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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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Best answer
92 votes
92 votes

Answer should be (B).

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

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

$= 4KB$.

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

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11 votes
11 votes

Answer is (A)

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 by
10 votes
10 votes
B)

 

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

 

Sender window size= min(Wcongestion,Wadverstised)

                                =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.
 
Answer:

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