THE NEWSVENDOR AND APPLICATIONS 2 T HE N EWSVENDOR M ODEL 3 ON ELL - - PowerPoint PPT Presentation

OPERATİONS & LOGİSTİCS MANAGEMENT İN AİR TRANSPORTATİON

PROFESSOR DAVİD GİLLEN (UNİVERSİTY OF BRİTİSH COLUMBİA ) & PROFESSOR BENNY MANTİN (UNİVERSİTY OF WATERLOO)

Air Transportation Systems and Infrastructure Strategic Planning Module 7-8 : 12 June 2014 Istanbul Technical University Air Transportation Management M.Sc. Program

THE NEWSVENDOR AND APPLICATIONS

2

THE NEWSVENDOR MODEL

3

O’NEİLL’S HAMMER 3/2 WETSUİT TİMELİNE

Nov Dec Jan Feb Mar Apr May Jun Jul Aug Generate forecast

• f demand and

submit an order to TEC Receive order from TEC at the end of the month Spring selling season Left over units are discounted

• Marketing’s forecast for sales is 3200 units.

4

O’NEİLL’S HAMMER 3/2 WETSUİT TİMELİNE

• Marketing’s forecast for sales is 3200 units.
• 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.

• How many units shall O’Neill order?!
• O’Neill is facing a “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.

5

FORECASTİNG: DEMAND DİSTRİBUTİON

• Marketing's forecast is a single number.
• However, demand is more likely to follow some distribution.
• Traditional distributions from statistics include, e.g., the normal,

gamma, Poisson distributions

6

50 100 150 200 Probability Demand NORMAL GAMMA 1 2 3 4 5 6 7 8 9 10 Probability Demand POISSON

FORECASTİNG: DEMAND DİSTRİBUTİON

• How can we uncover the underlying distribution?
• We can look at historical forecast performance at O’Neill

7

1000 2000 3000 4000 5000 6000 7000 1000 2000 3000 4000 5000 6000 7000 Forecast Actual demand .

Forecasts and actual demand for surf wet-suits from the previous season

• The idea: use historical actual to forecast ratio (the A/F ratio)

from past observations.

• Calculate average and standard deviation of A/F ratio.
• Set the mean of the normal distribution to
• Set the standard deviation of the normal distribution to
• Recall:

FORECASTİNG: DEMAND DİSTRİBUTİON

Forecast demand Actual ratio A/F 

Forecast ratio) A/F (Expected demand actual Expected   Forecast ratios) A/F

• f

deviation (Standard demand actual

• f

deviation Standard  





 

N i i

x

1 2

) ( N 1  

8

• Marketing’s forecast for sales is 3200 units. Hence:
• 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.

FORECASTİNG: DEMAND DİSTRİBUTİON

3192 3200 9975    . demand actual Expected 1181 3200 369    . demand actual

• f

deviation Standard

Product description Forecast Actual demand Error A/F Ratio JR ZEN FL 3/2 90 140

• 50

1.5556 EPIC 5/3 W/HD 120 83 37 0.6917 JR ZEN 3/2 140 143

• 3

1.0214 WMS ZEN-ZIP 4/3 170 156 14 0.9176

… … … … …

ZEN 3/2 3190 1195 1995 0.3746 ZEN-ZIP 4/3 3810 3289 521 0.8633 WMS HAMMER 3/2 FULL 6490 3673 2817 0.5659 Average 0.9975 Standard deviation 0.3690

Note: Full data in hidden slide 9

DİSTRİBUTİON AND DENSİTY FUNCTİONS

50 100 150 200 Probability Demand 0.00 0.20 0.40 0.60 0.80 1.00 50 100 150 200 Probability Demand

Distribution function Density function

The probability of a particular

• utcome occurring

The probability the outcome will be a particular value or smaller

11

NORMAL DİSTRİBUTİON

• Characterized by 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 some quantity – N (, ) describes the demand distribution – Prob{demand is Q or lower} = Prob{the outcome of a standard normal is z or lower}, where – Look up z in the Standard Normal Distribution Function Table to get the desired probability or use Excel: Prob{} = normsdist(z)

• r

Q z Q z         

50 100 150 200 Probability Demand

Q

12

THE STANDARD NORMAL DİSTRİBUTİON

Standard Normal Distribution Function Table (continued), F(z )

z 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 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

• Q: What is the probability the outcome of a standard

normal will be z = 0.28 or smaller?

