Question

Consider a variation of this slide

Swapping (1) Time Operating Operating Operating Operating Operating Operating Operating system system system system system stem System (b) (d) (e) (a) (c) (g)

Assume that the memory of this computer is of 1Gbytes and OS takes 200 Mbytes in the beginning or the bottom like in part (a) of the figure.

Assume that processes A, B, C, D, E, F, and G each takes 100M, 120M, 140M, 160M, 180M, 200M, and 300M bytes to run.

(a)(5%) Note a way to represent that A runs, B runs, C runs, and B is swapped out is like below in Figure 2.

(figure 2)

Show how it looks with A runs, B runs, C runs, B out, D runs, A out, and E runs with a picture similar to Figure 2.

(b)(7%) It is apparent that all 7 processes can not run concurrently in this memory since the memory they use together is 100+120+140+160+180+200+300 = 1200 M > 800 M.

What is the maximum number n of processes that can run concurrently? Explain! We know n < 7, and n > 1.

(c)(7%) With n processes running in memory (so that any n+1 processes take more than 800 M bytes memory), find the combination that utilizes the most of the memory. Note there are at least two different ways of running n processes in the memory.

(d)(7%) For two processes running in the memory, we notice that with A and B of 100 + 120 = 220M, this is the combination using the least amount of memory; and with F and G of 200 + 300 = 500M bytes, this is the combination using the most amount of memory. If the sum of two processes can not be more than 400M bytes, which combination uses the most amount of memory <= 400M bytes? Find out that.

Answer 1

We have memory in a Computer is 1GB that is 1024MB.

Initially OS took 200M in the 1024MB, so now we have 824MB.

A process can take – 100MB

B process can take – 120MB

C process can take – 140MB

D process can take – 160MB

E process can take – 180MB

F process can take – 200MB

G process can take – 300MB

Totally we have 7 processes that can take 1200MB. But we have only 824MB space remaining in memory.

A)

B,C)

We have 824 MB remaining after OS took 200MB.

With A + B + C + D + E = 700M. and we left with 124 MB. If any process can added, it will cross 824 MB. That is why we can’t accommodate either F or G. so we can run 5 processes currently.

Or else we can make F can accommodate, by not adding A.

B + C + D + E + F = 800MB. And we left with just 24MB.

Or else we can make G can accommodate, by not adding E.

A + B + C + D + G = 820 MB. And we left with 4MB, this is optimal utilization of memory.

In 3 cases, we can run only 5 processes. But in 3^rd way, we can fully utilize the memory.

D)

We have 3 different ways to run n processes (that is 5 processes).

One is A + B + C + D + E = 700MB. Here the combination that utilizes the most of the memory is D and E, which is 160 + 180 = 340 MB.

Second way is B + C + D + E + F = 800MB. Here the combination that utilizes the most of the memory is D and E, which is 180 + 200 = 380 MB.

Third way is A + B + C + D + G = 820 MB. Here the combination that utilizes the most of the memory is D and G, which is 160 + 300 = 460 MB.