Download List six access technologies

CEN 4500C Computer Networks Fundamentals
Instructor: Prof. A. Helmy
Homework 1: Introduction to Computer Networks and the Internet
Date Assigned: Sept 3rd, 2007. Due Date: Sept 18th, 2007 (beginning of lecture)
Total (max) points: 140
R4. List six access technologies. Classify each one as residential access, company access,
or mobile access. <6 points>
1. Dial-up modem over telephone line: residential; 2. DSL over telephone line:
residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched
Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example,
WAP): mobile [each technology 1 point]
R5. Is HFC transmission rate dedicated or shared among users? Are collisions possible in
a downstream HFC channel? Why or why not? <4 points>
HFC bandwidth is shared among the users. [2 points] On the downstream channel, all
packets emanate from a single source, namely, the head end. Thus, there are no collisions
in the downstream channel (although there may be collisions on the upstream channel!).
[2 points]
R7. What is the transmission rate of Ethernet LANs? For a given transmission rate, can
each user on the LAN continuously transmit at that rate? <4 points>
Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps. [2
points] For an X Mbps Ethernet (where X = 10, 100, 1,000 or 10,000), a user can
continuously transmit at the rate X Mbps if that user is the only person sending data. If
there are more than one active user, then each user cannot continuously transmit at X
Mbps. [2 points]
R11. What advantage does a circuit-switched network have over a packet-switched
network? <4 points>
A circuit-switched network can guarantee a certain amount of end-to-end bandwidth for
the duration of a call. Most packet-switched networks today (including the Internet)
cannot make any end-to-end guarantees for bandwidth.
R12. Why is it said that packet switching employs statistical multiplexing? Contrast
statistical multiplexing with the multiplexing that takes place in TDM. <6 points>
In a packet switched network, the packets from different sources flowing on a link do not
follow any fixed, pre-defined pattern. [3 points] In TDM circuit switching, each host gets
the same slot in a revolving TDM frame. [3 points]
R15. Suppose users share a 2Mbps link. Also suppose each user requires 1Mbps when
transmitting, but each user transmits only 20 percent of the time. <12 points>
a. When circuit switching is used, how many users can be supported? <2 points>
b. For the remainder of this problem, suppose packet switching is used. Why will
there be essentially no queuing delay before the link if two or fewer users transmit
at the same time? Why will there be a queuing delay if three users transmit at the
same time? <4 points>
c. Find the probability that a given user is transmitting. <2 points>
d. Suppose now there are three users. Find the probability that at any given time, all
three users are transmitting simultaneously. Find the fraction of time during which
the queue grows. <4 points>
15. a) 2 users can be supported because each user requires half of the link bandwidth.
b) Since each user requires 1Mbps when transmitting, if two or fewer users transmit
simultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of
the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three
users transmit simultaneously, the bandwidth required will be 3Mbps which is more than
the available bandwidth of the shared link. In this case, there will be queuing delay before
the link.
c) Probability that a given user is transmitting = 0.2
d) Probability that all three users are transmitting simultaneously =
Since the queue grows when all the users are transmitting, the fraction of time during
which the queue grows (which is equal to the probability that all three users are
transmitting simultaneously) is 0.008.
R16. Consider sending a packet from a source host to a destination host over a fixed route.
List the delay components in the end-to-end delay. Which of these delays are constant
and which are variable? <4 points>
The delay components are processing delays, transmission delays, propagation delays,
and queuing delays. All of these delays are fixed, except for the queuing delays, which
are variable. [Notes: one can argue that processing delays maybe variable if the load on
the machine/router is variable. Also, one can argue that transmission delays are variable
in media where the link capacity is variable (as in wireless 802.11)]
R19. Suppose Host A wants to send a large file to Host B. The path from Host A to Host
B has three links, of rates R1=500kbps, R2=2Mbps, and R3=1Mbps. <8 points>
a. Assuming no other traffic in the network, what is the throughput for the file
transfer. <2 points>
b. Suppose the file is 4 million bytes. Roughly, how long will it take to transfer the
file to Host B? <2 points>
c. Repeat (a) and (b), but now with R2 reduced to 100kbps. <4 points>
19. a) 500 kbps
b) 62.5 seconds
c) 100kbps; 312.5 seconds
R23. What are the five layers in the Internet protocol stack? What are the principal
responsibilities of each of these layers? <10 points>
The five layers in the Internet protocol stack are – from top to bottom – the application
layer, the transport layer, the network layer, the link layer, and the physical layer.
The principal responsibilities are: [also outlined in Section 1.5.1 in the book.]
- Application layer: to exchange packets of information (messages) with other
applications that are distributed over various end systems. Examples include
email (SMTP), file transfer (FTP), web (HTTP) and DNS, among others.
