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On a TCP connection, current congestion window size is Congestion Window = 4 KB. The window size advertised by the receiver is Advertise Window = 6 KB. The last byte sent by the sender is LastByteSent = 10240 and the last byte acknowledged by the receiver is LastByteAcked = 8192. The current window size at the sender is

1. 2048 bytes
2. 4096 bytes
3. 6144 bytes
4. 8192 bytes
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https://gateoverflow.in/32145/find-the-size-of-senders-window-if

What is the difference in both of them ??????????

Ans should be (B)

Current Sender window $=\text{min(Congestion Window, Advertised Window)}$

$=\text{min}(4KB, 6KB)$

$= 4KB$.

answered by (81k points) 5 17 37
edited
Sender's window size = minimum(Congestion window, Receiver window).

Therefore, Sender's window size =min(4kb,6kb)

=4kb

=4096 bytes.

Rest of the data is of no use. They are there to create confusion .
@Bikram sir then there is no significance of unaknowledge bytes here right?
yes, it have No significance .
To be a bit specific-
Window size is $4 \times 1024 = 4096$ Bytes. That is from $8193$ to $8193+4096-1(=12288)$.

Where Byte number $8193$ to $10240$ are "$\text{Sent but not acked}$",

and Byte number $10241$ to $12288$ are "$\text{Not sent yet}$" (will send according to advertized reciever winow size).
so nicely described bro cleared all my doubts.. thanks...:)
last byte acknowledge is $8192$. it means sequence number of first Byte sent by sender is $8192$ not $8193$.
@reena_kandari Last byte acknowledged by the receiver is 8192. it means it is already acknowledged by Receiver and Receiver is expecting sequence number 8193 , So First byte sequence number of next window is 8193 not 8192 which is sent by Sender but not yet acknowledged by Receiver .
yes true..thanks for correcting!!
just for the concept I got this doubt. Last byte ack by receiver means sequence number receiver expecting next right? so the next sequence number from sender must be 8193 and last byte of this window is  8193+2^10 is that so?

Current Sender window = min (Congestion Window, Advertised Window)= min(4KB, 6KB)= 4KB

Unacknowledged Bytes= 10240 - 8192 = 2048 Bytes =2 KB

Since 2KB data is unacknowledged it can send only (Current window- Unacknowledged Bytes)= 4KB- 2KB =2KB = 2048 Bytes

edited
doubt @Arjun sir
answer should be 4096 bytes only. Please clear it any expert advice is needed.??? when sender window size is asked, it will obviously include outstanding bytes.
sender may not send more than 2 kb until some acks are not received. Right??

So this will be current sender window size or min of( cwnd and rwnd) will be current  sender window size??
somebody please explain properly,is answer 4KB or 2KB???has the sender sent 4KB only?can it not be the sender sent 6KB and 2 KB in not acknowledged?or is ot thatat sender has sent 4KB only as it is current congestion window and out of that 2KB has not been acknowledged?? please check it!

To calculate Current Window Size a sender should follow below steps:

1. Calculate Unacknowledged Bytes = LastByteSent - LastByteAcked

Unacknowledged Bytes = 10240 - 8192 = 2048 Bytes

Receiver Available Window Size = 6KB - 2KB = 4KB

3. Calculate Sender Current Window Size = min(Congestion Window Size, Receiver Available WindowSize)

Sender Current Window Size = min( 4KB, 4KB) = 4KB

• The current window size at the sender is = $\large\color{maroon}{4}$KB

edited
Please suggest if anything wrong !
B)

Since Receiver window size is 6KB and N/W congestion window size is 4KB so sender has to send only 4KB window.

=min(4KB,6KB)

=4KB

10240 B is last byte that is sent and last byte that is acknowledged is 8192.So free space in window will be 2048 B but question is just asking you Ws which will be 4KB.

answered by (1k points) 1 1 1

The current sender window size is 4KB  = 4096
See the following for more details:

Consider the following example, see sender window size is 9.

answered by (43.6k points) 1 1 1
Its more of a doubt
At the start of transmission, the data that can be sent =min(4kb,6kb)=4kb

So 4096 bytes are sent
then the congestion window size doubles(if I  assume the connection is in slow phase)=8192kb

so next max data that can be send=min(8kb,6kb)=6kb

total data sent uptil now=4096+6144=10240 bytes
and the data is acknowledged upto 8192 bytes.
So 2048 bytes are unacknowledged and currently only 4096 bytes can be sent which is hence the current window..is it right?

-acknowledgement receive is of 8192 so it gt unmarked and next 8193 got marked(selected) to the 10240 which is last byte send by the sender(not congestion).

10240-8193=2047+1(8193 itself) =2048

sender window size =min(con window,adver window) is applicable before transmision after transmision window size changes according to the acknowledgements recieved at sender side
edited
C, 6144 as new window size will reach to max, which is receiver window size.
answered by (7k points) 1 1
how explain

Sender has transferred 4KB( as it is his window size) and amongst that he has received the acknowledgement of 2048 bytes and waiting for ack of remaining 2048 byte. Hence he has increased the window size to 6KB(4KB+2KB) and will maintain that even if he receives the rest of the acks as the max sender window size will be receiver window size.

but until it acknowleges will the window slide?

+1 vote