can also be writtenr − c

F (x∗

) = (r − c) + (c − s)

r − c > 0 is the marginal profit when x <

c − s > 0 is the marginal loss when x > w

Choose x∗ so that the fraction of time you d

too much is marginal profit

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NewsvendorExample 1

r = 1, cProblem = .25, s = 0, µw = 10

x* x* vs. r

86

90

94

98

102

106

110

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NewsvendorExample 2

r =Problem 1, c = .75, s = 0, µw = 10

x* x* vs. r

105

101

97

93

89

85

81

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NewsvendorExample 1

r = 1, cProblem = .25, s = 0, µw = 10

x* x* vs. c

130

80

90

100

110

120

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NewsvendorExample 1

r = 1, cProblem = .25, s = 0, µw = 10

x* x* vs. s

130

110

114

118

122

126

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NewsvendorExample 2

r =Problem 1, c = .75, s = 0, µw = 10

x* x* vs. s

121

117

113

109

105

101

97

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NewsvendorExample 1

r = 1, cProblem = .25, s = 0, µw = 10

x* x* vs. Mean

240

40

80

120

160

200

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NewsvendorExample 2

r = 1, cProblem = .75, s = 0, µw = 10

x* x* vs. Mean

30

70

110

150

190

230

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NewsvendorExample 1

r = 1, cProblem = .25, s = 0, µw = 10

x* x* vs. Std

114

112

110

108

106

104

102

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NewsvendorExample 2

r = 1, cProblem = .75, s = 0, µw = 10

x* x* vs. Std

88

90

92

94

96

98

100

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�

NewsvendorWhy is x∗ linear in µw

Problem

r − cF (x∗) =

r − s

Assume demand w is N (µw, σw). Then

w − µwF (w) = Φ

σw

where Φ is the standard normal cumulative d

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�


Problem

Therefore

� x∗ − µw r − c

Φ =σw r − s

or

� x∗ − µw r − c= Φ−1σw r − s

or

r − cx∗ = µw + σwΦ

−1

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Problem

Note that

Φ−1(k) > 0 if k > .5

and

Φ−1(k) < 0 if k < .5

Therefore r − c• x∗ increases with σw if > .5, and

r − sr − c

•

x∗ decreases with σw if < .5.r − s

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EOQ

• Economic Order Quantity

• Motivation: economy of scale in ordering.

• Tradeoff:

⋆ Each time an order is placed, the compa

fixed cost in addition to the cost of the go⋆ It costs money to store inventory.

A i

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Assumptions

EOQ

•

No randomness.

• No delay between ordering and arrival of g

•

No backlogs.• Goods are required at an annual rate of λ

year. Inventory is therefore depleted at the

units per year.• If the company orders Q units, it must pay

the order. s is the ordering cost , c is the u

•

It costs h to store one unit for one year. h

P bl

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Problem

EOQ

•

Find a strategy for ordering materials that

minimize the total cost.

• There are two costs to consider: the order

and the holding cost.

S i

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ScenarioEOQ

• At time 0, inventory level is 0.

•

Q units are ordered and the inventory leve

instantaneously to Q.

• Material is depleted at rate λ.

•

Since the problem is totally deterministic, wwait until the inventory goes to zero before

next. There is no danger that the inventory

zero earlier than we expect it to.

S i

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ScenarioEOQ

•

Because of the very simple assumptions, w

assume that the optimal strategy does not

over time.

• Therefore the policy is to order Q each tim

inventory goes to zero. We must determineoptimal Q.

S i

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ScenarioEOQ

I n v e n t o r y

Q

jumpdecrease at rate λ

T 2T

T = Q/λ (years)

F l ti

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Formulation

EOQ

• The number of orders in a year is 1/T = λ

Therefore, the ordering cost in a year is sλ

• The average inventory is Q/2. Therefore t

average inventory holding cost is hQ/2.

•

Therefore, we must minimizehQ sλ

C = +2 Q

over Q.

F l ti

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Formulation

EOQ

ThendC h sλ

= − = 0dQ 2 Q2

or

2sλ

Q∗ =h

E amples

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Examples

EOQ

In the following graphs, the base case is

•

λ = 3000• s = .001

• h = 6

Note that

Q∗ = 1

Examples

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Examples

EOQ

0

10

20

30

40

50

0 2 4 6 8

o s t

s=.01s=.001

s=.0001

Examples

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Examples

EOQ

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 2000 4000 6000 8000 1

Q *

Examples

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Examples

EOQ

0

0.5

1

1.5

2

2.5

3

3.5

0 0 002 0 004 0 006 0 008

Q

*

Examples

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Examples

EOQ

0

1

2

3

4

5

6

7

8

0 2 4 6 8

Q *

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More Issues

• Random delivery times and amounts.

