Tag: Sliding Window Protocol

Stop and Wait ARQ | Go back N | SR Protocol

Flow Control Protocols-

 

In computer networking, there are various flow control protocols-

 

 

Sliding Window Protocol-

 

  • Sliding window protocol allows the sender to send multiple frames before needing the acknowledgements.
  • It is more efficient.

 

Read More- Sliding Window Protocol

 

Implementations-

 

Various implementations of sliding window protocol are-

  1. Stop and Wait ARQ
  2. Go back N
  3. Selective Repeat

 

In this article, we will compare these sliding window protocols.

 

Comparison Table-

 

Stop and Wait ARQGo back NSelective RepeatRemarks
Efficiency1 / (1+2a)N / (1+2a)N / (1+2a)Go back N and Selective Repeat gives better efficiency than Stop and Wait ARQ.
Window SizeSender Window Size = 1

Receiver Window Size = 1

Sender Window Size = N

Receiver Window Size = 1

Sender Window Size = N

Receiver Window Size = N

Buffer requirement in Selective Repeat is very large.

If the system does not have lots of memory, then it is better to choose Go back N.

Minimum number of sequence numbers required2N+12 x NSelective Repeat requires large number of bits in sequence number field.
Retransmissions required if a packet is lostOnly the lost packet is retransmittedThe entire window is retransmittedOnly the lost packet is retransmittedSelective Repeat is far better than Go back N in terms of retransmissions required.
Bandwidth RequirementBandwidth requirement is LowBandwidth requirement is high because even if a single packet is lost, entire window has to be retransmitted.

Thus, if error rate is high, it wastes a lot of bandwidth.

Bandwidth requirement is moderateSelective Repeat is better than Go back N in terms of bandwidth requirement.
CPU usageLowModerateHigh due to searching and sorting required at sender and receiver sideGo back N is better than Selective Repeat in terms of CPU usage.
Level of difficulty in ImplementationLowModerateComplex as it requires extra logic and sorting and searchingGo back N is better than Selective Repeat in terms of implementation difficulty.
AcknowledgementsUses independent acknowledgement for each packetUses cumulative acknowledgements (but may use independent acknowledgements as well)Uses independent acknowledgement for each packetSending cumulative acknowledgements reduces the traffic in the network but if it is lost, then the ACKs for all the corresponding packets are lost.
Type of TransmissionHalf duplexFull duplexFull duplexGo back N and Selective Repeat are better in terms of channel usage.

 

Conclusions-

 

  • Go back N is more often used than other protocols.
  • SR protocol is less used because of its complexity.
  • Stop and Wait ARQ is less used because of its low efficiency.
  • Depending on the context and resources availability, Go back N or Selective Repeat is employed.
  • Selective Repeat and Stop and Wait ARQ are similar in terms of retransmissions.
  • Go back N and Selective Repeat are similar in terms of efficiency if sender window sizes are same.
  • SR protocol may be considered as a combination of advantages of Stop and Wait ARQ and Go back N.
  • SR protocol is superior to other protocols but because of its complexity, it is less used.

 

Important Notes-

 

Note-01:

 

Protocols at data link layer like HDLC (Low level protocols) use Go back N.

This is because-

  1. Bandwidth is high
  2. CPU is very busy doing routing job
  3. Error rate is low since out of order packets are not possible in wired medium

 

Note-02:

 

Protocols at transport layer like TCP (High level protocols) use selective repeat.

 

PRACTICE PROBLEMS BASED ON SLIDING WINDOW PROTOCOLS-

 

Problem-01:

 

If the bandwidth between the sender and receiver is sufficient, CPU and buffers are moderate, then which flow control protocol would you suggest to use?

 

Solution-

 

The suggested protocol would be Go back N.

 

Problem-02:

 

If the bandwidth between the sender and receiver is moderate, CPU and buffers are sufficient, then which flow control protocol would you suggest to use?

 

Solution-

 

The suggested protocol would be Selective Repeat.

 

To gain better understanding about sliding window protocols comparison,

Watch this Video Lecture

 

Next Article- Miscellaneous Problems On Flow Control Protocols

 

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Watch video lectures by visiting our YouTube channel LearnVidFun.

Flow Control Protocols | Practice Problems

Flow Control Protocols-

 

In computer networking, there are various flow control protocols-

 

 

 

In this article, we will discuss practice problems based on these flow control protocols.

