Description

Course Number: EE450
Homework 3
Assignment 3
Chapter 6
Problem 18
Problem 19
At t = 245 bit times, both A and B detect collision and start transmitting a jam signal. Because the length of a jam signal is 48 bits, at t = 245 + Tp + Tt = 245 + 245 + 48 = 538 bit times, both A and B finish receiving the whole jam signal, and then A detects the channel is idle. After that, A must wait for a IFS time, that is, 96 bit times before transmitting, so at t = 538 + 96 = 634 bit times, A starts transmitting. At t = 634 + Tp = 634 + 245 = 879 bit times, the first bit of A’s retransmission reaches B. After t = 245 + 48 = 293 bit times, B finishes transmitting a jam signal, and it needs to wait Kb * 512 = 1 * 512 bit times of backoff time and 96 bit times of IFS time. Therefore, at t = 293 + 512 + 96 = 901 bit times, B schedules its retransmission. Because of 879 < 901, B refrains from transmitting at its scheduled time.
Problem 22
I assign MAC address and IP address to clients needed as follow:
Clients MAC/IP
A AA-AA-AA-AA-AA-AA/192.168.1.101
Router(interface for A) BB-BB-BB-BB-BB-BB/192.168.1.1
Router(interface for F) CC-CC-CC-CC-CC-CC/192.168.3.1
F DD-DD-DD-DD-DD-DD/192.168.3.101
(i)
Source MAC address: AA-AA-AA-AA-AA-AA
Destination MAC address: BB-BB-BB-BB-BB-BB
Source IP address: 192.168.1.101
Destination IP address: 192.168.3.101
(ii)
Source MAC address: AA-AA-AA-AA-AA-AA
Destination MAC address: BB-BB-BB-BB-BB-BB
Source IP address: 192.168.1.101
Destination IP address: 192.168.3.101
(iii)
Source MAC address: CC-CC-CC-CC-CC-CC
Destination MAC address: DD-DD-DD-DD-DD-DD
Source IP address: 192.168.1.101
Destination IP address: 192.168.3.101
Problem 23
Problem 24
There are three departments, and each hub only allows one collision domain, which means the maximum throughput for each hub is 100 Mbps. However, two servers still keep the same: they have their own collision domain, and each collision domain’s maximum throughput is 100 Mbps. Thus, the maximum aggregate throughput is 300 + 100 + 100 = 500 Mbps.
Problem 25
All clients, 11 hosts and 2 servers, share a collision domain, so the maximum aggregate throughput is 100 Mbps.
Problem 26
Event Switch Table Links
Before (i) empty
B sends a frame to E Flood the frame to all links except the one with B
After (i) 1. Know B MAC address
Before (ii) 1. Know B MAC address
E replies with a frame to B Forward the frame to B link
After (ii) 1. Know B MAC address
2. Know E MAC address
Before (iii) 1. Know B MAC address
2. Know E MAC address
A sends a frame to B Forward the frame to B link
After (iii) 1. Know B MAC address
2. Know E MAC address
3. Know A MAC address
Before (iv) 1. Know B MAC address
2. Know E MAC address
3. Know A MAC address
B replies with a frame to A Forward the frame to A link
After (iv) 1. Know B MAC address
2. Know E MAC address
3. Know A MAC address
(i) Because the switch doesn’t know which port is for E, so it floods the frame to all links.
(ii) Because the switch knows which port is for B, so it only forwards the frame to B link.
(iii) Because the switch knows which port is for B, so it only forwards the frame to B link.
(iv) Because the switch knows which port is for A, so it only forwards the frame to A link.
Chapter 7
Problem 5
a. Two stations, each associated with a different AP, are in the same channel. It means two stations share the media and bandwidth. For the shared media, two stations will receive each other’s frame, but they will not process the received frame, because the MAC address is not specifically addressed to them. For the shared bandwidth, the data rate of two stations will be affected. Plus, it is possible a collision will occur. Thus, CSMA/CA or RTS/CTS takes into place.
b. If two stations, both associated with different AP, are in different channels, there will not be shared media, shared bandwidth, and collision. Two stations can transmit as they want if there is no other station.