THE NEWSVENDOR AND APPLICATIONS 2 T HE N EWSVENDOR M ODEL 3 ON - - PowerPoint PPT Presentation
N EWSVENDOR & R EVENUE M ANAGEMENT P ROFESSOR D AVID G ILLEN (U NIVERSITY OF B RITISH C OLUMBIA ) & P ROFESSOR B ENNY M ANTIN (U NIVERSITY OF W ATERLOO ) Istanbul Technical University Logistic Management in Air Transport Air
NEWSVENDOR & REVENUE MANAGEMENT
PROFESSOR DAVID GILLEN (UNIVERSITY OF BRITISH COLUMBIA) & PROFESSOR BENNY MANTIN (UNIVERSITY OF WATERLOO)
Logistic Management in Air Transport Module 7-8 17 December 2015 Istanbul Technical University Air Transportation Management M.Sc. Program
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Nov Dec Jan Feb Mar Apr May Jun Jul Aug Generate forecast
submit an order to TEC Receive order from TEC at the end of the month Spring selling season Left over units are discounted
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c = $110 per suit
– This is also called the salvage value.
– Order too much and inventory is left over at the end of the season – Order too little and sales are lost.
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gamma, Poisson distributions
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50 100 150 200 Probability Demand NORMAL GAMMA 1 2 3 4 5 6 7 8 9 10 Probability Demand POISSON
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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
from past observations.
Forecast demand Actual ratio A/F
Forecast ratio) A/F (Expected demand actual Expected Forecast ratios) A/F
deviation (Standard demand actual
deviation Standard
N i i
x
1 2
) ( N 1
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standard deviation 1181 to represent demand for the Hammer 3/2 during the Spring season.
3192 3200 9975 . demand actual Expected 1181 3200 369 . demand actual
deviation Standard
Product description Forecast Actual demand Error A/F Ratio JR ZEN FL 3/2 90 140
1.5556 EPIC 5/3 W/HD 120 83 37 0.6917 JR ZEN 3/2 140 143
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
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
The probability the outcome will be a particular value or smaller
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that has mean = 0 and standard deviation = 1.
– 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)
Q z Q z
50 100 150 200 Probability Demand
Q
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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
normal will be z = 0.28 or smaller?
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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
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– our demand model is a normal distribution with mean 3192 and standard deviation 1181
c = $110 per suit
– This is also called the salvage value.
– What are the costs associated with too much/too little?
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– 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
– 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
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– Expected loss on the Qth unit = Co · F(Q)
the chance of underage: – Expected gain on the Qth unit = Cu · (1-F(Q))
the expected benefit from
while the expected loss of
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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
equals the expected gain on the Qth unit:
satisfy all demand (i.e., demand is Q or lower) equals the critical ratio.
Q F C Q F C
u
1 ) (
u
C C C Q F ) ( 80 . 80 20 80
u
C C C
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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
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
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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
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– If Q = 3000, then to satisfy all demand, demand must be ≤ 3000.
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
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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
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not satisfied:
– Stockout probability = 1 – F(Q) = 1 –0.4364 = 56.36%
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!
the probability all demand is satisfied (the in-stock) need not be 99%
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{0,10,20,…,190,200} with probabilities as per graph
demand, D:
– If D ≤ Q, lost sales = 0 – If D = 130, lost sales = D – Q = 10
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
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500 1000 1500 2000 2500 3000 3500 1000 2000 3000 4000 5000 Expected Lost Sales Order quantity
– Find its z-statistic. – Look up in the Standard Normal Loss Function: L(-0.16)=0.4840
=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
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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
left Expected value Salvage
Sales Expected Cost
Profit Expected
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– There is a single order/production/replenishment opportunity. – Demand is uncertain. – There is a “too much” vs. “too little” challenge
profit.
stock probability or fill rate.
u
C C C Q F ) (
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With material partially adapted from Netessine and Shumsky; teaching note on Yield Management; Metin’s class notes; Phillips (2005); Boyd (2007) 30
– technique to maximize revenue by matching fixed supply with uncertain demand – A very successful application of the critical ratio
– Airlines:
– Hotels:
– car rental:
– commercials
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– Southwest – THE Low-Fare Airline
– Day-to-day adjusting of prices to address demand realization and updating of expectations
– A technique to maximize revenue by matching fixed supply with uncertain demand
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industry in the US was
published and controlled by Civil Aeronautics Board
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– 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
– 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.
had great people, tremendous value, terrific growth. We did a lot of things
automation issues.”
