ECE 407: Introduction to Computer Communications
Homework #7  Solutions
Problem 5 Solution:
If we divide 10011 into 10101010100000 we get 1011011100, with a remainder of R = 0100.
Problem 8 Solution:
Problem 10 Solution:
a. Let
R
A
is the throughput of node A when there is no other transmission.
A’s average throughput =
R
A
·
p
A
(1

p
B
).
The total eﬃciency =
p
A
(1

p
B
) +
p
B
(1

p
A
)
b. When
p
A
= 2
p
B
,
Node A’s average throughput =
R
A
·
2
p
B
(1

p
B
).
Node B’s average throughput =
R
B
·
p
B
(1

2
p
B
).
Although
R
A
=
R
B
, node A’s average throughput is not twice as large as that of node B. To
make this happen,
p
A
(1

p
B
) = 2
p
B
(1

p
A
)
p
A

p
A
p
B
= 2
p
B

2
p
A
p
B
p
A
=
2
p
B
1+
p
B
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documentc. Assume that average throughput of each node when the others are not transmitting is R.
A’s average throughput =
R
·
2
p
(1

p
)
N

1
Any other node’s throughput =
R
·
p
(1

2
p
)(1

p
)
N

2
Problem 14 Solution:
a.b. IP address and MAC address assignment
Interface
IP address
MAC address
A
192.168.1.1
111111111111
B
192.168.1.2
222222222222
R11(left)
192.168.1.3
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Eun
 IP address, Subnetwork, MAC address, average throughput

Click to edit the document details