COMPUTER NETWORKS

NET2201 COMPUTER NETWORKS
Tutorial 1
TUTORIAL QUESTIONS FOR CHAPTER 1: COMPUTER NETWORKS AND THE INTERNET
1. Compare and contrast packet switching and circuit switching. (8 marks)
Both packet switching and circuit switching are used in the network core.
Packet switching
Circuit switching
Requirement on end-to-end resource reservation
No. A source uses resources (e.g., bandwidth) along a path from source host to destination host in an on-demand manner.
Yes. A source host must reserve resources along a path (called circuit or dedicated end-to-end connection) from a source host to a destination host. The resource must be reserved for the entire duration of the communication session.
Performance Efficiency
Higher. There is more sharing of link capacity.
Lower. Reserved resources may not be fully utilized, and may not be sufficient. Underutilized reserved resources cannot be used for other packets.
Packets need to wait at queue?
Yes. There is variable and unpredictable delay, and not suitable for real-time services.
No. Reserved resource provides guaranteed constant rate.
Complexity
Circuit switching has a higher complexity since reservation requires end-to-end signaling protocol.
Cost Efficiency
Packet switching has a higher cost efficiency, whereby cost is lower since there is no reservation.
Popularity
Packet switching is more popular.
2. Do you agree that a traffic intensity of greater than 0.8 indicates congestion? Explain your answer. (4 marks)
Yes. Traffic intensity is calculated by La/R, where L represents packet length, a represents packet arrival rate, and R represents transmission rate. As traffic intensity approaches 1, the packet length L and packet arrival rate a increase, and the transmission rate R reduces. So, the queue size increases without bound and queuing delay approaches infinity, which are an indication of congestion.
3. This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and B, connected by a
single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B.
a. Express the propagation delay, dprop, in terms of m and s. (1 mark)
𝒅𝒑𝒓𝒐𝒑=π’Žπ’”
b. Determine the transmission time of the packet, dtrans, in terms of L and R.
(1 mark) 𝒅𝒕𝒓𝒂𝒏𝒔=𝑳𝑹
c. Ignoring processing and queuing delays, obtain an expression for the end-to-end delay. (1 mark)
π’…π’†πŸπ’†=𝒅𝒑𝒓𝒐𝒑+𝒅𝒕𝒓𝒂𝒏𝒔=π’Žπ’”+𝑳𝑹 seconds
d. Suppose Host A begins to transmit the packet at time t = 0. At time t = dtrans, where is the last bit of the packet? (1 mark)
The bit has just left Host A.
e. Suppose dprop is greater than dtrans. At time t = dtrans, where is the first bit of the packet? (1 mark)
The first bit is in the link and has not reached Host B.
f. Suppose dprop is less than dtrans. At time t = dtrans, where is the first bit of the packet? (1 mark)
The first bit has reached Host B.
Host A
Host A
Host B
Host B
π‘‘π‘π‘Ÿπ‘œπ‘
π‘‘π‘‘π‘Ÿπ‘Žπ‘›π‘ 
First bit
First bit
Host A
Host A
Host B
Host B
π‘‘π‘π‘Ÿπ‘œπ‘
π‘‘π‘‘π‘Ÿπ‘Žπ‘›π‘ 
First bit
First bit
g. Suppose s = 2.5 Γ— 108, L = 120 bits, and R = 56 kbps. Find the distance m so that
dprop equals dtrans. (2 marks)
𝒅𝒑𝒓𝒐𝒑 = 𝒅𝒕𝒓𝒂𝒏𝒔
π’Ž
𝒔
=
𝑳
𝑹
π’Ž =
𝑳
𝑹
Γ— 𝒔
π’Ž =
𝟏𝟐𝟎
πŸ“πŸ”πŸŽπŸŽπŸŽ
Γ— 𝟐. πŸ“ Γ— πŸπŸŽπŸ–
π’Ž = πŸ“πŸ‘πŸ” km
4. Consider the following scenario:
Suppose that there are M client-server pairs. Denote RS, RC and R for the rates of the
server links, client links, and network link. Assume all other links have abundant
capacity and that there is no other traffic in the network besides the traffic generated by
the M client-server pairs. Derive general expression for throughput in terms of RS, RC
and R and M. (3 marks)
Throughput = min {𝑹𝒔, 𝑹𝒄, 𝑹
𝑴
}
M
M
5. Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose, the propagation speed over the link is 2.5 Γ— 108 meters/sec.
