Homework 3 Answers

Jeffrey Absher



24.  (15 pts) Ex. 15, 16 and 18 in Chapter 8.
15) Is the size of the ARP packet fixed? Explain.
No, The ARP packet size must vary because it contains 2 Hardware/MAC addresses in it and 2 different protocol addresses in it. Depending on the datalink and network protocol used the size addresses vary
16) Is the size of the RARP packet fixed? Explain
No, The RARP packet size must vary because it contains 2 Hardware/MAC addresses in it and 2 different protocol addresses in it. Depending on the datalink and transport protocol used the size addresses vary.
18)What is the size of a RARP packet when the protocol is IP and the hardware is Ethernet?
28 bytes = 8 bytes of header info plus12 bytes hardware +8 bytes network data


25.  (10 pts) Ex. 40 and 41 in Chapter 11.
40) A client uses UDP to send data to a server. The data is (are) 16 bytes. Calculate the efficiency of this transmission at the UDP level (ration of useful bytes to total bytes)?
0.66=2/3 = 16/24
41) Redo the previous exercise calculating the efficiency of the transmission at the IP level. Assume no options for the IP header.
0.36 = 4/11=16 / (24 + 20)


26.   (10 pts) Ex. 37 in Chapter 12.
37)TCP sends a segment at 4:30:20. It does not receive an acknowledgement. At 4:30:25, it retransmits the previous segment. It receives an acknowledgement at 4:30:27. What is the new value for RTT according to Karn’s algorithm if the previous RTT was four seconds?
Four Seconds (no change)


27.   (5 pts) Ex. 40 in Chapter 12
If HLEN is 0111, how many bytes of option are included in the segment?
8 bytes


28.   (10 pts) Ex. 43 in Chapter 12.
43)TCP opens a connection using an initial sequence number of 14,534. The other party opens the connection with and ISN of 21732. Show the three TCP segments during the connection establishment.
See attached sheet.


29.   (20 pts) Ex. 47 and 48 in Chapter 12.
47) A TCP connection is using a window size of 10,000 bytes and the previous acknowledgement number was 22,001. It receives a segment that acknowledges bytes 24,001. Draw a diagram to show the situation of the window before and after.
See attached sheet.
48) Redo exercise 47 if the receiver has changed the window size to 11,000.
See attached sheet


30.   (10 pts) Why would we want to impose a maximum packet lifetime for a transport layer connection? Why is it necessary to decouple the buffering from the acknowledgments (or flow control from the acknowledgments) in the transport layer?
If transport layer connections do not have a maximum packet lifetime, a late segment from a pervious connection may arrive and be mistaken for a segment from the current session due to the unreliability of the network layer and the reuse/wrapping of connection/sequence numbers.

Buffering and acknowledgements must be decoupled because unlike the datalink layer where frames are transmitted in sequence and acks are waited on prior to transmission of the next frame (“stop and wait”) the transport layer may be sending data over many datalinks and stop-and-wait would be extremely inefficient. Instead the sliding window is used for flow control allowing the sender to send data and wait for acks. This allows segment to arrive out of order (which couldn’t happen with stop and wait) and allows the sender to keep sending as long as it can “on credit” until the limit (window size) is reached.