Consider a singleline singleserver waiting line system Suppo

Consider a single-line, single-server waiting line system. Suppose that there is only physical space for 2 units in the system (one in line and one being served). The arrival rate is 80 people per hour. The manager, Johnny Three Toes, has the choice among workers of different speeds to be the server. He wants to save money by hiring the slowest server that will result in an average number of units in the system (L) equal to 2.

a) Using Johnny Three Toes’ strategy, what will the server’s service rate per hour () need to be?

Answer: mu = 120

b) What is the probability of having 3 or more units in the system?

Solution

Given,

Arrival rate = a = 80 / hour

Let, Service rate = S people per hour

The average number of units in the system asper queuing theory formula for M/M/1 type queue system

= a^2/( S x ( S – a )) + a/ S

= ( a^2 + a x S - a^2) / (S x ( S -a ) )

= axS/Sx (S -a)

= a/( s-a)

It is mentioned that the average number of units in the system = 2

Therefore,

a/(s-a) = 2

Or, a = 2s – 2a

Or, 2S = 3.a

Or. S = 1.5 a = 1.5 x 80 = 120

SERVER’S SERVICE RATE PER HOUR = 120 PERSONS

Probability of having 3 or more units in the system

= 1 – Probability of less than 3 units in the system

= 1 – ( Probability of ZERO unit in the system + probability of 1 unit in the system + Probability of 2 units in the system )

Probability of Zero unit in the system =Po = 1 – a/ s = 1 – 80/120 = 1 – 0.666 = 0.334

Probability of 1 unit in the system = P1 = ( a/s) x Po = 80/120 x Po = 0.666 x 0.334 = 0.2224

Probability of 2 units in the system = P2 = ( a/s)^2 x Po = 0.666 x 0.666 x 0.334 = 0.1481

Therefore , Po + P1 + P2 = 0.334 + 0.2224 + 0.1481 = 0.7045

Thus,

Probability of having 3 or more units in the system = 1 – 0.7045 = 0.2955

PROBABILITY OF 3 OR MORE UNITS IN THE SYSTEM = 0.2955