• A: The answer is 0.6103, or 61.03%

13

THE STANDARD NORMAL DİSTRİBUTİON

Standard Normal Distribution Function Table (continued), F(z )

z 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 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

• Q: For what z is there a 70.19% chance that the outcome
• f a standard normal will be that z or smaller?
• A: the answer is z = 0.53

14

O’NEİLL’S HAMMER 3/2 WETSUİT TİMELİNE

• Marketing’s forecast for sales is 3200 units.

– our demand model is a normal distribution with mean 3192 and standard deviation 1181

• 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.

• How many units shall O’Neill order?!
• O’Neill is facing a “too much/too little problem”:

– What are the costs associated with too much/too little?

15

“TOO MUCH” AND “TOO LİTTLE” COSTS

• Co = overage cost

– The consequence of ordering one more unit than what you would have ordered had you known demand. – 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. – For the Hammer 3/2 Cu = Price – Cost = p – c = 190 – 110 = 80

16

“TOO MUCH” AND “TOO LİTTLE” COSTS

• Idea: balance the risk vs. benefit of an additional unit
• Ordering one more unit increases the chance of overage …

– Expected loss on the Qth unit = Co · F(Q)

• … but the benefit/gain of ordering one more unit is the reduction in

the chance of underage: – Expected gain on the Qth unit = Cu · (1-F(Q))

• As more units are ordered,

the expected benefit from

• rdering one unit decreases

while the expected loss of

• rdering one more unit increases!

17

10 20 30 40 50 60 70 80 90 800 1600 2400 3200 4000 4800 5600 6400 Expected gain or loss . Qth unit ordered Expected gain Expected loss

MAXİMİZE EXPECTED PROFİT

• Expected profit is maximized when the expected loss on the Qth unit

equals the expected gain on the Qth unit:

• Rearrange terms in the above equation:
• 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

   

Q F C Q F C

u

• 

   1 ) (

u

• u

C C C Q F   ) ( 80 . 80 20 80    

u

• u

C C C

18

HAMMER 3/2’S ORDER QUANTİTY

• The critical ratio is 0.80
• Find the critical ratio inside the Standard Normal Distribution

Function Table:

– 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 :

4196 1181 85 . 3192          z Q

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

19

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

20

IN-STOCK PROBABİLİTY

• All demand is satisfied if demand is the order quantity, Q,
• r smaller.

– If Q = 3000, then to satisfy all demand, demand must be ≤ 3000.

• The distribution function tells us the probability demand is

Q or smaller!

– What is the in-stock probability if the order quantity is Q = 3000? – The z-statistic is: – Using the Std. Normal Distribution Function Table, we have F(-0.16) = 0.4364 – That is, if 3000 units are ordered, then there is a 43.64% chance all demand will be satisfied.

16 . 1181 3192 3000         Q z

21

IN-STOCK PROBABİLİTY

• Now, suppose that we wish to find the order quantity,

Q, that satisfies are required in-stock probability

– Consider the Hammer 3/2. What is Q if a 99% in-stock probability is required? – Find the z-statistic: from the Standard Normal Distribution Function Table we have F(2.33) = 0.9901. Hence z = 2.33. – Convert the z-statistic into an order quantity: Q =  + z ·  = 3192 + 2.33 · 1181 = 5944

22

OTHER MEASURES OF SERVİCE PERFORMANCE

• The stockout probability is the probability some demand is

not satisfied:

– Stockout probability = 1 – F(Q) = 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%

23

EXPECTED LOST SALES

• Suppose demand can be one
• f the following values

{0,10,20,…,190,200} with probabilities as per graph

• Suppose Q = 120
• What is expected lost sales?
• It depends on the actual

demand, D:

