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

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".

Step-by-step 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} * \text{round} - \text{trip time})}{(\text{segment size window })}`

`=(10^9 * 0.06)/(576 * 8)\\\\=(10^9 *6)/(576 * 8* 100)\\\\=(10^7 * 1)/(96 * 8)\\\\=(10^7 * 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} = (10^9 * 0.06)/(16,000 * 8)`

`= (10^9 * 0.06)/(16,000 * 8)\\\\ = (10^9 * 6)/(16,000 * 8 * 100)\\\\ = (10^4 * 3)/(16 * 4)\\\\ = (30000)/(64 )\\\\=468.75`

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