 

PRACTICE PROBLEMS BASED ON FLOW CONTROL PROTOCOLS-

 

Problem-01:

 

In what protocols is it possible for the sender to receive an acknowledgement for a packet that falls outside its current window?

  1. Stop and Wait
  2. Selective Repeat
  3. Go back N
  4. All of the above

 

Solution-

 

  • Delayed Acknowledgements fall outside the current window.
  • They may occur in any of the flow control protocols and received by the sender.

 

Thus, correct option is (D).

 

Problem-02:

 

On a wireless link, the probability of packet error is 0.2. A stop and wait protocol is used to transfer data across the link. The channel condition is assumed to be independent from transmission to transmission. What is the average number of transmission attempts required to transfer 100 packets?

  1. 100
  2. 125
  3. 150
  4. 200

 

Solution-

 

Method-01:

 

Given-

  • Probability of packet error = 0.2
  • We have to transfer 100 packets

 

Now,

  • When we transfer 100 packets, number of packets in which error will occur = 0.2 x 100 = 20.
  • Then, these 20 packets will have to be retransmitted.
  • When we retransmit 20 packets, number of packets in which error will occur = 0.2 x 20 = 4.
  • Then, these 4 packets will have to be retransmitted.
  • When we retransmit 4 packets, number of packets in which error will occur = 0.2 x 4 = 0.8 ≅ 1.
  • Then, this 1 packet will have to be retransmitted.

 

From here, average number of transmission attempts required = 100 + 20 + 4 + 1 = 125.

Thus, Option (B) is correct.

 

Method-02:

 

REMEMBER

If there are n packets to be transmitted and p is the probability of packet error, then-

Number of transmission attempts required

= n + np + np2 + np3 + …… + ∞

= n / (1-p)

 

Substituting the given values, we get-

Average number of transmission attempts required = 100 / (1-0.2) = 125.

Thus, Option (B) is correct.

 

Problem-03:

 

Compute the fraction of the bandwidth that is wasted on overhead (headers and retransmissions) for a protocol on a heavily loaded 50 Kbps satellite channel with data frames consisting of 40 bits header and 3960 data bits. Assume that the signal propagation time from the earth to the satellite is 270 msec. ACK frames never occur. NAK frames are 40 bits. The error rate for data frames is 1% and the error rate for NAK frames is negligible.

  1. 1.21 %
  2. 2.12 %
  3. 1.99 %
  4. 1.71 %

 

Solution-

 

Consider 100 frames are being sent. Then, we have-

 

Useful Data Sent-

 

Since each frame contains 3960 data bits, so while sending 100 frames,

Useful data sent

= 100 x 3960 bits

= 396000 bits

 

Useless Data Sent / Overhead-

 

In general, overhead is due to headers, retransmissions and negative acknowledgements.

 

Now,

  • The error rate for data frames is 1%, therefore out of 100 sent frames, error occurs in one frame.
  • This causes the negative acknowledgement to follow which causes the retransmission.

 

So, we have-

  • Overhead due to headers = 100 x 40 bits = 400 bits.
  • Overhead due to negative acknowledgement = 40 bits.
  • Overhead due to retransmission = 40 bits header + 3960 data bits = 4000 bits.

 

From here,

Total overhead

= 400 bits + 40 bits + 4000 bits

= 8040 bits

 

Calculating Efficiency-

 

Efficiency (η) = Useful data sent / Total data sent

 

Here,

  • Useful data sent = 396000 bits
  • Total data sent = Useful data sent + Overhead = 396000 bits + 8040 bits = 404040 bits

 

Substituting the values, we get-

Efficiency (η)

= 396000 bits / 404040 bits

= 0.9801

 

Calculating Bandwidth Utilization-

 

Bandwidth Utilization

= Efficiency x Bandwidth

= 0.9801 x 50 Kbps

= 49.005 Kbps

 

Calculating Bandwidth Wasted-

 

Bandwidth wasted

= Bandwidth – Bandwidth Utilization

= 50 Kbps – 49.005 Kbps

= 0.995 Kbps

 

Calculating Fraction of Bandwidth Wasted-

 

Fraction of bandwidth wasted

= Wasted Bandwidth / Total Available Bandwidth

= 0.995 Kbps / 50 Kbps

= 0.0199

= 1.99 %

 

Thus, Option (C) is correct.

