Question

1 Jobs arrive to a file server consisting of a CPU and two disks (disk 1 and disk 2). With probability 13/20, a job goes from CPU to disk ĺ, and with probability 6/20, a job goes from CPU to disk 2. With probability 1/20, a job is finished after its CPU operation and is immediately replaced by another job. There are always 3 jobs in the system. The mean time to complete the CPU operation is .039 second The mean time to complete the disk 1 operation is .18 second, and the mean time to complete the disk 2 operation is .26 second a Determine the steady-state distribution of the num b What is the average number of jobs at CPU? Disk c What is the probability that CPU is busy? Disk 1? d What is the average number of jobs completed per ber of jobs at each part of the system 1? Disk 2? Disk 2? second by CPU? Disk 1? Disk 2?

Answer 1

Answer:

Jobs arrive to a file server consisting of a CPU, disk 1 and disk 2. A job goes from CPU to disk 1 with probability 13/20 and a job goes from CPU to disk 2 with probability 6/20. A job is finished after its CPU operation with probability 1/20 and the job is immediately replaced by another job. There are always 3 jobs in the system. The mean time to complete CPU operation, disk1 operation and disk 2 operations is 0.039 second, 0.18 second and 0.26 second respectively.

The manufacturing and computer systems where there is a constant number of jobs present are called closed queuing networks. Let P be the probability that a job will go to server j after completing service at station i. It is assumed that the service time at server j follows an exponential distribution with parameter u, The system has s servers and at all times exactly N jobs are present. Let i, be the arrival rate to system / Thus izs i-1 Since jobs never leave the system, so for each i, Let n, be the number of jobs present at server i. So

Thus for any state n, its steady state probability πΝ (n) is determined from the following formula. G(N) Here G(N)- Y,

The set of equations formed for the given problem are 13 207 20 The steady-state distribution of the number of jobs at each part of the system is to be determined. There are an infinite number of solutions to this system. Arbitrarily choosing 1 yields the solution and 13 20 20

In cells G10:G13, compute C (k) p,k 0,1,2,3 In cells G10:110, enter C, (0)-l,i= 1,2,3 Now copy from H11 to H11:113 the formula This implements the recursion C,(k)-C-I(k)pC(k-1) Thus we can find the value of G10-102.185 from the value of C, (3) in the cell 13

The excel worksheet is given below LambdI MI phoi 0.05 0.039 0.65 0.18 03 026 2 0 1 128205128 489316239604700855 2 16436554919 3134086 262907261 3 210725063 718501149 102.185568

We can now generate all possible system states efficiently by starting with n,0and listing those states in order of increasing values of n2. Then we increase n, to 1and list all states in increasing values of n2 Once we have n, 3, we will have listed all states. To efficiently generate all possible states, we copy down from B25 the formula IF(D24-0,B24+1,B24) This formula increments n,by 1 if n 0, which is same as having n,-3-n, otherwise the formula keeps n, constant.

Then copy down from C25 the formula IF(B25-B24-1,0,C24+1) This formula makes n, Oif we have just increased the value of n, otherwise the formula increments the value of n, by 1.

Finally from D25, copy down the formula 3-B25-C25 This ensures that n3-n-n,

In cells E24 E33, use the formula in equation (1) to compute the steady state probability for each state by copying from E24 to E25 E33 the formula (SGS8AB24) (SHS8AC24) (SIS8AD24)SIS13

The screen shot of the excel worksheet is given below

履 Jobs at Jobs at Disk Disk 3 Probability Jobs at CPU 23 24 25 26 27 28 29 3 0.01503329 2 0.04704864 10.14724483 0 0.46082177 2 0.01670366 1 0.05227627 0 0.16360536 1 0.01855962 0 0.05808474 0 0.0206218 30 31 32

a. Now the steady state distribution of the number of jobs at each part of the system is to be determined. Use the SUMIF function in excel First compute in cell H25, the probability of 0 parts at machine 1 with the formula SUMIF (SBS24:SBS33,G25,E24:E33) This formula adds up every number in column E (which contains state probabilities) for the rows and 3 parts at CPU In a similar way, compute the steady state probability distribution for disk 1 and disk 2.

The table showing the steady-state distribution of the number of jobs at each part of the system is given below. Prob CPU 0 067014854 1 023258529 2 00766437 3 0.0206218 Disk 1 0 007091838 1 0.15740966 2 031085019 3 046082177 Dik 2 0 0.70313368 1 021808072 2 00637523 3 001503329

b. The mean number of jobs present at CPU can be computed using the formula Average number of jobs =Σ. (Probability ofijobs at CPU) i=1

In cell K31, the mean number of jobs at CPU is computed using the formula SUMPRODUCT (G25:G28,H25:H28) Similarly, the mean number of jobs at disk 1 and disk 2 are computed in cells K32 and K33 respectively.

Thus the average number of jobs at CPU is 0.46 jobs the average number of jobs at Disk 1 is 2.16 jobs and the average number of jobs at Disk 2 is 0.39 jobs

C. e the probability that CPU is busy, we subtract from 1 the probability that CPU has 0 jobs. This is computed in cell L31 using the formula 1-H25 Similarly the probability that disk 1 is busy are computed by subtracting from 1 the probability that disk 1 has 0 jobs and the probability that disk 2 is busy are computed by subtracting from 1 the probability that disk 2 has 0 jobs. The computations are done in cells L32 and L33.

Thus the CPU is busy 32.9% of the time: Disk 1 is busy 92.9% of the time and Disk 3 is busy 96% thetime 29.6% of the time

d. To determine the average number of jobs completed per second by CPU, multiply the probability that CPU is busy by the service rate of CPU. Similarly the average number of jobs completed per second by disk 1 and disk 2 are computed. The computations are done in cells M31 M33

The screen shot of the excel worksheet is given below Mean Mean Completions Number Number Prob busy per second Machine 1 0.44773944 0.32985146 0.0128642 Machine 2 2.16157536 0.92908162 0.1672347 Machine 3 0.39068521 0.296866320.0771852

Thus the CPU on an average completes 0.012 jobs/second: Disk 1 on an average completes o.167j jobs second and Disk 2 on an average completes O.0771 İobs/second

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