CS 321
Details on the exam itself:
Review for Final
 Mon, Dec 12, 4:00 - 6:00 pm
 Open book, open notes exam. Bring these items with you.
 Bring a calculator if you wish to use one.
 Covers chapters 6 through 11 in the text and the supplementary items that were added. (The added items included topics such as error detection, error correction, the spanning tree algorithm, Nyquist formula, Shannon-Hartley law, Dijkstra's algorithm, digital signal encoding with schemes such as
B8ZS, etc.) The various course projects are not directly included in the exam.
Review Questions:
1) Show the encoding of the bit string 1 1 0 1 0 1 1 1 0 0 1 using Bipolar-AMI.
2) Show the encoding of the bit string 1 1 0 1 0 1 1 1 0 0 1 using Manchester encoding.
3) Show the encoding of the bit string 0 1 0 0 0 0 0 0 0 0 0 1 using the B8ZS scheme.
4) Run the spanning tree algorithm on the following. S indicates a switch, H indicates a host. Assume that all of the switches have the default priority of 32,768. The number inside each switch is an
(abbreviated) MAC address. The numbers on the links are the costs.
H H H H H H H H H H
55
S
100
100
80
S
100
100
12
S
19
19
73
S
100
30
S
19
20
S
100 18
S
100

5) In a communications link with essentially no noise, what is the channel capacity (in bps, Kbps, or
Mbps) if there are 4 signal levels and the bandwidth is 2000 Hz?
6) Suppose that the spectrum of a noisy communications channel runs from 1 to 1.5 MHz, so that the bandwidth is 0.5 MHz. Also suppose that the decibel version of the signal-to-noise ratio, SNR dB
, is
21. What is the channel capacity (in bps, Kbps, or Mbps)?
7) Run Dijkstra's algorithm on the graph below to find the least cost path from a to z. Each link has the same cost in both directions. Show each step (where you circle one new vertex) in a new drawing.
Remember to record both the cost and the path at each vertex. b 4 e
3 5
1
7
1 f a d z
3
4
2 g
2
1
2 c
5 h
8) Use our Hamming code to show the bits used to add error correction to the ASCII letter given by the
7 bits 0 1 1 0 1 1 1. Then flip the next to the last bit of the data and show how this would be located and corrected.
9) Calculate the CRC (also called a FCS) for the data string 1 1 0 1 1 1 0 0 when the divisor is 1 0 0 1 1.
Then suppose that when we transmit the data string followed by the FCS, the first (leftmost) data bit gets flipped. Show how this is detected.

Answers:
1) Show the encoding of the bit string 1 1 0 1 0 1 1 1 0 0 1 using Bipolar-AMI.
1 1 0 1 0 1 1 1 0 0 1
You could also have started at the negative voltage and then alternated between positive and negative for the
1 bits.
2) Show the encoding of the bit string 1 1 0 1 0 1 1 1 0 0 1 using Manchester encoding.
1 1 0 1 0 1 1 1 0 0 1
3) Show the encoding of the bit string 0 1 0 0 0 0 0 0 0 0 0 1 using the B8ZS scheme.
0 1 0
0
0
0
0
0
0
+
0
-
0
0
0
-
0
+
0 1
V V
Refer to your handout on encoding schemes for the details on the encodings used in questions 1, 2, and 3.

4) First we choose the root switch. Since the priority numbers are all the same, it is the one with the lowest MAC address.
H H H H H H H H H H
55
S
100 80
S
100 73
S
100 18
S
100
100 root switch
12
S
19
19
100
30
S
100
19
20
S
Next, we figure out the root port costs and use these to determine the root port for each switch (other than the root switch).
H H H H H H H H H H
55
S
RPC
100
RP
RPC
200
100
100
RPC
200 root switch
80
S
RPC
100
RP
100
12
S
RPC
219
100
19
RPC
200
RP
RPC
19
73
S
RP RPC
RPC
119
100
300
RPC
219
100
RPC
219
RPC
RP
119
100
18
S
RPC
400
30
S
19
RP
RPC
19
38
RPC
19
RPC
38
20
S

Then we find the designated port costs and use these to determine the designated port for each link. All ports on the root switch are automatically designated ports, so don't bother to figure out costs there. Also, since a root port cannot be a designated port, that helps in finding some of the DP's without looking at costs. When the 2 ends of a link have the same DPC's, choose as DP the port on the switch with the lower ID number.
H H H H H H H H H H
55
S
RP
DPC
100
100
100
DP
DPC
100
DP root switch
80
S
RP
100
DP
DP
DPC
100
100
DPC
119
12
S
DP 19
73
S
RP
DPC
119
RP
DP
100
DP
30
S
100 DP
DPC
119
100
RP
18
S
DP
19
RP
DPC
DP
19
DPC
19
19
20
S
Finally, we block each port that is not a DP or RP. Each blocked port is marked with an X below.
H H H H H H H H H H
55
S
X
100 DP 80
S
DP 100
X
73
S
X
100 DP 18
S
RP
RP
100
DP 100
DP root switch
12
S
DP
DP
19
19
RP DP
RP
RP
DP
100
DP
30
S
X
19
20
S
100
RP

