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>U</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
