Tanenbaum NAK and Retransmission

Question

Compute the fraction of the bandwidth that is wasted on overheads(headers and retransmission ) for protocol 6 on a heavily loaded 50 kbps satellite channel with data frames consisting of 40 headers 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.The sequence numbers are 8 bits.

Comments

given 26 minute for this question but not solved

how weak my concept

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 Ayush Upadhyaya What is the answer ??

I am getting 2.97%

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Check my approach @ srestha mam

Answer 1

No, There won't be any retransmissions on time out. It is SR with NAK.

Reason: See here, in this question , the transmission is said to be successful if there is no NAK( negatively acknowledged ). And it is clearly stated that ACK (+ve ACK) frames never occur. That is even if the whole window (every packet) is successfully received, the receiver won't send anything. Thus accordingly sender must be configured not to retransmit on time out . Actually , time out here signifies successful transmission.

now my solution :

the fraction of overhead in transmission

i.e 1.9%

Comments

Very nice explanation and solution. Thanx a million.

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@erh,

here in total data transmitted = 100*4000 + 40 + 4000

100*4000 is for frames,40 is for NAK and 4000 is for retransmitted packet..right??

and in overhead- 100*40 + 40+ 4000

here,100*40 for headers, 40 is for NAk or header ??and last 4000 is for retransmitted frame due to error,but where is retransmitted frame due to NAk??

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40 is for NAK

retransmitted frame due to error and retransmitted frame due to NAK , must be same !

how can it be different? Because NAK tells the sender about errornous frame. 4000 is the retransmitted due to Nak

Secondly, https://gateoverflow.in/57517

Answer 2

this ratio is the fraction of overhead in transmission

i.e 1.9%

Comments

would there be error on retransmitted data as well ? should we have to consider it for retransmitted data as well ? if so how long we have to do like that ?

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In SR are there any explicit acknowledgements or not??

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Sender window size= 7 frames here because of bandwidth constraint. if we calculate the no of frames we can transmit in one transmission and propagation delay it is comming 1+540*50/4000=7.

So just because of 7 bits sequence number how can we assume that sender window size= 2^7=128.

???

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@sushmita , you are right window size = min { max possible sequence number , 1+2a }

Considering that 100<128 is creating confusion, here i never said window size is 128 or 100 , but 100 is valid and permissible sequence number for frames.

Regarding ,Explicit acknowledgements: are there any implicit acknowledgements? Acknowledgements are always explicitly sent by the receiver .

Positive Acknowledgement: the receiver explicitly notifies the sender which packets, messages, or segments were received correctly. Positive Acknowledgement therefore also implicitly informs the sender which packets were not received and provides detail on packets which need to be retransmitted.

Source:https://en.wikipedia.org/wiki/Retransmission_(data_networks)

Negative Acknowledgment (NACK): the receiver explicitly notifies the sender which packets, messages, or segments were received incorrectly and thus may need to be retransmitted (RFC 4077).

Similarly , not receiving any ACK implicitly tells the sender the frames are received successfully (in above scenario).

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@pC, yes there would be errors on retransmitted data .

To account that case i propose-

$Total Transmissions =\mathbf{n/ 1-p}$

where p is error rate and n number of frames to be transmitted. p=0.01 then

$Total \, Transmissions =\mathbf{n/ 1-0.01}$

$=\mathbf{100n/ 99}$

then its better to take n as a multiple of 99.

i.e for n=99 total transmissions = 100

then $overhead \, ratio = \left ( (40*99)+ 4000 +40 \right ) / \left ( \left ( 100 * 4000 \right ) + 40 \right )$

$= 0.019998$

P.S : n could be 198, 297.....then respective total transmissions will be 200 , 300 and respective retransmission  will be 2, 3.... .ie #NAK received 2 , 3.... respectively, but still overhead ratio = 0.019998

i think this solution is more accurate then what i have provided, but i need someone to  verify it . Tell me if i am wrong somewhere.

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u mean not receiving any NAK implicitly means that the frames were successfully transmitted?

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Thanx for the clear explanation.

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So if we consider error in retransmitted data too, have u calculated the no of retransmissions using infinite GP series??

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Why seq number bits given?? I mean how to use it as per your explaination?

Sender window size is 7 frames but why considering n=100??

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Even if we use 128, the answer would be same as this change will be accounted both in denominator and numerator in the wasted Bandwidth ratio.

Extra 2  frames need to be retransmitted (1% error in 128 frames =1.28)

Answer 3

According to my understanding 1kbps is lost

2% is lost

50 kbps of 2% is 1 kbps is LOST

Comments

128 is maximum window size. it is not necessary to send all frames at a time ...

we can assume 100 frames also..

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@gshivam u r rt..

@kapil as we r calculating efficency it will be better considering all...what u say?

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@asu, yes we can consider.

actually, they are using SR to give a upper bound on the number of frames send...

what if i send 1000 frames, then efficiency will be different

so, not to be confused with what value we take, just to get a simple idea , taking 100 frames is also correct...

with 128 , it will be more correct..

Answer 4

Error Rate = 1\100 ie, so out of  100 frames only 1 frame is retransmited.

 Header = 40 bits., so  For 100 Frames = 4000 bits(40*100)

 Total useful Data = 3960 *100 = 396000 bits

Now one Frame is Retransmitted = 3960(data) + 40 bits(header) => 4000 bits

Also NAK is sent = 40 bits

Overhead = Headers + Retranmissions + NAK / Total Data

=> 4000 + 4000 + 40 / 396000 + 4000 + 4000 + 40

=> 0.0198

so in percentage % => 0.0198 * 100 => 1.98 %

Comments

@Manas-What would be total transmissions for 128 frames?

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See I did like this. Let me know if I am in the correct direction.

The sequence number bits $m=8$

Protocol is Selective repeat, so $W_s=128$

Now since this protocol, unlike GBN, allows out of order packets, NAK will be generated only for corrupted packets.

The error rate for a data packet=1%.

So, the number of NAK generated(hence the number of retransmission needed)=$\lceil 0.01 \times 128 \rceil=2$

So, in total sender transmitted 130 frames.

Total data transmitted=$(130 \times 4000)+(2 \times 40)(NAK)=520080$

Overhead data sent=$(128 \times 40)+(2 \times 4000)(retransmissions)+(2 \times40)(NAK)=13200$

Bandwidth wasted=$\frac{13200}{520080}=2.53\,percent$

I am getting another answer :(

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@ayush first of all NAK generated  = floor(0.01*128) = 1

so only 1 retransmission needed

Now caclculate u will get the right answer