Answer to question a)
The “working set” is short hand for “parts of memory that the present algorithm is utilizing” and is dictated by which parts of memory the CPU simply happens to get to.
The main point is to find the pages within and outside of the working set. The inexact time the page was last utilized and the R (Referenced) bit. The vacant white rectangle symbolizes alternate fields not required for this algorithm, for example, the page outline number, the insurance bits, and the M (Modified) bit.
Answer to question b)
The memory administration in the working framework is a basic usefulness, which enables the portion of memory to the procedures for execution and deallocates the memory when the procedure is never again required. In this article, we will talk about two memory administration plans paging and segmentation. On the off chance that we discuss the essential contrasts between the paging and segmentation it is, a page is a settled sized block though, a segment is a variable-sized block.
In a joined paging/segmentation framework a client address space is separated into various segments at the watchfulness of the software engineer. Each segment is thusly separated into various settled size pages which are equivalent long to a fundamental memory outline. On the off chance that a segment is not as much as a page long, the segment involves only one page. From the programmer`s perspective, a legitimate address still comprises of a segment number and a segment offset. From the system`s perspective, the segment offset is seen as a page number and page offset for a page inside the predefined segment.
Answer to question c)
1, 2, 3, 4, 1, 2, 5, 3, 4, 1, 3, 2, 5, 6, 4, 3, 7, 6, 3, 4
FIFO
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|
1 |
2 |
3 |
4 |
1 |
2 |
5 |
3 |
4 |
1 |
3 |
2 |
5 |
6 |
4 |
3 |
7 |
6 |
3 |
4 |
|
0 |
1 |
1 |
1 |
4 |
4 |
4 |
5 |
5 |
5 |
1 |
1 |
1 |
1 |
6 |
6 |
6 |
7 |
7 |
7 |
7 |
|
1 |
2 |
2 |
2 |
1 |
1 |
1 |
3 |
3 |
3 |
3 |
3 |
5 |
5 |
5 |
3 |
3 |
3 |
3 |
4 |
|
|
2 |
3 |
3 |
3 |
2 |
2 |
2 |
4 |
4 |
4 |
2 |
2 |
2 |
4 |
4 |
4 |
6 |
6 |
6 |
||
|
|
* |
* |
Page Hit – 2
Page Fault – 20-2 = 18
LRU
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|
1 |
2 |
3 |
4 |
1 |
2 |
5 |
3 |
4 |
1 |
3 |
2 |
5 |
6 |
4 |
3 |
7 |
6 |
3 |
4 |
|
0 |
1 |
1 |
1 |
4 |
4 |
4 |
5 |
5 |
5 |
1 |
1 |
1 |
5 |
5 |
5 |
3 |
3 |
3 |
3 |
3 |
|
1 |
2 |
2 |
2 |
1 |
1 |
1 |
3 |
3 |
3 |
3 |
3 |
3 |
6 |
6 |
6 |
7 |
7 |
7 |
4 |
|
|
2 |
3 |
3 |
3 |
2 |
2 |
2 |
4 |
4 |
4 |
2 |
2 |
2 |
4 |
4 |
4 |
6 |
6 |
6 |
||
|
|
* |
* |
Page Hit – 2
Page Fault – 20-2 = 18
OPTIMAL
|
|
1 |
2 |
3 |
4 |
1 |
2 |
5 |
3 |
4 |
1 |
3 |
2 |
5 |
6 |
4 |
3 |
7 |
6 |
3 |
4 |
|
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
2 |
5 |
6 |
6 |
6 |
6 |
6 |
6 |
6 |
|
1 |
2 |
2 |
2 |
2 |
2 |
5 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
3 |
|
|
2 |
3 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
7 |
7 |
7 |
7 |
||
|
|
* |
* |
* |
* |
* |
* |
* |
* |
* |
Page Hit – 9
Page Fault – 20-9 = 11
Question 2: Inter-process Communication
Answer to question a)
Answer to question b)
|
Allocation |
Max |
Available |
|||||||
|
A |
B |
C |
A |
B |
C |
A |
B |
C |
|
|
P0 |
1 |
2 |
2 |
9 |
8 |
8 |
2 |
2 |
2 |
|
P1 |
1 |
1 |
2 |
4 |
3 |
3 |
|||
|
P2 |
2 |
1 |
1 |
2 |
1 |
7 |
|||
|
P3 |
2 |
2 |
1 |
3 |
3 |
2 |
|||
|
P4 |
2 |
2 |
2 |
7 |
7 |
7 |
Need Matrix
|
A |
B |
C |
|
|
P0 |
8 |
6 |
6 |
|
P1 |
3 |
2 |
1 |
|
P2 |
0 |
0 |
6 |
|
P3 |
1 |
1 |
1 |
|
P4 |
5 |
5 |
5 |
P0 à need <= available
866<= 222
Not execute
P1à need <= available
321 <= 222
Not execute
P2à need <=available
006<=222
Not execute
P3 à need <= available
111<=222
Execute the process
Now available = allocation + available = 221+222 = 443
P4 à need <= available
555<= 443
Not execute
P0 à need <= available
866<= 443
Not execute
P1à need <= available
321 <= 443
Execute the process
Now available = 112+443 = 555
P2à need <=available
006<=555
Not execute
P4 à need <= available
555<= 555
Process execute
Now available = 222+555 = 777
P0 à need <= available
866<= 777
Not execute
P2à need <=available
006<=777
Process execute
Now available = 211+777 = 988
