Computer Networks · TEE · papers
Q3(a) · Alternating-bit state transitions
2023-11-22 · AI-only Sem V variant; incomplete recovered scan
Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.
Q3(a). Use the printed rdt3.0 alternating-bit sender/receiver: S0 sends data0; S3 accepts ACK0; S5 sends data1; S8 accepts ACK1; S2/S7 are timeout retransmissions; S1/S6 ignore corrupt or wrong ACKs. Receiver R0 accepts new data0 and sends ACK0; R2 accepts new data1 and sends ACK1; R3/R1 handle corrupt or duplicate data while waiting for 0/1 respectively. Complete the global transition sequence S0, R0, t1, t2, t3, t4 that delivers two messages without errors, including the second ACK at the sender. Justify each transition. [5 marks]
Answer
Given
Two messages; no DATA or ACK errors; original FSM transition labels in the question.
To find
The global interleaved transition sequence, including the final ACK.
Formula and why
Send DATA → receive and acknowledge DATA → receive matching ACK → advance the sequence bit.
Each successful ACK permits the sender to send the next new message.
Solution
The sequence bit labels the outstanding DATA packet. ACK0 confirms data0 and ACK1 confirms data1 in this printed FSM; these are not TCP cumulative byte numbers. An ACK loss does not advance the sender. A duplicate data packet is acknowledged again but not delivered twice.
| Send0 | → | ACK0 | → | Send1 | → | ACK1 |
| Order | Transition | Action |
|---|---|---|
| 1 | S0 | Send data0; start timer |
| 2 | R0 | Accept data0; deliver; send ACK0 |
| 3 | S3 | Accept ACK0; stop timer |
| 4 | S5 | Send data1; start timer |
| 5 | R2 | Accept data1; deliver; send ACK1 |
| 6 | S8 | Accept ACK1; stop timer |
Required sequence: S0, R0, S3, S5, R2, S8. For the four placeholders: t1=S3, t2=S5, t3=R2, t4=S8.