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February 12, 2018 | Author: Anonymous | Category: Math, Statistics And Probability, Normal Distribution
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Matching Supply with Demand: An Introduction to Operations Management Gérard Cachon ChristianTerwiesch All slides in this file are copyrighted by Gerard Cachon and Christian Terwiesch. Any instructor that adopts Matching Supply with Demand: An Introduction to Operations Management as a required text for their course is free to use and modify these slides as desired. All others must obtain explicit written permission from the authors to use these slides.

Slide ‹#›

The Newsvendor Model

Slide ‹#›

O’Neill’s Hammer 3/2 wetsuit

Slide ‹#›

Hammer 3/2 timeline Generate forecast of demand and submit an order to TEC

Spring selling season

Nov Dec Jan

Feb Mar Apr May Jun

Receive order from TEC at the end of the month



Jul Aug

Left over units are discounted

Marketing’s forecast for sales is 3200 units.

Slide ‹#›

Newsvendor model implementation steps 

Generate a demand model:  Determine a distribution function that accurately reflects the possible demand outcomes, such as a normal distribution function.



Gather economic inputs:  Selling price, production/procurement cost, salvage value of inventory



Choose an objective:  e.g. maximize expected profit or satisfy an in-stock probability.



Choose a quantity to order.

Slide ‹#›

Demand models

Slide ‹#›

What is a demand model?

GAMMA

POISSON

NORMAL Probability



A demand model specifies what demand outcomes are possible and the probability of these outcomes. Traditional distributions from statistics can be used as demand models:  e.g., the normal, gamma, Poisson distributions

Probability



0

50

100 Demand

150

200

0 1 2 3 4 5 6 7 8 9 10

Demand Slide ‹#›

Distribution and density functions? 

The density function tells you the probability of a particular outcome occurring.  e.g. the probability demand will be exactly 5 units.



The distribution function tells you the probability the outcome will be a particular value or smaller.  e.g. the probability demand will be 5 or fewer units.



Both can be used to describe a demand model but we will most often work with distribution functions.



Density functions can take all kinds of shapes (they don’t always have to look like a bell curve)



Distribution functions always start low and increase towards 1.0

Slide ‹#›

Distribution and density functions 

Two ways to describe the same demand model:

Distribution function

Density function

1.00

0.60

Probability

Probability

0.80

0.40 0.20 0.00 0

50

100 150 Demand

200 Slide ‹#›

0

50

100 Demand

150

200

Normal distribution – details 

 

All normal distributions are characterized by two parameters, mean = m and standard deviation = s All normal distributions are related to the standard normal that has mean = 0 and standard deviation = 1. For example:  Let Q be some quantity, and (m, s) the parameters of the normal distribution used to model demand.  Prob{demand is Q or lower} = Prob{the outcome of a standard normal is z or lower}, where

z 

Qm

s

or Q  m  z  s

Look up z in the Standard Normal Distribution Function Table to get the desired probability or use Excel: Prob{the outcome of a standard normal is z or lower} = normsdist(z)

Slide ‹#›

The Standard Normal Distribution Function Table Quantities on the outside Standard Normal Distribution Function Table (continued), F (z ) z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.00 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.8413

0.01 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438

0.02 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.8461

0.03 0.5120 0.5517 0.5910 0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238 0.8485

0.04 0.5160 0.5557 0.5948 0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264 0.8508

0.05 0.5199 0.5596 0.5987 0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.8531

0.06 0.5239 0.5636 0.6026 0.6406 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.8554

0.07 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577

Probabilities in the inside Slide ‹#›

0.08 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599

0.09 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621

The Standard Normal Distribution Function Table  



How to start with a quantity and find the corresponding probability? Q: What is the probability the outcome of a standard normal will be z = 0.28 or smaller?  Look for the intersection of the fourth row (with the header 0.2) and the ninth column (with the header 0.08) because 0.2+0.08 = 0.28, which is the z we are looking for. A: The answer is 0.6103, which can also be written as 61.03%

Standard Normal Distribution Function Table (continued), F (z ) z 0.0 0.1 0.2 0.3 0.4 0.5