P0 à need <= available
866<= 988
Process execute
Now available = 1 2 2 + 9 8 8 = 10 10 10
Process sequence {P3, P1, P4, P2, P0}
The system is in safe state
Answer to question c)
|
#include<stdio.h> #include<stdlib.h> int mutex=1,full=0,empty=3,x=0; int main() { int n; void producer(); void consumer(); int wait(int); int signal(int); printf(“n1.Producern2.Consumern3.Exit”); while(1) { printf(“nEnter your choice:”); scanf(“%d”,&n); switch(n) { case 1: if((mutex==1)&&(empty!=0)) producer(); else printf(“Buffer is full!!”); break; case 2: if((mutex==1)&&(full!=0)) consumer(); else printf(“Buffer is empty!!”); break; case 3: exit(0); break; } } return 0; } int wait(int s) { return (–s); } int signal(int s) { return(++s); } void producer() { mutex=wait(mutex); full=signal(full); empty=wait(empty); x++; printf(“nProducer produces the item %d”,x); mutex=signal(mutex);} void consumer() { mutex=wait(mutex); full=wait(full); empty=signal(empty); printf(“nConsumer consumes item %d”,x); x–; mutex=signal(mutex); } |
Question 3: Files and File Systems
Answer to question a)
The File Allocation Table, FAT, utilized by DOS is a variety of linked allocation, where every one of the connections are put away in a different table toward the start of the disk. The advantage of this approach is that the FAT table can be reserved in memory, extraordinarily enhancing irregular access speeds.
Answer to question b)
In contiguous allocation, files are relegated to contiguous regions of auxiliary stockpiling. A client determines ahead of time the size of the region expected to hold a file to be made. In the event that the coveted measure of contiguous space isn’t accessible, the file can’t be made.
Answer to question c)
Numerous apparently basic I/O tasks are really made out of sub-activities. For instance, erasing a file on an I-node based system (extremely this implies erasing the last connect to the I-node) requires expelling the passage from the catalog, setting the I-node on the free rundown, and putting the file blocks on the free rundown. Since I/O tasks can command the time required for finish client forms, extensive exertion has been used to enhance the execution of these activities. All broadly useful file systems utilize a (non-request) paging algorithm for file storage (read-just systems, which frequently utilize contiguous allocation, are the significant special case). Files are broken into settled size pieces, called blocks that can be scattered over the disk. Note that in spite of the fact that this is paging, it isn’t called paging (and might not have an express page table).
In reality, it is more muddled since different improvements are performed to endeavor to have successive blocks of a solitary file put away sequentially on the disk. This is examined beneath.
Note that every one of the blocks of the file are put away on the disk, i.e., it isn’t request paging.
One can envision systems that do use request paging-like algorithms for disk block storage. In such a system just a portion of the file blocks would be put away on disk with the lay on tertiary storage (some sort of tape). Maybe NASA does this with their enormous datasets. A district of kernel memory is committed to monitoring the free blocks. One bit is alloted to each block of the file system. The bit is 1 if the block is free.
Answer to question d)
FCFSThat’s First Come First Serve, In that wording only it was saying like which one will come first it will serve. So
Wait time of each process is as follows −
Process
Wait Time: Service Time – Arrival Time
P0
0 – 0 = 0
P1
5 – 1 = 4
P2
8 – 2 = 6
P3
16 – 3 = 13
Average Wait Time: (0+4+6+13) / 4 = 5.75
SSFIn SSTF (Shortest Seek Time First), asks for having shortest seek time are executed first. Along these lines, the seek time of each demand is figured ahead of time in line and after that they are planned by their computed seek time. Accordingly, the demand close to the disk arm will get executed first. SSTF is surely a change over FCFS as it diminishes the normal reaction time and builds the throughput of framework.
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