•

Order lead time.

• Vanishing inventory.

• Combinations of these issues and random

ordering/setup costs.

Make to Stock Qu

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Base Stock Make-to-Stock Qu

Policy

Usual assumptions:

• Demand is random.

•

Inventory is held (unlike newsvendor problem).

• No ordering cost, batching, etc.

Policy• Try to keep inventory at a fixed level S .

Issue

• Stockout: a demand occurs when there is no stock

Make to Stock Qu

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Policy

Issues

• What fraction of the time will there be stoc

how much demand occurs during stockout

• How much inventory will there be, on the a

•

How much backlog will there be, on the av

Make to Stock Qu

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Policy

Production Dispatch Material

Raw SupplierFrom

Factory

Floor

FG

Bac

Material Flow Information Flow

• When an order arrives (generated by “Demand”),

⋆ an item is requested from finished goods invento⋆ an item is delivered to the customer from the finis

inventory (FG) or, if FG is empty, backlog is incre

and⋆ a signal is sent to “Dispatch” to move one item fr

Make to Stock Qu

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Policy


Raw SupplierFrom

Factory

Floor

FG

Bac


• There is some mechanism for ordering raw materia

suppliers.

• We do not consider it here.

• We assume that the raw material buffer is never em

Make to Stock Qu

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Policy


Raw SupplierFrom

Factory

Floor

FG

Bac


• Whenever the factory floor buffer is not empty and

production line is available, the production line take

from the factory floor buffer and starts to work on it

• When the production line completes work on a par

finished part in the finished goods buffer or, if there

it sends the part to the customer and backlog is red

S Make-to-Stock Qu

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PolicyQ


Raw SupplierFrom

Factory

Floor

F

Ba


• Q(t) = factory floor inventory at time t.

•

I (t) =

⋆ the number of items in FG, if this is nonnegative and thereor

⋆ – (the amount of backlog), if backlog is non-negative and F

• There are no lost sales. I (t) is not bounded from below. It cnegative value

B S k Make-to-Stock Qu

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PolicyQ


Raw SupplierFrom

Factory

Floor B


• Assume Q(0) = 0 and I (0) = S . Then Q(0) + I (0) = S .

• At every time t when a demand arrives,

⋆ I (t) decreases by 1 and Q(t) increases by 1 so Q(t) + Ichange.

• At every time t when a part is produced,

⋆ I (t) increases by 1 and Q(t) decreases by 1 so Q(t) + Ichange.


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PolicyQ


Raw SupplierFrom

Factory

Floor B


• Also, Q(t) ≥ 0 and I (t) ≤ S .

• Assume the factory floor buffer is infinite.

• Assume demands arrive according to a Poisson pr

rate parameter λ.

• Assume the production process time is exponentia

distributed with rate parameter µ.


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Policy

• Assume λ < µ.

Therefore, in steady state,

prob(Q = q) = (1 − ρ)ρq, q ≥ 0

where ρ = λ/µ < 1.

• What fraction of the time will there be stoc

That is, what is prob(I ≤ 0)?

B St k Make-to-Stock Qu

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Policy

∞

prob(I ≤ 0) = prob(Q ≥ S ) = (1 − ρ)ρq

q=S

∞ ∞ = (1 − ρ) ρq = (1 − ρ)ρS ρq = (1 − ρ)ρS

1q=S q=0

soprob(I ≤ 0) = ρS

• How much demand occurs during stockout periods?

ρS λ


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Policy

•

How much inventory will there be, on the average?

on what you mean by “inventory”.)

ρ•

EQ =1 − ρ

• EI = S − EQ is not the expected finished goods

is the expected inventory/backlog.

