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OPSM 405 Service Management

Ko ç Un iversity. OPSM 405 Service Management. Class 16: Yield management: overbooking. Zeynep Aksin zaksin @ku.edu.tr. Reservation System. Forecasting. Overbooking Levels. Discount Allocation. Yield Management System. current demand cancellations. cancellation rate estimates. future

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OPSM 405 Service Management

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  1. Koç University OPSM 405 Service Management Class 16: Yield management: overbooking Zeynep Aksin zaksin@ku.edu.tr

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

  3. Dealing with cancellationsOverbooking control total cost Basic Problem: E[Rev] $ opportunity cost of unsold seats overbooking costs 0 capacity overbooking limit (BL) #seats sold

  4. Overbooking Two basic costs: • Stock outs 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 0 5 5 1 10 15 2 20 35 3 15 50 4 15 65 5 10 75 6 5 80 7 5 85 8 5 90 9 5 95 10 5 100

  7. Overbooking Approach 1: Using Averages In Table 9.1 the average number of no-shows is calculated by 0x0.05 + 1x0.10 + 2x0.20 + 3x0.15 +…+ 10x0.05 = 4.05. Take up to four overbookings.

  8. Overbooking Approach 2: Spreadsheet Analysis

  9. 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 • Cs = $120, Co = $50.00 • Co/(Cs + Co) = 29.% • Overbook 2 rooms • Table 9.1: Hotel California No-Show Experience • No-Shows % of Experiences Cumulative % of • Experiences • 0 5 5 • 1 10 15 • 2 20 35 29%

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

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

  13. 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”

  14. Example: Bulvar Palas Bulvar Palas No-Show Experience: (Daily) No-shows Probability P[no show] Cum. Prob. P[no show<x] 0 0.07 0.00 1 0.19 0.07 2 0.22 0.26 3 0.16 0.48 4 0.12 0.64 5 0.10 0.76 6 0.07 0.86 7 0.04 0.93 8 0.02 0.97 9 0.01 0.99

  15. 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

  16. 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?

  17. 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

  18. Spreadsheet approach

  19. 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

  20. Likya World • 105/(300+105)=0.2592 • Look up in a standard normal table to obtain z=-0.645 • So number of seats to overbook= 20-0.645*10=13.5 • Alternatively use NORMINV(0.2592,20,10)

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

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