OPSM 501: Operations Management Week 10: Supply Chain contracts Newsvendor Koç University Graduate School of Business MBA Program Zeynep Aksin zaksin@ku.edu.tr.
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OPSM 501: Operations Management
Week 10:
Supply Chain contracts
Newsvendor
Koç University Graduate School of BusinessMBA Program
Zeynep Aksinzaksin@ku.edu.tr
2
Hamptonshire Express
Anna has a degree from journalism & operations research
She has started a daily newspaper in her hometown She used a leased PC: lease cost $10 per day A local printer prints newspapers at 0.20 per copy Sales the next day between 6 am and 10 am Newsstand rental $30 per day Express sold to customers at $1 per copy Copies not sold by 10 am are discarded Anna estimates daily demand to be distributed
N(500,100)
4
Ordering Level and Profits in Vertically Integrated Channel
h=1; Anna sells to market directly:i* = 584; E[Profit] = $331.33; Fill rate
98%
5
Improving demand through effort
After 6 weeks of operation, Anna thinks she can improve demand by adding a profile section
Experiments indicate that demand is a function of time she invests in preparing the section
She thinks D=500 +50 h
6
Question 2
How many hours should she invest daily in the creation of the profile section? Assume the opportunity cost of her time is $10 per hour.
Compare optimal profits to previous scenario
7
Optimal Level of Effort in Vertically Integrated Channel
Demand potential increases by 50 Expected profit increases by 0.8*50
h
h
h h+1
0 1 40
1 2 16.56
2 3 12.71
3 4 10.71
4 5 9.44
)1(*50*0.8 hh
i* = 684E[Profit] = 371.33
8
Delegating sales to Ralph
Anna is really busy, so asks Ralph to take-over the retailing portion of her job.
Ralph agrees to run the newsstand from 6 AM to 10 AM and pay the daily rent of $30
He estimates demand the next day based on viewing a copy of the paper the previous night at 10 PM
He buys the papers from Anna at $0.8 per copy Ralph is responsible for unsold copies at the end of the
day
9
Question 3
Assuming h=4 try to determine the optimal stocking quantity for Ralph?
Why is this quantity different than the one in Question 2? Now vary h in spreadsheet 3c which calculates the
optimal newsboy quantity for the differentiated channel, i.e. to maximize Ralph’s profit.
How would changing the transfer price from the current value of 0.8 impact Ann’s effort level and Ralph’s stocking decision? (Spreadsheet 3d)
Compare an integrated (centralized) firm to a differentiated (decentralized) one.
10
Ordering Level and Profits in Differentiated Channel
Case 1. h=4; Anna sells to market directly:i* = 684; E[Profit] = $371.33; Fill rate
98%
Case 2. h=4; Anna sells thru Ralph:i* = 516; E[Total Profit] = $322
Anna makes $260Ralph makes $ 62Fill rate 84%
Why??
11
Effect of Transfer (Wholesale) Price in Differentiated Channel
Breakdown of total profits (h=4)
0
50
100
150
200
250
300
350
400
transfer price
$ralph
anna
12
Optimal Effort in Decentralized Channel
Optimal effort level for Anna is h=2 (and not 4).
h=2 h=4
Stocking quantity: $487 $516
Anna’s profit: $262 $260
Ralph’s profit: $56 $ 62
Total profit: $318 $322
Fill rate: 83% 84%
Why??
13
Inefficiencies in a Differentiated Channel
Supplier chooses w, retailer chooses i* Retail ignores +ve effect of stocking one more
unit on supplier Supplier ignores +ve effect of cutting
wholesale price/increasing effort on retailer Supplier prices above marginal cost/exerts
low effort Retailer stocks less Supply chain profits shrink
14
Contracts
Specifies the parameters within which a buyer places orders and a supplier fulfills them
Example parameters: quantity, price, time, quality Double marginalization: buyer and seller make
decisions acting independently instead of acting together; both of them make a margin on the same sale – gap between potential total supply chain profits and actual supply chain profits results
Buyback contracts can be offered that will increase total supply chain profit
15
Returns policies
Rationale: set buyback price b so that (retailer cost structure
= supply cost structure)
Supplier can use both w and b Supplier is bundling insurance with the
good
sr
sc
br
bw
17
Reasons for return policies
Supplier is less risk averse than retailers
Supplier has a higher salvage value Safeguarding the brand Signalling information Avoiding brand switching
18
Costs of Return Policies
Extra transportation and handling Extra depreciation Getting the return rate wrong Retailer incentives
19
The case of books
Returns as in Hamptonshire Express… …However publisher has no control of return
quantities No control of inventory-shelf arrangements No control over private-label No control of retail price
Key difference: power has shifted from publisher to retailer
20
Video sales
Hollywood: video rentals and sales $20B business, and largest source of revenue
Rentals slipping– Competition from direct services– Customer dissatisfaction (20% cannot rent
video they want on a typical trip) What’s the problem? Bad forecasting?
