There are 10 vehicles in a queue when an attendant opens a toll booth. Vehicles arrive at the booth at a rate of 4 per minute. The attendant opens the booth and improves the service rate over time following the function u (t) = 1.1 +.30t, where u (t) is in vehicles per minute and t is in minutes. I need to find the maximum queue length. I included my solution below for the total delay.

u (t) = 4 when t = (4-1.1) / 0.3 = 9.67mins

(9.67 to t) ? (1.1+0.3t) dt = 10

? 1.1t + 0.15t2 ] (9.67 to t) = 10

? t = 11.97mins

So, delay = 11.97 - 9.67 = 2.30mins

Question

Engineering

Iyana Michael

13 March, 05:15

There are 10 vehicles in a queue when an attendant opens a toll booth. Vehicles arrive at the booth at a rate of 4 per minute. The attendant opens the booth and improves the service rate over time following the function u (t) = 1.1 +.30t, where u (t) is in vehicles per minute and t is in minutes. I need to find the maximum queue length. I included my solution below for the total delay.

u (t) = 4 when t = (4-1.1) / 0.3 = 9.67mins

(9.67 to t) ? (1.1+0.3t) dt = 10

? 1.1t + 0.15t2 ] (9.67 to t) = 10

? t = 11.97mins

So, delay = 11.97 - 9.67 = 2.30mins

+3

Answers (1)

Know the Answer?

Itzel Middleton · Answer 1

as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.

Explanation:

4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum