Solution

Multi-level queue scheduling divides the ready queue into multiple queues, according to various categories of processes, and favours short-term and I/O-bound processes.

Solution

Minimum waiting time can be 0 as the process may be given the CPU immediately after its arrival. Maximum wait time is when the process gets last the CPU ie after executing n-1 process for b time each. So maximum wait time is (n-1)b

Solution

Implementation of the second-chance algorithm in the form of a queue may be inefficient as

it incurs the cost of moving the pages with reference bit 1 at the end of the queue. Therefore,

the pages are moved around in the queue unnecessarily. A better approach is to have a circular queue instead of a general queue. The circular queue will store all the pages with their status bits as before, but now, there is a hand (a pointer) that indicates the next page in the circular queue. This circular queue is known as a clock.

Solution

Seek time

Solution

The SSTF algorithm may be best suitable for batch-processing systems where high throughput and average response time are required. However, it may not be efficient for interactive and online systems where high response time and its predictability are required.

Solution

Resoruce allocation graph (RAG) is used to represent resource allocation graphically.

Solution

In FCFS the seek time will be high and throughput will be low. The algorithm may be good

in performance when the load on the disk is low and the requests are uniformly distributed.

However, as the load on the disk increases, FCFS provides poor performance as it increases the average response time. However, the FCFS will provide low variance of response time due to random seek, that is, the arriving disk requests cannot get ahead of already pending requests

Solution

The property considers the execution of an algorithm with two different page-frame sizes, say m and n where n < m. At an instant of time, all the pages that are in the memory when the page-frame size is n would also be in the memory when the page frame size is m. In other words, the set of pages when the page-frame size is n is a subset of the set of pages when the page-frame size is m. The implication of stack property is that the number of page faults in page-frame size n is more compared to that of m.

Solution

Solution

Solution

Solution

Since each page table entry is 4 bytes and each page contains 4 Kbytes, then a one-

age page table would point to 1024 = 2^10 pages, addressing a total of 2^10 × 2^12 =

2^22 bytes. The address space however is 2^64 bytes. Adding a second layer of page

tables, the top page table would point to 2^10 page tables, addressing a total of 2^32

bytes. Continuing this process, we need 6 levels to fit last page table in one page.

Solution

TLBHit[TLB+M] + (1-TLBHIT)[TLB+(1-p)[2M]+p(pfs)]

=0.85*(110) + 0.15*(10+0.98*200 + 0.02*2*1000000)

=6124.4ns

Solution

The above is similar to sequential code x=x+1; x=y+1. Therefore, x=2, y=1 after execution. Note that the code is run only once as no loop is given.

Solution

S2, S5, S4, S1, S3. Please refer to steps involved in page fault servicing.

Solution

P1Working set = {0, 1, 2, 10}

P2 Working set = {0, 100, 101, 102, 103, 104}

P3 Working set = {0, 1, 2, 3, 4, 5}

Therefore,

Working set size of P1 = 4

Working set size of P2 = 6

Working set size of P3 = 6

Total memory required = 4 + 6 + 6 = 16 frames

Solution

In B tree, the deletion of key causes either no change in the height or the height of the tree will be decreased by 1.

Solution

Solution

Combination of prime and non prime attributes should be taken as non prime only. Hence, in {wx->yz, xy->z} the FD xy->z is transitive dependency.

Solution

(a, a, c) - (a,c) = 0

Solution

The problems may occur when the transactions are executed concurrently are: Lost update, Dirty read, Incorrect summary problem etc.

Solution

The SQL LIKE clause is used to compare a value to similar values using wildcard operators. There are two wildcards used in conjunction with the LIKE operator.

• The percent sign (%)
• The underscore (_)

The percent sign represents

Solution

In D->C both C and D are non prime attributes. So, it is a transitive dependency but not partial dependency. Hence, D->C is in 2NF but not in 3NF

Solution

Shared locks exist when two transactions are granted read access. One transaction gets the shared lock on data and when the second transaction requests the same data it is also given a shared lock.

Solution

Solution

Blocking factor (Index) =512/13=39

No. of blocks in 1stlevel =10,000/39=257

No. of blocks in 2nd level =7

Solution

R2/R1 gives minimum of zero rows as there is no row in R2 for each row of R1.

R2/R1 gives max of X2/X1 rows as there is a row in R2 for each row of R1.

Solution

3

Solution

• Each entity in X is participating in one entity in relation.
• Each entity in Y is participating in one or more entities in relation.

We can merge X and Relation as one table:

Solution

n*Bp+(n-1)<=k512

n*8+(n-1)<=12512

8n+12n-12<=512

⎡26.2⎤=26

Solution

Order of Internal node & leaf node are 26 and 22.

New orders are

Internal node =26 x 0.9=23

Leaf node =22 x 0.9=19

Root : 1 node, 23 pointers

Level 1: 23 nodes, 23 x 23=529

Level 2: 529 x 23=12167

Level 3: 12167 x 19=231173

Solution

Solution