Inefficient replenishment?
21
Revenue Sharing
Reduce wholesale price in return for a share of revenues
Encourages retailers to stock more $60 a tape
– $3/rental – rent each tape 20 times to break even
$9 a tape, studio receives half revenue– $3/rental – rent each tape 6 times to break
even Retailer stocks more
22
Revenue sharing
When does it work?– marginal cost of increasing inventory low– administrative burden low– for price-sensitive products
23
The Impact of Revenue Sharing
Blockbuster Instituted the “Go Home Happy” marketing initiative
Results– Store traffic went up– Market share 4th quarter 98 = 26%– Market share 2nd quarter 99 = 31%– Revenue in 2nd quarter 99: +17% from 98– Cash flow in 2nd quarter 99: +61% from 98
Supply
Sources:plantsvendorsports
RegionalWarehouses:stocking points
Field Warehouses:stockingpoints
Customers,demandcenterssinks
Production/purchase costs
Inventory &warehousing costs
Transportation costs Inventory &
warehousing costs
Transportation costs
Supply Chain Management: the challenge
Global optimization– Conflicting Objectives– Complex network of facilities– System Variations over time
Managing uncertainty– Matching Supply and Demand– Demand is not the only source of uncertainty
Recall Zara’a Approach to Demand uncertainty
Expected demand Actual demand
Small batches
Excess stock and unmet demand are avoided by stopping production when market saturates
Flu vaccine example
Each year’s flu vaccine is different: can’t produce ahead or keep from last year
Flu vaccine production requires growing strains: there is a lead time
Factories have limited capacity Demand is uncertain Need to commit to production before flu season starts Result: frequent shortage of vaccine or left overs at the
end of the season
Historical forecast performance at O’Neill
0
1000
2000
3000
4000
5000
6000
7000
0 1000 2000 3000 4000 5000 6000 7000
Forecast
Act
ual d
eman
d
.
Forecasts and actual demand for surf wet-suits from the previous season
Empirical distribution of forecast accuracy
Empirical distribution function for the historical A/F ratios.
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75
A/F ratio
Prob
abili
ty
Product description Forecast Actual demand Error* A/F Ratio**JR ZEN FL 3/2 90 140 -50 1.56EPIC 5/3 W/HD 120 83 37 0.69JR ZEN 3/2 140 143 -3 1.02WMS ZEN-ZIP 4/3 170 163 7 0.96HEATWAVE 3/2 170 212 -42 1.25JR EPIC 3/2 180 175 5 0.97WMS ZEN 3/2 180 195 -15 1.08ZEN-ZIP 5/4/3 W/HOOD 270 317 -47 1.17WMS EPIC 5/3 W/HD 320 369 -49 1.15EVO 3/2 380 587 -207 1.54JR EPIC 4/3 380 571 -191 1.50WMS EPIC 2MM FULL 390 311 79 0.80HEATWAVE 4/3 430 274 156 0.64ZEN 4/3 430 239 191 0.56EVO 4/3 440 623 -183 1.42ZEN FL 3/2 450 365 85 0.81HEAT 4/3 460 450 10 0.98ZEN-ZIP 2MM FULL 470 116 354 0.25HEAT 3/2 500 635 -135 1.27WMS EPIC 3/2 610 830 -220 1.36WMS ELITE 3/2 650 364 286 0.56ZEN-ZIP 3/2 660 788 -128 1.19ZEN 2MM S/S FULL 680 453 227 0.67EPIC 2MM S/S FULL 740 607 133 0.82EPIC 4/3 1020 732 288 0.72WMS EPIC 4/3 1060 1552 -492 1.46JR HAMMER 3/2 1220 721 499 0.59HAMMER 3/2 1300 1696 -396 1.30HAMMER S/S FULL 1490 1832 -342 1.23EPIC 3/2 2190 3504 -1314 1.60ZEN 3/2 3190 1195 1995 0.37ZEN-ZIP 4/3 3810 3289 521 0.86WMS HAMMER 3/2 FULL 6490 3673 2817 0.57* Error = Forecast - Actual demand** A/F Ratio = Actual demand divided by Forecast
Normal distribution tutorial
All normal distributions are characterized by two parameters, mean = and standard deviation =
All normal distributions are related to the standard normal that has mean = 0 and standard deviation = 1.
