Example - INFORMS 

Recall: D  Normal(5000,2000) 



We have lookup tables for Standard Normal distribution ( =0, =1)  Convert to Standard Normal!



F (Q)  P ( D  Q) 

4  0.444 9

D Q  P ( D  Q )  P ( D    Q   )  P    P( Z  z )  ( z )     Safety Standard normal 

Z

zstock

Expected demand

Find z from the lookup table  Q =  z + 

Standard Normal Distribution - 1 Density function  ( z ) 

e



z2 2

2 Symmetric around the mean Area under the entire curve  1  P(Z  )

1

Standard Normal Distribution - 2 Density function  ( z ) 

e



z2 2

2 Symmetric around the mean Area under the entire

We will use ɸ(z) and F(z) interchangeably

curve  1  P(Z  )

F(z)=P(Z≤z)

z

Standard Normal Distribution - 2 Density function  ( z ) 

e



z2 2

2 Symmetric around the mean Area under the entire

We will use ɸ(z) and F(z) interchangeably

curve  1  P(Z  )

F(z)=P(Z≤z)

F(-z)=P(Z≤-z)=1-F(z)

-z

z

2

Example - INFORMS P( Z  z )   ( z )  Area under the curve  0.444  z  -0.14 Q*   z    2000(0.14)  5000 Q*  4720

z = - 0.14

Example - INFORMS Why is Q* co=50 z=0.21

Q* Q*

Standard deviation

Standard deviation

Example – Fashion Bags Input    

Unit cost c = $28.50 Selling price p = $150 Salvage value s = $20 Cost of inventory = $0.40 for each dollar tied up in inventory at the end of the season

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Example – Fashion Bags Input    

Unit cost c = $28.50 Selling price p = $150 Salvage value s = $20 Cost of inventory = $0.40 for each dollar tied up in inventory at the end of the season

Computed input:   

Holding cost per bag cu = co =

F (Q) 

h=

cu  cu  co

Example – Fashion Bags Input    

Unit cost c = $28.50 Selling price p = $150 Salvage value s = $20 Cost of inventory = $0.40 for each dollar tied up in inventory at the end of the season

Computed input:   

Holding cost per bag cu = p – c = $121.5 co = c – s + h = $19.9

F (Q) 

h = (0.40)(28.50) = $11.4

cu 121.5   0.86 cu  co 121.5  19.9

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Example – Fashion Bags – Normally distributed demand Demand ~ Normal(   150,   20) F (Q)  0.86  z  1.08

Area under the curve= Critical ratio = 0.86

 Q*  z    (20)(1.08)  150 Q *  172

z = 1.08

Example – Fashion Bags – Uniformly distributed demand 

Demand Uniform between 50 and 250  =150 

Same expected demand as in Normal distribution

0.86  (Q * 50) Q*  222

1  200

Uniform density

Area under the curve = Critical ratio = 0.86

1/200

Q 250

50 Q*= 222

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Example – Fashion Bags 

Even though both the Normal and the Uniform distributions have the same mean (=150), why did we get different quantities?  

Normal distribution  Q*=172 Uniform distribution  Q*=222

Example – Fashion Bags 

Even though both the Normal and the Uniform distributions have the same mean (=150), why did we get different quantities?  

Normal distribution  Q*=172 Uniform distribution  Q*=222

Because of the variance (equivalently, standard deviation) and the shape of the distribution !!!  

Normal  =20 Uniform  =57.7

Normal

Uniform

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Class exercise Georgia Tech bookstore must decide how many 2005 calendars to order. Each calendar costs the bookstore $2 and is sold for $4.50. After January 1, any unsold calendars are returned to the publisher for a refund of 75 cents per calendar. The number of calendars sold by January 1 follows the probability distribution below. How many calendars should be ordered to maximize expected profit? # of calendars sold Probability

100

150

200

250

300

0.30

0.20

0.30

0.15

0.05

Class exercise Q: number of calendars ordered D: demand D≤Q  Cost = 2Q – 4.5D - 0.75(Q-D) = 1.25Q - 3.75D D≥Q  Cost = 2Q – 4.5Q = -2.5Q cu = 4.5 – 2 = $2.5 co = 2 – 0.75 = $1.25 Critical ratio = (2.5)/(2.5+1.25) = 2/3  Q=? # of calendars sold

100

150

200

250

300

Probability

0.30

0.20

0.30

0.15

0.05

F(Q)=P(D≤Q)

0.30

0.50

0.80

0.95

1.00

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Summary - Newsvendor  



Single period Depending on the relationship between the cost of shortage or excess inventory, we may order more or less than expected demand Optimal order quantity  

Higher variability may cause an increase or a decrease in the optimal order quantity 

As  increases, Q* will deviate more from the mean

Inventory Control - Demand Variability Constant/Stationary Variable/Non-Stationary Uncertainty Stochastic Deterministic



increases as shortage cost increases decreases as holding cost increases

Economic Order Quantity (EOQ) – Tradeoff between fixed cost and holding cost

Lot size/Reorder level (Q,R) or (s,S) models – Tradeoff between fixed cost, holding cost, and shortage cost

Aggregate Planning – Planning for capacity levels given a forecast

Materials Requirements Planning (MRP)

Very difficult problem!

Newsvendor – single period

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