• We want to know

I if I > 0E (I +) = E = E (I |I > 0)pro

h i


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PolicyS

E (I |I > 0) = i prob(I = i|I > 0)i=1

S

i prob(I = iS

i prob(I = i ∩ I > 0) i=1= = S prob(I > 0) i=1

prob(I = j j=1

S −1

(S − q) prob(Q = q)q=0

=S −1

prob(Q = s)

s=0


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Base Stock Make to Stock Qu

Policy

S −1 S −1

S prob(Q = q) − q prob(Q =q=0 q=0

=S −1

prob(Q = q′)q ′=0

S −1 S −1

q prob(Q = q) (1 − ρ) qρq

q=0 q=0= S − = S − =

S −

1 S −

1prob(Q = q′) (1 − ρ) ρq


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Policy

ThereforeS −1

qρq

q=0

E (I |I > 0) = S − S −1

ρq

q=0

and S −1 S −1

E (I +) = S (1 − ρ) ρq − (1 − ρ) qq=0 q=0

S −1

= S prob(Q < S) − (1 − ρ) qρq


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Policy

• How much backlog will there be, on the av

•

This time, we are looking for −E (I −), whe

I if I ≤ 0

E (I −

) = E = E (I |I ≤ 0)0 otherwise−∞

E (I |I ≤ 0) = i prob(I = i|I ≤i 0


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Policy−∞

i prob(I−∞ prob(I = i ∩ I < 0) i=0= i =

−∞prob(I < 0) i=0prob(I

j=0∞

(S − q) prob(Q = q)

q=S =∞

prob(Q = s)s=S

where, because Q = S − I , we substitute q = S −


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Policy

∞ ∞

S prob(Q = q) − q prob(Q =q=S q=S

=∞

prob(Q = q′)′q =S

∞ S −1

q prob(Q = q) EQ − q prq=S q=0

= S − = S −∞ S −

1prob(Q = q′) 1 −

prob


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Policy

S −1

EQ − (1 − ρ) qρq

q=0= S −S −1

1 − (1 − ρ) ρq

q=0so

E (I −) = E (I |I ≤ 0)prob(I ≤ 0

( +) ( )


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Policyprob I pos

0.2

0.4

0.6

0.8

1

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

λ


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10


PolicyEIposEIneg

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

-10

-20

-30

-40

λ


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Policy

0.2

0.4

0.6

0.8

1

prob I pos, lambda=.6 prob I pos, lambda=.9

0

0 5 10 15 20 25 30 35

S


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Policy

-10

-5

0

5

10

15

20

25

30

35

40

0 5 10 15 20 25 30 35

EIpos, lambda=.6EIneg, lambda=.6EIpos, lambda=.9EIneg, lambda=.9

S


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Base Stock a e to Stoc Qu

Policy Optimization of

We have analyzed a policy but we haven’t completely

How do we select S ?

One possible way: choose S to minimize a function texpected costs due to finished goods inventory (E (I

backlog (E (I −)). That is, minimize

EC (I ) = E (hI + − bI −) = hE (I +) − bE (

where h is the cost of holding one unit of inventory founit and b is the cost of having one unit of backlog fo


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Base Stock Q


cost

−b

h


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Base Stock Q


5

10

15

20

25

30

35

40

Cost, lambda=.6Cost, lambda=.9

0

0 5 10 15 20 25 30 35

S


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Base StockPolicy Optimization of

Problem: Find S ∗ to minimize hE (I +) − bE

Solution

1

: S

∗

= ⌊S ˜⌋ or ⌊S

˜⌋ + 1, where

S ̃+1 hρ =

h + b

Note: ρS ̃+1 = prob(Q ≥ S ̃+ 1) = prob(I


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That is, we choose S so thath

prob(I < 0) =h + b

Look familiar? This is the solution to the newsvendor problem ifI = x − w,h = c − s, and b = r − c because

r −F (x∗) = prob(w ≤ x∗) = prob(I ≥ 0) =

(r − c) +so

r − c c − sprob(I < 0) = 1 − =

(r − c) + (c − s) (r − c) + (

c − s is the cost of one more unit of inventory when x > w, ie I


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profit

r−c

s−c

w

Newsvendor profit function


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Questions:

•

How does the problem change if we considfloor inventory? How does the solution cha

• How does the problem change if we consid

finished goods inventory space ? How doesolution change?

•

How else could we have selected S ?

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Q, R Policy

• Fixed ordering cost, like in EOQ problem

• Random demand, like in newsvendor or ba

problems.

• Policy: when the inventory level goes down

a quantity Q.• Hard to get optimal R and Q; use EOQ an

stock ideas.

Inventory

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InventoryPosition

Inventory position

Q

Lead time

Inventory

Q/ λ

Material ordered Material arrives

The current order must satisfy demand until the next

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