Question

In order to reduce the waiting time further, a traffic light was installed to replace the yield sign. Assuming the departure process after the light was installed remained uniform (deterministic). It was found that the average waiting time in queue after the traffic light was installed was 10 sec/veh. What is the average departure rate (in vph) from the traffic light if the average arrival rate remains the same?

Answer 1

Answer:

Explanation:

(a). from the question we have that the rate of arrival and departure is given thus,

departure rate D  = 250 vph

arrival rate R = 200 vph

therefore utilization rate U becomes = arrival rate / departure rate = 200/250

U = 0.8

we are asked to compute the average waiting time in queue which becomes

average waiting time = R/D(D-R)

i.e. average waiting time = 200/250(250-200) = 0.016 hours/veh =

= 57.6 sec/veh

(ii) the average time spent in the system;

= 1/D-R = 1/250-200 = 0.02hrs/veh = 72 sec/veh.

(iii). the average queue length at this stop sign becomes;

R²/D(D-R) = 200²/250(250-200) = 3.2

(b). the average waiting time in queue is to be calculated, and it is given as

= U /2D (1-U) = 0.8/2*250(1-0.8) = 28.8 sec/veh

(ii). the average time spent in the system is given thus as;

average time= 2-U /2D(1-U) = 2 - 0.8 / 2*250(1-0.8) = 43.2 sec/veh

(iii). average queue length at this stop sign

average length =  U²/ 2(1-U) = 0.8²/2(1-0.8) = 1.6

(c). to reduce the waiting time further, a traffic light was installed to replace the yield sign, given that the average waiting time in queue after the traffic light was installed was  found to be 8 sec/veh.

therefore the average waiting time to stay in the queue is;

8 sec/veh = 0.0022 hr/veh

we already know that U = R/D

given that R = 200 vph

D = ?

from the equation above

U = 200/ D .........(2)

substituting value gives

200/D = 0.0044 D(D-200/D)

D = 333.5 veh/hr

the departure rate gives 333.5 vph

cheers, i hope this helps