Stochastic Models
Inventory
Control
Inventory Forms
 Form
 Raw
Materials
 Work-in-Process
 Finished Goods
Inventory Function
 Safety
Stock
 inventory
held to offset the risk of unplanned
demand or production stoppages
 Decoupling
Inventory
 buffer
inventory required between adjacent
processes with differing production rates
 Synchronized Production
 In-transit
(pipeline)
 materials
moving forward through the value chain
 order but not yet produced/received
Inventory Function

Decoupling Inventory
buffer
inventory required between adjacent processes
with differing production rates
Synchronized Production

In-transit (pipeline)
materials
moving forward through the value chain
order but not yet produced/received
 Cycle
Inventory
 orders
in lot size not equal to demand
requirements to lower per unit purchase costs
Inventory Function

In-transit (pipeline)
materials moving forward through the value chain
 order but not yet produced/received


Cycle Inventory

orders in lot size not equal to demand requirements
to lower per unit purchase costs
 Seasonal
Inventory
 produce
in low demand periods to meet the needs
in high demand periods
 anticipatory - produce ahead of planned
downtime
Inventory Costs
 Item
Cost (C)
 Order Cost (S)
 Process
 Holding
Setup Costs
Costs (H)
 Function
of time in inventory, average inventory
level, material handling, utilities, overhead, ...
 Often calculated as a % rate of inventory cost (iC)
 Stockout
 reflects
or Shortage Costs (s)
costs associated with lost opportunity
Economic Order Quantity
Q
D
r
1
T
L
Q = reorder quantity
D = demand rate
T = inventory cycle
time
order
arrives
r = reorder point
L = leadtime
Economic Order Quantity
Q
Avg.
Inventory
r
T
L
Q = reorder quantity
D = demand rate
T = inventory cycle
time
r = reorder point
L = leadtime
Inventory Cost
TRC = Total Relevant Cost
= Order Cost + Holding Costs
 S  1 HQT = cost per cycle
2
TCU = Total Relevant Cost per Unit Time
= TRC/T
 S T  1 2 HQ
Cost per unit Time
But,
T = length of a cycle
 QD
SD
TCU 

Q
1
2
HQ
Cost per Unit Time
Total Cost per Unit Time
30
25
20
Cost
Order
15
Hold
TCU
10
5
0
500
1,000
1,500
Order Q
2,000
Cost per Unit Time
Total Cost per Unit Time
30
25
20
Cost
Order
15
Hold
TCU
10
5
0
500
1,000
1,500
2,000
Order Q
Note: that minimal per unit cost occurs when
holding cost = order cost
(per unit time)
Economic Order Quantity
Find min TCU
TCU
 SD
( Q
0
Q
Q
2
0  -SDQ 
Q*
2SD

H
2SD
iC
1
2
1
HQ
)
2
H
Example
The monthly demand for a product is 50 units.
The cost of each unit is $500 and the holding
cost per month is estimated at 10% of cost. It
costs $50 for each order made. Compute the
EOQ.
Sol:
Q* 
2*50*50
= 10
.1*500
Optimal Inventory Cost
SD
Recall TCU 

Q
TCU*

SD
Q*

1
2
1
HQ
*
HQ
2
SD
2SD
1
*
 2H
TCU 
2SD
H
H
Optimal Inventory Cost
TCU *  2HSD  2iCSD
Example:
TCU *  2*.1*500*50*50
= $500 per month
Orders per year
N = number of orders per year
D
N *
Q
HD
2S
Example: D = 50 / month, Q* = 10
D = 50 x 12 = 600 / yr.
H = .1x12x500 = $600 / unit-yr
N
600*600
2*50
= 60
Cycle Time
T = cycle time
*
Q
T

D
2S
HD
Example: D = 50 units/month, Q* = 10
10
T

50
2*50 = .2 months = 6 days
50*50
Reorder Point
L = lead time
r = reorder point
= inventory depleted in time L
= L*D
Example: Lead time for company is 2 days. Demand
is 50 units per month or 1.67 units/day.
r = 2*1.67 = 3.33
Reorder at 4 units
Lead Time
Example 2: Suppose our lead time is closer to 8
days.
r = 8*1.67 = 13.33
but, recall we only order 10 units at a time
r = 13.33 - 10 = 3.33
Example (cont.)
Reorder at 4 units 1 cycle ahead.
T
10
4
time
L
reorder
order
arrives
Sensitivity
Recall that
Q*
2SD
H
Now suppose we deviate by p amount so that
Q = Q*(1+p). What affect does this have on
total cost? Let
PCP = Percentage Cost Penalty
TCU ( Q)  TCU ( Q * )
PCP 
x100
*
TCU ( Q )
Senstivity (cont.)
SD
Recall TCU 

