Flow Control Protocols-

In computer networking, there are various flow control protocols-

• Stop and Wait Protocol
• Stop and Wait ARQ
• Go back N Protocol
• Selective Repeat Protocol

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:

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 (T_t) = 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 T_p is more, time out will occur and retransmission will take place but if T_p 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-

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. 2^n-1
3. 2ⁿ-1
4. 2^n-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,

So, we have-

2ⁿ = W + W

2ⁿ = 2W

W = 2^n-1

Therefore, maximum window size possible of sender and receiver = 2^n-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 (T_p) = 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 (T_t)

= 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 = T_p / T_t

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-

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 = 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 = 2^n-1

Point-02:

In SR protocol,

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

Point-03:

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:

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:

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:

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:

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-

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

To gain better understanding about Selective Repeat ARQ,

Watch this Video Lecture

Next Article- Practice Problems On Selective Repeat Protocol

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