Question

Using 5-bit sequence numbers, what is the maximum size of the send and receive
windows for each of the following protocols?
a. Stop-and-Wait
b. Go-Back-N
c. Selective-Repeat

Answer 1

Explanation:

a. for sender=1

for receiver=1

b. for sender= 2^m-1

= 2^5-1

=32-1

=31

for receiver=1

C. for sender=2^(m-1)

=2^(5-1)

=2^4

=16

for receiver=16

Answer 2

Answer:

Sequence numbers 0 to 31 can be generated using a five-bit sequence number. The Nth packet's sequence number is (N mod 32). This indicates that the sequence number (101 mod 32) or 5 is included in the 101st packet.

Explanation:

1- Stop and wait protocol is also called as a one-bit sliding window protocol.

The maximum size of the sender and receiver window is 1.

2- The maximum sender window size in go-back n is (2^n)-1, n is the sequence number.  

=  2^5 -1

= 32-1

= 31

The maximum receiver window size is 1.

3- In selective repeat arq, the maximum window size is (2^n)/2 where n is sequence number, the sequence number here is 5.

                                                     so, 2^5/2

                                                        = 32/2

                                                        = 16  

Therefore, the maximum size of send and receive windows for each of the protocols given:

• Stop‐and‐Wait ARQ 1 1
• Go‐Back‐N ARQ 32 - 1 = 31 1

• Selective‐Repeat ARQ 32/2 = 16 32/2 = 16

Find more like this here:

https://brainly.com/question/18762484

#SPJ1