Computer networks placement model papers

Computer networks placement model papers

Q1. A full duplex link is used to send data from host X to host Y where X and Y are using sliding window protocol for flow control. The size of send and receive window is 5 packets each. Consider the specifications given below:

Data packets sent only from X to Y are all 1000byte long,
Transmission time is 50μs
Acknowledge packets sent from Y to X are very small
Transmission time of acknowledge packets is negligible.
The propagation delay over the link is 200μs.
In this communication, the maximum achievable throughput is

1. 11.11 x 10⁶ bps
2. 16.11 x 10⁶ bps
3. 21.21 x 10⁶ bps
4. 7.24 x 10⁶ bps
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Q2. The distance between station A and station B = X km.

Length of all frames = L bits.
The propagation delay per kilometer = d s.
The channel capacity = C b/s.
Processing delay = negligible.
Protocol used = sliding window protocol.
For maximum utilization, the minimum number of bits for the sequence number field in a frame =

1. | log₂ ((2XdC + 2L) / L)|
2. | log₂ ((2XdC + L) / 2L)|
3. | log₂ ((2XdC + L) / L)|
4. | log₂ (2XdC / L)|
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Q3. Let P be the client process and S be the server process. P need to make a TCP connection to S. S after executing system calls socket (), bind () and listen () in that order it is preempted. P executes system call socket () followed by system call connect () in order to connect to S. The system call accept () is not executed by S. The event that could take place is

1. System call connect () returns in a core dump
2. System call connect () returns an error
3. System call connect () blocks
4. System call connect () returns successfully.
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For Q4 and Q5 refer to the data given below:
The network given below has 6 routers A1 to A6 and is connected with the links having weights.



Q4. The distance vector based routing algorithm is used by all the routers to update their routing tables. Each router starts with its routing table initialized to contain an entry for each neighbor with the weight of the respective connecting link. In the network after all the routing tables gets stabilized the number of links that will never be used for carrying any data

1. 1
2. 5
3. 6
4. 4
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Q5. Consider a case where the weights of unused links in Q4 are changed to 2. Until all routing tables stabilize, the distance vector algorithm is used again. Now how many links will remain unused?

1. 5
2. 4
3. 2
4. 1
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Q6. Which one of the following statement is true for a layer-4 firewall which is a device that can look at all protocol headers up to the transport layer?

1. It cannot block TCP traffic from a specific user on a multiple-user system during 9.00pm and 5.00pm
2. It cannot stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address.
3. It cannot block entire HTTP traffic during 9.00 pm and 5.00pm
4. It cannot block all ICMP traffic.
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Q7. In HTML, language elements permit certain actions other than describing the structure of the web document. Pure HTML pages (without any server or client side scripting) does not support

1. Automatically redirecting to another page upon download.
2. Displaying the client time as part of the page
3. Refreshing the page automatically after a specified interval.
4. Embed web objects from different sites into the same page.
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Q8. We have source computer S and destination computer D. S transmits a file to D over a network of two routers A1, A2 and three links X1, X2, X3. X1 connects S to A1, X2 connects A1 to A2 and X3 connects A2 to D. The size of file is 10⁶ bits. Each link is of 100 km in length and the speed at which signal travel over each link is 10⁸ m/s. Link bandwidth on each link is 1Mbps. The file that is to be transmitted is broken down into 1000 packets each of size 1000 bits. In transmitting the file from S to D the total sum of transmission and propagation delay is

1. 3000ms
2. 2005ms
3. 1050ms
4. 1005ms
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Q9. An internet service provider (ISP) has 254.248.128.0/20 CIDR-based IP address available with it. The ISP wants to divide and give this chunk of addresses to different organizations and keep some with itself. So ISP decided to give half of this chunk of addresses to A, Quarter of it to B and retained remaining with itself. The valid allocation of addresses to A and B are

1. 245.248.128.0/21 and 245.248.128.0/22
2. 245.248.136.0/21 and 245.248.128.0/22
3. 245.248.132.0/22 and 245.248.128.0/23
4. 245.248.136.0/24 and 245.248.134.0/21
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Q10. Consider TCP’s additive increase multiplicative decrease (AIMD) algorithm. Consider an instance where the window size at the start of the slow start phase is 2MSS and threshold at the start of the first transmission is 8MSS. What will be the congestion window size at the end of tenth transmission if we assume that time-out occurs during the fifth transmission?

1. 7MSS
2. 20MSS
3. 11 MSS
4. 16MSS
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Q11. Encryption is done at A and decryption is done at B. A adds a digital signature s to message S using public key cryptography, encrypts and then sends it to B for decryption. The sequence of keys that is used for the operations is

1. Encryption: A’s private key followed by B’s public key;
Decryption: B’s private key followed by A’s public key.
2. Encryption: A’s public key followed by B’s private key;
Decryption: B’s public key followed by A’s private key.
3. Encryption: A’s private key followed by B’s private key;
Decryption: A’s public key followed by B’s private key.
4. Encryption: A’s private key followed by B’s private key;
Decryption: A’s public key followed by B’s public key.
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ANSWER: 1. Encryption: A’s private key followed by B’s public key;
Decryption: B’s private key followed by A’s public key.