– If D ≤ Q, lost sales = 0 – If D = 130, lost sales = D – Q = 10

• Expected lost sales =

10·Prob{D = 130} + 20·Prob{D = 140} + … + 80·Prob{D = 200}

0,01 0,02 0,03 0,04 0,05 0,06 0,07 0,08 0,09 0,1 20 40 60 80 100 120 140 160 180 200

Loss of 10 units Loss of 40 units

24

500 1000 1500 2000 2500 3000 3500 1000 2000 3000 4000 5000 Expected Lost Sales Order quantity

EXPECTED LOST SALES: HAMMER 3/2S

• Suppose O’Neill orders 3000 Hammer 3/2s.
• How many sales will be lost on average?

– Find its z-statistic. – Look up in the Standard Normal Loss Function: L(-0.16)=0.4840

• r, in Excel: L(z)

=Normdist(z,0,1,0)-z·(1-Normsdist(z))

– Evaluate lost sales:

16 . 1181 3192 3000         Q z 572 4840 . 1181 ) ( Sales Lost Expected      z L 

Everything is lost “Nothing” is lost

25

OTHER MEASURES: HAMMER 3/2S

Expected sales =  - Expected lost sales = 3192 – 572 = 2620 Expected Left Over Inventory = Q - Expected Sales = 3000 – 2620 = 380 Note: the above equations hold for any demand distribution

       

000 , 202 $ 380 20 $ 2620 80 $          inventory

• ver

left Expected value Salvage

• Cost

Sales Expected Cost

• Price

Profit Expected

26

0.00 0.10 0.20 0.30 0.40 0.50 0.60 2 4 6 8 10 Q Density function probability 0.625 1.25 2.5 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 10 20 30 Q Density function probability 5 10 20

OTHER DİSTRİBUTİONS: POİSSON

• Characteristics:

– Defined only by its mean (standard deviation = square root(mean)) – Does not always have a “bell” shape, especially for low demand. – Discrete distribution function: only non-negative integers – Good for modeling demands with low means (e.g., less than 20) – If the inter-arrival times of customers are exponentially distributed, then the number of customers that arrive in a given interval of time has a Poisson distribution.

Here are six Poisson density functions with means 0.625, 1.25, 2.5, 5,10 and 20

27

POİSSON DİSTRİBUTİON: EXAMPLE

• EcoTea plans to sell a gift basket of Tanzanian teas through its specialty

stores during the Christmas season.

• Basket price is $55, purchase cost is $32 and after the holiday season left
• ver inventory will be cleared out at $20.
• Estimated demand at one of its stores is Poisson with mean 4.5.
• One order is made for the season.
• What is the optimal order quantity?

– The critical ratio: Co=32-20=12; Cu=55-32=23; Cu / (Cu + Cp) = 23 / 35 = 0.6571 – Look up in Table (or in Excel =POISSON.DIST(S,4.5,TRUE)): order 5 baskets

• If they order 6 baskets,

what is their lost sales?

• From the loss function

table, they can expect to lose 0.32 in sales.

Poisson Distribution Function Table Mean S

…

4.25 4.5 0.01426 0.01111 1 0.07489 0.06110 2 0.20371 0.17358 3 0.38621 0.34230 4 0.58012 0.53210 5 0.74494 0.70293 6 0.86169 0.83105 Poisson Loss Function Table Mean S … 4.25 4.5 4.25000 4.50000 1 3.26426 3.51111 2 2.33915 2.57221 3 1.54286 1.74579 4 0.92907 1.08808 5 0.50919 0.62019 6 0.25413 0.32312

28

NEWSVENDOR MODEL SUMMARY

• The model can be applied to settings in which …

– There is a single order/production/replenishment opportunity. – Demand is uncertain. – There is a “too much” vs. “too little” challenge

• The critical ratio trades off the overage and underage costs:
• The corresponding order quantity maximizes expected

profit.

