1. OPSM 405 Service Management
Koç University
OPSM 405 Service Management
Class 16:
Yield management: overbooking
Zeynep Aksin

2. Yield Management System
Reservation System
current demand
cancellations
Forecasting
cancellation rate estimates
future
demand
estimates
Overbooking Levels
overbooking levels
Discount Allocation
fare class allocations

3. Dealing with cancellations Overbooking control
total cost
Basic Problem:
E[Rev] $
opportunity cost of unsold seats
overbooking costs
capacity
overbooking
limit (BL)
#seats sold

4. Overbooking Two basic costs: Stock outs Overage
customers have a reservation and there are no rooms left
Overage
customers denied advance reservation and rooms are unoccupied

5. Example: Hotel California
Stock outs: 0.8 x $150 = $120
Overage: $50

6. Table 9.1: Hotel California No-Show Experience
No-Shows % of Experiences Cumulative % of
Experiences

7. Overbooking Approach 1: Using Averages
In Table 9.1 the average number of no-shows is calculated by 0x x x x0.15 +…+ 10x0.05 = 4.05.
Take up to four overbookings.

8. Overbooking Approach 2: Spreadsheet Analysis

9. Book more guests until:
Overbooking Approach 3: Marginal Cost Approach
Book more guests until:
E(cost of dissatisfied customer) = E(cost of empty room)
Cost of dissatisfied customer * Probability that there are fewer no-shows than overbooked rooms =
Cost of empty room * Probability that there are more no-shows than overbooked rooms

10. Hotel California Co/(Cs + Co) = P(Overbook  No Shows) Hotel Data 29%
Overbook 2 rooms
Table 9.1: Hotel California No-Show Experience
No-Shows % of Experiences Cumulative % of
Experiences
29%

11. Overbooking: Marginal analysis
n-th cust shows up?
+$r p
yes no
1-p
Seat available for
n-th cust.? $0
yes
Accept n-th request? no
n-th cust shows up?
yes p
-$s
yes no
1-p no $0 $0
N(n,p) is binomial r.v. with parameters n,p

12. Accept if ...

13. Example

15. Dynamic Overbooking Overbooking Time to Event Event Occurs
Reservations Start

16. Overbooking over time %Capacity 100% 90 days to departure o
booking limit
reservations with overbooking
100%
reservations without overbooking 90
days to departure o

17. Bulvar Palas The contribution of each room is 40YTL per night.
If a guest holding a reservation is turned away owing to overbooking, then other costs are incurred:
Arrangements with a nearby hotel
Penalties associated with lost good will
Management estimates this cost as 100YTL per guest “walked”

18. Bulvar Palas No-Show Experience: (Daily)
Example: Bulvar Palas
Bulvar Palas No-Show Experience: (Daily)
No-shows Probability P[no show] Cum. Prob. P[no show<x]

19. Marginal analysis How much can I overbook?
Overbook too few 40YTL, P(no show>x)
Overbook too many 100YTL, P(no show<x)
Keep overbooking as long as
40*P(no show>x) > 100*P(no show<x) or
P(no show<x) < 40/(40+100)=0.286
Overbook 2 rooms based on no-show distribution

20. Example
The Ozhas bus company is currently assessing its Istanbul-Adana run. The number of customers that do not show up after making a reservation are uniformly distributed from 1 to 10. Tickets costs are 45YTL, and if a particular bus run is full, a passenger with a reservation is given passage on a rival company’s bus at a cost of 75YTL. Using the averages method, what should Ozhas’s overbooking policy be?

21. Averages method
Using the averages method, the average number of no shows is calculated by:
0(0.0)+1(0.1)+2(0.1)+3(0.1)+4(0.1)+5(0.1)+6(0.1)+7(0.1)+8(0.1)+9(0.1)+10(0.1)
= 5.5

22. Spreadsheet approach No shows Probability 1 2 3 4 5 6 7 $0 ($75)
$45
Cs =
$75
Number of Reservations Overbooked
No shows
Probability 1 2 3 4 5 6 7 $0
($75)
($150)
($225)
($300)
($375)
($450)
($525)
0.1
($45)
($90)
($135)
($180)
($270)
($315) 8
($360) 9
($405) 10
Total cost
($248)
($203)
($170)
($149)
($140)
($143)
($158)
($185)

23. Likya World
Number of customers who book a night and fail to show up is Normally distributed with mean 20 and standard deviation 10
Bumping a customer costs 300 YTL
If room is not sold, hotel loses revenue of 105 YTL

24. Likya World
105/( )=0.2592
Look up in a standard normal table to obtain z=-0.645
So number of seats to overbook=
*10=13.5
Alternatively use NORMINV(0.2592,20,10)

25. Obtaining the Probability
Standardized Normal Probability Table (Portion) Z
.00
.01
.02 s
= 1
0.0
.50000
.50399
.50798 Z : : : :
2.0
.97725
.97784
.97831
.97725 m
= 0
2.0 Z
2.1
.98214
.98257
.98300 z
Probabilities in body