Sample Online Test

Each block size = 128 Bytes
Disk block address = 8 Bytes
∴ Each disk can contain = 128/8 = 16 addresses
Size due to 8 direct block addresses: 8×128
Size due to 1 indirect block address: 16×128
Size due to 1 doubly indirect block address: 16×16×128
So maximum possible file size:
8×128+16×128+16×16×128 = 1024+2048+32768 = 35840 Byte = 35 kBytes

1. AcquireLock(L){
2. while (Fetch_And_Add(L,1))
3. L = 1;
}
4. ReleaseLock(L){
5. L = 0;
6. }
Let P and Q be two concurrent processes in the system currently executing as follows
P executes 1,2,3 then Q executes 1 and 2 then P executes 4,5,6 then L = 0 now Q executes 3 by which L will be set to 1 and thereafter no process can set L to zero, by which all the processes could starve

The completion order for round robin is given below
At time t = 2
P₁ will move to ready queue, so status of ready queue will be:

                     ↑ front

At time t = 3
P₃ enters the system so it will be enqueued.

                     ↑ front

At time t = 4
When quantum of P₂ is done
We need to enqueue P₂ into ready queue & next process to be executed which will be taken from front of ready queue i.e. P₁. Ready queue:

                     ↑ front

So using round robin scheduling, the completion order is P₁,P₃,P₂

Effort person per month= a.(KDSI)^B
KDSI = kilo LOC
= 2.8 X (40)^1·20
= 2.8 X 83.6511
= 234.22 person per month

Method used by P1 & P2 enters into critical section when S1 = S2 & S1! = S2 respectively, since both are opposite conditions so mutual exclusion is there, but if P1's while loop true then it will always be true & in P2 if while loop true it would run infinitely so no. progress. Hence (A) is correct option.

In FIFO page replacement policy, the pages which entered first are replaced first so for first 100 accesses, 100 page faults will occur. Now in memory there are last 4 pages 97, 98, 99 & 100th page. So during reverse access there 4 pages would not create any page fault & other 97 page faults. So total 100 + 96 = 196 page faults. Hence (A) is correct option.

I. SRIF scheduling may cause starvation since the processes which require large CPU burst periods would have to wait.
II. Generally preemptive scheduling doesn’t cause starvation but in some cases, it may cause starvation when one process doesn’t block due to I/O so others have to wait.
III. It is quite obvious that due to preemption involvement round orbit is better than FCFS, in case of response time.
All are true
Hence (D) is correct option.

Let us see what will happen initially S₀ = 1, S₁ = 0 S₂ = 0 so P₁ & P₂ will wait & P₀ runs & make S₀ = 0 wait decrements semaphore by 1 & release increments if by 1. So after P₀ 's 1^st run S₁ = 1 & S₂ = 1 ‘0’ is printed Now P₁ & P₂ can run & make S₁ = 0, S₂ = 0 but both increments S₀ = 1 again. So P₀ again starts and print '0' so this would continue again & again. So '0' will be printed at least twice. Hence (A) is correct option.

Allotment of resources would be successful to all cases but in option (B) where n = 21 & k = 12. Let use take i = 10 So for P₁₀ only R₁₀ can be allotted whereas for P₁₁ odd case (else). R₂₁₋₁₁ i.e R₁₀ can only be allotted. This is an conflict. Hence (B) is correct option.

Total resources = 3 2 3 2

All process computed without deadlock Hence (A) is correct option.

Belady’s anomaly is a properly of FIFO page replacement policy in which the no. of page faults increases even if the no. of page frames are increased in some cases. Hence (A) is correct option.

A page table is in main memory which mop a page frame to a virtual memory page. So essential content is page frame number Hence (B) is correct option.

Arranging in order.


 

Total = 16 + 14 + 1 + 4 + 5 + 3 + 1 + 2 + 2 + 71
= 119 ms
Hence (B) is correct option.

Single level page table is not preferred since it requires the no. of page table enteries equal to the no. of virtual memory addresses, but a multilevel page table has smaller no. of enteries so reduce the size of page table needed. Hence (B) is correct option.