5) In a communications link with essentially no noise, what is the channel capacity (in bps, Kbps, or
Mbps) if there are 4 signal levels and the bandwidth is 2000 Hz?
Use the Nyquist formula: C = 2 B log
2
(M), where B is bandwidth and M is the number of signalling levels. Note that log
2
(4) = 2 since 2 is the power you put on the base, 2, to get 4.
C = 2(2000) log
2
(4) = 4000(4) = 16,000 bps = 16 Kbps
6) Suppose that the spectrum of a noisy communications channel runs from 1 to 1.5 MHz, so that the bandwidth is 0.5 MHz. Also suppose that the decibel version of the signal-to-noise ratio, SNR dB
, is
21. What is the channel capacity (in bps, Kbps, or Mbps)?
First find the regular SNR:
SNR dB
= 10 log
10
(SNR)
21 = 10 log
10
10
2.1
= SNR
(SNR)
2.1 = log
10
(SNR)
SNR = 125.89
Then use the Shannon-Hartley law:
C = B log
2
(1 + SNR)
C = 0.5 * 10 6 * log
2
(1 + SNR)
C = 0.5 * 10
6
* log
2
(126.89) = 0.5 * 10
6
* log
10
(126.89) / log
10
(2)
C = 0.5 * 10
6
* 2.1034 / .3010 = 3,494,000 = 3.494 Mbps
7) Run Dijkstra's algorithm on the graph below to find the least cost path from a to z. Each link has the same cost in both directions. Show each step (where you circle one new vertex) in a new drawing.
Remember to record both the cost and the path at each vertex. b
3(a)
4 e
3
5 a
1 d
7
1 f z
3
0 2 2
4 g
1 2
4(a) c
5 h

a
0 b
3(a)
4
7(a,b) e
3
4
4(a) c
1 d
4(a,b)
7
2
1 f g
2
1
5 h
5
3
2 z
At the next step either d or c could be circled: b
3(a)
4
7(a,b) e
3
1 1 f a d
4(a,b)
7
0 2 2
4 g
1
4(a) c
5
11(a,b,d) 3
5 h
2 z

a
0
3
4
4(a) b
3(a)
1
2
4 d
4(a,b)
7(a,b) e
1
7
5
1 f
11(a,b,d) 3
2 g
5(a,c)
2 c
5 h
9(a,c) z a
0
3
4
4(a) b
3(a)
1
2
4 d
4(a,b)
7(a,b) e
1
7
5
1 f
7(a,c,g) 3
2 g
5(a,c)
2 c
5 h
9(a,c) z
At the next stage we can either circle e or f:

a
0
3
4
4(a) b
3(a)
1
2
4 d
4(a,b)
7(a,b) e
1
7
5
1 f
7(a,c,g) 3
2 g
5(a,c)
2 c
5 h
9(a,c) z
12(a,b,e) a
0
3
4
4(a) b
3(a)
1
2
4 d
4(a,b)
7(a,b) e
1
7
5
1 f
7(a,c,g) 3
2 g
5(a,c)
2 c
5 h
9(a,c) z
10(a,c,g,f)

a
0
3
4
4(a) b
3(a)
1
2
4 d
4(a,b)
7(a,b) e
1
7
5
1 f
7(a,c,g) 3
2 g
5(a,c)
2 c
5 h
9(a,c) z
10(a,c,g,f)
There was no change in the cost or path info at z at the above step. In the last picture we would simply circle z, which is trivial, so we leave that out.
Final result: The lowest cost path from a to z that we found was a, c, g, f, z and it had cost 10.
8) Use our Hamming code to show the bits used to add error correction to the ASCII letter given by the
7 bits 0 1 1 0 1 1 1. Then flip the next to the last bit of the data and show how this would be located and corrected. bit # data bits
1
2
3
4
0 check bits
1
1
0
5
6
7
1
1
0
8
9
10
11
1
1
1
1
Thus the string that we transmit would be 1 1 0 0 1 1 0 1 1 1 1.
Flipping the next to the last bit gives 1 1 0 0 1 1 0 1 1 0 1.
The error is detected and corrected as follows:

bit # received bits new check bits
1
2
1
1
0
1
6
7
8
3
4
5
1
0
1
0
0
1
0
1
9
10
11
1
0
1
The fact that the new check bits are not all zeros indicates an error.
Writing the new check bits in reverse order gives the binary number 1010, which is ten in decimal.
Flip bit 10 and the data is corrected! The entire string is now 1 1 0 0 1 1 0 1 1 1 1. Taking out the 4 check bits leaves us with the original data: 0 1 1 0 1 1 1.
9) Calculate the CRC (also called a FCS) for the data string 1 1 0 1 1 1 0 0 when the divisor is 1 0 0 1 1.
Then suppose that when we transmit the data string followed by the FCS, the first (leftmost) data bit gets flipped. Show how this is detected.
A 5-bit divisor yields a 4-bit FCS, so begin by adding 4 zeros to the right end of the data string.
Then divide (modulo 2) by the divisor.
1 1 0 0 0 0 0 0
---------------------------------------
1 0 0 1 1 | 1 1 0 1 1 1 0 0 0 0 0 0
1 0 0 1 1
--------------
1 0 0 0 1 Remember to use exclusive OR, not ordinary subtraction! 1 0 0 1 1
--------------
1 0 0 0 0
1 0 0 1 1
---------------
1 1 0 0 0
As long as the leading 1 matches, the divisor goes into it.
1 0 0 1 1
--------------
1 0 1 1 The 4-bit remainder is the FCS.
The transmitted string is thus 1 1 0 1 1 1 0 0 1 0 1 1, the original data followed by the FCS.
The received string, with the flipped bit, is 0 1 0 1 1 1 0 0 1 0 1 1
We check for error(s) by doing another division:

1 0 1 0 0 1 1
---------------------------------------
1 0 0 1 1 | 0 1 0 1 1 1 0 0 1 0 1 1
1 0 0 1 1
--------------
1 0 0 0 0
1 0 0 1 1
--------------
1 1 1 0 1
1 0 0 1 1
--------------
1 1 1 0 1
1 0 0 1 1
--------------
1 1 1 0

Not all zeros indicates an error!