- Transport (or TCP/UDP) layer: to break down the messages into segments at one
end (the sender) and reassemble them into messages again (at the other, receiving,
end). TCP provides connection-oriented service, flow control (to match the speed
of senders and receivers), congestion control (to slow down senders when the
network is congested/overloaded) and reliable delivery (through retransmissions).
The transport layer operates end-to-end.
- Network (or IP) layer: it ‘routes’ the segments (after breaking them down into IP
datagrams) through the network [segmentation occurs at the sender, reassembly
at the receiver]. The network layer runs the routing protocols that calculate the
best (or shortest) paths to get the packets from any point in the network to any
other point, based on IP addresses. These protocols also re-calculate routes after
failures, and sometime provide load balancing by routing excess traffic over
lightly loaded paths. Routing in the Internet occurs hop-by-hop in a store-andforward fashion. The network layer provides unicast (point-to-point) or multicast
(multi-point to multi-point) routing. A routing protocol operates at a domain
(multiple-links) level.
- Link layer: breaks down the datagrams into frames and performs link-specific
functions/services, including identification of the MAC layer addresses
(sender/receiver) for that hop. Some link layers offer broadcast, multicast or
unicast communications. Some offer reliable delivery (on a per-hop or per-link
basis) as in WiFi (802.11) via retransmissions or redundancy codes [this is
different from the end-to-end reliable TCP service]. For shared media (as in
Ethernet, WiFi) the medium access control (MAC) protocol is particularly
important at the link layer. Each hop/link the datagram traverses may have a
different link layer.
- Physical
layer:
moves
bits
on
links.
Bit/signal
encoding,
modulation/demodulation, transmission and reception on the wire (twisted pair,
fiber) or on air (radio signals in wireless) are all services provided by the physical
layer.
R25. Which layers in the Internet protocol stack does a router process? Which layers does
a link-layer switch process? Which layers does a host process? <6 points>
Routers typically process layers 1 through 3 (physical, link and network layers). [Note:
Some modern routers sometimes act as firewalls or caching components, and process
layer four as well.] [2 points]
Link layer switches process layers 1 through 2 (physical and link layers). [2 points] Hosts
process all five layers. [2 points]
Q. How would you describe the overall structure of the Internet? <6 points>
What is the advantage of having such a structure? (Mention at least 2 items) <6 points>
For the first part, on the structure of the Internet:
It is a loose hierarchy from an organizational/provider or routing point of view. There are
several answers.
Provider hierarchy: various tiers of providers (tier 1 being the backbone providers and
tier 3 closer to local providers) are structured into provider-customer or peering
relationships. The hierarchy is loose (not strict) in that tier 2 (or 3) providers can provide
peering relationships without going through backbone providers (tier1).
Routing hierarchy: there are 2 levels of hierarchy for routing at the autonomous system
(AS) (or domain) level; the intra-domain routing and the inter-domain (border gateway
[BGP]) routing protocols. BGP establishes peering relationships between border routers
at different domains, while the intra-domain routing defines the routing protocol running
within a domain.
Also, a transit-stub topology structure exists in the Internet, where transit networks
(representing providers) with powerful backbone routers (connecting very high speed
links) with low out degree, and stub networks are towards the edges with highly clustered
(high degree) networks, connecting slower links leading to hosts.
Any one of the above structures is a satisfactory answer.
For the second part on the advantages:
The hierarchical structure has multiple advantages:
1- It allows each tier (or domain/autonomous system) to run its own internal
protocols and provides uniformity by providing standard protocols and
relationships between tiers (or domains)
2- It shields the effects of change in one tier/domain from directly affecting other
tiers/domains by keeping the effect local
3- It provides more flexibility in adding or removing domains or ISPs than a nonhierarchical (flat) structure
2 of these points should be sufficient
Q. What is the advantages and disadvantages of having a layered protocol architecture for
the Internet? (mention at least 3 advantages and 2 disadvantages) <10 points>
Is it true that the change in any of the layers does not affect the other layers? (support
your answer/arguments with examples) <4 points>
Advantages:
Allows an explicit structure to identify relationships between various pieces of the
complex Internet structure, by providing a reference model for discussion.
Provides a modular design that facilitates maintenance, updating/upgrading of protocols
and implementations (by various vendors) at the various layers of the stack.
Supports a flexible framework for future advances and inventions (such as mobile or
sensor networks).
Disadvantages: overhead of headers, redundancy of functions (sometimes not needed)
[such as reliability as the transport layer and the link layer, or routing at the network layer
and some link layer protocols (such as ATM)]
It is true in many cases that the change in one layer does not affect the change in the other
layers, but not always.
Examples of change that did not affect the other layers: change from FDDI to token ring,
to Ethernet at the MAC layer.
Examples of change that affected other layers: wireless vs. wired (performance of TCP
and routing degraded drastically). Introduction of 802.11 for wireless and ad hoc
networks (a change in the physical and MAC layers), does affect in a major way routing
at the network layer and the transport layers. In that case, many of the protocols needed
re-design.
