311_Session18_RevenueManagement

311_Session18_RevenueManagement - Operationsmanagement

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  Operations management Session 18:  Revenue Management Tools
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Operations Management 2 RM: A Basic Business Need What are the basic ways to improve profits? Profits Profits $ Reducing Cost Increasing Revenue Revenue Management
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 Session 18 Operations Management 3 Elements of Revenue  Management Pricing and market segmentation Capacity control Overbooking Forecasting  Optimization
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 Session 18 Operations Management 4 Pricing: How does it work? Objective:  Maximize revenue Example (Monopoly):  An airline has the following demand  information: Price Demand 0 ? 50 150 100 120 150 90 200 60 250 30 0 20 40 60 80 100 120 140 160 0 50 100 150 200 250 300 price d = (3/5)(300-p)
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 Session 18 Operations Management 5 Pricing: How does it work? What is the price that the airline should charge to  maximize revenue?  Note that this is equivalent  to determining how many seats the airline should  sell. The revenue depends on price, and is:   Revenue = price * (demand at that price) r(p) = p * d(p)        = p * (3/5) * (300 – p)        = (3/5) * (300p – p 2 ) We would like to choose the price that maximizes revenue.
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 Session 18 Operations Management 6 Finding the price that maximizes  revenue. 0 2000 4000 6000 8000 10000 12000 14000 16000 50 100 150 200 250 price Revenue is maximized when the price per seat is $150, meaning 90 seats are sold.
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 Session 18 Operations Management 7 Finding the price that maximizes  revenue. r(p) = p*d(p) = (3/5)*(300p-p 2 ) r’(p)=0 implies (3/5)(300-2p)=0 or p=150 Pricing each seat at $150 maximizes revenue. d(150)=(3/5)*(300-150)=90 This means we will sell 90 seats.
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 Session 18 Operations Management 8 What if the airline only holds 60  people? Then, r(d) = p(d)*d = 300d-(5/3)d 2 . First note that actually, revenue = price * min(demand, capacity). Second note that it is equivalent to think in terms of price or demand; i.e., d(p) = (3/5)*(300-p) implies p(d) = 300-(5/3)d. Is it possible we would want to sell less than 60 seats? To answer this question, plot revenue as a function of demand.
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 Session 18 Operations Management 9 Capacity Limit R(D) = p(D)*D = 300D-(5/3)D 2 . 0 2000 4000 6000 8000 10000 12000 14000 16000 0 20 40 60 80 100 120 140 160 demand It is obvious from the graph that revenue is maximized when 90 seats are sold (demand is 90), as we found originally. It is also clear that revenue is increasing from 0 to 90. We should sell as many as possible when capacity is less than 90. Conclusion: sell 60 seats at price p(60)=300-(5/3)*60=200.
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 Session 18 Operations Management 10 Pricing to Maximize Revenue:   The General Strategy Write revenue as a function of price. Find the price that maximizes the revenue function. Find the demand associated with that price.
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311_Session18_RevenueManagement - Operationsmanagement

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