Question

Although slow start with congestion avoidance is an effective technique for coping with congestion, it can result in long recovery times in high-speed networks, as this problem demonstrates.
A) Assume a round-trip delay of 60 msec (about what might occur across the continent) and a link with an available bandwidth of 1 Gbps and a segment size of 576 octets. Determine the window size needed to keep the pipeline full and the a worst case estimate of the time it will take to reach that window size after a timeout occurs on a new connection using Jacobson’s slow start with congestion avoidance approach.

B) Repeat part (a) for a segment size of 16 kbytes.

Answer 1

Answer:

The answer to this question can be defined as follows:

In option A: The answer is "13020".

In option B: The answer is "468 Segments".

Explanation:

Given:

The value of round-trip delay= 60 m-second

The value of Bandwidth= 1Gbps

The value of Segment size = 576 octets

window size =?

Formula:

$\text{Window size =} \frac{(\text{Bandwidth} \times \text{round} - \text{trip time})}{(\text{segment size window })}$

                     $=\frac{10^9 \times 0.06}{576 \times 8}\\\\=\frac{10^9 \times6}{576 \times 8\times 100}\\\\=\frac{10^7 \times 1}{96 \times 8}\\\\=\frac{10^7 \times 1}{768}\\\\=13020.83$

So, the value of the segments is =13020.833 or equal to 13020

Calculating segments in the size of 16 k-bytes:

$\text{Window size} = \frac{10^9 \times 0.06}{16,000 \times 8}$

                    $= \frac{10^9 \times 0.06}{16,000 \times 8}\\\\ = \frac{10^9 \times 6}{16,000 \times 8 \times 100}\\\\ = \frac{10^4 \times 3}{16 \times 4}\\\\ = \frac{30000}{64 }\\\\=468.75$

The size of 16 k-bytes segments is 468.75 which is equal to 468.