GATE IT 2005 | Question: 73

Question

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:

• A. $2048$ bytes
• B. $4096$ bytes
• C. $6144$ bytes
• D. $8192$ bytes

Comments

https://gateoverflow.in/32145/find-the-size-of-senders-window-if

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

please tell before gate :)

---

There is no difference.Both are same :)

---

What wiil be the window size at receiver?

---

In this question, we don't have to think about AIMD algo right ?

---

Current window size is 4KB

---

Is last byte acknowledged by receiver is same as last byte received by receiver ?

---

@Harpuneet No it’s not

---

The Actual window size from the client side will not calculate the rwnd size on its own, it will just take the minimum of what is advertised to it, whatever data is there in the recvbuffer is not it’s concern. The min of (18,20) is 18, so 18 is the answer

---

How to find the free space in sender window size

---

In geeks for geeks the answer is given to be 6144Kb. Here is the link –https://www.geeksforgeeks.org/gate-gate-it-2005-question-73/. What is the correct theory for this question??

---

the last line of the question has so significance in calculating the answer,

current window size is min of congestion window and advertisment window.

---

REC ACK WAS 8192 THEN SENDER SEND SEGMENT WHOSE LAST BIT WAS 10240 HENCE SIZE OF SEGMENT SENT IS (10240-8192=2048=2KB) AFTER THIS SENDER INC HIS WINDOW EXPONENTIALLY HENCE AFTER THIS HIS WINDOW SIZE WOULD BE 4KB HENCE 4096

---

On this question there is very good and deep explanation done by Sachin Sir on one of his CN marathon video of TCP https://www.youtube.com/watch?v=ZFk-SJk7pp4&list=PLIPZ2_p3RNHgz_uIqO3iGfZETeqNAyVUH&index=13&t=12011s  at time 2:55:43 approx

Answer 1

Answer should be (B).

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

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

$= 4KB$.

Comments

To add to the already informative comment section here:

The amount of unacknowledged bytes at the sender can't exceed the Sender Window Size.

That is,

$\text{LastByteSent} – \text{LastByteAcked} \ ≤ \ \text{min{cwnd,rwnd}} $

 

As evident by the details in the question: $10240-8192=2048 \leq 4096$

---

@JashanArora can i say the unack bytes at the sender is = lastbytesent - lastbyteacked?

---

USABLE SPACE IS 2KB LEFT BUT THE TOTAL WINDOW SIZE IS STILL 4KB

Answer 2

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

Comments

Please suggest if anything wrong !

---

Super answer

Answer 3

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

Comments

Receiver available window size is actually receiver advertised window size, it's not total or maximum receiver window size.

In other words receiver advertised window size tells space free at receiver's buffer.

Please correct me if I am wrong.

---

Yes I also think the above explanation on that part is wrong. Total window size would be 8KB = unacknowledged size of window 2KB + Advertised Window size given as 6KB.

But according to the question, it doesn't ask for window size of the receiver instead it asks for "current window size at sender" i.e. advertisement window size according to sender's perspective. This according to TCP's congestion control method, we apply minimum of advertisement window size and congestion window size.

Even I m not completely sure about this above explanation, pls let me know if I m correct or wrong.

---

@Ankush Kundaliya perfect explanation 

Answer 4

 

Comments

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.

 

---

by free space you mean the bytes which haven't been sent yet in the window. right?

Answer 5

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

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

Answer 6

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?