The given scheduling definition takes two parameters, one is dynamically assigned process priority and the other is ‘T’ time unit to re-evaluate the process priorities. This dynamically assigned priority will be deciding processes order in ready queue of round robin algorithm whose time quantum is same as ‘T’ time units. As all the processes are arriving at the same time, they will be given same priority but soon after first ‘T’ time burst remaining processes will get higher priorities

x ⊕ y = x2 + y2 = y2 + x2 = y ⊕ x

∴ commutative

Not associative, since, for example

(1 ⊕ 2) ⊕ 3 ≠ 1 ⊕ (2 ⊕ 3)

RAM chip size = 1k × 8 [1024 words of 8 bits each]

RAM to construct = 16k × 16

Number of chips required = (16k × 16)/(1k × 8) = 16 × 2 [16 chips vertically with each having 2 chips
horizontally]

So to select one chip out of 16 vertical chips, we need 4 x 16 decoder.

Available decoder is – 2 x 4 decoder

To be constructed is 4 x 16 decoder

So we need 5, 2 x 4 decoder in total to construct 4 x 16 decoder.

MBR ← PC

MAR ← X

PC ← Y

Memory ← MBR

Which one of the following is a possible operation performed by this sequence?

PC content is stored in memory via MBR and PC gets new address from Y. It represents a function call (routine), which is matching with interrupt service initiation

No Explaination.

Clock period=Maximum stage delay+ overhead (Buffer) =10+1=11 ns

Assume FI-1, DI-2, FO-3, EI-4, WO-5

So number of clocks required to complete the program is = 15 clocks and time taken is = 15
×11 ns=165 ns.

Coupling = number of external links/number of modules = 2/2

Cohesion of a module = number of internal links/number of methods

Cohesion of M1 = 8/4; Cohesion of M2 = 6/3; Average cohesion=2

After moving method m to M2, graph will become

Coupling = 2/2

Cohesion of M1 = 6/3; Cohesion of M2 = 8/4; Average cohesion=2

Clustered index is built on ordering non key field and hence if the index is clustered then the data records of the file are organized in the same order as the data entries of the index.

42797 KB = (42797 × 1024)/512 = 85594 sectors

Starting is (1200,9,40) contains total 24 + (6×64)=408 sectors

Next, 1201, --------, 1283 cylinders contains total 1024×83 = 84992 sec tors

each cylinder contains 16×64 = 1024 sec tors)

Total = 408+84992 = 85400 sec tors

The required cylinder number is (1284) which will contain the last sector of the file