1. Data Communications: Homework 1 Chapter 1 –Problem 9 The number of cables for each type of network is: a. Mesh: n (n – 1) / 2 b. Star: n c. Ring: n – 1 d. Bus: one backbone and n drop lines –Problem 16 a. Cable links: n (n – 1) / 2 = (6 × 5) / 2 = 15 b. Number of ports: (n – 1) = 5 ports needed per device –Problem 18 This is a LAN. The Ethernet hub creates a LAN as we will see in Chapter 13.

2. Data Communications: Homework 1 Chapter 1 –Problem 22 –Problem 29 List at least 5 RFCs

3. Data Communications: Homework 1 Chapter 2 –Problem 16 a. Route determination: network layer b. Flow control: data link and transport layers c. Interface to transmission media: physical layer d. Access for the end user: application layer –Problem 18 a. Communication with user’s application program: application layer b. Error correction and retransmission: data link and transport layers c. Mechanical, electrical, and functional interface: physical layer d. Responsibility for carrying frames between adjacent nodes: data link layer

4. Data Communications: Homework 1 Chapter 2 –Problem 20 –Problem 21

5. Data Communications: Homework 1 Chapter 2 –Problem 22 22. If the corrupted destination address does not match any station address in the network, the packet is lost. If the corrupted destination address matches one of the stations, the frame is delivered to the wrong station. In this case, however, the error detection mechanism, available in most data link protocols, will find the error and discard the frame. In both cases, the source will somehow be informed using one of the data link control mechanisms discussed in Chapter 11. –Problem 26 26. advantages and disadvantages of combining the session, presentation, and application layer into single application layer –Problem 28 28. which layer is responsible for translation, encryption, compression(presentation layer) in the internet model –Problem 30 30. find all of network layer models proposed in the OSI model, explain the differences between them

6. Data Communications: Homework 1 Chapter 3 –Problem 28 The signal is nonperiodic, so the frequency domain is made of a continuous spectrum of frequencies as shown in Figure 3.4.

7. Data Communications: Homework 1 Chapter 3 –Problem 42 We can approximately calculate the capacity as a. C = B × ( SNRdB /3) = 20 KHz × ( 40 /3) = 267 Kbps b. C = B × ( SNRdB /3) = 200 KHz × ( 4 /3) = 267 Kbps c. C = B × ( SNRdB /3) = 1 MHz × ( 20 /3) = 6.67 Mbps

8. Data Communications: Homework 1 Chapter 3 –Problem 46 We have (bit length) = (propagation speed) × (bit duration) The bit duration is the inverse of the bandwidth. a. Bit length = (2 × 108 m) × [(1 / (1 Mbps)] = 200 m. This means a bit occupies 200 meters on a transmission medium. b. Bit length = (2 × 108 m) × [(1 / (10 Mbps)] = 20 m. This means a bit occupies 20 meters on a transmission medium. c. Bit length = (2 × 108 m) × [(1 / (100 Mbps)] = 2 m. This means a bit occupies 2 meters on a transmission medium.

9. Data Communications: Homework 1 Chapter 4 –Problem 22 The data rate is 100 Kbps. For each case, we first need to calculate the value f / N. We then use Figure 4.6 in the text to find P (energy per Hz). All calculations are approximations. a. f /N = 0/100 = 0 → P = 1.0 b. f /N = 50/100 = 1/2 → P = 0.5 c. f /N = 100/100 = 1 → P = 0.0 d. f /N = 150/100 = 1.5 → P = 0.2

10. Data Communications: Homework 1 Chapter 4 –Problem 24 a. The output stream is 01010 11110 11110 11110 11110 01001. b. The maximum length of consecutive 0s in the input stream is 21. c. The maximum length of consecutive 0s in the output stream is 2.

11. Data Communications: Homework 1 Chapter 4 –Problem 28 a. In a lowpass signal, the minimum frequency is 0. Therefore, we can say fmax = 0 + 200 = 200 KHz → fs = 2 × 200,000 = 400,000 samples/s The number of bits per sample and the bit rate are nb = log21024 = 10 bits/sample N = 400 KHz × 10 = 4 Mbps b. The value of nb = 10. We can easily calculate the value of SNRdB SNRdB = 6.02 × nb + 1.76 = 61.96 c. The value of nb = 10. The minimum bandwidth can be calculated as BPCM = nb × Banalog = 10 × 200 KHz = 2 MHz