1 / 2

Answers of Exercise 11

Answers of Exercise 11. 1. Explain similarities and differences between UDP and TCP protocols Answer: See the lecture note. 2. The following is a TCP header in hexadecimal format.    05320017 00000001 00000000 501007FF 00000000 a. What are the source and destination port numbers?

felice
Download Presentation

Answers of Exercise 11

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Answers of Exercise 11 1. Explain similarities and differences between UDP and TCP protocols Answer: See the lecture note. 2. The following is a TCP header in hexadecimal format.    05320017 00000001 00000000 501007FF 00000000 a. What are the source and destination port numbers? Source port: 0532h=1330, Destination port: 0017h=23b. What is the sequence number? 00000001c. What is the acknowledgement number? 00000000d. What is the length of the header? 20 octets or bytese. What is is the type of the segment? Acknowledgement messagef. What is the window size? 07FF(h)=2047 bytes 3. A computer uses TCP to send a data to the other computer. The data is 100 bytes. Calculate the efficiency (ratio of useful bytes to total bytes) first at the TCP level (no optional field), then at the IP level (no option field), and finally at Ethernet link layer ( no option field), respectively. Answer: TCP level: 100/120 = 83.3% IP level: 100/140 = 71.4% Ethernet level: 100/166 = 60.2%

  2. Answers of Exercise 11 4. Corrupted and lost segments are overcome using acknowledge and retransmission technique in TCP. In the adaptive retransmission technique, timeout setting is very important and use the algorithm RTO(n+1)= β RTT(n+1) where RTT(n+1)= α RTT(n) + (1-α)RTTn. Suppose α =0.8, β=1.5 and RTT(0) = 500ms, calculate RTO(1) ~ RTO (5) for RTT1=600ms, RTT2=400ms, RTT3 (lost), and RTT4=700ms. Answer: The algorithm to calculate RTO RTO(n+1)=1.5RTT(n+1) and RTT(n+1)= 0.8RTT(n)+0.2RTTn when the nth datagram is acknowledged RTO(n+1)=2.0xRTO(n) and RTT(n+1)=RTT(n) when the nth datagram is lost n 0 1 2 3 4 RTTn 0 600 400 - 700 RTT(n+1) 500 520 496 496 537 RTO(n+1) 750 780 744 1488 805 5. Suppose two programs use TCP to establish a connection, communicate, terminate the connection, and then open a new connection. Further suppose a FIN message sent to shut down the first connection is duplicated and delayed until the second connection has nee established. If a copy of the old FIN is delivered, will TCP terminate the new connection? Why? Answer: TCP will not terminate the new connection since each connection will have a unique sequence number to be distinguished from other connections.

More Related