# Assume that the bandwidth for a TCP connection is 1048560 bits /sec. Let α be the value of RTT in milliseconds (rounded off to the nearest integer) af

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Assume that the bandwidth for a TCP connection is 1048560 bits /sec. Let α be the value of RTT in milliseconds (rounded off to the nearest integer) after which the TCP window scale option is needed. Let β be the maximum possible window size with window scale option. Then the values of α and β are

1. 63 milliseconds, 65535 x 214
2. 63 milliseconds, 65535 x 216
3. 500 milliseconds, 65535 x 214
4. 500 milliseconds, 65535 x 216

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Correct Answer - Option 3 : 500 milliseconds, 65535 x 214

Concept:

RTT (Round trip time) is the length of time it takes for a signal to be sent plus the length of time it takes for an acknowledgement of that signal to be received.

Bandwidth delay product: It gives the maximum amount of data that can be transmitted by the sender at a given time before waiting for acknowledgement.

It is measured in terms of RTT × Bandwidth.

Explanation:

In TCP, sequence number is limited to 16 bits i.e. TCP allows scaling of windows when bandwidth delay product is greater than 65,535.

Here, it is given that bandwidth = 1048560 bits/sec

RTT = α ms

Maximum possible window size = β

Bandwidth delay product = 1048560 × ∝

65535 B = 1048560 × α

α = 500 milliseconds

When we do scaling, window size increases from 64KB to 1 GB i.e. from 216 B to 230 B.

A 14-bit shift count is used in TCP header.

i.e. from 65535 to 65535 × 214 Bytes