Question

A large insurance company maintains a central computing system that contains a variety of information about customer accounts. Insurance agents in a six-state area use telephone lines to access the customer information database. Currently, the company's central computer system allows three users to access the central computer simultaneously. Agents who attempt to use the system when it is full are denied access; no waiting is allowed. Management realizes that with its expanding business, more requests will be made to the central information system. Being denied access to the system is inefficient as well as annoying for agents. Access requests follow a Poisson probability distribution, with a mean of 40 calls per hour. The service rate per line is 18 calls per hour.
Reqiured:
a. What is the probability that 0, 1, 2, and 3 access lines will be in use?
b. What is the probability that an agent will be denied access to the system?
c. What is the average number of acess lines in use?

Answer 1

Answer:

Part A:

$P_0=0.13298\\P_1=0.29547\\P_2=0.328320\\P_3=0.24321$

Part B:

$P_3=0.24321$

Part C:

$Average\ Number:=1.6817$

Step-by-step explanation:

Given data:

Mean =μ= 40 calls

Service Rate=λ=18 calls

Solution:

Formula:

$P_i=\frac{\frac {(\frac{\mu}{\lambda}) ^{i} }{i!} }{\sum \frac {(\frac{\mu}{\lambda}) ^{i} }{i!}}$

Where i is access lines

Part A:

Probability of 0 access line, i=0:

$P_0=\frac{\frac {(\frac{40}{18}) ^{0} }{0!} }{\frac {(\frac{40}{18})^{0} }{0!}+\frac {(\frac{40}{18})^{1} }{1!}+\frac {(\frac{40}{18})^{2} }{2!}+\frac {(\frac{40}{18})^{3} }{3!}}$

$P_0=\frac{1}{1+2.222+2.4691+1.8290}$

$P_0=0.13298$

Probability of 1 access line, i=1:

$P_1=\frac{\frac {(\frac{40}{18}) ^{1} }{1!} }{\frac {(\frac{40}{18})^{0} }{0!}+\frac {(\frac{40}{18})^{1} }{1!}+\frac {(\frac{40}{18})^{2} }{2!}+\frac {(\frac{40}{18})^{3} }{3!}}$

$P_1=\frac{2.222}{1+2.222+2.4691+1.8290}\\P_1=0.29547$

Probability of 2 access line, i=2:

$P_2=\frac{\frac {(\frac{40}{18}) ^{2} }{2!} }{\frac {(\frac{40}{18})^{0} }{0!}+\frac {(\frac{40}{18})^{1} }{1!}+\frac {(\frac{40}{18})^{2} }{2!}+\frac {(\frac{40}{18})^{3} }{3!}}$

$P_1=\frac{2.4691}{1+2.222+2.4691+1.8290}\\P_2=0.328320$

Probability of 3 access line, i=3:

$P_3=\frac{\frac {(\frac{40}{18}) ^{3} }{3!} }{\frac {(\frac{40}{18})^{0} }{0!}+\frac {(\frac{40}{18})^{1} }{1!}+\frac {(\frac{40}{18})^{2} }{2!}+\frac {(\frac{40}{18})^{3} }{3!}}\\P_3=\frac{1.8290}{1+2.222+2.4691+1.8290}\\P_3=0.24321$

Part 2:

Only 3 users can access simultaneously, denied Probability is P_3(4th access line)

$P_3=0.24321$

Part 3:

Average Number:

$Average\ Number:\frac{\mu}{\lambda} (1-P_3)\\Average\ Number:\frac{40}{18} (1-0.24321)\\\\Average\ Number:=1.6817$