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store
is known
– firm can differentiate and price-discriminate among customers
customer segments
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– 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
– 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:
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– Your guests either book a room in advance…. – Or, wait till the last minute – So you charge two rates:
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
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– number of last minute customers is normally distributed with mean 75 and standard deviation 25 – Advance booking: $200/night – Late booking: $500/night
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
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– $500 · E[high fare sales] + $200 · Booking limit
– Expected sales = Expected demand - Lost sales
– Expected lost sales = σ·L(z)
– L(z)=f(z)-z ·(1- F(z)) =Normdist(z,0,1,0)-z·(1-Normsdist(z))
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– Expected lost sales = σ·L(0.253)=25·0.285 = 7.13
– Expected sales = Expected demand - Lost sales = 75 – 7.13 = 67.87
– Expected left over inventory = Q - Expected sales = 82 – 67.87 = 14.13.
– $500 · Exp. sales + $200 · Booking limit = $57,535. – Note: without RM:
Excel
– More than 200? – How many more?
– 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:
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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
Critical Value Overbooking Limit
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we can charge two prices. That is, in the calculation of Cu we use the conservative value of $200.
rooms for late arrivals.
Assume compensation is $350, then we can overbook 5 rooms.
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– “Accept all Y-class passengers as long as you have an empty seat, but accept no more than 70 Q-class passengers”
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many seats to reserve for Y-class
# of Y-class Passengers … 29 30 31 32 33 34 35 36 … Prob. … 51% 50% 49% 47% 44% 40% 35% 30% … Expected marginal revenue … $255 $250 $245 $235 $220 $200 $175 $150 …
a University of Waterloo graduate.
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– Assume b{6}=10 and b{5,6}=20 – After selling 10 rooms to class 5:
– Assume b{6}=10, b{5}=10 – After selling 10 rooms to class 5:
requests for class 5...
b1 b2 b3 b4 b5 b6
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– 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
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
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b1 b2 b3 bn-1 bn
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}
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b1 b2 b3 bn-1 bn
yn yn-1 yn-2 y2 y1 y0 bn+1 bn bn-1 b3 b2 b1
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flexibility where supply flexibility is not possible.
– Single decision in the face of uncertainty. – Underage and overage penalties.
– 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.)
– seasonality, special events, changing fares and truncation of demand data.
– Different aircrafts, ability to move rental cars around.
– One-way vs. round-trip tickets. – Saturday-night stay requirement. – Non-refunds. – Advanced purchase requirements.
– Not all customers using a given piece of capacity (a seat on a flight leg, a room for one night) are equally valuable.
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– Full fare: $1,000 – Discount: $400 – Deep discount: $100
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– Full fare: $1,000 – 30-50 seats – Discount: $400 – 70-100 seats – Deep discount: $100 – 100-150 seats
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Nested!
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Full fare: Discount: Deep discount not necessary: can always sell all seats
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$200 · (# of stranded)2
two people costs $800, three people, $1,800, etc.
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determined:
– Full fare: $1100, $1000, and $900 – Discount: $440, $400, and $360 – Reflecting +/- 15% deviations in demand.
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50 100 150 200 250 300 350 400 450 500 10 20 30 40 50 60 70 80 90 Airfare ($) Days out ATL-LAS BOI-DEN BUF-RSW CVG-LGA
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– n classes – The vector b of booking limits is given – Set vector B to 0 – While Remaining Capacity =b1-B1>0 do
(note, m<0 implies a cancellation)
– End while
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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)
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) 66
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)
4th class: B4-B5=1; 5th class: B5=0.
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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.
– The difference between making and losing money is measured by a handful of passengers.
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– 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
pricing segmentation contributes to revenues:
– Fare simplification removes restrictions, resulting in reduced segmentation of demand
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A SOLUTION TO THE MULTI-LEG CUSTOMER: BUCKETS
booking limits for the O’Hare-JFK leg, one for each fare class.
valuable, so you could have six booking limits,
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 75
ANOTHER SOLUTION TO MULTI-LEGS: BID PRICES
– 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
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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