For example:– Let Q be the order quantity, and (, ) the parameters of the normal demand
forecast.– Prob{demand is Q or lower} = Prob{the outcome of a standard normal is z or
lower}, where
– (The above are two ways to write the same equation, the first allows you to calculate z from Q and the second lets you calculate Q from z.)
– Look up Prob{the outcome of a standard normal is z or lower} in the Standard Normal Distribution Function Table.
orQ
z Q z
11-34
-
0.0020
0.0040
0.0060
0.0080
0.0100
0.0120
0.0140
0.0160
0.0180
0 25 50 75 100 125 150 175 200
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.018
-100 -75 -50 -25 0 25 50 75 100
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
-4 -3 -2 -1 0 1 2 3 4
Converting between Normal distributions
Start with = 100,= 25, Q = 125
Center the distribution over 0 by subtracting the mean
Rescale the x and y axes by dividing by the standard deviation
125
100125
Q
z
11-35
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:
Set the mean of the normal distribution to
Set the standard deviation of the normal distribution to
Using historical A/F ratios to choose a Normal distribution for the demand forecast
Forecast
demand Actual ratio A/F
Forecast ratio A/F Expected demand actual Expected
Forecast ratios A/F of deviation Standard
demand actual of deviation Standard
11-36
O’Neill’s Hammer 3/2 normal distribution forecast
3192320099750 . demand actual Expected
118132003690 . demand actual of deviation Standard
O’Neill should choose a normal distribution with mean 3192 and standard deviation 1181 to represent demand for the Hammer 3/2 during the Spring season.
Product description Forecast Actual demand Error A/F RatioJR ZEN FL 3/2 90 140 -50 1.5556EPIC 5/3 W/HD 120 83 37 0.6917JR ZEN 3/2 140 143 -3 1.0214WMS ZEN-ZIP 4/3 170 156 14 0.9176
… … … … …ZEN 3/2 3190 1195 1995 0.3746ZEN-ZIP 4/3 3810 3289 521 0.8633WMS HAMMER 3/2 FULL 6490 3673 2817 0.5659Average 0.9975Standard deviation 0.3690
11-37
Empirical vs normal demand distribution
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0 1000 2000 3000 4000 5000 6000
Quantity
Prob
abili
ty
.
Empirical distribution function (diamonds) and normal distribution function withmean 3192 and standard deviation 1181 (solid line)
11-38
Costs
Production cost per unit (C): $80 Selling price per unit (S): $125 Salvage value per unit (V): $20 Fixed production cost (F): $100,000 Q is production quantity, D demand
Profit =Revenue - Variable Cost - Fixed Cost + Salvage
Best Solution
Find order quantity that maximizes weighted average profit.
Question: Will this quantity be less than, equal to, or greater than average demand?
What to Make?
Question: Will this quantity be less than, equal to, or greater than average demand?
Average demand is 13,100 Look at marginal cost Vs. marginal profit
– if extra jacket sold, profit is 125-80 = 45– if not sold, cost is 80-20 = 60
So we will make less than average
Scenarios
Scenario One:– Suppose you make 12,000 jackets and demand ends
up being 13,000 jackets.– Profit = 125(12,000) - 80(12,000) - 100,000 = $440,000
Scenario Two:– Suppose you make 12,000 jackets and demand ends
up being 11,000 jackets.– Profit = 125(11,000) - 80(12,000) - 100,000 + 20(1000) = $
335,000
44
Scenarios and their probabilitiesDemand
Pro
du
ctio
n q
uan
tity
8000 10000 12000 14000 16000 1800011% 11% 28% 22% 18% 10%
5,000 $125,000.00 $125,000.00 $125,000.00 $125,000.00 $125,000.00 $125,000.00 $125,000
5,500 $147,500.00 $147,500.00 $147,500.00 $147,500.00 $147,500.00 $147,500.00 $147,500
