COMP90007 Internet Technologies Semester 2, 2021 Assignment 1 Sample Solutions
Question 1 (1 point)
Compared to OSI, the hybrid model removes presentation and session layers, and include protocols used in practice.
Compared to TCP/IP, the hybrid model clearly splits physical and data link layer to avoid having a crowded link layer division. The hybrid model also distinguishes the concept of protocol, service and interfaces, which was not done in TCP/IP model.
Question 2 (1 point)
Amount of data: 5 * 1280 * 720 * 3 * 8 = 110,592,000 bits
(1) over a 256kbps modem
Transmission delay = 110,592,000/(256*1000) = 432s Propagation delay = 1500/200,000 = 0.0075s
Latency = 432 + 0.0075 = 432.0075s
(2) over a 1Gbps broadband link
Transmission delay = 110,592,000/10^9 = 0.110592s Propagation delay = 1500/200,000 = 0.0075s Latency = 0.110592 + 0.0075 = 0.118092s
Question 3 (1 point)
Using Shannon’s theorem, we have: B x log(1+S/N)
SNR = 20dB, S/N = 100
Max data rate = 16kHz x log2(1+100) = 16kHz x 6.658 = 106.528 kbps
Using Nyquist’s theorem, we have: 2B x log2 V Maxdatarate=2x16kHzxlog2 8=2x16kHzx3=96kbps
The max data rate is 96kbps, as Nyquist’s limit is the bottleneck.
Question 4 (1 point)
The check bits are at position 1, 2 and 4.
The data to send is _ _1_000, if odd parity is used,
P1 + P3 + P5 + P7 = ?+1+0+0 P1 = 0 P2 + P3 + P6 + P7 = ?+1+0+0 P2 = 0 P4 + P5 + P6 + P7 = ?+0+0+0 P4 = 1 The data to send is 0011000
Question 5 (1 point)
You are expected to answer this question relating the flow graph to the service primitives like LISTEN, CONNECT, ACCEPT, etc. and explain the results observed for selected service primitives.
As this question is open-ended, please check the comments on your answers (will be released soon with your grade) for more details.