Question

Camile owns Crunch Code, a company that provides quick programming solutions. Clients send Crunch projects via their web page and Crunch bangs out the needed code as quickly as possible. Camile has five programmers who do all of the coding. On average, a project arrives once every 4.1 hours, with a standard deviation of 4.2 hours. Each project is assigned to one programmer and that programmer takes on average 18.6 hours to complete each project with a standard deviation of 18.6 hours.
Required:
How many uncompleted projects does Crunch Code have in the system on average at any given time?

Answer 1

The average number of uncompleted projects at any given time is equal to 11.955.

Given the following data:

• Number of programmers, m = 5.

• Interarrival time, a = 4.1 hours.

• Standard deviation of interarrival time = 4.2 hours.

• Processing time, p = 18.6 hours.

• Standard deviation of processing time = 18.6 hours.

How to calculate the average number of uncompleted projects.

First of all, we would determine the squared coefficient of variation (SCV) of both the interarrival time and processing time as follows;

For interarrival time:

$SCV = (\frac{\alpha _a}{a} )^2\\\\SCV = (\frac{4.2}{4.1} )^2\\\\SCV = (1.024 )^2$

SCV = 1.05.

For processing time:

$SCV = (\frac{18.6}{18.6} )^2\\\\SCV = (1.0 )^2$

SCV = 1.0.

For utilization:

$U=\frac{p}{ma} \\\\U=\frac{18.6}{5 \times 4.1}$

U = 0.907.

Next, we would determine the average number of projects that are waiting for a programmer by using this formula:

$T_q=\frac{p}{m} \times \frac{u^{\sqrt{(2(m+1))-1}} }{1-u} \times\frac{C_a^2 + C_p^2}{2} \\\\T_q=\frac{18.6}{5} \times \frac{0.907^{\sqrt{(2(5+1))-1}} }{1-0.907} \times \frac{1.05^2 +1.0^2}{2} \\\\T_q=3.72 \times 7.7785\times 1.05125\\\\T_q=30.42$

The number of projects that are being programmed is given by:

$I_p =mu\\\\I_p=5 \times 0.907\\\\I_p=4.535$

The number of customers that are waiting to be serviced is given by:

$I_q=\frac{T_q}{a} \\\\I_q=\frac{30.42}{4.1} \\\\I_q=7.42$

Now, we can determine the average number of uncompleted projects at any given time as follows:

$I=I_q+I_p\\\\I=7.42+4.535$

I = 11.955.

Read more on standard deviation here: https://brainly.com/question/4302527