Suppose that a queueing system has two servers, an exponential interarrival time distribution with a mean of 2 hours, and an exponential service-time distribution with a mean of 2 hours for each server. Furthermore, a customer has just arrived at 12:00 noon. (a) What is the probability that the next arrival will come (i) before 1:00 P.M., (ii) between 1:00 and 2:00 P.M., and (iii) after 2:00 P.M.?

Respuesta :

Answer:

What is the probability that the next arrival will come (i) before 1:00 P.M = 0.3935

between 1:00 and 2:00 P.M. = 0.2387

after 2:00 P.M= 0.3679

Step-by-step explanation:

Let T be inter arrival time. Tn exponential = λ = 2

Si be service time. i = 1, 2. SIn exponential= ∪ = [tex]\frac{1}{2}[/tex] ,

Arrival of first customer: 12:00

(a) What is the probability that the next arrival will come (i) before 1:00 P.M., (ii) between 1:00 and 2:00 P.M., and (iii) after 2:00 P.M.?

i) P(T < 1) = F(1) = 1 - ε[tex]^-{\frac{1}{2}}[/tex] *[tex]^{1}[/tex] = 0.3935

ii) P(1 < T < 2) = F(2) - F(1) = (1 -ε[tex]-\frac{1}{2} * 2[/tex]) -  (1 -ε[tex]-\frac{1}{2} * 1[/tex])

= 0.2387

iii) P( T > 2) = ε[tex]-\frac{1}{2} * 2[/tex] = 0.3679