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TABLE 35.5 State Probabilities for Example 35.5 | ||||
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Number of Jobs at | ||||
Terminals | CPU | Disk A | Disk B | Probability |
0 | 0 | 0 | 3 | 0.021 |
0 | 0 | 1 | 2 | 0.031 |
0 | 0 | 2 | 1 | 0.047 |
0 | 0 | 3 | 0 | 0.070 |
0 | 1 | 0 | 2 | 0.010 |
0 | 1 | 1 | 1 | 0.016 |
0 | 1 | 2 | 0 | 0.023 |
0 | 2 | 0 | 1 | 0.005 |
0 | 2 | 1 | 0 | 0.008 |
0 | 3 | 0 | 0 | 0.003 |
1 | 0 | 0 | 2 | 0.063 |
1 | 0 | 1 | 1 | 0.094 |
1 | 0 | 2 | 0 | 0.141 |
1 | 1 | 0 | 1 | 0.031 |
1 | 1 | 1 | 0 | 0.047 |
1 | 2 | 0 | 0 | 0.016 |
2 | 0 | 0 | 1 | 0.094 |
2 | 0 | 1 | 0 | 0.141 |
2 | 1 | 0 | 0 | 0.047 |
3 | 0 | 0 | 0 | 0.094 1.000 |
Without the table, the probability of one job at disk A can be computed as follows:
Similarly, probabilities of 0, 2, and 3 jobs at disk A can be shown to be 0.383, 0.211, and 0.070. Using these values, the average queue length at disk A can be shown to be 0.97 and the variance of the queue length is 0.87.
The system throughput is
The device utilizations are
UCPU = XDCPU = 0.264 × 0.78 = 0.206
UA = XDA = 0.264 × 2.34 = 0.618
UB = XDB = 0.264 × 1.56 = 0.412
The average number of jobs at the devices are
The device reponse times are
The system response time is
We can check the computation by computing the average number of jobs in the system:
N = X(R + Z) = 0.264(6.694 + 4.68) = 3
The complete convolution algorithm is summarized in Box 35.1.
Box 35.1 Convolution Algorithm
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