0.00 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915

0.01 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950

0.02 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985

0.03 0.5120 0.5517 0.5910 0.6293 0.6664 0.7019

Slide ‹#›

0.04 0.5160 0.5557 0.5948 0.6331 0.6700 0.7054

0.05 0.5199 0.5596 0.5987 0.6368 0.6736 0.7088

0.06 0.5239 0.5636 0.6026 0.6406 0.6772 0.7123

0.07 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157

0.08 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190

0.09 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224

The Standard Normal Distribution Function Table  



How to start with a probability and find the corresponding quantity? Q: For what z is there a 70.19% chance that the outcome of a standard normal will be that z or smaller?  Find the probability inside the table.  Add the row and column headers for that probability A: the answer is z = 0.53

Standard Normal Distribution Function Table (continued), F (z ) z 0.0 0.1 0.2 0.3 0.4 0.5

0.00 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915

0.01 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950

0.02 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985

0.03 0.5120 0.5517 0.5910 0.6293 0.6664 0.7019

Slide ‹#›

0.04 0.5160 0.5557 0.5948 0.6331 0.6700 0.7054

0.05 0.5199 0.5596 0.5987 0.6368 0.6736 0.7088

0.06 0.5239 0.5636 0.6026 0.6406 0.6772 0.7123

0.07 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157

0.08 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190

0.09 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224

The Standard Normal Distribution Function Table Standard Normal Distribution Function Table (continued), F (z ) z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

 

0.00 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.8413

0.01 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438

0.02 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.8461

0.03 0.5120 0.5517 0.5910 0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238 0.8485

0.04 0.5160 0.5557 0.5948 0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264 0.8508

0.05 0.5199 0.5596 0.5987 0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.8531

0.06 0.5239 0.5636 0.6026 0.6406 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.8554

0.07 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577

0.08 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599

0.09 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621

Q: What is the probability that the outcome of a standard normal is 0.05 or lower? Q: For what z is there a 0.67 probability that the outcome of the standard normal is that z or lower? Slide ‹#›

The Standard Normal Distribution Function Table Standard Normal Distribution Function Table, F (z ) z -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0.0   

-0.09 0.2148 0.2451 0.2776 0.3121 0.3483 0.3859 0.4247 0.4641

-0.08 0.2177 0.2483 0.2810 0.3156 0.3520 0.3897 0.4286 0.4681

-0.07 0.2206 0.2514 0.2843 0.3192 0.3557 0.3936 0.4325 0.4721

-0.06 0.2236 0.2546 0.2877 0.3228 0.3594 0.3974 0.4364 0.4761

-0.05 0.2266 0.2578 0.2912 0.3264 0.3632 0.4013 0.4404 0.4801

-0.04 0.2296 0.2611 0.2946 0.3300 0.3669 0.4052 0.4443 0.4840

-0.03 0.2327 0.2643 0.2981 0.3336 0.3707 0.4090 0.4483 0.4880

-0.02 0.2358 0.2676 0.3015 0.3372 0.3745 0.4129 0.4522 0.4920

-0.01 0.2389 0.2709 0.3050 0.3409 0.3783 0.4168 0.4562 0.4960

0.00 0.2420 0.2743 0.3085 0.3446 0.3821 0.4207 0.4602 0.5000

Q: What is the probability that the outcome of a standard normal is -0.38 or lower? Q: For what z is there a 0.40 probability that the outcome of the standard normal is that z or lower? Q: For what z is there a 0.75 probability that the outcome of the standard normal is greater than z? Slide ‹#›

The Newsvendor Model: Develop a demand model

Slide ‹#›

Historical forecast performance at O’Neill

.

7000 6000

Actual demand

5000 4000 3000 2000 1000 0 0

1000

2000

3000

4000

5000

6000

7000

Forecast

Forecasts and actual demand for surf wet-suits from the previous season Slide ‹#›

Empirical distribution function of forecast accuracy



Start by evaluating the actual to forecast ratio (the A/F ratio) from past observations.

