Answer:
a) 0.36
b) 0.3
c) Yes
Step-by-step explanation:
Given:
Probability of no traffic delay in one period, given no traffic delay in the preceding period = P(No_Delay) = 0.9
Probability of finding a traffic delay in one period, given a delay in the preceding period = P(Delay) = 0.6
Period considered = 30 minutes
a)
Let A be the probability that for the next 60 minutes (two time periods) the system will be in the delay state:
As the Probability of finding a traffic delay in one period, given a delay in the preceding period is 0.6 and one period is considered as 30 minutes.
So probability that for the next two time periods i.e. 30*2 = 60 minutes, the system in Delay is
P(A) = P(Delay) * P(Delay) = 0.6 * 0.6 = 0.36
b)
Let B be the probability that in the long run the traffic will not be in the delay state.
This statement means that the traffic will not be in Delay state but be in No_Delay state in long run.
Let C be the probability of one period in Delay state given that preceding period in No-delay state
:
P(C) = 1 - P(No_Delay)
= 1 - 0.9
P(C) = 0.1
Now using P(C) and P(Delay) we can compute P(B) as:
P(B) = 1 - (P(Delay) + P(C))
= 1 - ( 0.6 + 0.10 )
= 1 - 0.7
P(B) = 0.3
c)
Yes this assumption should be questioned for this traffic problem because it implies that traffic will be in Delay state for the 30 minutes and just after 30 minutes, it will be in No_Delay state. However, traffic does not work like this in general and it makes this scenario unrealistic. Markov process model can be improved if probabilities are modeled as a function of time instead of being presented as constant (for 30 mins).
