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A context switch from a process $P_{old}$ to a process $P_{new}$ consists of the following steps:

1. Step I:saving the context of $P_{old}$;
2. Step II: running the scheduling algorithm to pick $P_{new}$;
3. Step III: restoring the saved context of $P_{new}$.

Suppose Steps I and III together take $T_0$ units of time. The scheduling algorithm takes $nT_1$ units of time, where $n$ is the number of ready-to-run processes. The scheduling policy is round-robin with a time slice of $10$ms. Compute the CPU utilization for the following scenario: $k$ processes become ready at almost the same instant in the order $P_1, P_2, . . . , P_k;$ each process requires exactly one CPU burst of $20$ms and no I/O burst.

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CPU Utilization = 10*k / [10*k + k*(T0 + k*T1)]

by (43 points)
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Can u give detailed solution.. I think answer should be $\frac{2\times K\times10}{2\times 10\times k + (T+KT1)(2k-1))}$ because burst time is 20ms

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