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.)

Case 1:  Let us assume 1 car takes AMB route, 3 cars take ANB route
Then travel time of AMB will be A-M + M-B = 6 + 20 = 26
Then travel time of ANB will be A-N + N-B = (20+2) + (6+3*2) = 34

Now, one car(A) travelling on ANB broke the rule and decided to move on AMB route. then

Case 2: Let us assume 2 cars take AMB route, 2 cars take ANB route
Then travel time of AMB will be A-M + M-B =(6+3) + (20+0.9) = 29.9
Then travel time of ANB will be A-N + N-B = (20+1) + (6+3) = 30

Since the car A reduced its time from 34 to 29.9, the case-1 route is not optimal, hence case 1 is invalid.
Now, one car(B) travelling on ANB broke the rule and decided to move on AMB route. then, case

Case 3:  Let us assume 3 cars take AMB route, 1 car take ANB route

Then travel time of AMB will be A-M + M-B = (6+3*2) + (20+0.9*2) = 33.8
Then travel time of ANB will be A-N + N-B = (20) + (6) = 29

Now, one car(C) travelling on AMB broke the rule and decided to move on AMNB route. then
Case 4:  Let us assume 2 cars take AMB route, 1 car take AMNB route and other take ANB route. 

Then travel time of AMB will be A-M + M-B = (6+3*2) + (20+0.9) = 32.9
Then travel time of AMNB will be A-M + M-N + N-B = (6+3*2) + (7) + (6+3) = 28
Then travel time of ANB will be A-N + N-B = (20) + (6+3) = 29

Since this car(C) reduced its time from 33.8 to 28, the route is not optimal, hence case 3 is invalid.
Since this car(B) reduced its time from 30 to 28, the route is not optimal, hence case 2 is invalid.
Now, one car(D) travelling on AMB broke the rule and decided to move on AMNB route. then

Case 5:   Let us assume 1 car takes AMB route, 2 cars take AMNB route and other takes ANB route.

Then, the portion A-M will be travelled by 3 cars, M-B by one car, M-N by 2 cars, A-N by 1 car and N-B by 3 cars. 

Then, travel time of AMB will be A-M + M-B = (6+3*2) + (20) = 32
Then, travel time of AMNB will be A-M + M-N + N-B = (6+3*2) + (7+1) + (6+3*2) = 32
Then, travel time of ANB will be A-N + N-B = (20) + (6+3*2) = 32

It is clear that the car D reduced its time from 32.9 min to 32 min if it broke the rule. Hence, case 4 is invalid.
In this arrangement of case 5, no car can improve their travel time by changing their path
Hence, the optimal allocation will be to order 2 cars on A-M-N-B route.