P5. 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 that the propagation speed is s m/s. Host A is to
send a packet of size L bits to host B. <18 points>
a. Express the propagation delay, dprop, in terms of m and s. <3 points>
b. Determine the transmission time of the packet, dtrans, in terms of L and R. <3
points>
c. Obtain an expression for the end-to-end delay (ignore queuing and processing
delays). <3 points>
d. Suppose A begins to transmit the packet at time t=0. At time t=dtrans, where is the
last bit of the packet? <2 points>
e. Suppose dprop is greater than dtrans. At time t=dtrans, where is the first bit of the
packet? <2 points>
f. Suppose dprop is less than dtrans. At time t=dtrans, where is the first bit of the
packet? <2 points>
g. Suppose s=2.5·108 m/s, L=100 bits, and R=28kbps. Find the distance m so that
dprop equals dtrans. <3 points>
a) d m s prop = / seconds.
b) d L R trans = / seconds.
c) d (m/ s L / R) end to end = + − − seconds.
d) The bit is just leaving Host A.
e) The first bit is in the link and has not reached Host B.
f) The first bit has reached Host B.
g) Want
P24. In modern packet-switched networks, the source host segments long, applicationlayer messages (for example, an image or a music file) into smaller packets and sends the
packets into the network. The receiver then reassembles the packets back into the original
message. We refer to this process as message segmentation. Figure 1.24 illustrates the
end-to-end transport of a message with and without message segmentation. Consider a
message that is 7.5106 bits long that is to be sent from source to destination in Figure 1.24.
Suppose each link in the figure is 1.5Mbps. Ignore propagation, queuing, and processing
delays. <14 points>
a. Consider sending the message from source to destination without message
segmentation. How long does it take to move the message from the source host to
the first packet switch? Keeping in mind that each switch uses store-and-forward
packet switching, what is the total time to move the message from source host to
destination host? <3 points>
b. Now suppose that the message is segmented into 5,000 packets, with each packet
being 1,500 bits long. How long does it take to move the first packet from source
host to the first switch? When the first packet is being sent from the first switch to
the second switch, the second packet is being sent from the source host to the first
switch. At what time will the second packet be fully received at the first switch?
<3 points>
c. How long does it take to move the file from source host to destination host when
message segmentation is used/ Compare this result with your answer in part (a)
and comment. <4 points>
d. Discuss the drawbacks of message segmentation. <4 points>
a) Time to send message from source host to first packet switch =
With store-and-forward switching, the total time to move message from source host to
destination host = 5sec× 3 hops = 15sec
b) Time to send 1st packet from source host to first packet switch = .
Time at which 2nd packet is received at the first switch = time at which 1st packet is
received at the second switch = 2×1msec = 2 msec
c) Time at which 1st packet is received at the destination host = 1msec× 3 hops = 3 msec .
After this, every 1msec one packet will be received; thus time at which last (5000th)
packet is received =3 msec+ 4999*1msec = 5.002sec . It can be seen that delay in using
message segmentation is significantly less (almost 1/3rd).
d) Drawbacks:
i. Packets have to be put in sequence at the destination.
ii. Message segmentation results in many smaller packets. Since header size is usually the
same for all packets regardless of their size, with message segmentation the total amount
of header bytes is more.
R26. What is the difference between a virus and a worm? <6 points>
a) Virus: Requires some form of human interaction to spread. Classic example: E-mail
viruses.
b) Worms: No user replication needed. Worm in infected host scans IP addresses and port
numbers, looking for vulnerable processes to infect.
R27. How can a distributed denial of service (DDoS) attack be created? <6 points>
In DDoS attack, attackers make services (server, bandwidth) unavailable to legitimate
users by overwhelming the network or server with bogus traffic and excess requests.
First, the attacker(s) select the target to be attacked. Then, they break into vulnerable
hosts in the network via malware to create a botnet. Finally, the compromised hosts send
packets or requests to/towards the target (server).
R28. Alice and Bob send packets to each other over a network, while Trudy is positioned
such that she can capture all the packets sent by Alice and Bob and send whatever she
wants to Bob and Alice. List at least three of the malicious things Trudy can do. <6
points>
i. Trudy can pretend to be Bob to Alice (and vice-versa) and partially or completely
modify the message(s) being sent from Bob to Alice. For example, she can easily change
the phrase “Alice, I owe you $1000” to “Alice, I owe you $10,000”.
ii. Trudy can pretend to be Bob at a later time via record-and-playback, resending
passwords and potentially gaining access to authorized services (that only Bob should
access).
iii. Furthermore, Trudy can even drop the packets that are being sent by Bob to Alice
(and vise-versa), even if the packets from Bob to Alice are encrypted.
[or other…]