 

Problem-04:

 

Consider 1 Mbps error free line. The maximum frame size is 1000 bits. New packets are generated about 1 sec apart. The time out interval is 10 msec. If the ack timer is eliminated. How many times the average message be transmitted?

 

  1. Only once
  2. Twice
  3. Thrice
  4. Can’t say

 

Solution-

 

  • Transmission delay (Tt) = L / B = 1000 bits / 106 bits per sec = 1 msec.
  • After packet is put on the link, the time out timer is started which is 10 msec long.
  • The next packet is transmitted after 1 sec = 1000 msec.
  • If no acknowledgement is received within 10 msec, the packet will be retransmitted.
  • We have been asked how many times the average message be transmitted i.e. how many retransmissions are possible.
  • Retransmission occurs or not depends on the propagation delay (Tp).
  • If Tp is more, time out will occur and retransmission will take place but if Tp is less, then there will be no time out.
  • Since propagation delay (Tp) is not given in the question, therefore we can not say anything.

 

Thus, Option (D) is correct.

 

Problem-05:

 

What is the effect on line utilization if we increase the number of frames for a constant message size?

 

  1. Lower line efficiency
  2. Higher line efficiency
  3. No change in line efficiency
  4. No relation between line efficiency and frame size

 

Solution-

 

In both the following cases, line utilization remains the same-

  • Whether the entire message is sent as a single entity
  • Or the entire message is divided into frames and then frames are sent.

This is because line contains the same amount of data in both cases.

 

So,

  • If the number of frames are increased by dividing the message, there is no change in line efficiency.
  • The line efficiency remains the same.

 

Thus, Option (C) is correct.

 

Next Article- Flow Control Methods | Single Parity Check

 

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Selective Repeat Protocol | Practice Problems

Selective Repeat Protocol-

 

Before you go through this article, make sure that you have gone through the previous article on Selective Repeat Protocol.

 

We have discussed-

  • Sliding Window Protocols allow the sender to send multiple frames before needing acknowledgements.
  • Selective Repeat is an implementation of a sliding window protocol.

 

In this article, we will discuss practice problems based on selective repeat protocol.

 

 

PRACTICE PROBLEMS BASED ON SELECTIVE REPEAT PROTOCOL-

 

Problem-01:

 

The maximum window size for data transmission using the selective repeat protocol with n bit frame sequence numbers is-

  1. 2n
  2. 2n-1
  3. 2n-1
  4. 2n-2

 

Solution-

 

We know-

  • With n bits, total number of sequence numbers possible = 2n.
  • In SR Protocol, sender window size = receiver window size = W (say)

 

For any sliding window protocol to work without any problems,

 

Min Available Sequence Numbers

= Sender window size + Receiver window size

 

So, we have-

2n = W + W

2n = 2W

W = 2n-1

Therefore, maximum window size possible of sender and receiver = 2n-1

Thus, Option (B) is correct.

 

Problem-02:

 

In SR protocol, suppose frames through 0 to 4 have been transmitted. Now, imagine that 0 times out, 5 (a new frame) is transmitted, 1 times out, 2 times out and 6 (another new frame) is transmitted.

At this point, what will be the outstanding packets in sender’s window?

 

  1. 341526
  2. 3405126
  3. 0123456
  4. 654321

 

Solution-

 

In SR Protocol, only the required frame is retransmitted and not the entire window.

 

Step-01:

 

Frames through 0 to 4 have been transmitted-

4 , 3 , 2 , 1 , 0

 

Step-02:

 

0 times out. So, sender retransmits it-

0 , 4 , 3 , 2 , 1

 

Step-03:

 

5 (a new frame) is transmitted-

5 , 0 , 4 , 3 , 2 , 1

 

Step-04:

 

1 times out. So, sender retransmits it-

1 , 5 , 0 , 4 , 3 , 2

 

Step-05:

 

2 times out. So, sender retransmits it-

2 , 1 , 5 , 0 , 4 , 3

 

Step-06:

 

6 (another new frame) is transmitted-

6 , 2 , 1 , 5 , 0 , 4 , 3

 

Thus, Option (B) is correct.

 

Problem-03:

 

The selective repeat protocol is similar to Go back N except in the following way-

 

  1. Frame Formats are similar in both the protocols
  2. The sender has a window defining maximum number of outstanding frames in both the protocols
  3. Both uses piggybacked acknowledgements where possible and does not acknowledge every frame explicitly.
  4. Both uses piggyback approach that acknowledges the most recently received frame

 

Solution-

 

Also Read- Go back N Protocol

 

Option (A)-

 

  • Both the protocols use the same frame formats because both are sliding window protocols.
  • The variation occurs only in the coding and implementation.