• The firm can determine other service measure such as in-

stock probability or fill rate.

u

• u

C C C Q F   ) (

29

REVENUE MANAGEMENT

With material partially adapted from Netessine and Shumsky; teaching note on Yield Management; Metin’s class notes; Phillips (2005); Boyd (2007) 30

REVENUE MANAGEMENT

• What is Revenue Management

– technique to maximize revenue by matching fixed supply with uncertain demand – A very successful application of the critical ratio

• Examples

– Airlines:

• AA credits RM for increasing revenue by $500 million per year.
• Delta airlines has attributed a $300 million gain.

– Hotels:

• Marriot Hotel attributes $100 million in additional revenue to RM

– car rental:

• National Car Rental used RM to rescue itself from the bank of bankruptcy

– commercials

31

THE DİFFERENT ASPECTS OF PRİCİNG

• Strategic pricing

– Southwest – THE Low-Fare Airline

• Operational pricing

– Day-to-day adjusting of prices to address demand realization and updating of expectations

• Revenue Management

– A technique to maximize revenue by matching fixed supply with uncertain demand

32

BRİEF HİSTORY: AİRLİNES

• Until 1978, the aviation

industry in the US was

• regulated. Fares were

published and controlled by Civil Aeronautics Board

33

BRİEF HİSTORY: AİRLİNES

• Deregulation in 1978
• PeopleExpress established circa 1980

– Charged 70% less than major airlines – Fast growth over 1980-1984: targeting underserved Leisure market – In 1984 entered AA’s core routes: Newark-Chicago and New Orleans-LA

• In January 1985, AA’s counterattack: Ultimate Super Saver

– Discount fare if reserving 2 weeks before departure and staying over a Saturday night. – Restrictions on the number of seats available – Leisure passengers used USS, Business travelers paid full fare.

• September 1985, on the verge of bankruptcy, PeopleExpress’s CEO: “We

had great people, tremendous value, terrific growth. We did a lot of things

• right. But we didn’t get our hands around the yield management and

automation issues.”

34

BRİEF HİSTORY: AİRLİNES

• Airline Reservation Before Computers

35

REVENUE MANAGEMENT: WHEN TO USE IT

• Fixed inventory or capacity that is expensive or impossible to

store

• Inventory/capacity committed to a customer before all demand

is known

• Different customer segments exist

– firm can differentiate and price-discriminate among customers

• Same unit of inventory or capacity can satisfy different

customer segments

36

CUSTOMER SEGMENTATİON

• Leisure Passengers vs. Business Passengers

– Leisure passengers are highly price sensitive – They book earlier – They are more flexible in choosing departure/arrival times – More likely to stay over for Saturday night

• Can further segment within the segment
• Other segmentations:

– Government employees, senior citizens, children, frequent flyers, etc. – International pricing: Same ticket at different prices in different countries. (what is the logic here?) – Distribution channel based segmentation: Travel agent in person vs. Travel agent software/website vs. Airline’s website. – City-of-origin pricing:

• Hotels: Kamaaina rates for Hawaii residents
• Theme Parks: Disney’s resident rates for Orlando residents

37

HOTEL EXAMPLE

• You manage a hotel with 200 rooms

– Your guests either book a room in advance…. – Or, wait till the last minute – So you charge two rates:

• Advance booking: Bargain rate
• Late booking: Premium rate
• Should the hotel fill up rooms with advance

customers? – Why? – How many rooms should be set aside for last- minute customers? – What information do you need?

Capacity: 200 rooms

These X rooms are protected for high paying customers 200-X is the booking limit for low paying customers

Note: allocation is virtual. Room # 168 can serve both classes 38

39

HOTEL EXAMPLE

• Assume

– number of last minute customers is normally distributed with mean 75 and standard deviation 25 – Advance booking: $200/night – Late booking: $500/night

• How many rooms shall be protected?

Cu= Co= P=Cu/(Cu+Co)= Z= QLB= 500-200=300 200 300/500=0.6 0.253 75+0.253*25=81.3 =>82

40

HOTEL EXAMPLE

• How many high-fare travelers will be rejected?

– Expected lost sales = σ·L(0.253)=25·0.285 = 7.13

• How many high-fare travelers will be accommodated?

– Expected sales = Expected demand - Lost sales = 75 – 7.13 = 67.87

• How many rooms will remain empty?