6,000 $170,000.00 $170,000.00 $170,000.00 $170,000.00 $170,000.00 $170,000.00 $170,000
6,500 $192,500.00 $192,500.00 $192,500.00 $192,500.00 $192,500.00 $192,500.00 $192,500
7,000 $215,000.00 $215,000.00 $215,000.00 $215,000.00 $215,000.00 $215,000.00 $215,000
7,500 $237,500.00 $237,500.00 $237,500.00 $237,500.00 $237,500.00 $237,500.00 $237,500
8,000 $260,000.00 $260,000.00 $260,000.00 $260,000.00 $260,000.00 $260,000.00 $260,000
8,500 $230,000.00 $282,500.00 $282,500.00 $282,500.00 $282,500.00 $282,500.00 $276,725
9,000 $200,000.00 $305,000.00 $305,000.00 $305,000.00 $305,000.00 $305,000.00 $293,450
9,500 $170,000.00 $327,500.00 $327,500.00 $327,500.00 $327,500.00 $327,500.00 $310,175
10,000 $140,000.00 $350,000.00 $350,000.00 $350,000.00 $350,000.00 $350,000.00 $326,900
10,500 $110,000.00 $320,000.00 $372,500.00 $372,500.00 $372,500.00 $372,500.00 $337,850
11,000 $80,000.00 $290,000.00 $395,000.00 $395,000.00 $395,000.00 $395,000.00 $348,800
11,500 $50,000.00 $260,000.00 $417,500.00 $417,500.00 $417,500.00 $417,500.00 $359,750
12,000 $20,000.00 $230,000.00 $440,000.00 $440,000.00 $440,000.00 $440,000.00 $370,700
12,500 -$10,000.00 $200,000.00 $410,000.00 $462,500.00 $462,500.00 $462,500.00 $366,950
13,000 -$40,000.00 $170,000.00 $380,000.00 $485,000.00 $485,000.00 $485,000.00 $363,200
13,500 -$70,000.00 $140,000.00 $350,000.00 $507,500.00 $507,500.00 $507,500.00 $359,450
14,000 -$100,000.00 $110,000.00 $320,000.00 $530,000.00 $530,000.00 $530,000.00 $355,700
14,500 -$130,000.00 $80,000.00 $290,000.00 $500,000.00 $552,500.00 $552,500.00 $340,400
15,000 -$160,000.00 $50,000.00 $260,000.00 $470,000.00 $575,000.00 $575,000.00 $325,100
15,500 -$190,000.00 $20,000.00 $230,000.00 $440,000.00 $597,500.00 $597,500.00 $309,800
16,000 -$220,000.00 -$10,000.00 $200,000.00 $410,000.00 $620,000.00 $620,000.00 $294,500
16,500 -$250,000.00 -$40,000.00 $170,000.00 $380,000.00 $590,000.00 $642,500.00 $269,750
Average Profit
Expected Profit
Expected Profit
Expected Profit
$0
$100,000
$200,000
$300,000
$400,000
8000 12000 16000 20000
Order Quantity
Pro
fit
Expected Profit
Expected Profit
$0
$100,000
$200,000
$300,000
$400,000
8000 12000 16000 20000
Order Quantity
Pro
fit
Expected Profit
Expected Profit
$0
$100,000
$200,000
$300,000
$400,000
8000 12000 16000 20000
Order Quantity
Pro
fit
Important Observations
Tradeoff between ordering enough to meet demand and ordering too much
Several quantities have the same average profit Average profit does not tell the whole story
Question: 9000 and 16000 units lead to about the same average profit, so which do we prefer?
Key Insights from this Model
The optimal order quantity is not necessarily equal to average forecast demand
The optimal quantity depends on the relationship between marginal profit and marginal cost
Fixed cost has no impact on production quantity, only on whether to produce or not
As order quantity increases, average profit first increases and then decreases
As production quantity increases, risk increases. In other words, the probability of large gains and of large losses increases
Example
Mean demand=3.85 How much would you order?
Demand Probability
1 0.10
2 0.15
3 0.20
4 0.20
5 0.15
6 0.10
7 0.10
Total 1.00
Single Period Inventory Control
Economics of the Situation Known:1. Demand > Stock --> Underage (under stocking) Cost
Cu = Cost of foregone profit, loss of goodwill
2. Demand < Stock --> Overage (over stocking) Cost
Co = Cost of excess inventory
Co = 10 and Cu = 20 How much would you order? More
than 3.85 or less than 3.85?
