What is the minimum number of bits you need for the sequence number to achieve the above maximum transmission rate?

Maximum window size = 1 + 2*a where a = Tp/Tt

Minimum sequence numbers required = 1 + 2*a.

All the packets in the current window will be given a sequence number. Number of bits required to represent the sender window = ceil(log2(1+2*a)).

Step-by-step explanation: But sometimes number of bits in the protocol headers is pre-defined. Size of sequence number field in header will also determine the maximum number of packets that we can send in total cycle time. If N is the size of sequence number field in the header in bits, then we can have 2N sequence numbers.

Window Size ws = min(1+2*a, 2N)

If you want to calculate minimum bits required to represent sequence numbers/sender window, it will be ceil(log2(ws)).

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680