 

Option (B)-

 

  • In both the protocols, sender has a window which defines the maximum number of outstanding frames.

 

Option (C)-

 

  • Both the protocols use piggybacked acknowledgements wherever possible.
  • Sending acknowledgements along with the data are called as piggybacked acknowledgements.
  • But Go back N protocol uses cumulative acknowledgements and does not acknowledge every frame explicitly.
  • On the other hand, Selective repeat protocol acknowledges each frame independently.

 

Option (D)-

 

  • Both the protocols use piggyback approach.
  • Go back N acknowledges the most recently received frame by sending a cumulative acknowledgement which includes the acknowledgement for previous packets too if any.
  • On the other hand, Selective Repeat protocol acknowledges all the frames independently and not only the recently received frame.

 

Thus, Options (C) and (D) are correct.

 

Problem-04:

 

Consider a 128 x 103 bits/sec satellited communication link  with one way propagation delay of 150 msec. Selective Retransmission (repeat) protocol is used on this link to send data with a frame size of 1 KB. Neglect the transmission time of acknowledgement. The minimum number of bits required for the sequence number field to achieve 100% utilization is ________ .

 

Solution-

 

Given-

  • Bandwidth = 128 x 103 bits/sec
  • Propagation delay (Tp) = 150 msec
  • Frame size = 1 KB

 

Now,

  • To achieve 100% utilization, efficiency must be 100%.
  • Efficiency is 100% when sender window size is optimal i.e. 1+2a

 

Calculating Transmission Delay-

 

Transmission delay (Tt)

= Frame size / Bandwidth

= 1 KB / (128 x 103 bits per sec)

= (1 x 210 x 8 bits) / (128 x 103 bits per sec)

= 64 msec

 

Calculating Value of ‘a’-

 

a = Tp / Tt

a = 150 msec / 64 msec

a = 2.34

 

Calculating Optimal Sender Window Size-

 

Optimal sender window size

= 1 + 2a

= 1 + 2 x 2.34

= ⌈5.68⌉

= 6

 

Calculating Number Of Sequence Numbers Required-

 

In SR Protocol, sender window size and receiver window size are same.

So, sender window size = receiver window size = 6

 

Now,

For any sliding window protocol, minimum number of sequence numbers required

= Sender window size + Receiver window size

= 6 + 6

= 12

 

Calculating Bits Required in Sequence Number Field-

 

To have 12 sequence numbers,

Minimum number of bits required in sequence number field

= ⌈log2(12)⌉

= 4

 

Thus,

  • Minimum number of bits required in sequence number field = 4
  • With 4 bits, number of sequence numbers possible = 16
  • We use only 12 sequence numbers and rest 4 remains unused.

 

Next Article- Comparison Table Of Sliding Window Protocols

 

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Sliding Window Protocol | Practice Problems

Sliding Window Protocol-

 

Before you go through this article, make sure that you have gone through the previous article on Sliding Window Protocol.

 

We have discussed-

  • Sliding window protocol is a flow control protocol.
  • It allows the sender to send multiple frames before needing the acknowledgements.
  • Go back N and Selective Repeat are the implementations of sliding window protocol.

 

In this article, we will discuss practice problems based on sliding window protocol.

 

PRACTICE PROBLEMS BASED ON SLIDING WINDOW PROTOCOL-

 

Problem-01:

 

A 3000 km long trunk operates at 1.536 Mbps and is used to transmit 64 byte frames and uses sliding window protocol. If the propagation speed is 6 μsec / km, how many bits should the sequence number field be?

 

Solution-

 

Given-

  • Distance = 3000 km
  • Bandwidth = 1.536 Mbps
  • Packet size = 64 bytes
  • Propagation speed = 6 μsec / km

 

Calculating Transmission Delay-

 

Transmission delay (Tt)

= Packet size / Bandwidth

= 64 bytes / 1.536 Mbps

= (64 x 8 bits) / (1.536 x 106 bits per sec)

= 333.33 μsec

 

Calculating Propagation Delay-

 

For 1 km, propagation delay = 6 μsec

For 3000 km, propagation delay = 3000 x 6 μsec = 18000 μsec

 

Calculating Value Of ‘a’-

 

a = Tp / Tt

a = 18000 μsec / 333.33 μsec

a = 54

 

Calculating Bits Required in Sequence Number Field-

 

Bits required in sequence number field

= ⌈log2(1+2a)⌉

= ⌈log2(1 + 2 x 54)⌉

= ⌈log2(109)⌉

= ⌈6.76⌉

= 7 bits

 

Thus,

  • Minimum number of bits required in sequence number field = 7
  • With 7 bits, number of sequence numbers possible = 128
  • We use only (1+2a) = 109 sequence numbers and rest remains unused.

