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Tough Apps: Travelling Games


1. If x >= 3 and y <= 4, then setting the price at $300 is best for the hotels, though only x people will get rooms. Otherwise, if x + y >= 3, then charge $200 to get $800 for the two hotels. Otherwise, set the hotel prices to $100.

2. No. You capture no more customers than if you set the price to $300 and receive less revenue.

3. We already saw in the Warm-Ups that you would charge $100 if x = y = z = 2 to guarantee at least $300 in revenue. Any other price would give a lesser guarantee (though the possibility of more). If x >= 5, then each hotel may as well charge $300 as each will get at least two customers at that price. If x <= 4 and x + y >= 5, then $200 is the best price (because charging $300 would guarantee you only one customer and charging $200 gurantees at least two).

4. There are no x, y, z for which it is worthwhile to charge something between $200 and $300 unless you know the prices of the other. If, on the other hand, you knew the other hotel was charging $300 and x = 3 and y = 5, then you could charge say $280 and get all the $300 customers.

5. If x >= 3, then a price of $300 will guaranteee $900. If x + y >= 2, then $200 will guarantee $600. Your competitor can't charge even $200 unless x + y >= 5.

6. With monopoly power, you know all customers will come to you, so you can set the price through direct calculation. In a competitive market, you have to assume the worst case. If you can convince customers to come to you first, then you can charge the most per room.

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