a. Calculate the bandwidth-delay product, R Γ— dprop. (2 marks)
𝑹×𝒅𝒑𝒓𝒐𝒑 =π‘ΉΓ—π’Žπ’” =πŸΓ—πŸπŸŽπŸ”Γ—πŸΓ—πŸπŸŽπŸ•πŸ.πŸ“Γ—πŸπŸŽπŸ–
=πŸπŸ”πŸŽ,𝟎𝟎𝟎 bits
b. What does the bandwidth-delay product indicates? (2 marks)
𝑹×𝒅𝒑𝒓𝒐𝒑=[π’ƒπ’Šπ’•π’”π’”π’†π’„π’π’π’…π’”Γ—π’”π’†π’„π’π’π’…π’”]=π’ƒπ’Šπ’•π’”
So, it represents the number of bits in the link.
c. Consider sending a file of 800,000 bits from host A to host B. What is the maximum number of bits that will be in the link at any given time? (2 marks)
Since πŸπŸ”πŸŽ,𝟎𝟎𝟎<πŸ–πŸŽπŸŽ,𝟎𝟎𝟎, there are 160,000 bits in the link.
6. Referring to Question 5, but now with a link of R = 1 Gbps.
a. Calculate the bandwidth-delay product, R Γ— dprop. (2 marks)
𝑹×𝒅𝒑𝒓𝒐𝒑=π‘ΉΓ—π’Žπ’” =πŸΓ—πŸπŸŽπŸ—Γ—πŸΓ—πŸπŸŽπŸ•πŸ.πŸ“Γ—πŸπŸŽπŸ–
=πŸ–Γ—πŸπŸŽπŸ• bits
b. Consider sending a file of 800,000 bits from host A to host B. What is the maximum number of bits that will be in the link at any given time? (2 marks)
Since πŸ–Γ—πŸπŸŽπŸ•>πŸ–πŸŽπŸŽ,𝟎𝟎𝟎, there are 800,000 bits in the link.
Source: Chapter 1, Computer Networking – A Top-Down Approach, Kurose and Ross.
TUTORIAL QUESTIONS FOR CHAPTER 2: APPLICATION LAYER
1. Explain the differences between client-server architecture and peer-to-peer architecture. (8 marks)
Client-server architecture
Peer-to-peer architecture
Characteristics
Server is always on host.
Peers are not always-on hosts.
Server has permanent IP address.
Function
Server service requests from clients.
Peers request services from other peers, then provide services to other peers.
Advantages
β€’ Peer-to-peer architecture has higher self-scalability since new peers provide services to other peers.
β€’ Peer-to-peer architecture has lower cost since expensive servers are not necessary.
2. Why is a centralized DNS server impractical? (3 marks)
Centralized DNS has the following disadvantages:
β€’ Single point of failure
β€’ High traffic volume
β€’ High delay: DNS server may be far from DNS client
β€’ High maintenance: Large record for all hostname
3. Do you agree that recursive query is more efficient than iterated recursive query? Justify your answer. (4 marks)
No, iterated recursive query is more efficient. In recursive query, the burden of name resolution is put on the contacted name server, and so there is a heavy load at upper levels of hierarchy.
4. Consider the following scenario, for which there is an institutional network connected to the Internet and assume that the access link is of 15 Mbps. Suppose that the average object size is 850,000 bits and that the average request rate from the institution’s browsers to the origin servers is 16 requests per second. Also suppose that the amount of time it takes from when the router on the Internet side of the access link forwards an HTTP request until it receives the response is 3 seconds on average. Model the total average response time as the sum of the average access delay (that is, the delay from Internet router to institution router) and the average Internet delay. For the average access delay, use Ξ” / (1 β€’ Ξ”βˆ™Ξ²), where Ξ” is the average time required to send an object over the access link and Ξ² is the arrival rate of objects to the access link.
a. Find the total average response time. (4 marks)
Given
𝑹=πŸπŸ“ Mbps
𝑳=πŸ–πŸ“πŸŽπŸŽπŸŽπŸŽ bits
𝜷=πŸπŸ” requests/s
Internet delay = 3 s
Transmission delay Ξ”=𝑳𝑹
Access delay = Ξ”πŸβˆ’Ξ”πœ·
Total response delay = Internet delay + access delay + LAN delay
Ξ”=πŸ–.πŸ“Γ—πŸπŸŽπŸ“πŸ.πŸ“Γ—πŸπŸŽπŸ•=𝟎.πŸŽπŸ“πŸ”πŸ•s
Access delay = Ξ”πŸβˆ’Ξ”πœ·=𝟎.πŸŽπŸ“πŸ”πŸ•πŸβˆ’(𝟎.πŸŽπŸ“πŸ”πŸ•)(πŸπŸ”)=𝟎.πŸ”πŸπŸπ’”
Total response delay = 3 + 0.611 = 3.611s
b. Now suppose a cache is installed in the institutional LAN. Suppose the miss rate is 0.4. Find the total response time. (4 marks)
During miss, which happens to 40% of the requests, the 𝜷 is reduced by 40%.