 

Problem-02:

 

Compute approximate optimal window size when packet size is 53 bytes, RTT is 60 msec and bottleneck bandwidth is 155 Mbps.

 

Solution-

 

Given-

  • Packet size = 53 bytes
  • RTT = 60 msec
  • Bandwidth = 155 Mbps

 

Calculating Transmission Delay-

 

Transmission delay (Tt)

= Packet size / Bandwidth

= 53 bytes / 155 Mbps

= (53 x 8 bits) / (155 x 106 bits per sec)

= 2.735 μsec

 

Calculating Propagation Delay-

 

Propagation delay (Tp)

= Round Trip Time / 2

= 60 msec / 2

= 30 msec

 

Calculating Value of ‘a’-

 

a = Tp / Tt

a = 30 msec / 2.735 μsec

a = 10968.921

 

Calculating Optimal Window Size-

 

Optimal window size

= 1 + 2a

= 1 + 2 x 10968.921

= 21938.84

 

Thus, approximate optimal window size = 21938 frames.

 

Problem-03:

 

A sliding window protocol is designed for a 1 Mbps point to point link to the moon which has a one way latency (delay) of 1.25 sec. Assuming that each frame carries 1 KB of data, what is the minimum number of bits needed for the sequence number?

 

Solution-

 

Given-

  • Bandwidth = 1 Mbps
  • Propagation delay (Tp) = 1.25 sec
  • Packet size = 1 KB

 

Calculating Transmission Delay-

 

Transmission delay (Tt)

= Packet size / Bandwidth

= 1 KB / 1 Mbps

= (210 x 8 bits) / (106 bits per sec)

= 8.192 msec

 

Calculating Value of ‘a’-

 

a = Tp / Tt

a = 1.25 sec / 8.192 msec

a = 152.59

 

Calculating Bits Required in Sequence Number Field-

 

Bits required in sequence number field

= ⌈log2(1+2a)⌉

= ⌈log2(1 + 2 x 152.59)⌉

= ⌈log2(306.176)⌉

= ⌈8.25⌉

= 9 bits

 

Thus,

  • Minimum number of bits required in sequence number field = 9
  • With 9 bits, number of sequence numbers possible = 512.
  • We use only (1+2a) sequence numbers and rest remains unused.

 

Problem-04:

 

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?

  1. 7.69 x 106 Bps
  2. 11.11 x 106 Bps
  3. 12.33 x 106 Bps
  4. 15.00 x 106 Bps

 

Solution-

 

Given-

  • Sender window size = Receiver window size = 5
  • Packet size = 1000 bytes
  • Transmission delay (Tt) = 50 μs
  • Propagation delay (Tp) = 200 μs

 

Calculating Bandwidth-

 

We know,

Transmission delay = Packet size / Bandwidth

 

So, Bandwidth

= Packet Size / Transmission delay (Tt)

= 1000 bytes / 50 μs

= (1000 x 8 bits) / (50 x 10-6 sec)

= 160 Mbps

 

Calculating Value of ‘a’-

 

a = Tp / Tt

a = 200 μsec / 50 μsec

a = 4

 

Calculating Optimal Window Size-

 

Optimal window size

= 1 + 2a

= 1 + 2 x 4

= 9

 

Calculating Efficiency-

 

Efficiency (η)

= Sender window size / Optimal window size

= 5 / 9

= 0.5555

= 55.55%

 

Calculating Maximum Achievable Throughput-

 

Maximum achievable throughput

= Efficiency (η) x Bandwidth

= 0.5555 x 160 Mbps

= 88.88 Mbps

= 88.88 x 106 bps or 11.11 x 106 Bps

 

Thus, Option (B) is correct.

 

Problem-05:

 

Station A uses 32 byte packets to transmit messages to station B using a sliding window protocol. The round trip delay between A and B is 80 msec and the bottleneck bandwidth on the path between A and B is 128 Kbps. What is the optimal window size that A should use?

