There are two airlines with connecting ights from city A to city D. Each connec- tion has a half hour layover. Each airline has connecting ights so that if you miss one connection there are later options. Airline One has a set of ights that leaves at 8:00 and arrives in city D at 20:30, if all goes well. Airline Two has a set of ights that leaves at 8:00 and arrives in City D at 20:00, if all goes well, so Airline Two appears to be the better option. However, you study the airlines and discover that Airline One might be more liable to arrive on time at the intermediate cities, and that it may be more likely to arrive at D on time since the standard deviation is smaller. Flight times are normally distributed.

A(8:00, 4, 0.25) -> B(12:30, 4, 0.25) -> C(17:00, 3.5, 0.25) -> D

B(13:00, 4, 0.25) -> C(17:30, 3.5, 0.25) -> D

C(18:00, 3.5, 0.25) -> D

A(8:00, 3.5, 0.5) -> E(12:00, 4, 0.5) -> F(16:30, 3.5, 0.5) -> D

E(12:30, 4, 0.5) -> F(17:00, 3.5, 0.5) -> D

F(17:30, 3.5, 0.5) -> D

Airline Two Flights

Write a program to simulate the journey from A to D, for each airline, using the data above. Generate a large number of trials for each airline (maybe 10,000 would be enough), compute the arrival time for the rst ight at B or E, and decide which second ight you will be able to take. Compute the arrival time at C or F, decide which third ight you will be able to take, and compute the average arrival time at D for both options. If you arrive at B or E too late to take the second ight, count the trip as having been stranded, and the same for arriving at C or F too late for the third connecting ight. Compare arrival times at city D for the two options. For each option, Airline One and Airline Two, display average arrival time and probability of being stranded. You may assume departing ights are at gates next to arrivals, so that if a ight arrives within one minute of departure for the next time, the second ight may be used. Assume a minimum ight time of - 3, and a maximum time of + 4. If ight times fall out side these limits, change them to the max or min.