Computer networks miscellaneous
- Consider a TCP client and a TCP server running on two different machines. After completing data transfer, the TCP client calls close to terminate the connection and a FIN segment is sent to the TCP server. Server-side TCP responds by sending an ACK which is received by the client-side TCP. As per the TCP connection state diagram (RFC 793), in which state does the client-side TCP connection wait for the FIN from the server-side TCP?
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Correct Option: D
- The values of parameters for the Stop-and-Wait ARQ protocol are as given below :
Bit rate of the transmission channel = 1 Mbps.
Propagation delay from sender to receiver = 0.75 ms.
Time to process a frame = 0.25 ms.
Number of bytes in the information frame = 1980.
Number of bytes in the acknowledge frame = 20.
Number of overhead bytes in the information frame = 20.
Assume that there are no transmission errors. Then, the transmission efficiency (expressed in percentage) of the Stop-and-Wait ARQ protocol for the above parameters is _______ (correct to 2 decimal places).
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Given data: B = 1 Mbps , Tprocess = 0.25 ms, TPt = 0.75 ms LIF = 1980 bytes , L(Overhead) = 20 bytes , LAF = 20 bytes
Efficiency (η) = ?
As we know that,Transmission time of data = Data size Bandwidth So , TX = L = ( LIf + LAf ) = ( 1980 + 20 ) bytes B B 1 × 106( bits / sec )
(∵ 1 byte = 8 bit)= 2000 × 8 (bits) 106( bits / sec )
TX = 16 m sec
Propagation time = 0.75 m sec.Transmission time of TACK = LAF = 20 × 8 bits B 106
∴ TACK = 0.16 m sec.
Then transmission efficiency of stop and wait ARQ.⇒ TX = 16 (TX + TACK + 2TPt + TProcess) (16 0.16 + 2 × 0.75 + 0.25) = 16 = 0.8933 = 89.33% 17.91
Hence answer is 89.33%Correct Option: B
Given data: B = 1 Mbps , Tprocess = 0.25 ms, TPt = 0.75 ms LIF = 1980 bytes , L(Overhead) = 20 bytes , LAF = 20 bytes
Efficiency (η) = ?
As we know that,Transmission time of data = Data size Bandwidth So , TX = L = ( LIf + LAf ) = ( 1980 + 20 ) bytes B B 1 × 106( bits / sec )
(∵ 1 byte = 8 bit)= 2000 × 8 (bits) 106( bits / sec )
TX = 16 m sec
Propagation time = 0.75 m sec.Transmission time of TACK = LAF = 20 × 8 bits B 106
∴ TACK = 0.16 m sec.
Then transmission efficiency of stop and wait ARQ.⇒ TX = 16 (TX + TACK + 2TPt + TProcess) (16 0.16 + 2 × 0.75 + 0.25) = 16 = 0.8933 = 89.33% 17.91
Hence answer is 89.33%
- Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 106 bytes/ sec. A user on host A sends a file of size 103 bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?
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I Transmission
1000 + 100 = 1100 bytes (transmitted from A at a time)
In 1 second → 106 bytes∴ For 1100 bytes ⇒ 1100 × 8 = 8.8 m sec.(transmission time) 106
From A, R1 and R2 it takes T1 = 8.8 + 8.8 + 8.8 = 26.4 m sec. (to reach B)
II Transmission
100 + 100 = 200 bytes (transmitted 10 times from A)For 200 bytes ⇒ 200 × 8 = 1.6 m sec. 106
From A, transmission time to entire packet = 1.6 × 10 = 16 m sec.
From A, R1 and R2 it takes T2 = 16 + 1.6 + 1.6 = 19.2 m sec. (to reach B)
III transmission
50 + 100 = 150 bytes transmitted 20 times from A.For 150 bytes ⇒ 150 × 8 = 1.2 m sec. 106
From A, transmission time for entire packet = 1.2 × 20 = 24 m sec.
From A, R1 and R2 it takes T3 = 24 + 1.2 + 1.2 = 26.4 m sec. (to reach B)
∴ T1 = T3 = 26.4 m sec.
T2 = 19.6 m sec. < T3Correct Option: D
I Transmission
1000 + 100 = 1100 bytes (transmitted from A at a time)
In 1 second → 106 bytes∴ For 1100 bytes ⇒ 1100 × 8 = 8.8 m sec.(transmission time) 106
From A, R1 and R2 it takes T1 = 8.8 + 8.8 + 8.8 = 26.4 m sec. (to reach B)
II Transmission
100 + 100 = 200 bytes (transmitted 10 times from A)For 200 bytes ⇒ 200 × 8 = 1.6 m sec. 106
From A, transmission time to entire packet = 1.6 × 10 = 16 m sec.
From A, R1 and R2 it takes T2 = 16 + 1.6 + 1.6 = 19.2 m sec. (to reach B)
III transmission
50 + 100 = 150 bytes transmitted 20 times from A.For 150 bytes ⇒ 150 × 8 = 1.2 m sec. 106
From A, transmission time for entire packet = 1.2 × 20 = 24 m sec.
From A, R1 and R2 it takes T3 = 24 + 1.2 + 1.2 = 26.4 m sec. (to reach B)
∴ T1 = T3 = 26.4 m sec.
T2 = 19.6 m sec. < T3
- An IP datagram of size 1000 bytes arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is 100 bytes. Assume that the size of the IP header is 20 bytes. The number of fragments that the IP datagram will be divided into for transmission is __________ .
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Therefore the no. of fragments that are transferred in this case is 13.Correct Option: A
Therefore the no. of fragments that are transferred in this case is 13.
- Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a web page from a remote server, assuming that the host has just been restarted.
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When a web browser requests any web page from a remote server then that request (URL address) is mapped to IP Address using DNS, then TCP synchronisation occurs. Finally HTTP verifies that the address exists in the web server or not.
Correct Option: C
When a web browser requests any web page from a remote server then that request (URL address) is mapped to IP Address using DNS, then TCP synchronisation occurs. Finally HTTP verifies that the address exists in the web server or not.
