Computer Networks is crucial subject in computer science engineering stream. Here I am listing multiple-choice questions (MCQs) in the field of networking. These MCQs are previous years questions of GATE ( Graduate Aptitude Test in Engineering ) exam in India. But questions are good if you want to brush up your knowledge in the subject.
Consider a network with 6 routers R1 and R6 connected with links having weights as shown in the following diagram.
Question 1 and 2 are based on above statement
Q1. All the routers use the distance vector based routing algorithm to update their routing tables. Each 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?(Gate 2010)
Q2. 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?(Gate 2010)
Q3. Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same netmask N. Which of the values of N given below should not be used if A and B should belong to the same network? (Gate 2010)
Q4. Which one of the following is not a client-server application?(Gate 2010)
- Internet chat
- Web browsing
Q5. One of the header fields in an IP datagram is the Time-to-Live (TTL) field. Which of the following statements best explains the need for this field? (Gate 2010)
- It can be used to prioritize packets
- It can be used to reduce delays
- It can be used to optimize throughput
- It can be used to prevent packet looping
Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25 ms. Frames are to be transmitted into to maximally pack them in transit (within the link).
Question 6 and 7 are based on above statement
Q6. What is the minimum number of bits (l) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.(Gate 2009)
- l = 2
- l = 3
- l = 4
- l = 5
Q7. Suppose that the sliding window protocol is used with the sender window size of 2l, where l is the number of bits identified in the earlier part and acknowledgments is always piggybacked. After sending 2l frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time)(Gate 2009)
- 16 ms
- 18 ms
- 20 ms
- 22 ms
Q8. LetG(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect odd number of bits in error?(Gate 2009)
- G(x) contains more than two terms
- G(x) does not divide 1 + xk , for any K not exceeding the frame length
- 1 + x is a factor of G(x)
- G(x) has an odd number of terms
Q9.While opening a TCP connection, the initial sequence number is to be derived using a time-of-day (ToD) clock that keeps running even when the host is down. The low order 32 bits of the counter of TOD clock is to be used for the initial sequence numbers. The clock counter increments once per millisecond. The maximum packet lifetime is given to be 64s.
Which one of the choices given below is closet to the minimum permissible rate at which sequence numbers used for packets of a connection can increase?(Gate 2009)
Q10. In the RSA public key cryptosystem, the private and the public keys are (e, n) and (d, n) respectively, where n = p* and p and q are large primes. Besides, n is public and p and q are private. Let M be an integer such that 0 < M < nand φ(n) = (p − 1) (q − 1). Now consider the following equations.
I. M' = Me mod n
M = (M')d mod n
II. ed ≡ 1mod n
III.ed ≡ 1modφ(n)
IV. M' = Memod φ(n)
M = (M')dmod φ(n)
Which of the above equations correctly represent RSA cryptosystem? (Gate 2009)
- I and II
- I and III
- II and IV
- III and IV
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Disclaimer: This tutorial is for educational purpose only. Individual is solely responsible for any illegal act.