Computer networks miscellaneous
- In a RSA cryptosystem, a participant A uses two prime numbers p = 13 and q =17 to generate her public and private keys. If the public key of A is 35, then the private key of A is ________.
-
View Hint View Answer Discuss in Forum
As given data P = 13, q = 17, e = 35, d = ?
As per RSA Algorithm.
n = P × q = 13 × 17 = 221
By using RSA formula ,
φ(n) = (P – 1) (q – 1) = 12 × 16 = 192.
To find out the private key of A, then take Modulas operater for RSA formula.
(d × e) mod φ(n) = 1
(d × 35) mod (192) = 1
d = 11, Hence, answer is 11.Correct Option: B
As given data P = 13, q = 17, e = 35, d = ?
As per RSA Algorithm.
n = P × q = 13 × 17 = 221
By using RSA formula ,
φ(n) = (P – 1) (q – 1) = 12 × 16 = 192.
To find out the private key of A, then take Modulas operater for RSA formula.
(d × e) mod φ(n) = 1
(d × 35) mod (192) = 1
d = 11, Hence, answer is 11.
- Which of the following functionalities must be implemented by a transport protocol over and above the network protocol?
-
View Hint View Answer Discuss in Forum
The transport layer of TCP/IP model is very similar to the transport layer of OSI model. The transport layer is more commonly known as Host to Host protocol layer as it is responsible to setup and maintain (end to end communication) between two hosts. Hence, the transport protocols maintain the end to end connectivity.
Correct Option: D
The transport layer of TCP/IP model is very similar to the transport layer of OSI model. The transport layer is more commonly known as Host to Host protocol layer as it is responsible to setup and maintain (end to end communication) between two hosts. Hence, the transport protocols maintain the end to end connectivity.
- How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits and one parity bit?
-
View Hint View Answer Discuss in Forum
Serial communication link (baud) = 9600 = 9600 = 800 (start bit + data bit + end bit + parity bit) 1 + 8 + 2 + 1 12
Hence, 800 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission.Correct Option: B
Serial communication link (baud) = 9600 = 9600 = 800 (start bit + data bit + end bit + parity bit) 1 + 8 + 2 + 1 12
Hence, 800 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission.
- Two hosts are connected via a packet switch with 107 bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 10000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is __________.
-
View Hint View Answer Discuss in Forum
Transmission time for 5000 bits = Time for 500 bits to travel = 5000 = 50 microseconds 107
So, after 50 microseconds the packet is completely transmitted. But the packet must reach at the other and this happens exactly after propagation delay which is 20 microseconds. So, the switch receives packet after 50 + 20 = 70 microseconds.
According to question, switch forwards after 35 microseconds.
So, after 70 + 35 = 105 microseconds, the switch forwards the packet and at 105 + 70 = 175 microseconds the receiver receives the last bit of data.Correct Option: A
Transmission time for 5000 bits = Time for 500 bits to travel = 5000 = 50 microseconds 107
So, after 50 microseconds the packet is completely transmitted. But the packet must reach at the other and this happens exactly after propagation delay which is 20 microseconds. So, the switch receives packet after 50 + 20 = 70 microseconds.
According to question, switch forwards after 35 microseconds.
So, after 70 + 35 = 105 microseconds, the switch forwards the packet and at 105 + 70 = 175 microseconds the receiver receives the last bit of data.
- A link has a transmission speed of 106 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgement has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the oneway propagation delay(in milliseconds) is _______.
-
View Hint View Answer Discuss in Forum
Given, B = 106 bps
L = 1000 bytes
h = 25%
Tp = ?In stop-and-wait, η = 1 1 + 2a ⇒ 1 = 1 = 1 + 2a = 4 4 1 + 2a 2a = 3 ; a = 3 2 Tx = L = 8 × 103 = 8 ms B 106 Tp = 3 ; 2Tp = 3T Tx 2
2Tp = 24 ms
Tp = 12 ms
Correct Option: B
Given, B = 106 bps
L = 1000 bytes
h = 25%
Tp = ?In stop-and-wait, η = 1 1 + 2a ⇒ 1 = 1 = 1 + 2a = 4 4 1 + 2a 2a = 3 ; a = 3 2 Tx = L = 8 × 103 = 8 ms B 106 Tp = 3 ; 2Tp = 3T Tx 2
2Tp = 24 ms
Tp = 12 ms
