CS421/S03 Chapter 14 (7/e) 1. Problem 14.1
Solutions to Homework Four
a. The number of clusters M = 16. The total available bandwidth in each direction is 20 MHz. Therefore, the total number of channels in each way is 20 MHz/30 kHz = 666, and each
CS 421/S03
Homework Four Due time (07/28/03 in class)
Textbook: Data and Computer Communications, W. Stallings, 7 th edition. Chapter 14 Problem 14.1 Problem 14.4 Chapter 15 Problem 15.4 Problem 15.6 Problem 15.8 Chapter 16 Problem 16.1 Problem 16.
CS 421/S03
Homework Three Due time (07/21/03 in class)
Textbook: Data and Computer Communications, W. Stallings, 7 th edition. Chapter 10 Problem 10.1 Problem 10.4 Problem 10.7 Chapter 11 Problem 11.5 Problem 11.7 Chapter 12 Problem 12.14 Problem 1
CS 421/S03
Chapter 3 Problem 3.14 Problem 3.16 Problem 3.19 Problem 3.20 Problem 3.21 Chapter 4 Problem 4.2 Chapter 5
Homework One Due time (06/23/03 in class)
Problem 5. 7 Problem 5. 12 (assume r = 0.5) Problem 5.14 Problem 5.16 Problem 5.20 End.
CS 421/S03
Homework Two Due time (06/30/03 in class)
Textbook: Data and Computer Communications, W. Stallings, 7 th edition. Chapter 6 Problem 6.9 Problem 6.12 Chapter 7 Problem 7.2 Problem 7.9 Problem 7.12 Problem 7.15 Chapter 8 Problem 8.9 Proble
CS421/S03 Chapter 10 (7/e) 1. Problem 10.1
Solutions to Homework Three
Each telephone makes 0.5 calls/hour at 6 minutes each. Thus a telephone occupies a circuit for 3 minutes per hour. Twenty telephones can share a circuit (although this 100% util
CS421/S03 Chapter 6 (7/e) 1. Problem 6.9
Solutions to Homework Two
a. We have: Pr [single bit in error] = 10-3 Pr [single bit not in error] = 1 10-3 = 0.999 Pr [8 bits not in error] = (1-10-3)8 = 0.992 Pr [at least one error in frame] = 1-(1-10-3)
CS 421/S03
Chapter 3 (7/e) 1. Problem 3.14
Solutions to Homework One
N = 10log k + 10log T + 10log B = -228.6 + 10log 104 + 10log107 (dBW) = -228.6 + 40 + 70 = - 118.6 (dBW) 2. Problem 3.16 Use Nyquist's formula C = 2Blog2M. (a) C = 9600 bps, log2M