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

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

@akhllnadhpc I have some doubts:

In the question they have given 270 msec propagation time and 8 bit sequence number.These parameters will not be used?

what are NAK Frames?

Why only data is considered as useful transmission and not headers?

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First i need @arjun sir to varify this solution is correct or not .
NAK means Negative ACK , In Any transmission only Data bits is considered useful . We need only Data . dont we ?

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protocol 6 is selective repeat ARQ and sequence number bits are given.

so, we can get maximum sender window size , then why use 100 frames ..?

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how u know protocol 6=sr????

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@asu,

protocol 4 -> stop and wait ARQ

protocol 5 -> Go back N ARQ

protocol 6 -> selective repeat ARQ..

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any reference plz

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@asu https://fc562c112ad07a0225f1fb629f0b18424f556af7.googledrive.com/host/0B0UAwnKSXFy7dEp5MFdVbE1YMVU/2014/lec3.pdf

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@aakhil...this question is copied from a pdf...and in that pdf protocol 6 refers to SR ARQ....and 8 bit sequence number means 256 sequence nuber possibe..so sender windows size will be 128....

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If we consider 128 frames here, since sequence number is 8 bits then

1% loss will mean 1.28 Frames which will be considered 2 frames for retransmission then for this how many NAK will be send 1 or 2?

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@asu SIR, I don't know whether my solution is correct or not . I just give a try.
If you can plz solve it for me .
Need some experts or @arjun Sir , to varify it or rather solve it :D

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your solution is correct...as protocol 6 refers to sr protocol .so instead of taking 100 packets u can take 128 and everything is correct

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@asu sir...

In selective repeat we send individual acknowledgements. So when we will take 128 frames then 1 % error rate will mean 1.28 frames ⋍2 frames....So in that case we need to send 2 NAK and we will consider 80 bits in place of 40 bits ( FOR 100 frames we send 1 NAK=40bits) ....am i right?

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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