Advanced Computer Networks
Final Exam
Cynthia Michele Vincent
California State University Fullerton, Computer Science Department
[email protected]
CPSC 558, Section 1, Fall 02
December 19, 2002
Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
1. Consider a network path that consists of four 10 Mbps Ethernet links connected by 3 storeand-forward switches. Assume that each switch introduces 10 ms propagation delay and no
queuing delay.
a) Calculate the path’s one-way latency for sending a 2 KB message (2 pt)
b) Calculate the effective bandwidth for sending a 2 KB message across the path (3 pt).
c) Calculate the effective bandwidth for sending a sequence of 2 KB messages if a 30byte ACK has to be received before it can send next packet. (5 pt)
Answer for (a):
Note: I used 1 KB = 1000 for simplicity. If I used 1 KB = 1024, the latency is still approximate
also 11.6 ms.
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Final Exam, CPSC 558
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Cynthia Michele Vincent
Answer for (b):
Note: I used 1 KB = 1000 for simplicity. If I used 1 KB = 1024, the transfer time, 0.1016, would
be approximately the same, but the throughput is slightly bigger -- 161,198 bit /sec.
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Final Exam, CPSC 558
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Answer for (c)
Note: I used 1 KB = 1000 for simplicity from the very beginning. So, you might find my
answers slightly different.
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Final Exam, CPSC 558
December 19, 2002
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Cynthia Michele Vincent
2. What 2 scaling problems the internet is experiencing? Briefly describe subnet and
supernet (CIDR) concepts and explain how each addresses these two scaling problem. (10 pt)
The two scaling problems the internet is experiencing are:

Growth of backbone routing tables as more network numbers need to be stored.

Potential for 32-bit address space to run out.
Subnetting is used to break one large group into several smaller networks. These subnets
can be distributed throughout an enterprise, resulting in less waste and better logical organization.
Subnetting provides a simple way to reduce the total number of network numbers that are assigned.
The idea is to take a single IP network number and allocate the IP addresses with that network
number to several physical networks which are referred to as subnets.
Subnetting helps solve our scalability problems in two ways:

It improves our address assignment efficiency by letting us not use up an entire C or
class B address every time we add a new physical network.

It helps aggregate information. A complex collection of physical networks can be
made to look like a single network, so that the amt of info that routers need to store
can be reduced.
Classless Interdomain Routing (CIDR) is a form of IPv4 that routers use that ignores class. In
a classful system, a router determines the class of an address and then identifies the network and
host octets based on that class. With CIDR, a router uses a bitmask to determine the network and
host portions of an address, which are no longer restricted to using an entire octet.
CIDR helps solve IPv4's scalability problems and efficiency by providing the following:

The replacement of classful addressing with a more flexible and less wasteful
classless scheme.

