The average number of uncompleted projects at any given time is equal to 11.955.
Given the following data:
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:
[tex]SCV = (\frac{\alpha _a}{a} )^2\\\\SCV = (\frac{4.2}{4.1} )^2\\\\SCV = (1.024 )^2\\\\[/tex]
SCV = 1.05.
For processing time:
[tex]SCV = (\frac{18.6}{18.6} )^2\\\\SCV = (1.0 )^2\\\\[/tex]
SCV = 1.0.
For utilization:
[tex]U=\frac{p}{ma} \\\\U=\frac{18.6}{5 \times 4.1}[/tex]
U = 0.907.
Next, we would determine the average number of projects that are waiting for a programmer by using this formula:
[tex]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[/tex]
The number of projects that are being programmed is given by:
[tex]I_p =mu\\\\I_p=5 \times 0.907\\\\I_p=4.535[/tex]
The number of customers that are waiting to be serviced is given by:
[tex]I_q=\frac{T_q}{a} \\\\I_q=\frac{30.42}{4.1} \\\\I_q=7.42[/tex]
Now, we can determine the average number of uncompleted projects at any given time as follows:
[tex]I=I_q+I_p\\\\I=7.42+4.535[/tex]
I = 11.955.
Read more on standard deviation here: https://brainly.com/question/4302527
