1
$\begingroup$

If holes are half as large as processes, the fraction of memory wasted in holes is:

  1. $2/3$
  2. $1/2$
  3. $1/3$
  4. $1/5$

My attempt:

Somewhere it explained as:

Imagine processes as squares. If holes are also squares of half the side dimension of the processes (i.e. linear dimension is the largeness yardstick), then process area $= 4$ and hole area $= 1$. Then ratio of hole to total $= 1/5$ and you have your answer $(4)\space 1/5$ with a bunch of caveats.

Sorry, I didn't get the given solution, why processes as squares?

Can you explain it, please?

$\endgroup$
2
  • $\begingroup$ I don't think you want the side length to be half, rather the area. After all, there are no actual squares in memory $\endgroup$ Sep 18 '16 at 14:46
  • $\begingroup$ Yes, right @HagenvonEitzen. $\endgroup$
    – ً ً
    Sep 18 '16 at 14:48
2
$\begingroup$

I would think of memory as linear, not planar, because of the way it is addressed. Then you have as many holes as processes, so the fraction of holes is $\frac {\frac 12}{\frac 12+1}=\frac 13$

$\endgroup$
1
  • $\begingroup$ Thanks for nice explanation. $\endgroup$
    – ً ً
    Sep 18 '16 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.