21. Which
of the following services is not provided by wireless access point in 802.11
WLAN?
(A) Association (B) Disassociation
(C) Error
Correction (D) Integration
Answer:
C
Explanation:
22. Which
of the following fields in IPv4 datagram is not related to fragmentation?
(A) Type of
service (B) Fragment
offset
(C) Flags (D) Identification
Answer:
A
23. Four
channels are multiplexed using TDM. If each channel sends 100 bytes/second and
we multiplex 1 byte per channel, then the bit rate for the link is ...............
(A) 400 bps (B) 800 bps
(C) 1600 bps (D) 3200 bps
Answer:
D
Explanation:
The
multiplexer is shown in the Figure.
Each frame carries
1 byte from each channel; the size of each frame, therefore, is 4 bytes, or 32
bits. Because each channel is sending 100 bytes/s and a frame carries 1 byte
from each channel, the frame rate must be 100 frames per second. The bit rate
is 100 × 32 = 3200 bps.
24. In
a typical mobile phone system with hexagonal cells, it is forbidden to reuse a
frequency band in adjacent cells. If 840 frequencies are available, how many
can be used in a given cell?
(A) 280 (B) 210
(C) 140 (D) 120
Answer:
A
Explanation:
Each cell
has six other adjacent cells, in a hexagonal grid. If the central cell uses
frequency group A, its six adjacent cells can use B, C, B, C, B, and C
respectively. In other words, only 3 unique cells are needed. So the answer is
840/3 = 280 frequencies.
25. Using
p=3, q=13, d=7 and e=3 in the RSA algorithm, what is the value of ciphertext
for a plain text 5?
(A) 13 (B) 21
(C) 26 (D) 33
Answer:
Marks to all
Explanation:
p=3, q=13,
d=7, e=3, M=5, C=?
C = Me mod n
n = p*q
= 3*13 = 39
C = 53 mod 39
= 8
Answer is 8.
26. A
virtual memory has a page size of 1K words. There are eight pages and four
blocks. The associative memory page table contains the following entries:
Which of the
following list of virtual addresses (in decimal) will not cause any page fault
if referenced by the CPU?
(A) 1024,
3072, 4096, 6144 (B) 1234,
4012, 5000, 6200
(C) 1020,
3012, 6120, 8100 (D) 2021,
4050, 5112, 7100
Answer:
C
Explanation:
The pages
which are not in main memory are:
Page
|
Address
|
Address
that will cause page fault
|
1
|
1K
|
1024-2047
|
3
|
3K
|
3072-4095
|
4
|
4K
|
4096-5119
|
6
|
6K
|
6144-7167
|
1020 will
not cause page fault (1024-2047)
3012 will
not cause page fault (3072-4095)
6120 will
not cause page fault (4096-5119)
8100 will
not cause page fault (6144-7167)
27. Suppose
that the number of instructions executed between page faults is directly
proportional to the number of page frames allocated to a program. If the
available memory is doubled, the mean interval between page faults is also
doubled. Further, consider that a normal instruction takes one micro second,
but if a page fault occurs, it takes 2001 micro seconds. If a program takes 60
sec to run, during which time it gets 15000 page faults, how long would it take
to run if twice as much memory were available?
(A) 60 sec (B) 30 sec
(C) 45 sec (D) 10 sec
Answer:
C
Explanation:
T = Ninstr
x 1µs + 15,000 x 2,000 µs = 60s
Ninstr
x 1µs = 60,000,000 µs – 30,000,000 µs = 30,000,000 µs
Ninstr
= 30,000,000
The number
of instruction between two page faults is
Ninstr /NPageFaults
= 30,000,000/15,000 = 2,000
If the mean
interval between page faults is doubled, the number of instruction between two
page faults is also doubled and is 4,000. Now the number of page faults is
30,000,000/4,000
= 7,500
T’ = 30,000,000 µs + 7,500 x 2,000 µs
=
30,000,000 µs + 15,000,000 µs = 45,000,000 µs = 45s
Doubling the
memory, doesn’t mean that the program runs twice as fast as on the first
system. Here, the performance increase is of 25%.
28. Consider
a disk with 16384 bytes per track having a rotation time of 16 msec and average
seek time of 40 msec. What is the time in msec to read a block of 1024 bytes
from this disk?
(A) 57 msec (B) 49 msec
(C) 48 msec (D) 17 msec
Answer:
B
Explanation:
Time in msec
to read a block of 1024 bytes (Access time or Disk Latency) = seek time
+ average rotational delay + transfer time
If there are
16384 bytes per track there are 1024/16384 tracks to be read for this block.
Seek time =
40 msec
Rotational
delay = 16 msec
Transfer
time = (sectors_read/sectors per rev.)
x rotational delay
= (1024/16384) x 16
= 1
average
rotational delay = rotational delay/2 = 16/2 = 8
access time = 40 + 8 + 1 = 49
29. A
system has four processes and five allocatable resources. The current
allocation and maximum needs are as follows:
Allocated
|
Maximum
|
Available
|
|
Process A
|
1 0 2 1 1
|
1 1 2 1 3
|
0 0 x 1 1
|
Process B
|
2 0 1 1 0
|
2 2 2 1 0
|
|
Process C
|
1 1 0 1 0
|
2 1 3 1 0
|
|
Process D
|
1 1 1 1 0
|
1 1 2 2 1
|
The smallest
value of x for which the above system in safe state is .............
(A) 1 (B) 3
(C) 2 (D) 0
Answer:
Marks to all
Explanation:
If
Process A’s Maximum need is 1 1 2 1 2 instead of 1 1 2 1 3, then answer will be
x=1
The needs
matrix is as follows:
0 1 0 0 1
0 2 1 0 0
1 0 3 0 0
0 0 1 1 1
If x is 0, available
vector will be 0 0 0 1 1, we have a deadlock immediately.
If x is 1, available
vector will be 0 0 1 1 1, now, process D can run to completion. When it is
finished, the available vector is 1 1 2 2 1.
Now A can
run to complete, the available vector then becomes 2 1 4 3 2.
Then C can
run and finish, return the available vector as 3 2 4 4 2.
Then B can
run to complete. Safe sequence D A C B.
30. In
Unix, the login prompt can be changed by changing the contents of the file
...............
(A) contrab (B) init
(C) gettydefs (D) inittab
Answer:
C
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