– Expected left over inventory = Q - Expected sales = 82 – 67.87 = 14.13.

• What is the expected revenue?

– $500 · Exp. sales + $200 · Booking limit = $57,535. – Note: without RM:

• all rooms sold in advance = $40,000
• all rooms kept for late arrivals = $37,500
• if 75 rooms kept (z=0): $57,513

OVERBOOKİNG

• You can easily fill up your hotel (200 rooms)
• Some people with reservations may not show up
• How many reservations should you take?

– More than 200? – How many more?

• What information do you need?

– number of no-shows (x) is normally distributed with mean 5 and standard deviation 2, advance room rate is $200. – How many rooms (y) should be overbooked? – Same setup:

• Single decision when the number of no-shows is uncertain.
• Underage cost if x>y (insufficient number of rooms overbooked).
• Overage cost if x<y (too many rooms overbooked).

41

Critical Values and Overbooking Limits

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 100 200 300 400 500 Cost of Compensation Critical Value 2 3 4 5 6 7 8 9 10 11 Overbooking Limit

OVERBOOKİNG: EXAMPLE

Critical Value Overbooking Limit

• The challenge is to estimate the ill-will and penalty cost.
• If no compensation (assume $1 processing fee) => overbook 11 rooms
• If overage cost is $200 => overbook 5 rooms
• If overage cost is $500 => overbook 4 rooms

42

• Nesting

– Assume b{6}=10 and b{5,6}=20 – After selling 10 rooms to class 5:

• b{6}=10 and b{5,6}=10
• So can accept 10 for either class 5 or class 6.
• Without nesting:

– Assume b{6}=10, b{5}=10 – After selling 10 rooms to class 5:

• b{6}=10 and b{5}=0
• So can accept 10 for class 6, but reject

requests for class 5...

b1 b2 b3 b4 b5 b6

BOOKİNG LİMİTS ARE NESTED

43

• How to segment standard rooms?

– Note: the differences in cost do not justify the price differential

b1 b2 b3 b4 b5 b6

Class Price Internet Free Fruit basket Newspaper Water (1 bottle/day) Bed choice: 2Q

• r K

1 $250 Free Y Y Y Y 2 $220 $5/d Y Y Y Y 3 $190 $5/d N Y Y Y 4 $160 $5/d N N Y Y 5 $130 $5/d N N $3/b Y 6 $100 $5/d N N $3/b N

BOOKİNG LİMİTS ARE NESTED

44

BOOKİNG LİMİTS ARE NESTED

b1 b2 b3 bn-1 bn

• Booking limit bi limits bookings for classes j={i,…,n}.
• Protection level yi protects future reservations or classes j={1,2, …, i}.

Booking limits Protection levels

{1,.,n} {1,.,n-1} {1,2,3} {1,2} {1} {n} {n-1,n} {3,.,n} {2,.,n} {1,.,n}

45

BOOKİNG LİMİTS ARE NESTED

b1 b2 b3 bn-1 bn

• Capacity = booking limits + protection levels
• Capacity = bi + yi-1

yn yn-1 yn-2 y2 y1 y0 bn+1 bn bn-1 b3 b2 b1

46

REVENUE MANAGEMENT SUMMARY

47

• Yield management and overbooking give demand

flexibility where supply flexibility is not possible.

• In the model used:

– Single decision in the face of uncertainty. – Underage and overage penalties.

• These are powerful tools to improve revenue:

– American Airlines estimated a benefit of $1.5 billion over 3 years. – National Car Rental faced liquidation in 1993 but improved via yield management techniques. – Delta Airlines credits yield management with $300 million in additional revenue annually (about 2% of year 2000 revenue.)

REVENUE MANAGEMENT CHALLENGES

• Demand forecasting.

– Wealth of information from reservation systems but there is seasonality, special events, changing fares and truncation of demand data.

• Dynamic decisions.
• Variable capacity:

– Different aircrafts, ability to move rental cars around.

• Group reservations.
• How to construct good “fences” to differentiate among customers?

– One-way vs. round-trip tickets. – Saturday-night stay requirement. – Non-refunds. – Advanced purchase requirements.