Incremental AnalysisProbability Probability Incremental
Incremental that incremental that incremental Expected
Demand Decision unit is not needed unit is needed Contribution
1 First 0.00 1.00 -10(0.00)+20(1.00)
=20
2 Second 0.10 0.90 -10(0.10)+20(0.90)
=17
3 Third 0.25 0.75 12.5
4 Fourth 0.45 0.55 6.5
5 Fifth 0.65 0.35 0.5
6 Sixth 0.80 0.20 -4
7 Seventh 0.90 0.10 -7
Co = 10 and Cu = 20
Generalization of the Incremental Analysis
ChancePoint
Stock n-1
DecisionPoint
Stock n
Base Case
nth unit needed
nth unit not needed
Pr{Demand n}
Pr{Demand n-1}
Cash FlowCu
-Co
0
Generalization of the Incremental Analysis
ChancePoint
Stock n-1
DecisionPoint
Stock n
Base Case
Expected Cash FlowCu Pr{Demand n} -Co Pr{Demand n-1}
Generalization of the Incremental Analysis
Order the nth unit ifCu Pr{Demand n} - Co Pr{Demand n-1} >= 0
or
Cu (1-Pr{Demand n-1}) - Co Pr{Demand n-1} >= 0
or
Cu - Cu Pr{Demand n-1} -Co Pr{Demand n-1} >= 0
or
Pr{Demand n-1} =< Cu /(Co +Cu)
Then order n units, where n is the greatest number that satisfies the above inequality.
Incremental Analysis
IncrementalDemand Decision Pr{Demand n-1} Order the unit?1 First 0.00 YES
2 Second 0.10 YES3 Third 0.25 YES4 Fourth 0.45 YES5 Fifth 0.65 YES6 Sixth 0.80 NO -7 Seventh 0.90 NO
Cu /(Co +Cu)=20/(10+20)=0.66
Order quantity n should satisfy:P(Demand n-1) Cu /(Co +Cu)< P(Demand n)
Order Quantity for Single Period, Normal Demand
Find the z*: z value such that F(z)= Cu /(Co +Cu)
Optimal order quantity is: *zQ
Transform
X = N(mean,s.d.) to
z = N(0,1)
z = (X - mean) / s.d.
F(z) = Prob( N(0,1) < z)
Transform back, knowing z*:
X* = mean + z*s.d.
The Standard Normal Distributionz 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.68790.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.83891.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.86211.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.93191.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.97061.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.97672.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.98172.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.98572.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.98902.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99162.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99362.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99522.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.99642.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.99742.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.99812.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.99863.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.99903.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.99933.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.99953.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997
F(z)
z0
Example
z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.68790.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.83891.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.86211.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.93191.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.97061.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.97672.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.98172.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.98572.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.98902.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99162.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99362.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99522.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.99642.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.99742.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.99812.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.99863.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.99903.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.99933.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.99953.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997
If we want tohave cum. probability of95%z=1.64
For demand:Mean=20std dev=10
Then:Q=20 + 1.64*10=36.4
Example: Anna’s first stocking decision
Cu = 1-0.2=0.8 and Co = 0.2; P ≤ 0.8 / (0.8 + 0.2) = .8
Z.8 = .84 (from standard normal table or using NORMSINV() in Excel)
therefore Anna needs 500 + .84(100) = 584 papers
Example: What about Ralph’s first stocking decision?
Anna sets h=4 D=N(500+50*2, 100) Cu = 1-0.8=0.2 and Co = 0.8; P ≤ 0.2 / (0.8 +
0.2) = .2
Z0.2 = - Z0.8 =-0.84 (from standard normal table or using NORMSINV() in Excel)
therefore Anna needs 600 - 0.84(100) = 516 papers
Example 2: Finding Cu and Co
A textile company in UK orders coats from China. They buy a coat from 250€ and sell for 325€. If they cannot sell a coat in winter, they sell it at a discount price of 225€. When the demand is more than what they have in stock, they have an option of having emergency delivery of coats from Ireland, at a price of 290.
The demand for winter has a normal distribution with mean 32,500 and std dev 6750.
How much should they order from China??
Example 2: Finding Cu and Co
A textile company in UK orders coats from China. They buy a coat from 250€ and sell for 325€. If they cannot sell a coat in winter, they sell it at a discount price of 225€. When the demand is more than what they have in stock, they have an option of having emergency delivery of coats from Ireland, at a price of 290.
The demand for winter has a normal distribution with mean 32,500 and std dev 6750.
How much should they order from China??
Cu=75-35=40Co=25F(z)=40/(40+25)=40/65=0.61z=0.28 q=32500+0.28*6750=34390
Example 3: Single Period Inventory Management Problem
Manufacturing cost=60TL,
Selling price=80TL, Discounted price (at the end of the season)=50TL
Market research gave the following probability distribution for demand.
Find the optimal q, expected number of units sold for this orders size, and expected profit, for this order size.
Demand Probability500 0.10600 0.2700 0.2800 0.2900 0.101000 0.101100 0.10
P(D<=n-1)00.10.30.50.70.80.9
Cu=20 Co=10P(D<=n-1)<=20/30=0.66
<=0.66 q=800
For q=800:E(units sold)=710E(profit)=13,300
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