Actual demand Product description Forecast JR ZEN FL 3/2 90 140 EPIC 5/3 W/HD 120 83 JR ZEN 3/2 140 143 WMS ZEN-ZIP 4/3 170 163 HEATWAVE 3/2 170 212 JR EPIC 3/2 180 175 WMS ZEN 3/2 180 195 ZEN-ZIP 5/4/3 W/HOOD 270 317 WMS EPIC 5/3 W/HD 320 369 EVO 3/2 380 587 JR EPIC 4/3 380 571 WMS EPIC 2MM FULL 390 311 HEATWAVE 4/3 430 274 ZEN 4/3 430 239 EVO 4/3 440 623 ZEN FL 3/2 450 365 HEAT 4/3 460 450 ZEN-ZIP 2MM FULL 470 116 HEAT 3/2 500 635 WMS EPIC 3/2 610 830 WMS ELITE 3/2 650 364 ZEN-ZIP 3/2 660 788 ZEN 2MM S/S FULL 680 453 EPIC 2MM S/S FULL 740 607 EPIC 4/3 1020 732 WMS EPIC 4/3 1060 1552 JR HAMMER 3/2 1220 721 HAMMER 3/2 1300 1696 HAMMER S/S FULL 1490 1832 EPIC 3/2 2190 3504 ZEN 3/2 3190 1195 ZEN-ZIP 4/3 3810 3289 WMS HAMMER 3/2 FULL 6490 3673 * Error = Forecast - Actual demand ** A/F Ratio = Actual demand divided by Forecast

Slide ‹#›

Error* A/F Ratio** -50 1.56 37 0.69 -3 1.02 7 0.96 -42 1.25 5 0.97 -15 1.08 -47 1.17 -49 1.15 -207 1.54 -191 1.50 79 0.80 156 0.64 191 0.56 -183 1.42 85 0.81 10 0.98 354 0.25 -135 1.27 -220 1.36 286 0.56 -128 1.19 227 0.67 133 0.82 288 0.72 -492 1.46 499 0.59 -396 1.30 -342 1.23 -1314 1.60 1995 0.37 521 0.86 2817 0.57

Using historical A/F ratios to choose a normal distribution for the demand model 



Start with an initial forecast generated from hunches, guesses, etc.  O’Neill’s initial forecast for the Hammer 3/2 = 3200 units. Evaluate the A/F ratios of the historical data:

A/F ratio  

Actual demand Forecast

Set the mean of the normal distribution to

Expectedactualdemand Expected A/F ratio  Forecast 

Set the standard deviation of the normal distribution to

Standarddeviationof actualdem and Standarddeviationof A/F ratios Forecast

Slide ‹#›

O’Neill’s Hammer 3/2 normal distribution forecast Product description JR ZEN FL 3/2 EPIC 5/3 W/HD JR ZEN 3/2 WMS ZEN-ZIP 4/3

Forecast Actual demand 90 140 120 83 140 143 170 156

Error -50 37 -3 14

A/F Ratio 1.5556 0.6917 1.0214 0.9176







… …

ZEN 3/2 ZEN-ZIP 4/3 WMS HAMMER 3/2 FULL Average Standard deviation

3190 3810 6490

1195 3289 3673

1995 521 2817

0.3746 0.8633 0.5659 0.9975 0.3690

Expectedactualdemand 0.9975 3200 3192

Standarddeviationof actualdemand 0.369  3200 1181 

O’Neill can choose a normal distribution with mean 3192 and standard deviation 1181 to represent demand for the Hammer 3/2 during the Spring season. Slide ‹#›

The Newsvendor Model: The order quantity that maximizes expected profit

Slide ‹#›

Hammer 3/2 economics 

 

O’Neill sells each suit for p = $190 O’Neill purchases each suit from its supplier for c = $110 per suit Discounted suits sell for v = $90  This is also called the salvage value.



O’Neill has the “too much/too little problem”:  Order too much and inventory is left over at the end of the season  Order too little and sales are lost.



Recall, our demand model is a normal distribution with mean 3192 and standard deviation 1181.

Slide ‹#›

“Too much” and “too little” costs 

Co = overage cost  The consequence of ordering one more unit than what you would have ordered had you known demand. Suppose you had left over inventory (you over ordered). Co is the increase in profit you would have enjoyed had you ordered one fewer unit.  For the Hammer 3/2 Co = Cost – Salvage value = c – v = 110 – 90 = 20 



Cu = underage cost  The consequence of ordering one fewer unit than what you would have ordered had you known demand. Suppose you had lost sales (you under ordered). Cu is the increase in profit you would have enjoyed had you ordered one more unit.  For the Hammer 3/2 Cu = Price – Cost = p – c = 190 – 110 = 80 

Slide ‹#›

Balancing the risk and benefit of ordering a unit

90 80

.



