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CSE - Questions and solutions set-2

Computer science and engineering(CSE) - Question & solution set-2
Question:-
Option (A)
50bytes
Option (B)
100 bytes
Option (C)
200bytes
Option (D)
300bytes
Correct Option:
C
Question Solution:
Total time taken will be 4km/2x108=20microsec

The size of packet will be 20microsecx107=200bytes
Question:-
Option (A)
120.4 ms
Option (B)
160.5 ms
Option (C)
165.5 ms
Option (D)
590.0 ms
Correct Option:
C
Question Solution:
For K stage pipeline time taken will be

T=(K+(n-1))t where t=td+5 and td is the maximum delay time =160nsec

So here t=165nsec

Here K=4 n=1000

So here T=(4+(1000-1))165nsec=165.5msec
Question:-
Option (A)
93 g
Option (B)
0
Option (C)
14 g
Option (D)
none of above
Correct Option:
C
Question Solution:
same as (a/b)*g
Question:-
Option (A)
Ambiguous
Option (B)
Non ambiguous
Option (C)
Information is not sufficient to decide
Option (D)
None
Correct Option:
C
Question:-
Option (A)
1.83
Option (B)
2
Option (C)
3
Option (D)
6
Correct Option:
D
Question Solution:
This pipelined process have 6 states, lets look at the no of cycles required to complete this

First instruction will be executed in 3 + 2 + 5 + 4 + 6 + 2 = 22 cycles but subsequent instructions will be executed in 6 instructions.

Though you can solved it by doing complicated maths looking at what will be the condition after 22, 23 .... cycles but normally pipelined stage with highest number of cycle determine the speed.

Good method to calculate is to break and see the problem

if there were only one pipeline with 3 cycles then time would be 3 cycles

if there were only two pipeline stage with 3 cycles and two cycles then time would be again 3 cycles(small thought will make you analyze this)

similary after 3rd stage - 5 cycles

similary after 4th stage - 5 cycles

similary after 5th stage - 6 cycles

similary after 6th stage - 6 cycles