# Computer Networks (Test 4)

## Gate Exam : Cs Computer Science And Information Technology

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Computer Networks
| Computer Networks |
Q.1
Match the problem domains in Group I with the solution technologies in Group II
Group I                                                                     Group II
(p) Services oriented computing                                      (1) Interoperability
(q) Heterogeneous communicating systems                    (2) BPMN
(R) Information representation                                         (3) Publish-find bind
(S) Process description                                                    (4) XML
A. P – 1, Q – 2, R – 3, S – 4
B. P – 3, Q – 4, R – 2, S – 1
C. P – 3, Q – 1, R – 4, S – 2
D. P – 4, Q – 3, R – 2, S – 1
Explaination / Solution:
No Explaination.

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Q.2
Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same net mask N. Which of the values of N given below should not be used if A and B should belong to the same network?
A. 255.255.255.0
B. 255.255.255.128
C. 255.255.255.192
D. 255.255.255.224
Explaination / Solution:
No Explaination.

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Q.3
Consider a network with 6 routers R1 to R6 connected with links having weights as shown in the following diagram

All the routers use the distance vector based routing algorithm to update their routing tables. Each router starts with its routing table initialized to contain an entry for each neighbour with the weight of the respective connecting link. After all the routing tables stabilize, how many links in the network will never be used for carrying any data?
A. 4
B. 3
C. 2
D. 1
Explaination / Solution:
No Explaination.

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Q.4
Consider a network with 6 routers R1 to R6 connected with links having weights as shown in the following diagram

Suppose the weights of all unused links in the previous question are changed to 2 and the distance vector algorithm is used again until all routing tables stabilize. How many links will now remain unused?
A. 0
B. 1
C. 2
D. 3
Explaination / Solution:
No Explaination.

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Q.5
In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are
A. Last fragment, 2400 and 2789
B. First fragment, 2400 and 2759
C. Last fragment, 2400 and 2759
D. Middle fragment, 300 and 689
Explaination / Solution:

M= 0 – Means there is no fragment after this, i.e. Last fragment
HLEN=10 - So header length is 4×10=40, as 4 is constant scale factor
Fragment Offset = 300, that means 300×8 Byte = 2400 bytes are before this last fragment
So the position of datagram is last fragment
Sequence number of First Byte of Payload = 2400 (as 0 to 2399 Sequence no are used)
Sequence number of Last Byte of Payload = 2400+360-1=2759

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Q.6
Determine the maximum length of cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s
A. 1
B. 2
C. 2.5
D. 5
Explaination / Solution:

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Q.7
Consider the following sequence of micro–operations
MBR ← PC
MAR ← X
PC ← Y
Memory ← MBR
Which one of the following is a possible operation performed by this sequence?
A. Instruction fetch
B. Operand fetch
C. Conditional branch
D. Initiation of interrupt service