The various answers to the question are:
- To answer 90% of calls instantly, the organization needs four extension lines.
- The average number of extension lines that will be busy is Four
- For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.
<h3>How many extension lines should be used if the company wants to handle 90% of the calls immediately?</h3>
a)
A number of extension lines needed to accommodate $90 in calls immediately:
Use the calculation for busy k servers.
![$$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$$](https://tex.z-dn.net/?f=%24%24P_%7Bj%7D%3D%5Cfrac%7B%5Cfrac%7B%5Cleft%28%5Cfrac%7B%5Clambda%7D%7B%5Cmu%7D%5Cright%29%5E%7Bj%7D%7D%7Bj%20%21%7D%7D%7B%5Csum_%7Bi%3D0%7D%5E%7Bk%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B%5Clambda%7D%7B%5Cmu%7D%5Cright%29%5E%7Bt%7D%7D%7Bi%20%21%7D%7D%24%24)
The probability that 2 servers are busy:
The likelihood that 2 servers will be busy may be calculated using the formula below.
![P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$](https://tex.z-dn.net/?f=P_%7B2%7D%3D%5Cfrac%7B%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7B2%7D%7D%7B2%20%21%7D%7D%7B%5Csum_%7Bi%3D0%7D%5E%7B2%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7Bt%7D%7D%7Bi%20%21%7D%7D%24%24%5Capprox%200.3425%24)
Hence, two lines are insufficient.
The probability that 3 servers are busy:
Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.
![P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$](https://tex.z-dn.net/?f=P_%7Bj%7D%3D%5Cfrac%7B%5Cfrac%7B%5Cleft%28%5Cfrac%7B%5Clambda%7D%7B%5Cmu%7D%5Cright%29%5E%7Bj%7D%7D%7Bj%20%21%7D%7D%7B%5Csum_%7Bi%3D0%7D%5E%7B2%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B%5Clambda%7D%7B%5Cmu%7D%5Cright%29%5E%7Bi%7D%7D%7Bi%20%21%7D%7D%24%20%5C%5C%5C%5C%24P_%7B3%7D%3D%5Cfrac%7B%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7B3%7D%7D%7B3%20%21%7D%7D%7B%5Csum_%7Bi%3D0%7D%5E%7B3%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7B1%7D%7D%7Bi%20%21%7D%7D%24%24%5Capprox%200.1598%24)
Thus, three lines are insufficient.
The probability that 4 servers are busy:
Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.
![P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$](https://tex.z-dn.net/?f=P_%7Bj%7D%3D%5Cfrac%7B%5Cfrac%7B%5Cleft%28%5Cfrac%7B%5Clambda%7D%7B%5Cmu%7D%5Cright%29%5E%7Bj%7D%7D%7Bj%20%21%7D%7D%7B%5Csum_%7Bi%3D0%7D%5E%7Bk%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B%5Clambda%7D%7B%5Cmu%7D%5Cright%29%5E%7Bt%7D%7D%7Bi%20%21%7D%7D%24%20%5C%5C%5C%5C%24P_%7B4%7D%3D%5Cfrac%7B%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7B4%7D%7D%7B4%20%21%7D%7D%7B%5Csum_%7Bi%3D0%7D%5E%7B4%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7B7%7D%7D%7Bi%20%21%7D%7D%24)
Generally, the equation for is mathematically given as
To answer 90% of calls instantly, the organization needs four extension lines.
b)
The probability that a call will receive a busy signal if four extensions lines are used is,
![P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$](https://tex.z-dn.net/?f=P_%7B4%7D%3D%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7B4%7D%7D%7B%5Csum_%7Bi%3D0%7D%5E%7B4%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7B1%7D%7D%7Bi%20%21%7D%7D%20%24%5Capprox%200.0624%24)
Therefore, the average number of extension lines that will be busy is Four
c)
In conclusion, the Percentage of busy calls for a phone system with two extensions:
The likelihood that 2 servers will be busy may be calculated using the formula below.
![P_{j}=\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}$$\\\\$P_{2}=\frac{\left(\frac{20}{12}\right)^{2}}{\sum_{i=0}^{2 !} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$](https://tex.z-dn.net/?f=P_%7Bj%7D%3D%5Cfrac%7B%5Cleft%28%5Cfrac%7B%5Clambda%7D%7B%5Cmu%7D%5Cright%29%5E%7Bj%7D%7D%7Bj%20%21%7D%24%24%5C%5C%5C%5C%24P_%7B2%7D%3D%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7B2%7D%7D%7B%5Csum_%7Bi%3D0%7D%5E%7B2%20%21%7D%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B20%7D%7B12%7D%5Cright%29%5E%7Bt%7D%7D%7Bi%20%21%7D%7D%24%24%5Capprox%200.3425%24)
For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.
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