ME FLT-1 q29

Question

please explain this anyone

Comments

answer?

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220.

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swap out or swap in time can be calculated for 200Kb program 200kb/2Mbps=0.1=100msec+10(latency)=110

now for any process to come into main memory in the worst case a 200kb program must be transferred to swap space so 110msec + and the required 200kb program is to be brought into the memory so 110msec

total=220msec...this should be the time quantum for effective utilization as these can be done by DMA during the processing of the process...i think

Answer 1

When Time quantum is greater than or equal to context switch time, in that case we say that CPU is being effectivel utilized.

Because, RR favours CPU bound processes and a CPU-bound process runs for all its time slice and is put back into the ready queue rather than I/O bound processes because an I/O bound process uses the CPU for a time less than the time quantum before it is blocked for an I/O . Hence, it is always a good approach to make TQ larger than the time required to handle the clock interrupt and dispatching but should not be much higher.

Now, how the context switch occurs for a program residing on a HD ? It is in the form of swap in and swap out

So, for better CPU utilization TQ >= Swap In + Swap Out

=> Swap In = Time taken to transfer the program + average latency

=> Swap In = 200kb / 2mbps + 10 ms

=> Swap In = 110 ms

Hence, Minimum TQ for effective CPU utilization = 110 * 2 = 220 ms.

Comments

why not greater than 220 then? why are we taking "minimum"

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220 is also good. It should be significantly larger and not much.

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I have a question,

CPU will not be doing effective work at the time of context switch right,,

because all previous registers values are getting getting saved in memory

new register are getting in,

in this switching time CPU is not utilised.....!

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Yes, exactly . We need to minimize this time.

Actually, the fraction of CS overhead is C / (TQ + C) . Now, we need to minimize this overhead hence, TQ>=C but it should be considerably large or atleast equal, otherwise it degenerates to FCFS.

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I think we should take 200 Kb as 200*1024 bits

and

2Mbps as $2 \times 10^6$ bits per second

ans should be 224.8

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@kapil...How the swap in time is exactly same as swap out time..