Ordering one more unit increases the chance of overage …  Expected loss on the Qth unit = Co x F(Q)  F(Q) = Distribution function of demand = Prob{Demand

F (Q) 

Cu Co  Cu



The ratio Cu / (Co + Cu) is called the critical ratio.



Hence, to maximize profit, choose Q such that the probability we satisfy all demand (i.e., demand is Q or lower) equals the critical ratio.



For the Hammer 3/2 the critical ratio is Slide ‹#›

Cu 80   0.80 Co  Cu 20  80

Hammer 3/2’s expected profit maximizing order quantity   z 0.5 0.6 0.7 0.8 0.9

The critical ratio is 0.80 Find the critical ratio inside the Standard Normal Distribution Function Table: 0.00 0.6915 0.7257 0.7580 0.7881 0.8159

0.01 0.6950 0.7291 0.7611 0.7910 0.8186

0.02 0.6985 0.7324 0.7642 0.7939 0.8212

0.03 0.7019 0.7357 0.7673 0.7967 0.8238

0.04 0.7054 0.7389 0.7704 0.7995 0.8264

0.05 0.7088 0.7422 0.7734 0.8023 0.8289

0.06 0.7123 0.7454 0.7764 0.8051 0.8315





0.07 0.7157 0.7486 0.7794 0.8078 0.8340

0.08 0.7190 0.7517 0.7823 0.8106 0.8365

0.09 0.7224 0.7549 0.7852 0.8133 0.8389

If the critical ratio falls between two values in the table, choose the greater z-statistic … this is called the round-up rule.  Choose z = 0.85 Convert the z-statistic into an order quantity :

Q  m  z s  3192 0.851181 4196 Slide ‹#›

The Newsvendor Model: Performance Measures

Slide ‹#›

Newsvendor model performance measures 

For any order quantity we would like to evaluate the following performance measures:  In-stock probability Probability all demand is satisfied  Stockout probability Probability some demand is lost  Expected lost sales The expected number of units by which demand will exceed the order quantity  Expected sales The expected number of units sold.  Expected left over inventory The expected number of units left over after demand (but before salvaging)  Expected profit 









Slide ‹#›

In-stock probability 

The in-stock probability is the probability all demand is satisfied.



All demand is satisfied if demand is the order quantity, Q, or smaller.  If Q = 3000, then to satisfy all demand, demand must be 3000 or fewer.



The distribution function tells us the probability demand is Q or smaller!



Hence, the In-stock probability = F(Q) = F(z)

Slide ‹#›

Evaluate the in-stock probability 

What is the in-stock probability if the order quantity is Q = 3000?



Evaluate the z-statistic for the order quantity : z 



Look up F(z) in the Std. Normal Distribution Function Table:  F(-0.16) = 0.4364



Answer :  If 3000 units are orders, then there is a 43.64% chance all demand will be satisfied.

Slide ‹#›

Qm

s



3000  3192  0.16 1181

Choose Q subject to a minimum in-stock probability 

Suppose we wish to find the order quantity for the Hammer 3/2 that minimizes left over inventory while generating at least a 99% in-stock probability.



Step 1:  Find the z-statistic that yields the target in-stock probability.  In the Standard Normal Distribution Function Table we find F(2.32) = 0.9898 and F(2.33) = 0.9901.  Choose z = 2.33 to satisfy our in-stock probability constraint.



Step 2:  Convert the z-statistic into an order quantity for the actual demand distribution.  Q = m + z x s = 3192 + 2.33 x 1181 = 5944

Slide ‹#›

Other measures of service performance 

The stockout probability is the probability some demand is not satisfied:  Some demand is not satisfied if demand exceeds the order quantity, thus…  Stockout probability = 1 – F(Q) = 1 – In-stock probability = 1 –0.4364 = 56.36%



The fill rate is the fraction of demand that can purchase a unit:  The fill rate is also the probability a randomly chosen customer can purchase a unit.  The fill rate is not the same as the in-stock probability! e.g. if 99% of demand is satisfied (the fill rate) then the probability all demand is satisfied (the in-stock) need not be 99% 

Slide ‹#›

Expected lost sales: graphical explanation 

  

 

Suppose demand can be one of the following values {0,10,20,…,190,200} Suppose Q = 120 Let D = actual demand What is expected lost sales? If D
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