TCP

Question 1
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= 109  bits/second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.
A
33
B
34
C
35
D
36
       Computer-Networks       TCP       Gate 2018
Question 1 Explanation: 
In TCP, Sequence number field is 32 bit, which means 232 sequence number per byte are possible. Whatever be the starting sequence number the possible number will be 232bytes
The process of using all the sequence number and repeating a previously used sequence number.
The time taken to wrap around is called wrap around time:
Minimum Time = Wrap around time = Total number of bits in sequence number / Bandwidth = 232* 8 / 109= 34.35 == 34 (closest integer)
Question 2
Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a webpage from a remote server, assuming that the host has just been restarted.
A
HTTP GET request, DNS query, TCP SYN
B
DNS query, HTTP GET request, TCP SYN
C
DNS query, TCP SYN, HTTP GET request
D
TCP SYN, DNS query, HTTP GET request
       Computer-Networks       TCP       GATE 2016 set-2
Question 2 Explanation: 
When a browser requests a web page from a remote server then that requests (URL address) will be mapped to IP address using DNS query, then TCP synchronization takes place after that HTTP verify whether it is existed in the web server or not.
Question 3
   
A
III only
B
I and III only
C
I and IV only
D
II and IV only
       Computer-Networks       TCP       GATE 2015 (Set-01)
Question 3 Explanation: 
S1: FALSE
The sequence number of the subsequent segment depends on the number of 8-byte characters in the current segment.
S2: TRUE
Depending on the value of α or Estimated RTT it may or may not be greater than 1.
S3: FALSE
It is the size of the receiver's buffer that's never changed. Receive Window is the part of the receiver's buffer that's changing all the time depending on the processing capability at the receiver's side and the network traffic.
S4: TRUE
The number of unacknowledged bytes that A sends cannot exceed the size of the receiver's window. But if it can't exceed the receiver's window, then it surely has no way to exceed the receiver's buffer as the window size is always less than or equal to the buffer size. On the other hand, for urgent messages, the sender CAN send it in even though the receiver's buffer is full.
Question 4
Assume that the bandwidth for a TCP connection is 1048560 bits/sec. Let  α be the value of RTT in milliseconds. (rounded off to the nearest integer) after which the TCP window scale option is needed. Let β be the maximum possible window size the window scale option. Then the values of α and β are
A
63 milliseconds, 65535×214
B
63 milliseconds, 65535×216
C
500 milliseconds, 65535×214
D
500 milliseconds, 65535×216
       Data-Structures       TCP       GATE 2015 -(Set-2)
Question 4 Explanation: 
TCP header sequence number field consist 16 bits. The maximum number of sequence numbers possible = 2^16 = 65,535.
The wrap around time for given link = 1048560 * α. The TCP window scale option is an option to increase the receive window size. TCP allows scaling of windows when wrap around time > 65,535.
==> 1048560 * α. > 65,535*8 bits
==> α = 0.5 sec = 500 mss
Scaling is done by specifying a one byte shift count in the header options field. The true receiver window size is left shifted by the value in shift count. A maximum value of 14 may be used for the shift count value. Therefore maximum window size with scaling option is 65535 × 2^14
.
Question 5
     
A
Only I is correct
B
Only I and III are correct
C
Only II and III are correct
D
All of I, II and III are correct
       Computer-Networks       TCP       GATE 2015(Set-03)
Question 5 Explanation: 
In TCP, as sender and receiver can send segments at the same time, It is FULL-DUPLEX. TCP can use selective ACK and TCP uses byte streams that is every byte is send using TCP is numbered.
Question 6
A
4,2,1,3
B
1,2,3,4
C
4,1,2,3
D
2,4,1,3
       Computer-Networks       TCP       GATE 2014(Set-01)
Question 6 Explanation: 
First of all the browser must now know what IP to connect to. For this purpose browser takes help of Domain name system (DNS) servers which are used for resolving hostnames to IP addresses. As browser is an HTTP client and as HTTP is based on the TCP/IP protocols, first it establishes a TCP connection with the web server and requests a web page using HTTP, and then the web server sends the requested web page using HTTP. Hence the order is 4,2,1,3.
Question 7
Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.
A
1100 to 1300
B
1101 to 1301
C
1102 to 1302
D
1103 to 1303
       Computer-Networks       TCP       GATE 2014(Set-01)
Question 7 Explanation: 
Given that at the time of Time Out, Congestion Window Size is 32KB and RTT = 100ms
When Time Out occurs, for the next round of Slow Start, Threshold = (size of Cwnd) / 2
It means Threshold = 16KB
Slow Start
2KB
1RTT
4KB
2RTT
8KB
3RTT
16KB ----------- Threshold reaches. So Additive Increase Starts
4RTT
18KB
5RTT
20KB
6RTT
22KB
7RTT
24KB
8RTT
26KB
9RTT
28KB
10RTT
30KB
11RTT
32KB
So, Total no. of RTTs = 11 → 11 * 100 = 1100
Question 8
A
(i) only
B
(i) and (ii) only
C
(i) and (ii) only
D
(i), (ii) and (iii)
       Computer-Networks       TCP       Gate 2014 Set -03
Question 8 Explanation: 
While an IP Datagram is transferring from one host to another host, TTL, Checksum and Fragmentation Offset will be changed.
Question 9
A
0.015/s
B
0.064/s
C
0.135/s
D
0.327/s
       Computer-Networks       TCP       2009
Question 9 Explanation: 
32 bits are used to represent a sequence number. So we have 232 different sequence number.
The maximum packet lifetime is given is given 64s.
Maximum data rate possible(bandwidth) to avoid the wraparound = 232/64 = 226 Byte/sec.
The clock counter increments once per milliseconds = That means when then counter increments next possible sequence number is generated. The packet lifetime is 64 seconds and after this 64 seconds next sequence number is come. So that means in this 64 seconds only 1 sequence number is generated.
Hence the minimum rate is = 1/64 = 0.015/sec.
Question 10
In the slow start phase of the TCP congestion control algorithm, the size of the congestion window
A
does not increase
B
increases linearly
C
increases quadratically
D
increases exponentially
       Computer-Networks       TCP       Gate-2008
Question 10 Explanation: 
In slow start phase, window size will grow exponentially. when the threshold is reached and congestion avoidance phase begins. In congestion avoidance phase, the window is increased linearly.
Question 11
 
A
S2 and S3 only
B
S1 and S4
C
S1 and S3
D
S2 and S4
       Computer-Networks       TCP       Gate 2008-IT
Question 11 Explanation: 
S1 → True.
S2 → False, because if after ACK client immediately sends data then everything goes on without worry.
S3 → False.
S4 → True.
There are 11 questions to complete.