Consider an instance of TCP's Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is

Question

Consider an instance of TCP's Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a time-out occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

(i) 8 MSS
(ii) 14 MSS
(iii) 7 MSS
(iv) 12 MSS

Accepted Answer

### Answer
**(iii) 7 MSS**

### Step-by-step solution
Initial values:
- $cwnd=2$ MSS
- $ssthresh=8$ MSS

During slow start, the window grows rapidly until the threshold:

| Transmission | cwnd (MSS) | Phase |
|---|---:|---|
| 1 | 2 → 4 | Slow start |
| 2 | 4 → 8 | Slow start; reaches threshold |
| 3 | 9 | Congestion avoidance |
| 4 | 10 | Congestion avoidance |
| 5 | 11 | Congestion avoidance; timeout occurs |

On a timeout, TCP sets the new threshold approximately to half of the current congestion window:

$ssthresh=11/2\approx5$ MSS

and resets $cwnd$ to **1 MSS**.

The next transmissions then grow as:

| Transmission | cwnd (MSS) |
|---|---:|
| 6 | 2 |
| 7 | 4 |
| 8 | 5 |
| 9 | 6 |
| 10 | **7** |

Therefore, the congestion window at the end of the tenth transmission is **7 MSS**.

**Final answer: (iii) 7 MSS.**

Answer

Step-by-step solution

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