Kurose and Ross Edition 6 Exercise 3 Question R15 (Page No 287)

Question

Suppose Host A sends two TCP segments back to back to Host B over a TCP connection. The first segment has sequence number 90; the second has sequence number 110.

1. How much data is in the first segment?

2. Suppose that the first segment is lost but the second segment arrives at B. In the acknowledgment that Host B sends to Host A, what will be the acknowledgment number?

Answer 1

The sequence number of the first Byte of first segment  = 90 and The sequence number of the first Byte of second segment  = 110

which implies, The sequence number of the last Byte of first segment  = 109.

(A) Therefore number of Bytes in the segment is (109-90+1 = 20)


(B) Ack number will be 90. since the first segment is lost

Comments

Hey! How you wrote $109$?

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First Byte of 2nd segment sequence number is 110. so the previous byte which will be the last byte of Ist segment will be (110-1 = 109).

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Thanks

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I have a doubt. Host A is sending back to back segments to Host A. If the first segment if lost, then also the second segment would have arrived at Host B, and it should have sent ACK 130(As the second segment had sequence no 110). So, why is it 90?