Access delay = Ξ”πŸβˆ’Ξ”πœ·=𝟎.πŸŽπŸ“πŸ”πŸ•πŸβˆ’(𝟎.πŸ’)(𝟎.πŸŽπŸ“πŸ”πŸ•)(πŸπŸ”)=𝟎.πŸŽπŸ–πŸ—π’”
Response time for 40% of the traffic is 3 + 0.089 = 3.089s
Total response time = 0.6 (0) + 3.089 (0.4) = 1.24s
5. Consider distributing a file of F = 15 Gbits to N peers. The server has an upload rate of us = 30 Mbps, and each peer has a download rate of di = 2 Mbps and an upload rate of u. For N = 10, 100, and 1,000 and u = 300 Kbps, 700 Kbps, and 2 Mbps, prepare a chart giving the minimum distribution time for each of the combinations of N and u for both client-server distribution and P2P distribution. (10 marks)
Given 𝑭=𝟏.πŸ“Γ—πŸπŸŽπŸπŸŽ
𝒖𝒔=πŸ‘Γ—πŸπŸŽπŸ• bps
π’…π’Š=πŸΓ—πŸπŸŽπŸ” bps 𝑡={𝟏𝟎,𝟏𝟎𝟎,𝟏𝟎𝟎𝟎} 𝒖={πŸ‘Γ—πŸπŸŽπŸ“,πŸ•Γ—πŸπŸŽπŸ“,πŸΓ—πŸπŸŽπŸ”}
For the client-server architecture, 𝑫π‘ͺ𝑺=𝐦𝐚𝐱 {𝑡𝑭𝒖𝒔,π‘­π’…π’Žπ’Šπ’} =𝐦𝐚𝐱 {𝟏.πŸ“Γ—πŸπŸŽπŸπŸŽπ‘΅πŸ‘Γ—πŸπŸŽπŸ•,𝟏.πŸ“Γ—πŸπŸŽπŸπŸŽπŸΓ—πŸπŸŽπŸ”} =𝐦𝐚𝐱 {πŸ“πŸŽπŸŽπ‘΅,πŸ•πŸ“πŸŽπŸŽ}
10
100
1000
πŸ‘Γ—πŸπŸŽπŸ“
7500
50000
500000
πŸ•Γ—πŸπŸŽπŸ“
7500
50000
500000
πŸΓ—πŸπŸŽπŸ”
7500
50000
500000
For the peer-to-peer architecture,
π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {𝑭𝒖𝒔,π‘­π’…π’Žπ’Šπ’,𝑡𝑭𝒖𝒔+Ξ£π’–π’Šπ‘΅π’Š=𝟏} =𝐦𝐚𝐱 {𝟏.πŸ“Γ—πŸπŸŽπŸπŸŽπŸ‘Γ—πŸπŸŽπŸ• ,𝟏.πŸ“Γ—πŸπŸŽπŸπŸŽπŸΓ—πŸπŸŽπŸ”,𝟏.πŸ“Γ—πŸπŸŽπŸπŸŽπ‘΅πŸ‘Γ—πŸπŸŽπŸ•+π’–π’Šπ‘΅} =𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,𝟏.πŸ“Γ—πŸπŸŽπŸπŸŽπ‘΅πŸ‘Γ—πŸπŸŽπŸ•+π’–π’Šπ‘΅}
For π’–π’Š=πŸ‘Γ—πŸπŸŽπŸ“ and 𝑡=𝟏𝟎, π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,πŸ’πŸ“πŸ’πŸ“.πŸ’πŸ“}=πŸ•πŸ“πŸŽπŸŽ
For π’–π’Š=πŸ‘Γ—πŸπŸŽπŸ“ and 𝑡=𝟏𝟎𝟎, π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,πŸπŸ“πŸŽπŸŽπŸŽ}=πŸπŸ“πŸŽπŸŽπŸŽ
For π’–π’Š=πŸ‘Γ—πŸπŸŽπŸ“ and 𝑡=𝟏𝟎𝟎𝟎, π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,πŸ’πŸ“πŸ’πŸ“πŸ’.πŸ“πŸ’}=πŸ’πŸ“πŸ’πŸ“πŸ’.πŸ“πŸ’
For π’–π’Š=πŸ•Γ—πŸπŸŽπŸ“ and 𝑡=𝟏𝟎, π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,πŸ’πŸŽπŸ“πŸ’.πŸŽπŸ“}=πŸ•πŸ“πŸŽπŸŽ
For π’–π’Š=πŸ•Γ—πŸπŸŽπŸ“ and 𝑡=𝟏𝟎𝟎, π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,πŸπŸ“πŸŽπŸŽπŸŽ}=πŸπŸ“πŸŽπŸŽπŸŽ
For π’–π’Š=πŸ•Γ—πŸπŸŽπŸ“ and 𝑡=𝟏𝟎𝟎𝟎, π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,πŸπŸŽπŸ“πŸ’πŸ•.πŸ—πŸ“}=πŸπŸŽπŸ“πŸ’πŸ•.πŸ—πŸ“
For π’–π’Š=πŸΓ—πŸπŸŽπŸ” and 𝑡=𝟏𝟎, π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,πŸ‘πŸŽπŸŽπŸŽ}=πŸ•πŸ“πŸŽπŸŽ
For π’–π’Š=πŸΓ—πŸπŸŽπŸ” and 𝑡=𝟏𝟎𝟎, π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,πŸ”πŸ“πŸπŸ.πŸ•πŸ’}=πŸ•πŸ“πŸŽπŸŽ
For π’–π’Š=πŸΓ—πŸπŸŽπŸ” and 𝑡=𝟏𝟎𝟎𝟎, π‘«π‘·πŸπ‘·=𝐦𝐚𝐱 {πŸ“πŸŽπŸŽ,πŸ•πŸ“πŸŽπŸŽ,πŸ•πŸ‘πŸ–πŸ—.πŸπŸ”}=πŸ•πŸ“πŸŽπŸŽ
10
100
1000
πŸ‘Γ—πŸπŸŽπŸ“
7500
25000
πŸ’πŸ“πŸ’πŸ“πŸ’.πŸ“πŸ’
πŸ•Γ—πŸπŸŽπŸ“
7500
15000
πŸπŸŽπŸ“πŸ’πŸ•.πŸ—πŸ“
πŸΓ—πŸπŸŽπŸ”
7500
7500
7500
Source: Chapter 2, Computer Networking – A Top-Down Approach, Kurose and Ross.
TUTORIAL QUESTIONS FOR CHAPTER 3: TRANSPORT LAYER
1. One of the problems with timeout-triggered retransmissions is that the time-out period can be relatively long. When a segment is lost, this long timeout period forces the sender to delay resending the lost packet, thereby increasing the end-to-end delay. Is there any remedy for this? (4 marks)
