A Philadelphia taxi driver gets new customers at rate 1/5 per minute. With proba

Question

A Philadelphia taxi driver gets new customers at rate 1/5 per minute. With probability 1/3 the person wants to go to the airport, a 20 minute trip. After waiting in a line of cabs at the airport for an average of 35 minutes, he gets another fare and spends 20 minutes driving back to drop that person off. Each trip to or from the airport costs the customer a standard $28 charge (ignore tipping). While in the city fares with probability 2/3 want a short trip that lasts an amount of time uniformly distributed on [2, 10] minutes and the cab driver earns an average of $1.33 a minute (i.e., 4/3 of a dollar). (a) In the long run how much money does he make per hour? (b) What fraction of time does he spend going to and from the airport (inducing the time spent in line there)? A Philadelphia taxi driver gets new customers at rate 1/5 per minute. With probability 1/3 the person wants to go to the airport, a 20 minute trip. After waiting in a line of cabs at the airport for an average of 35 minutes, he gets another fare and spends 20 minutes driving back to drop that person off. Each trip to or from the airport costs the customer a standard $28 charge (ignore tipping). While in the city fares with probability 2/3 want a short trip that lasts an amount of time uniformly distributed on [2, 10] minutes and the cab driver earns an average of $1.33 a minute (i.e., 4/3 of a dollar). (a) In the long run how much money does he make per hour? (b) What fraction of time does he spend going to and from the airport (inducing the time spent in line there)?

Explanation / Answer

a)

= (Probability that he gets an airport drop) * ((Total Earnings)/(Total time taken in minutes))

= (1/3) * ((28*2) / (20+20+35) )= =56/225=0.249

Then, Expected money he earns in airport trips per hour=0.249*60=$14.93

=(Probability that he gets a within city fare) * (Avg. money he earns in within city fares)

= (2/3) * 1.33 = 0.886

Expected money he earns in within city trips per hour = 0.886*60 = 53.2

b)

Fraction of time spent going to and from the airport (excluding the time spent in line there)

=((Probability of airport trip) * (Time taken to and fro airport))/

((Probability of airport trip) * (Avg. time taken in airport trips))+((Probability of city trip) *(Avg. time taken in city trips))

= (1/3)*(20+20)/(((1/3)*(20+35+20))+((2/3)*((2+10)/2))))

=40/(12+20+35+20)=40/87=0.46

Hence, he spends 46% of time in a day driving to and from the airport

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