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Answer:

The probability of all processes which are waiting for I/O $=(0.4)^6 = 0.004096$

$\therefore$ CPU Utilization $=1-0.004096 = 0.995904$

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In a multiprogrammed system, multiple programs are kept in memory and are executed by the CPU in a way that maximizes the use of the CPU and other resources. In a system with a degree of 6, there are six programs in memory at any given time.

If each process spends 40% of its time waiting for I/O, this means that the CPU is not being used for 40% of the time for each process. Since there are six programs, this means that the CPU is not being used for a total of 40% * 6 = 240% of the time.

Since the total time is 100%, this means that the CPU utilization in this system will be 100% - 240% = -140%. This is not possible, since utilization must be a non-negative value.

One possible explanation for this result is that the assumptions made in the question are not accurate. It is not possible for a process to spend more time waiting for I/O than the total amount of time that it has to execute. This means that the assumption that each process spends 40% of its time waiting for I/O is incorrect, and the correct value is likely lower than 40%.

Alternatively, the question might be asking for the percentage of time that the CPU is being used by at least one of the processes. In this case, the answer would be 100% - 40% = 60%. This means that the CPU is being used by at least one process for 60% of the time, and is idle for the remaining 40% of the time.

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admin asked Oct 24, 2019
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In the text it was stated that the model of Fig. $2-11(a)$ was not suited to a file server using a cache in memory. Why not? Could each process have its own cache?