• Multi-leg passengers/multi-day reservations for cars and hotels:

– Not all customers using a given piece of capacity (a seat on a flight leg, a room for one night) are equally valuable.

48

ADDİTİONAL SLİDES

49

BOOKİNG LİMİTS AND PROTECTİON LEVELS

50

BOOKİNG LİMİTS AND PROTECTİON LEVELS

51

BOOKİNG LİMİTS AND PROTECTİON LEVELS

• The updating algorithm

– n classes – The vector b of booking limits is given – Set vector B to 0 – While Remaining Capacity =b1-B1>0 do

• Suppose m requests for class i: x=[m,…,m,0,…0]

(note, m<0 implies a cancellation)

• If b≥B+x, accept request and set B:=B+x;
• else, reject the request

– End while

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EXAMPLE

• b=(b1, b2, b3, b4, b5)=(100, 73, 12, 4, 0).
• B=(B1, B2, B3, B4, B5)=(0, 0, 0, 0, 0)

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Adapted from Metin’s class notes Request x=(x1, x2, x3, x4, x5) condition Accept? Updating, B= 5 for C2 (5, 5, 0, 0, 0) b2-x2=68≥0 b1-x1=95≥0 Yes (5, 5, 0, 0, 0) 1 for C2 (1, 1, 0, 0, 0) b2-B2-x2=67 ≥ 0 b1-B1-x1=94 ≥ 0 Yes (6, 6, 0, 0, 0) 1 for C4 (1, 1, 1, 1, 0) b4-B4-x4=3≥0, b3-B3-x3=11≥0, b2-B2-x2=66≥0, b1-B1-x1=93≥0 Yes (7, 7, 1, 1, 0) 3 for C3 (3, 3, 3, 0, 0) b3-B3-x3=8≥0 , b2-B2-x2=63≥0, b1-B1-x1=90≥0 Yes (10, 10, 4, 1, 0) 4 for C4 (4, 4, 4, 4, 0) b4-B4-x4=-1<0, b3-B3-x3=4 ≥0, b2-B2-x2=59≥0 b1-B1-x1=86≥0 NO (10, 10, 4, 1, 0) 2 for C1 (2, 5, 0, 0, 0) b-B-x= (88, 63, 8, 3, 0)≥0 Yes (10, 10, 4, 1, 0) +(2, 0, 0, 0, 0)= (12, 10, 4, 1, 0)

EXAMPLE

Request x=(x1, x2, x3, x4, x5) condition Accept? Updating, B= 30 for C2 (30, 30, 0, 0, 0) b-B-x=(58, 33, 8, 3, 0) ≥ 0 Yes (12, 10, 4, 1, 0) +(30, 30, 0, 0, 0) =(42, 40, 4, 1, 0) 20 for C2 (20, 20, 0, 0, 0) b-B-x=(38, 13, 8, 3, 0) ≥ 0 Yes (42, 40, 4, 1, 0) +(20, 20, 0, 0, 0) =(62, 60, 4, 1, 0) 10 for C3 (10, 10, 10, 0, 0) b-B-x=(28, 3, -2, 3, 0) No (62, 60, 4, 1, 0) 6 for C3 (6, 6, 6, 0, 0) b-B-x=(32, 7, 2, 3, 0) ≥ 0 Yes (62, 60, 4, 1, 0) +(6, 6, 6, 0, 0) =(68, 66, 10, 1, 0) 3 for C2 (3, 3, 3, 0, 0) b-B-x=(29, 4, -1, 3, 0) No =(68, 66, 10, 1, 0) 6 for C2 (6, 6, 0, 0, 0) b-B-x=(26, 1, 2, 3, 0) ≥ 0 Yes (68, 66, 10, 1, 0) +(6, 6, 0, 0, 0) =(74, 72, 10, 1, 0) 1 for C1 (1, 1, 1, 0, 0) b-B-x=(25, 0, 1, 3, 0) ≥ 0 Yes (74, 72, 10, 1, 0) +(1, 1, 1, 0, 0) =(75, 73, 11, 1, 0) 54