  1. 20
  2. 40
  3. 160
  4. 320

 

Solution-

 

Given-

  • Packet size = 32 bytes
  • Round Trip Time = 80 msec
  • Bandwidth = 128 Kbps

 

Calculating Transmission Delay-

 

Transmission delay (Tt)

= Packet size / Bandwidth

= 32 bytes / 128 Kbps

= (32 x 8 bits) / (128 x 103 bits per sec)

= 2 msec

 

Calculating Propagation Delay-

 

Propagation delay (Tp)

= Round Trip Time / 2

= 80 msec / 2

= 40 msec

 

Calculating Value of ‘a’-

 

a = Tp / Tt

a = 40 msec / 2 msec

a = 20

 

Calculating Optimal Window Size-

 

Optimal window size

= 1 + 2a

= 1 + 2 x 20

= 41 which is close to option (B)

Thus, Option (B) is correct.

 

Next Article- Go Back N Protocol

 

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Selective Repeat | Sliding Window Protocol

Sliding Window Protocol-

 

Before you go through this article, make sure that you have gone through the previous article on Sliding Window Protocol.

 

The two well known implementations of sliding window protocol are-

 

 

  1. Go back N Protocol
  2. Selective Repeat Protocol

 

In this article, we will discuss about Selective Repeat protocol.

Learn about Go back N Protocol.

 

Selective Repeat Protocol-

 

Selective Repeat protocol or SR protocol is an implementation of a sliding window protocol.

The features and working of this protocol are explained in the following points-

 

Point-01:

 

In SR protocol, sender window size is always same as receiver window size.

 

In SR protocol,

  • Sender window size = Receiver window size
  • The size is of course greater than 1 otherwise the protocol will become Stop and Wait ARQ.
  • If n bits are available for sequence numbers, then-

Sender window size = Receiver window size = 2n/2 = 2n-1

 

Point-02:

 

SR protocol uses independent acknowledgements only.

 

In SR protocol,

  • Receiver acknowledges each frame independently.
  • As receiver receives a new frame from the sender, it sends its acknowledgement.

 

Point-03:

 

SR protocol does not accept the corrupted frames but does not silently discard them.

 

In SR protocol,

  • If receiver receives a frame that is corrupted, then it does not silently discard that frame.
  • Receiver handles the situation efficiently by sending a negative acknowledgement (NACK).
  • Negative acknowledgement allows early retransmission of the corrupted frame.
  • It also avoids waiting for the time out timer to expire at the sender side to retransmit the frame.

 

Point-05:

 

SR protocol accepts the out of order frames.

 

In SR protocol,

  • Consider receiver receives a frame whose sequence number is not what the receiver expects.
  • Then, it does not discard that frame rather accepts it and keeps it in its window.

 

Point-06:

 

SR protocol requires sorting at the receiver’s side.

 

In SR protocol,

  • Receiver window is implemented as a linked list.
  • When receiver receives a new frame, it places the new frame at the end of the linked list.
  • When the received frames are out of order, receiver performs the sorting.
  • Sorting sorts the frames in the correct order.

 

Point-07:

 

SR protocol requires searching at the sender’s side.

 

In SR protocol,

  • Receiver does not reject the out of order frames.
  • Receiver accepts the out of order frames and sort them later.
  • Thus, only the missing frame has to be sent by the sender.
  • For sending the missing frame, sender performs searching and finds the missing frame.
  • Then, sender selectively repeats that frame.
  • Thus, only the selected frame is repeated and not the entire window.
  • That is why, the protocol has been named as “Selective Repeat Protocol“.

 

Point-08:

 

SR protocol leads to retransmission of lost frames after expiry of time out timer.

 

In SR protocol,

  • Consider a frame being sent to the receiver is lost on the way.
  • Then, it is retransmitted only after time out timer expires for that frame at sender’s side.

 

Efficiency of SR Protocol-

 

Efficiency of any flow control protocol is given by-

 

Efficiency = Sender Window Size in Protocol / (1 + 2a)

 

In selective repeat protocol, if sender window size = N, then-

 

Efficiency of SR Protocol = N / (1 + 2a)

 

To gain better understanding about Selective Repeat ARQ,

Watch this Video Lecture

 

Next Article- Practice Problems On Selective Repeat Protocol

 

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