Question

CPU-on-Demand (CPUD) offers real-time high-performance computing services. CPUD owns 1 supercomputer that can be accessed through the Internet. Their customers send jobs that arrive, on average, every 5 hours. The standard deviation of the interarrival times is 3 hours. Executing each job takes, on average, 4 hours on the supercomputer and the standard deviation of the processing time is 4.8 hours.
Required:
How long does the customer have to wait to have the job completed?

Answer 1

Answer:

The waiting time is "17.778 hours".

Explanation:

The given values are:

Number of supercomputers with CPUD,

m = 1

Interarrival time's standard deviation,

sd(a) = 3 hours

Processing time,

p = 4 hours

Job's Interarrival time,

a = 5 hours

Processing time's standard deviation,

sd(p) = 4.8 hours

Now,

The coefficient of variation of interarrival time will be:

⇒  $CVa=\frac{sd(a)}{a}$

             $=\frac{3}{5}$

             $=0.6$

The coefficient of variation of interarrival time will be:

⇒  $CVp=\frac{sd(p)}{p}$

             $=\frac{4.8}{4}$

             $=1.2$

The utilization will be:

⇒  $u=\frac{p}{ma}$

       $=\frac{4}{1\times 5}$

       $=0.8$

The expected time of waiting will be:

⇒  $Tq=\frac{p}{m}\times \frac{u^{sqrt[2(m+1)] -1}}{(1-u)}\times \frac{CVa^2+CVp^2}{2}$

          $=\frac{4}{1}\times \frac{(0.8^{sqrt(2(1+1))} - 1)}{(1-0.8)}\times \frac{0.6^2+1.2^2}{2}$

          $=4\times \frac{0.8}{0.2}\times 0.9$

          $=17.778 \ hours$