Question 3) Sliding Window Protocol

Question

In what protocols it is possible for sender to receive an ACK for a packet that falls outside of its window . Please justify ur answer .

a)Stop and wait

b)SR

c) GBN

d) all


My approach

W_s = 3
t=0  sends packets 1,2,3
                                                       t=1 reveiver ACK 1,2,3
t=2 sender time out and resent 1,2,3
                                                       t=3 receiver receive duplicates and re ACK
t=4 sender receives ACK and advances Ws

t=5 sender receives ACK receiver sent at t=3 This is outside window

The above is true for GBN and SR .
What about Stop and Wait ?
Answer is given as Stop and wait
 

Comments

Ans- B and C.

Answer 1

For GBN and SR...

Suppose the sender has a window size of 3 and sends packets 1, 2, 3 at 0t. At 1t the receiver ACKS 1, 2, 3. At 2t the sender times out and resends 1, 2, 3. At 3t the receiver receives the duplicates and re-acknowledges 1, 2, 3. At 4t the sender receives the ACKs that the receiver sent at 1t and advances its window to 4, 5, 6. At 5t the sender receives the ACKs 1, 2, 3 the receiver sent at 2t. These ACKs are outside its window.

For stop and wait..

Let sender at t0 send 0th packet. At t1 received by receiver and send ack1 (ack for 0th packet and expecting packet 1). Eventually sender time out and send 0th packet again at receiver side it is discarded as expecting packet is 1. Ack1 has been send again.

When first ack1 is received sender sends packet1 and expecting ack0. Suppose after t time second ack1 arrives but expected ack  is ack0 so ack1 is discarded.

So it is also true for stop and wait protocol.

Correct me ...

Comments

Thanks for explanation

but if sender window size is [n]  ,and it expects ack[n+1]

[n+1] number is still not part of sender sliding window

after receiving ack[n+1] senders window slide ,

so why we cannot say in case of commulative acknowledgement it receive such acknowledgement no(which is not in window]

true for GBN,SR and stop and wait

true for

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best answer by papesh

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@papesh for stop and wait it is correct but i don’t think for SR and GO BACK-N you have given the correct logic.

Let’s suppose there are 5 packets to transfer.

FOR GOBACK-3:

The total sequence number possible is 4 (0-3).

→ currently my window has 1,2,3 packets and the sender has transmitted all the 3 packets and the receiver’s counter is initialized to 0. Packet 1 got acknowledged but let’s suppose it got delayed and the receiver’s counter incremented to 1. 

→ packet 2 got acknowledged and the sender receives it and the receiver’s counter incremented to 2. Since in GO-BACK acknowledgments are cumulative. So receiving packet 2’s acknowledgement means packet 1 and packet 2 both has been acknowledged by the receiver. Therefore the window slides.

→ Currently my sliding window has 3 4 5. Suppose the sender now receives the acknowledgement of packet 0.

Same way we can give for SR also.