Fast Retransmission. Sender receives triple duplicate ACK for packet n, and retransmit packet n+1.
2. Consider the following figure of TCP window size as a function of time. Assuming TCP Reno is the protocol experiencing the behavior. In all cases, you should provide a short discussion justifying your answer.
a. Identify the intervals of time when TCP slow start is operating. (2 marks)
[1,6], [23,26]
b. Identify the intervals of time when TCP congestion avoidance is operating.
(2 marks)
[6,16], [17,22]
c. After the 16th transmission round, is segment loss detected by a triple duplicate ACK or by a timeout? (2 marks)
Triple duplicate ACK
d. After the 22nd transmission round, is segment loss detected by a triple duplicate ACK or by a timeout? (2 marks)
Timeout
e. What is the initial value of ssthresh at the first transmission round? (2 marks)
32
f. What is the value of ssthresh at the 18th transmission round? (2 marks)
42/2 = 21
g. What is the value of ssthresh at the 24th transmission round? (2 marks)
(29+1)/2 = 15
h. During what transmission round is the 70th segment sent? Consider congestion avoidance starts at 6th transmission round. (4 marks)
Round
Window Size
Packet Number
1
1
1
2
2
2-3
3
4
4-7
4
8
8-15
5
16
16-31
6
32
32-63
7
33
64-96
So, the 70th segment is sent in Round 7
i. Assuming a packet loss is detected after the 26th round by the receipt of a triple duplicate ACK, what will be the values of the congestion window size and of ssthresh? (4 marks)
cwnd = 8/2 = 4, ssthresh = 4
j. Suppose TCP Tahoe is used (instead of TCP Reno), and assume that triple duplicate ACKs are received at the 16th round. What are the ssthresh and the congestion window size at the 17th round? (2 marks)
cwnd = 1, ssthresh = 42/2 = 21
k. Again suppose TCP Tahoe is used, and there is a timeout at 16th round. How many packets have been sent out from 17th round till 22nd round, inclusive?
(4 marks)
Round
Window Size
17
1
18
2
19
4
20
8
21
16
22
21
Total = 52
3. Suppose, TCP uses AIMD sending a large file from a host to another over a TCP connection that has no loss.
a. Suppose TCP uses AIMD for its congestion control without slow start. Assuming cwnd increases by 1 MSS every time a batch of ACKs is received and assuming approximately constant round-trip times, how long does it take for cwnd increase from 6 MSS to 12 MSS (assuming no loss event)? (4 marks)
RTT
Window Size
6
1
7
2
8
3
9
4
10
5
11
6
12
So, it takes 6 RTT.
b. What is the average throughput (in terms of MSS and RTT) for this connection up through time = 6 RTT? (4 marks)
Throughput = (6 + 7 + 8 + 9 + 10 +11) MSS/ 6 RTT = 8.5 MSS/ RTT
Source: Chapter 3, Computer Networking – A Top-Down Approach, Kurose and Ross.
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