سوال ۱۰۱ کنکور ارشد مهندسی کامپیوتر ۱۴۰۱

Setup

Each process must first be brought into main memory over the DMA channel, then it uses the CPU.

Transfer (I/O) time per process — moving $10\,\text{MB}$ at $5\,\text{MB/sec}$:

$t_{\text{io}} = \frac{10\,\text{MB}}{5\,\text{MB/sec}} = 2\,\text{sec}.$

CPU time per process:

$t_{\text{cpu}} = 200\,\text{ms} = 0.2\,\text{sec}.$

Because DMA transfers data without occupying the CPU, the loading of one process overlaps with the computation of others.

Multiprogramming utilization model

In the classic multiprogramming model, a process spends a fraction $p$ of its life-time waiting on I/O (here, waiting to be loaded by DMA):

$p = \frac{t_{\text{io}}}{t_{\text{io}} + t_{\text{cpu}}} = \frac{2}{2 + 0.2} = \frac{2}{2.2} \approx 0.91.$

With $n$ processes resident in memory at once, the probability that all of them are simultaneously blocked on I/O (leaving the CPU idle) is $p^{\,n}$, so

$\text{CPU utilization} = 1 - p^{\,n}.$

Take the limit

The key phrase is that the system can load many processes («فرایندهای زیادی») and run them concurrently with DMA, i.e. the degree of multiprogramming $n$ is large. Since $0 < p < 1$,

$\lim_{n \to \infty} \bigl(1 - p^{\,n}\bigr) = 1.$

Already for a modest $n$ the term $p^{\,n}$ collapses ($0.91^{10} \approx 0.39$, $0.91^{30} \approx 0.06$), pushing utilization toward $100\%$. With many overlapped processes the CPU is essentially never starved.

Hence the CPU utilization is closest to $\mathbf{100\%}$ — option 1.

Exam trick: The whole purpose of multiprogramming + DMA overlap is to hide I/O latency. When the problem says “many processes loaded concurrently,” read it as $n \to \infty$ in $1 - p^{\,n}$, so utilization $\to 100\%$ — do not compute the single-process ratio $0.2/2.2 \approx 9\%$, which would only apply to one job with no overlap.