Q
1
2
HQ
SD
TCU ( Q)  *
 1 2 HQ * (1  p)
Q (1  p)
Miracle 1 Occurs
TCU ( Q) 
1
2
1 

2SDH  (1  p) 

1
p



Sensitivity (cont.)
TCU *  2HSD
Recall
1
PCP =
2
1 

2SDH  (1  p) 

1
p



2HSD
1 

= 50  (1  p)  1 p 
 

100
2HSD
x 100
Sensitivity (cont.)
1 

PCP = 50  (1  p)  1 p 
 

Miracle 2 Occurs
 p2 
PCP  50

1
p
  
100
Example
Recall that Q* = 10. Suppose now that a minimum
order of 15 is introduced. Compute the
percentage cost penalty (increase).
15  10
p
 .5
10
50(.52 )
PCP 
 8.3
1 .5
Total relevant costs increase 8.3%
Example 2
Suppose demand forecast increases by 25% so
that D = 50(1.25) = 62.5. Then
TCU *  2 * 62.5 * 50 * 50  559
or TCU* increases by 11.8%
Shortages
Im
Q
r
time
Q-R
L
T
T1
T2
T1 = time inventory carried
H = holding cost
T2 = time of stockout
S = order cost
Im = max inventory level
p = cost per unit short per unit time
Inventory Costs
TCR = order + holding + shortages
S
1
HI
T

m
2
1
1
2
p( Q  Im ) T2
Miracle 3 Occurs
Q 
*
2DS p  H
H
p
R 
*
2DS
p
H p +H
Example
Suppose we allow backorders for our previous
example. We estimate that the cost of a backorder
is $1 per unit per day ($30 / month). Then
Q 
*
2*50*50
50
30 50
30
= 16.3 = 17 units
Production Model (ELS)
Im
P-D
1
T1
Q
D
P
P-D
S
H
Im
T2
T
= batch size order quantity
= demand rate
= production rate
= replenish rate during T1
= setup costs
= holding cost /unit-time
= max inventory level
time
Production Model (ELS)
Im
P-D
1
T1
T
T1
T2
Im
T2
T
time
= cycle length = T1 + T2 = Q/D
= length of production run = Q/P
= depletion time = Im/D
= max inventory level
= (P-D)T1 = (P-D)Q/P
Costs
TC = total costs per cycle = order + holding
1 H ( P  D) QT
S
2
P
TCU = Cost per unit time  TC / T
S 1 H ( P  D) Q
 
T 2
P
SD H ( P  D) Q


Q
2P
Optimal Q* (ELS)
SD H ( P  D)
TCU
0 2 
2P
Q
Q
Solving for Q,
Q 
*
2SDP
EOQ

H ( P  D)
1 DP
Max Inventory
Im = max inventory
= (P-D)T1
= (P-D)Q*/P
*
D
D

I m  Q  1    ELS 1  


P
P
Summary
Im
P-D
1
T1
Q* 
T2
T
2SDP
EOQ

H ( P  D)
1 DP
D

I m  Q  1    EOQ 1  D
P

P
*
time
Probabilistic Models
Im
R=B+LD
B
Q*
time
L
D
B
R
DL
= demand rate
= buffer stock
= reorder point = B+LD
= actual demand from time of order to
time of arrival
Probabilistic Model
Let = max risk level for out of stock condition
Idea:
we want to set a buffer level B so that the
probability of running out of stock is < .
 > P{out of stock}
= P{demand in DL > R}
= P{DL > B+LD}
Example
Prob: Suppose S=$100, H=$.02/day, L=2 days.
D = daily demand N(100, 10).
From EOQ model, Q* = 1,000 units
Find: Buffer level, B, such that probability of
out of stock < .05.
Solution
DL = demand for 2 days = D1 + D2
Question: D1 & D2 are identically independently
distributed normal variates with mean 
and standard deviation =10.
What can we say about the distribution of DL?
Prob. Review
Suppose we have a random variable XL given by
X = Y1 + Y2
Then
E[X] = E[Y1] + E[Y2]
2
x

