8.3k views

Station $A$ needs to send a message consisting of $9$ packets to Station $B$ using a sliding window (window size $3$) and go-back-n error control strategy. All packets are ready and immediately available for transmission. If every 5th packet that $A$ transmits gets lost (but no acks from $B$ ever get lost), then what is the number of packets that $A$ will transmit for sending the message to $B$?

1. $12$
2. $14$
3. $16$
4. $18$
edited | 8.3k views
–4
There is a  formula
N=1/(1-p)

where p -probability of failure

N-total number of packets(including the extra packets) to be sent for one data packet .

Can this be applied here ?

p=1/5 = 0.2

N=1/1-0.2 = 1/0.8

for 9 packets , 9/0.8 = 11.25 ~ 12
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thanks! now understood clearly
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This applies to stop and wait only..
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Some more variation in the given question

Q - The question asked above is same, only the  window size is 4 instead of 3. So now how many transmissions will be there?

Ans - 21

 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Notice -->  Increasing window size may perform poor if the given link is faulty.

Thank You  @Nashreen Sultana, @VS and @reena_kandari ji.

+1

This might help ...

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There is lot of confusion got created on the question https://gateoverflow.in/230305/slidind-window-protocol#c240121

Can someone clear that

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This strategy not work here

Total $16$ packet Transmission

Correct Answer: $C$

answered by (187 points)
edited
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why is 1,2,3,4 sent at once, only 1,2,3 should be sent at once(due to window size 3) then it recieves a cummulative ACK and then 4,5,6 should be sent from which 5 gets lost then recieve ACK for 4 and then 5,6,7 should be sent and so on. Shouldn't we consider the window size 3, given in the question and cummulative ACKs in GBN?

Since all packets are ready initially itself, we can assume a timeout is detected after all possible packets are sent. So, the sending happens as shown in figure (I draw the figure assuming 10 packets. For 9 packets answer will be 16).

answered by Veteran (408k points)
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When 4th packet gets lost then the ACKs received for 5th and 6th packet are discarded. Are those ACKs, actually NAKs or ACKs with seq. no as 4( asking for 4th packet)?? Also, does the timeout occurs for the fourth packet's ACK?
+1

@VS @Nashreen Sultana @reena_kandari for your question

The question asked above is same, only the window size is 4 instead of 3. So now how many transmissions will be there?

Yes , For window size 4 answer will be 21 not 23,26,20
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@Rajesh Pradhan Yes , Your statement is right . @Arjun Sir, There need small correction in Diagram however answer is right .


as per ur diagram from the beginning for the 5th and 6th numbered packet Receiver wont give Acknowledgement. Bcoz it will discard any out of order packet @time it is expecting for 4th numbered packet only.
Above statement is required for correction in Diagram .
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@RISHABH SHRIVAS, please Use small fonts rather than these large ones!! make the things more readable.

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Sorry for that . Done :)
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bad explanation...somehow ppl just find d answer widout clear explanation
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@Sunita Please explain it better :)

@Rishabh Thanks, I'll add.
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@Arjun Sir, Please give the solution considering SR protocol
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Hi, just a query here. Why are we considering individual ACKs rather than Cummulative ACKs
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@Arjun  sir,

@Tuhin Dutta  sir,

In GBN the timeout timer is maintained for First outstanding frames for Entire send window..eg

When Ws = 4,5,6 TO timer for 4 is maintained and window slides when ACK for 4 comes ..So if 5th one also lost with 4th [2 losses in same window ] ..it will be handled by retransmision of window 4,5,6 because TO timer for 4 expires ..

Am i right sir??

16 is the answer

Q doesn't say that only after 3rd packet B will send theacknowledgement for 1st packet from A
So we can say

"All packets are ready and immediately available for transmission"

packet1 from A - Tranmit1
acknowledgement (ack) for packet1(pckt) from B
packet2 from A - Transmit2
ack for pckt2 from B
packet3 from A - Transmit3
ack for pckt3 from B
packet4 from A - Transmit4
ack for pckt4 from B
packet5 from A - Transmit 5          (packet5 lost, so no ack from B for pckt5)
packet6 from A - Transmit 1          (go back n, so no ack for pckt6 also)
packet7 from A - Transmit 2          (go back n, so no ack for pckt7 also)
Timeout for packet5
packet5 from A - Transmit 3
ack for pckt5 from B
packet6 from A - Transmit 4
ack for pckt6 from B
packet7 from A - Transmit 5         (packet7 lost, so no ack for pckt7)
packet8 from A - Transmit 1         (go back n, so no ack for pckt8 also)
packet9 from A - Transmit 2         (go back n, so no ack for pckt9 also)
Timeout for packet7
packet7 from A - Transmit 3
ack for pckt7 from B
packet8 from A - Transmit 4
ack for pckt8 from B
packet9 from A - Transmit 5         (packet9 lost, so no ack for pckt9)
Timeout for packet9
packet9 from A - Transmit 1
ack for pckt9 from B

B now receives all 9 packets.
In total 16 transmits from station A

answered by Junior (533 points)
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why are we having timeout for 5th packet only after 7th packet ?
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What is the criteria for time out here?
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Yes! Please explan time out criteria ?
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window size = 3

so as per Q   5th packet is lost  implies till 4th packet transmission is fine and ACK is also received but 5th one is lost so current window status is 5th 6th 7th packet as window size is 3.  5th one lost so no ACK. so sender keeps waiting till timeout for ACK. After timeout retransmission takes place.

GBN receiver window size one till 5th is received nothing goes forward
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Okay.
@Neal
Means if window size is 'n' then wait for ack up to n packets and after that time out .

Hence 16 option C

answered by Boss (10.5k points)
I think the most intuitive way to get how Go back N work is to remember this 2 points and try it yourself

1. The window should slide ahead as we go on receiving ack and we will miss ack only for frame which is lost i.e frame 5 according to question

2. We should go back and send frame starting from frame no which was lost till the end of current window size.

Suppose if frame 5th was lot we should resent 5  6 7 because currently these 3 frames are in window.

PS: we should also count frames which were sent but discarded as out of order frame

Let's say we have 9 frames to send

1 2 3 4 5 6 7 8  9

We will send 1 then window will slide and currently 2 3 4 will be frame in the window then we will send 2 window will slide now currently 3 4 5 will be in window snapshot then we will send 3 and window will slide again and current snapshot will have 4 5 6 in window then next 4 will be sent current snapshot will be 5 6 7.

Now point 2 if 5th one is lost then what frames will be sent again well start from 5th and go to end of window that is 7 as window size is 3 so 5 6 7 will be sent again. Try to work it out by using above 2 points plus don't forget to slide the window ahead as you go on sending frame and receive ack. Remember 5 6 7 were sent before as 5 was not received 6 7 were discarded make sure you count then as well

If you get it right the sequence you will get at the end is

1 2 3 4 5 6 7 5 6 7 8 9 7 8 9 9

Good luck
answered by Active (1.2k points)

In GBN we send entire window again if any frame lost

answered by Active (3k points)
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In case of GBN when any packet is lost we send entire window again.

If the question says $SR$ protocol then we send only lost packet entire not  window.

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