Sliding-Window-Protocol

Question 1
Consider the sliding window flow-control protocol operating between a sender and a receiver over a full-duplex error-free link. Assume the following:
  • The time taken for processing the data frame by the receiver is negligible.
  • The time taken for processing the acknowledgement frame by the sender is negligible.
  • The sender has an infinite number of frames available for transmission.
  • The size of the data frame is 2,000 bits and the size of the acknowledgment frame is 10 bits.
  • The link data rate in each direction is 1 Mbps (=106bits per second).
  • One way propagation delay of the link is 100 milliseconds.
The minimum value of the sender’s window size in terms of the number of frames, (rounded to the nearest integer) needed to achieve a link utilization of 50% is ______.
A
51
Computer-NetworksSliding-Window-ProtocolGATE 2021 CS-Set-1Video-Explanation
Question 1 Explanation: 

Tt(packet) = L / B.W => 2000 bits / 10^6 bps = 2  x 10^-3 sec = 2 millisec


Tt(Ack) = L / B.W. => 10 bits / 10^6 bps = 10^-5 sec = 10^-2 millisec = 0.01 millisec


Tp = 100 millisec


Total time = Tt(packet) + 2 x Tp + Tt(Ack)

=> 2 + 2 x 100 + 0.01 = 202.01 millisec


Efficiency = 50 % = ½


Efficiency = Useful Time /  Total time


½ = n x Tt / Total time 

 

=> 2 x n x Tt =  Total time

=>2 x n x 2 = 202.01  

=> n = 202.01 / 4 => 50.50

 

For minimum, we have to take ceil, Hence size of window = 51

Question 2

Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 µs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 µs. What is the maximum achievable throughput in this communication?

A
7.69 × 106 bps
B
11.11 × 106 bps
C
12.33 × 106 bps
D
15.00 × 106 bps
Computer-NetworksSliding-Window-ProtocolGATE 2003
Question 2 Explanation: 
Given, Tt = 50 μs, Tp = 200 μs, L = 1000 bytes, N = 5,
Transmission rate , Tt = L / B.W
Therefore, B.W. = L / Tt = 1000 bytes/ 50 μs = 8000 bits / 50 μs=160 Mbps
Efficiency = N / 1 + 2a, where a = Tp / Tt
Efficiency = 5 * 50 / (50+400) = 250/450 = 5/9
Maximum achievable throughput = Efficiency * B.W = (5/9)*160 Mbps = 88.88 Mbps = = 11.11 x 106 bytes per second
*Actual option should be in bytes per second.
Question 3
Consider a network using the pure ALOHA medium access control protocol, where each frame is of length 1,000 bits. The channel transmission rate is 1 Mbps (= 106bits per second). The aggregate number of transmissions across all the nodes (including new frame transmissions and retransmitted frames due to collisions) is modelled as a Poisson process with a rate of 1,000 frames per second. Throughput is defined as the average number of frames successfully transmitted per second. The throughput of the network (rounded to the nearest integer) is _________. 
A
135
Computer-NetworksSliding-Window-ProtocolGATE 2021 CS-Set-2Video-Explanation
Question 3 Explanation: 

1 frames takes = Tt = L / B.w. => 1000 / 10^6 = 1 millisec

1000 frame Tt = 1000 x 1 millisec = 1 sec

In 1 sec, 1000 frames sends, which is 1 millisec per frame.

So, G = 1

 

Efficiency of Pure Aloha (η) = G x e-2G

where G = Number of requests per time slot willing to transmit.

e = Mathematical constant approximately equal to 2.718

So, η = 1 x 2.718(-2 x 1) = 0.1353

Therefore, In 1 sec1000 frames = 0.1353 x 1000 = 135.3(closest integer) =>135

 

Throughput =>  135

Question 4
Consider the three-way handshake mechanism followed during TCP connection established between hosts P and Q. Let X and Y be two random 32-bit starting sequence numbers chosen by P and Q respectively. Suppose P sends a TCP connection request message to Q with a TCP segment having SYN bit = 1, SEQ number = X, and ACK bit = 0. Suppose Q accepts the connection request. Which one of the following choices represents the information present in the TCP segment header that is sent by Q to P?
A
SYN bit = 1, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 0
B
SYN bit = 0, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 1
C
SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X+1, FIN bit = 0
D
SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X, FIN bit = 0
Computer-NetworksSliding-Window-ProtocolGATE 2021 CS-Set-2Video-Explanation
Question 4 Explanation: 

 

Q will send the SYN bit = 1 to the connection establishment.

Q Seq number will be Y different from X

ACK bit = 1 because sending the ACK

ACK number = X+1 (Next seq number id)

FIN bit = 0 (Because establishing the connection) 

Question 5

Suppose that it takes 1 unit of time to transmit a packet (of fixed size) on a communication link. The link layer uses a window flow control protocol with a window size of N packets. Each packet causes an ack or a nak to be generated by the receiver, and ack/nak transmission times are negligible. Further, the round trip time on the link is equal to N units. Consider time i > N. If only acks have been received till time i(no naks), then the goodput evaluated at the transmitter at time i(in packets per unit time) is

A
1 – N/i
B
i/(N + i)
C
1
D
1 – e(i/N)
Computer-NetworksSliding-Window-ProtocolGATE 2006-IT
Question 5 Explanation: 
Goodput is the application level throughout, i.e., the no. of useful information bits delivered by the network to a certain destination per unit of time.
So, successful delivery of packet can be assured if ack has been received for it.
So till time 'i' we would have transmitted 'i' packets but only (i - N) can be acknowledged as minimum time for a packet to get acknowledged is N (since RTT is N which is equal to the window size, there is no waiting for the sender).
So, successfully delivered packets = (i - N)
Time for transmission = i
Goodput = Successfully delivered data/Time
= (i - N)/i
= 1 - N/i
Question 6

A 1Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 × 108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.

A
120 bytes
B
60 bytes
C
240 bytes
D
90 bytes
Computer-NetworksSliding-Window-ProtocolGATE 2008-IT
Question 6 Explanation: 
Time to reach satellite = 36504000/3×108 = 0.121685
RTT = 4×Time to reach satellite (S1→S, S→S2, S2→S, S→S1)

∴ RTT = 0.48
Efficiency = N×Tt/Tt+2Tp
= N×Tt/Tt+RTT
0.25 = 127×Tt/Tt+0.48
0.25Tt + 0.25 × 0.48 = 127Tt
0.25 × 0.48 = 126.5Tt
0.25 × 0.48 × 106/126.5 = L
L = 952 bit ≈ 120 byte
There are 6 questions to complete.

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