A new one-way road is built from M to N. Each car now has three possible routes to travel from A to B: A-M-B, A-N-B and A-M-N-B. On the road from M to N, one car takes 7 minutes and each additional car increases the travel time per car by 1 minute. Assume that any car taking the A-M-N-B route travels the A-M portion at the same time as other cars taking the A-M-B route, and the N-B portion at the same time as other cars taking the A-N-B route.
How many cars would the police department order to take the A-M-N-B route so that it is not possible for any car to reduce its travel time by not following the order while the other cars follow the order? (Assume that the police department would never order all the cars to take the same route).___
A new one-way road is built from M to N. Each car now has three possible routes to travel from A to B: A-M-B, A-N-B and A-M-N-B. On the road from M to N, one car takes 7 minutes and each additional car increases the travel time per car by 1 minute. Assume that any car taking the A-M-N-B route travels the A-M portion at the same time as other cars taking the A-M-B route, and the N-B portion at the same time as other cars taking the A-N-B route.
How many cars would the police department order to take the A-M-N-B route so that it is not possible for any car to reduce its travel time by not following the order while the other cars follow the order? (Assume that the police department would never order all the cars to take the same route).
Following are the scenarios:
For a car to not be able to reduce its travel time by not following the order, all the cars cannot take the same route. So either two or three cars should go through A-M. If two cars go through M-B, one car can break the police order and go through MN and reach B in 9 + 7 + 12 = 28 minutes as compared to 29.9 minutes had both gone through A-M-B. If two cars go through A-M and one is directed to go through M-N, one of the cars which was directed to go through A-N can break the police order and go through A-M-B and save time as follows: Original time (A-N-B) = 21 + 12 = (three cars) = 33
New time = 12 (3 cars) + 20 .9 = 32.9
The police department cannot direct both cars to go through M-N as in that case all four cars would go through N-B.
In case three cars are directed to go through A-M, either one car can be directed through M-N or two cars can be directed through M-N.
If one car is directed through M-N, one of the two cars directed through M-B, can break the police order and go through M-N, and save time as shown.
Original time (A-M-B) = 12 (3 cars) + 20.9 = 32.9
New time (A-M-N-B) = 12 + 8 + 12 = 32 minutes.
Hence, two cars must be directed through M-N such that any car breaking the police order cannot reduce the travel time.
Following are the scenarios:
For a car to not be able to reduce its travel time by not following the order, all the cars cannot take the same route. So either two or three cars should go through A-M. If two cars go through M-B, one car can break the police order and go through MN and reach B in 9 + 7 + 12 = 28 minutes as compared to 29.9 minutes had both gone through A-M-B. If two cars go through A-M and one is directed to go through M-N, one of the cars which was directed to go through A-N can break the police order and go through A-M-B and save time as follows: Original time (A-N-B) = 21 + 12 = (three cars) = 33
New time = 12 (3 cars) + 20 .9 = 32.9
The police department cannot direct both cars to go through M-N as in that case all four cars would go through N-B.
In case three cars are directed to go through A-M, either one car can be directed through M-N or two cars can be directed through M-N.
If one car is directed through M-N, one of the two cars directed through M-B, can break the police order and go through M-N, and save time as shown.
Original time (A-M-B) = 12 (3 cars) + 20.9 = 32.9
New time (A-M-N-B) = 12 + 8 + 12 = 32 minutes.
Hence, two cars must be directed through M-N such that any car breaking the police order cannot reduce the travel time.
