Problem No. 1. At a local crisis center, three volunteers answer the phones to h

Question

Problem No. 1. At a local crisis center, three volunteers answer the phones to help solve residents' problems. Calls arrive randomly, at an average rate of 20 per eight hour day. The amount of time that a volunteer spends with a caller is exponentially distributed with and average of 40 minutes. If all the volunteers are busy, the callers are put on hold and they are assigned in the order of their arrival. a. Determine the fraction of time each volunteer is busy. b. On the average, how long is a caller put on hold ? c. On the average, how many callers are on hold ? d. How much time, on the average, a caller spends on phone ?

Explanation / Answer

Arrival rate, = 20 per 8-hr day = 20/8 = 2.5 per hour

Service rate, = (1/service time in hour) = 1/(40/60) = 1.5 per hour

Number of servers, s = 3 volunteers

a) Fraction of time each volunteer is busy = /(s) = 2.5/(3*1.5) = 0.556

b) This is a M/M/s queue model, with number of servers, s = 3

For s=0, (/)^s/s! = (2.5/1.5)^0/0! = 1

For s=1, (/)^s/s! = (2.5/1.5)^1/1! = 1.6667

For s=2, (/)^s/s! = (2.5/1.5)^2/2! = 1.3889

For s=3, (/)^s/s! = (2.5/1.5)^3/3! = 0.7716

(/)^s/s! (for s=0 to 2) = 1+1.6667+1.3889 = 4.0556

term 2 = ((/)^3/3!)/(1-/3) = 0.7716/(1-2.5/(3*1.5)) = 1.7361

P0 = 1/((/)^s/s!+term 2) = 1/(4.0556+1.7361) = 0.1727

Average time a caller is put on hold, Wq = P0*((/)^3/3!)*(/s)/(1-(/s))^2/ = 0.1727*0.7716*0.5556/(1-0.5556)^2/2.5 = 0.15 hours = 9 minutes

c) Average number of callers put on hold, Lq = Wq* = 0.15*2.5 = 0.375

d) Average time a caller spends on phone, W = Wq+1/ = 0.15+1/1.5 = 0.817 hours or 49 minutes

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.