Due date: Tuesday, Sept. 20, 1999

Weight: 5% of the final mark

1. Problem 2.1

1. [2 marks]

If two devices transmit at the same time, their signals will be on the medium at the same time, interfering with each other.

2. [4 marks] (The answer can be found in pp. 371. 4 marks can be given for any 4 of the following points.)

Advantages of a centralized scheme:

• It may afford great control over access for providing such things as priorities, overrides, and guaranteed capacity.
• It enables the use of relatively simple access logic at each station.
• It avoides problems of distributed coordination among peer entities.

• In creates a single point of failure; that is there is a point in the network that, if it fails, causes the entire network to fail.
• It may act as a bottleneck, reducing performance.

1. Problem 2.3 [4 marks]

　

Elapsed time = 1000 miles / 600 mph = 5 hours = 18,000 seconds.

Amount of data per diskette = 900 x 1024 x 8 = 6.55 x 10**6 bits

Number of diskettes = 16 x 2000 x 10 = 0.32 x 10**6

Data transfer rate = 6.55 x 0.32 x 10**12 / 18,000 = 116 Mbps.

　

　

　

2. Problem 2.7 [9 marks]

1. [4 marks]

(30 pictures/s) (480 x 500 pixels/picture) = 7.2 x 10⁶ pixels/s

Each pixel can take on one of 32 values and can therefore be represented by 5 bits: R = 7.2 x 10⁶ pixels/s x 5 bits/pixel = 36 x 10⁶ bps

　

2. [4 marks]

1. Problem 2.9 [4 marks]

C = W log₂ (1 + S/N)

W = 300 Hz

(S/N)_dB = 3

S/N = 10^0.3

C = 300 log₂ (1 + 10^0.3)

C = 300 log₂ (2.99) = 475 bps

　

2. Problem 2.10 (b) [3 marks]

9600 = 2W x 8

W = 600 Hz

　

4. Problem 5.1 [9 marks]

1. [3 marks, -2 marks if data bits is treated as 8 instead of 7]

There are 7 data bits, 1 start bit, 1.5 stop bits, and 1 parity bit

G = 7/(1 + 7 + 1 + 1.5 bits) = 7/10.5

Therefore, R = 0.67B

　

2. [3 marks]

Each frame contains 48 + 128 = 176 bits.

The number of characters is 128/8 = 16

The number of data bits is 16 x 7 = 112

Therefore, R = (112/176)B = 0.64B

　

3. [3 marks]