2
1
   2 cov( y1 , y2 )
2
2
If Y1 & Y2 are independent, then
x2   12   22
Solution (cont.)
Recall
DL = demand for 2 days = D1 + D2
~ N(L,L)
Then
L = E[DL] = E[D1] + E[D2] = 200
 
2
L
2
D1

2
D2
2
2
10
10


 200
 L  200  14.14
Solution (cont.)
DL ~ N(200, 14.14)
 < P{DL > B + LD}
= P{DL > B + L}
B
B
 Dl   l

 P

  P Z 

L
 L
 L

Solution (cont.)
Recall for our problem that =.05 and L=14.14.
Then,
B 

.05  P Z 

14.14 

=.05
0
Z=1.645
B
 1.645
14.14
B = 23.3
Summary
For D ~ N(100,10), L = 2 days, Q* = 1,000 units,
and = risk level = .05
DL ~ N(L=200, L=14.14)
B = ZL = 23.3
R = B + DL = B + L = 23.3 + 200 = 224
Optional Replacement
Im
R=B+LD
B
time
L
In the continuous review model, an order of Q* is made
whenever inventory level reaches the reorder point R.
We can also utilize periodic review systems with
variable order quantities. The two most common are
Optional Replacement (s,S)
P system
Optional Replacement
S
s
time
1
2
3
4
5
At t=1, inventory level is above minimum stock level s,
no order is made.
At t =2, inventory level is below s, order up to S
s = R = B+DL
S = Q*
P System
T
time
1
2
3
4
5
Order up to Target level T at each review interval P.
Let
DP+L = demand in review period + lead time
P+L = standard deviation of demand in period P+L
 = level of risk associated with a stockout
T = DP+L + ZP+L
Newsboy Problem
Often inventory for a single product is met only
once; e.g. News Stand (can’t sell day old papers)
Pet Rocks
Christmas Trees
If Q > D,
incur costs for Q but revenue only for D
If Q < D,
incur opportunity costs in form of lost sales
Newsboy (cont.)
Objective: Determine best order quantity which
maximizes expected profit
Payoff Matrix:
if Qi  D j
 PQi
Rij  
 PDi  L( Qi  D j ) if Qi  D j
Rij = payoff for order quantity Qi and demand level Dj
P = profit per unit sold
L = loss per unit not sold
Newsboy (cont.)
Expected Payoff:
m
EP ( Qi )   PD j Rij
j 1
where
EP(Qi) = expected payoff for order quantity Qi
PD j = probability of demand level j
Rij = payoff for order quantity Qi and demand level Dj
Example; Newsboy
Boy Scout troop 53 plans to sell Christmas trees to earn
money. Each tree costs the troop $10 and can be sold
for $25. They place no value on lost sales due to lack
of trees, L=0. Demand schedule is shown below.
Demand
100
120
140
160
180
200
P{demand}
0.10
0.15
0.25
0.25
0.15
0.10
Example (cont.)
Payoff Matrix
P = Profit = $25 - $10 per tree sold
L = Loss = $-10 per tree not sold
Order Q
100
120
140
160
180
200
100
1,500
1,300
1,100
900
700
500
120
1,500
1,800
1,600
1,400
1,200
1,000
Demand Level
140
1,500
1,800
2,100
1,900
1,700
1,500
160
1,500
1,800
2,100
2,400
2,200
2,000
180
1,500
1,800
2,100
2,400
2,700
2,500
200
1,500
1,800
2,100
2,400
2,700
3,000
Example (cont.)
Expected Payoff:
Order Q
100
120
140
160
180
200
0.1
100
150
130
110
90
70
50
0.15
120
225
270
240
210
180
150
Demand Level
0.25
140
375
450
525
475
425
375
0.25
160
375
450
525
600
550
500
0.15
180
225
270
315
360
405
375
0.1
200
150
180
210
240
270
300
Expected
Payoff
1,500
1,750
1,925
1,975
1,900
1,750
Order quantity has largest expected payoff of $1,975
order 160 trees