EXAMPLE

Request x=(x1, x2, x3, x4, x5) condition Accept? Updating, B= 1 for C5 (1, 1, 1, 1, 1) b-B-x=(24, -1, 0, 2, -1) No (75, 73, 11, 1, 0) 1 for C4 (1, 1, 1, 1, 0) b-B-x=(24, -1, 0, 2, 0) No (75, 73, 11, 1, 0) 1 for C3 (1, 1, 1, 0, 0) b-B-x=(24, -1, 0, 3, 0) No (75, 73, 11, 1, 0) 1 for C2 (1, 1, 0, 0, 0) b-B-x=(24, -1, 1, 3, 0) No (75, 73, 11, 1, 0) 25 for C1 (25, 0, 0, 0, 0) b-B-x=(0, 0, 1, 3, 0) ≥ 0 Yes (75, 73, 11, 1, 0) +(25, 0, 0, 0, 0) =(100, 73, 11, 1, 0)

• Note: now we have that b-B=(25, 0, 1, 3, 0)
• Only first class requests are accepted from now on!
• Note: now we have that b-B=(0, 0, 1, 3, 0)
• All classes are closed!
• Recover bookings: 1st class: B1-B2=27; 2nd class: B2-B3=62; 3rd class: B3-B4=10;

4th class: B4-B5=1; 5th class: B5=0.

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EXAMPLE: VİSUALİZATİON

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EXAMPLE: VİSUALİZATİON

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EXAMPLE: VİSUALİZATİON

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EXAMPLE: VİSUALİZATİON

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EXAMPLE: VİSUALİZATİON

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SOME U.S. AİRLİNE İNDUSTRY OBSERVATİONS

• Since deregulation (1979), 137 carriers have filed for
• bankruptcy. (2000 data)
• From 1995 - 1999 (the industry’s best 5 years ever), airlines

earned 3.5 cents on each dollar of sales:

– The US average for all industries is around 6 cents. – From 1990 – 1999, the industry earned 1 cent per $ of sales.

• Carriers typically fill 72.4% of seats and have a break-even load
• f 70.4%. (2000 data)

– The difference between making and losing money is measured by a handful of passengers.

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CHANGİNG FARE STRUCTURES

• Major shifts in airline pricing strategies since 2000

– Growth of low-fare airlines with relatively unrestricted fares – Matching by legacy carriers to protect market share and stimulate demand – Increased consumer use of internet search engines to find lowest available fare options – Greater consumer resistance to complex fare structures and huge differentials between highest and lowest fares offered

• Recent moves to “simplified” fares overlook the fact that

pricing segmentation contributes to revenues:

– Fare simplification removes restrictions, resulting in reduced segmentation of demand

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A SOLUTİON TO THE MULTİ-LEG CUSTOMER: BUCKETS

• With segment control, there are only three

booking limits for the O’Hare-JFK leg, one for each fare class.

• But an O’Hare-Heathrow customer may be more

valuable, so you could have six booking limits,

• ne for each fare-itinerary combination.
• But that leads to many booking limits, so group

fare-itineraries combinations with similar values into buckets:

O’Hare JFK Heathrow

Fare class O'Hare to JFK O'Hare to Heathrow Y $724 $1,610 M $475 $829 Q $275 $525 Bucket Itinerary Fare class O'Hare to Heathrow Y 1 O'Hare to Heathrow M O'Hare to JFK Y 2 O'Hare to Heathrow Q O'Hare to JFK M 3 O'Hare to JFK Q 63

ANOTHER SOLUTİON TO MULTİ-LEGS: BİD PRİCES

• Assign a bid price to each segment:
• A fare is accepted if it exceeds the sum of the bid prices on the segments it uses:

– For example, an O’Hare-JFK fare is accepted if it exceeds $290 – A O’Hare-Heathrow fare is accepted if it exceeds $290+$170 = $460

• The trick is to choose good bid-prices.

O’Hare JFK Heathrow

Fare class O'Hare to JFK O'Hare to Heathrow Y $724 $1,610 M $475 $829 Q $275 $525

O'Hare to JFK JFK to Heathrow Bid price $290 $170

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