Question

Operating System a. What is the meaning of the term busy waiting? What other kinds of waiting are there in an operating system? Can busy waiting be avoided altogether? Explain your answers. (5 pts) b. Explain why spinlocks are not appropriate for single-processor systems yet are often used in multiprocessor systems. (5 pts) c. Using the program shown below, explain what the output will be at lines X and Y. Assume that you have a pre-emptive priority system. (10 pts) #include #include #inculde #define SIZE 5 int nums[SIZE] = {0,1,2,3,4}; int main() { int i; pid_t pid; pid = fork(); if (pid ==0) { for (i =0; i < SIZE; i++) { nums[i] *= -i; printf(“CHILD: %d “,nums[i]); /*LINE X */ } } else if (pid > 0) { wait (NULL); for (i = 0; i < SIZE; i++) printf(“PARENT: %d “,nums[i]); /* LINE Y */ } return 0; } Part2 Consider a system with 3 physical frames of memory that is given the following page memory reference sequence (20 pts): 1, 3, 6, 7, 1, 3, 6, 7, 1, 3, 6, 7 What is the number of page faults that would occur for each of the following page replacement algorithms? a. An optimal page replacement algorithm (7 pts) b. LRU (7 pts) c. Does an optimal page replacement algorithm exist that does not require future knowledge for cyclical memory reference sequences such as shown above? If so, give a general algorithm. If not, explain why. (6 pts) 1) Consider a machine with a physical memory of 8 GB, a page size of 8 KB, and a page table entry size of 4 bytes. How many levels of page tables would be required to map a 46-bit virtual address space if every page table fits into a single page? Be explicit in your explanation. (5 points) 2) List the fields of a Page Table Entry in your scheme. (5 points) 3) Without a cache or TLB, how many memory operations are required to read or write a single 32-bit word? (5 points) 4) How much physical memory is needed for a process with three pages of virtual memory (for example, one code, one data, and one stack page)? (5 points)

Answer 1

Basically busy waiting is a scenario when a process looks for a condition repeatedly so it is busy to check the condition as well as it is waiting for it in this way a process will be in busy waiting condition. In other hand we can say when the processor is doing nothing but spending its clock cycle continuously and waiting for some event occurrence then the scenario will be called as busy waiting.

Busy waiting can't be avoided altogether. But some time it can be avoided but it will incurs some associated overhead like putting the process in sleep and wake up.

b). Implementing synchronisation primitive by disabling interrupt is not appropriate for single processor system because if a user level program will be able to disable interrupt then it can disable the timer interrupt also which will effect the context switching.

A semaphore can be initialized to the socket connection which is allowable. When connection is accepting accquire() will be called and after releasing the connection relese() will be called. If the system will reach the highest number of allowable connections then it will block the acquire() method until Atleast one connection is released.

c)

#include <sys/types.h>

#include <stdio.h>

#include <unistd.h>

#define SIZE 5

int nums[SIZE] = {0,1,2,3,4};

int main()

{

1. int i;

2. pid t pid;

3. pid = fork();

4. if (pid == 0) {

5. for (i = 0; i < SIZE; i++) {

6. nums[i] *= -i;

7. printf("CHILD: %d ",nums[i]); /* LINE X */

8. }

9. }

10. else if (pid > 0) {

11. wait(NULL);

12. for (i = 0; i < SIZE; i++)

13. printf("PARENT: %d ",nums[i]); /* LINE Y */

14. }

15. return 0;

16. }

Figure 3.35 What output will be at Line X and Line Y?

As per the same logic which we discussed in problem 1, the line 4 will be evaluated to true by the child process, and the lines 5,6,7,8 will be executed by the child process. The line 10 will be evaluated to true by the parent process, and the lines 11,12,13 will be executed by the parent process.

Initially nums[5] = {0,1,2,3,4}.

The same values will be passed to the child process.

By child process:

Line 5: It indicates that the loop (line 6, and line 7) will be evaluated for 5 times, for different values of i:

i = 0: nums[0] = nums[0] * -i.

• nums[0] = 0 * 0. /*0 will be printed*/

i = 1: nums[1] = nums[1] * -i.

• nums[1] = 1 * -1. /*-1 will be printed*/

i = 2: nums[2] = nums[2] * -i.

• nums[2] = 2 * -2. /*-4 will be printed*/

i = 3: nums[3] = nums[3] * -i.

• nums[3] = 3 * -3. /*-9 will be printed*/

i = 4: nums[4] = nums[4] * -i.

• nums[4] = 4 * -4. /*-16 will be printed*/

Therefore at Line X the output will be:

CHILD: 0 CHILD: -1 CHILD: -4 CHILD: -9 CHILD: -16

By parent process:

Line 10: It indicates that the loop (line 12, and line 13) will be evaluated for 5 times, after waiting the wait(NULL) system call once.

The wait() system call is used to wait for state changes in a child of the calling process, and obtain information about the child whose state has changed. That means this loop will wait till the child process finishes its execution.

After that, the loop will be evaluated, for different values of i:

i = 0: nums[0] value is 0.

i = 1: nums[1] value is 1.

i = 2: nums[2] value is 2.

i = 3: nums[3] value is 3.

i = 4: nums[4] value is 4.

Therefore at Line Y the output will be:

PARENT: 0 PARENT: 1 PARENT: 2 PARENT: 3 PARENT: 4

part 2:

a).

Table size=page size=2^13
Physical memory=2^33
Virtual address space=46 bit
Page entry size=4 bytes

So tolat entry in a page=page size/page entry size => VPN(Virtual Page no)=11 bit long

page offset=13 bit (as page size is 2^13)

So it can be devided only in this way=(11+11+11)+13
So, It should be a three level paging.