Enhanced route aggregation, also known as supernetting. CIDR helps solve our
scalability problems by aggregating routes. CIDR allows routers to group routes
together to minimize the quantity of routing information carried by the core routers.
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
3. TCP has a 32-bit sequence number and 16 bits for advertised window. Assuming a 100 ms
RTT and 200s maximum segment lifetime.
a) Identify two problems when the link bandwidth increases (2 pt)
b) At what bandwidth does each of problems become an issue (4 pt)
c) Describe how the internet addresses these two issues (4 pt)
The two problems when link bandwidth increases are:
1) Sequence Number wrapping
around, and 2) Advertised Window becomes smaller (not allowing it to fill larger delay X bandwidth).
At bandwidth 163.84 Mbps or higher, wrap around can occur. At bandwidth 5 Mbps, the
problem of Advertised Window becoming smaller becomes an issue.
The calculation for the problem of sequence number wrapping around is shown below.
Assuming 200s maximum segment lifetime:
1) Find what formula was used in the book. We do this by deriving the formula used in the
table on page 388 of texbook.
2) Solve for bandwidth.
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Final Exam, CPSC 558
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Cynthia Michele Vincent
The calculation for the problem of Advertised Window getting smaller is shown below.
Assuming 100ms RTT:
1) Find the inequality.
2) Solve for bandwidth.
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Final Exam, CPSC 558
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Internet addresses the problem of Sequence Number wrapping around too soon on a highspeed network by making TCP use the 32-bit timestamp to effectively extend the sequence number
space. TCP decides whether to accept or reject a segment based on a 64-bit identifier that has the
SequenceNum field in the low-order 32 bits and the timestamp in the high-order 32 bits. Since the
timestamp is always increasing, it serves to distinguish between two different incarnations of the
same sequence number. (Textbook, page 394)
Internet addresses the problem of Advertised Window becoming smaller by allowing TCP to
advertise a larger window, thereby allowing it to fill larger (delay X bandwidth) pipes that are made
possible by high-speed networks. This extension involves an option that defines a scaling factor for
the advertised window. This option allows the two sides of TCP to agree that the Advertised Window
field counts larger chunks. (Textbook, page 395)
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Final Exam, CPSC 558
December 19, 2002
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Cynthia Michele Vincent
4. Suppose we want to transmit the message 11001001 and protect it from error using CRC
polynomial X3 + 1,
a) Show the message it should be transmitted (4 pt).
b) If the leftmost bit got inverted due to noise during transmission, show the result of
receiver CRC calculation. Explain how would the receiver detect the error? (6 pt)
11001001 corresponds to the polynomial
M(x) = (1) x7 + (1) x6 + (0) x5 + (0) x4 + (1) x3 + (0) x2 + (0) x1 + (1) x0
M(x) = x7 + x6 + x3 + x0
Multiply M(x) by x3 [This is the xk]
That is, add 3 zeroes at the end of the message: T(x) = 11001001000
C(x) = x3 + 1 is 1 0 0 1
The remainder is shown below in the figure above: 00110
Subtract the remainder from T(x):
11001001000 – 110 = 11001001010
The answer 11001001010 would exactly be divisible by C(x). This is the message to be
transmitted. Let’s check if it is actually exactly divisible by C(x):
C(x) = 1 0 0 1
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Final Exam, CPSC 558
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Cynthia Michele Vincent
The recipient divides the received polynomial by C(x) and if the remainder is 0, concludes
there were no errors. We can see from the result in the above calculation that the remainder is 0.
Now, if the leftmost bit got inverted, we will have the following sequence of bits:
01001001010
The receiver will detect the error if he divides the received polynomial (01001001010) by C(x)
= 1001 and if the remainder is NOT 0, concludes there were errors. We can see from the above
calculation that the remainder is 0001 (not zero).
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
5. Suppose two learning bridges B1 and B2 form a loop as shown below and do not
implement spanning tree algorithm. Each bridge maintains a single table of (address,
interface) pairs.
a) What will happen if L sends to M? (3 pt).
b) Suppose after a short while, M replies to L. Give a sequence of events that leads to
one packet from M and one packet from L circuiting the loop in opposite directions (5
pt).
If L sends a frame to M. I will call the frame F.
F: L  M
// F = frame
Assume that B1 gets the frame F first. B1 knows that the frame F belongs to E1 LAN.
Therefore, B1 will create his table and enter: B1(L, E1)
Since B1 does not know the frame F’s destination, so he floods or copies the frame F to the
other E2 LAN. Now, B2 sees the frame that was copied. B2 knows that the frame belongs to E2
LAN. Therefore, B2 will create his table and enter: B2(L, E2)
Now, M replies to L. This means L sends a frame, which I will call frame G, to M.
G: M  L
// G = frame
Assume that B1 gets the frame G first. B1 knows that the frame G belongs to E2 LAN.
Therefore, B1 will create his table and enter: B1(M, E2).
Now, because B1 has already the information about L (from his table B1(L, E1), he takes the
frame to E1. Now, B2 sees that frame on E1. B2 will create his table and enter: B2(M, E1)
At this point, both frames will keep circulating:
When B1 sees a frame F (addressed to M) on E1, he refers to his table B1(M, E2). So, he forwards
the frame to E2.
When B1 sees a frame G (addressed to L) on E1, he refers to his table B1(L, E1). Because the
frame is already on E1, B1 discards the frame G.
When B1 sees a frame F (addressed to M) on E2, he refers to his table B1(M, E2). Because the
frame is already on E2, B1 discards the frame F.
When B1 sees a frame G (addressed to L) on E2, he refers to his table B1(L, E1). So, he forwards
the frame to E1.
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Final Exam, CPSC 558
December 19, 2002
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When B2 sees a frame F (addressed to M) on E1, he refers to his table B2(M, E1). Because the
frame is already on E1, B2 discards the frame F.
When B2 sees a frame G (addressed to L) on E1, he refers to his table B2(L, E2). So, he forwards
the frame to E2.
When B2 sees a frame F (addressed to M) on E2, he refers to his table B2(M, E1). So, he forwards
the frame to E1.
When B2 sees a frame G (addressed to L) on E2, he refers to his table B2(L, E2). Because the
frame is already on E2, B2 discards the frame G.
Notice that frame F keeps circulating (pink lines), and frame G also keeps circulating (blue lines).
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Final Exam, CPSC 558
December 19, 2002
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Cynthia Michele Vincent
6. Given the following extended LAN.
a) Indicate which paths are selected by the spanning tree algorithm. (4 pt)
b) What happens in the spanning algorithm if bridge B3 does not participate and drops all
spanning tree messages? (3 pt)
c) What happens in the spanning algorithm if bridge B3 does not participate and simply
forwards all spanning tree messages? (3 pt)
The paths selected by the spanning tree algorithm are shown below.
If B3 does not participate and drops all spanning tree messages, the downstream bridges will
not receive the configuration messages, and after waiting a specified period of time, they will all claim
to be the root, and the algorithm will kick in again to elect a new root and new designated bridge
(page 192-195 of textbook).
The algorithm is dynamic, so the bridges are always prepared to
reconfigure themselves into a new spanning tree should some bridge fail or does not participate. The
paths selected when B3 does not participate and drops all spanning tree messages are shown below.
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Final Exam, CPSC 558
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Cynthia Michele Vincent
If B3 does not participate and simply forwards all spanning tree messages -- B3 will not think
of him as a root. It will only forward all messages. The resulting figure (shown below) will include B3
and the figure will be just like the figure in (a).
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Final Exam, CPSC 558
December 19, 2002
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Cynthia Michele Vincent
7. Explain why an Ethernet has a minimal packet size of 64 bytes. Give an example to
illustrate how a smaller size of packet can be a problem (10 pt).
One reason why Ethernet has a minimal packet size of 64 bytes is to make it easier to
distinguish valid frames from garbage. 802.3 states that valid frames must be at least 64 bytes long,
form destination address to checksum. Another reason is to prevent a station from completing the
transmission of a short frame before the first bit has even reached the far end of the cable, where it
may collide with another frame.
An example is illustrated in Tanenbaum’s Computer Networks book on page 282, and is
shown below. At time 0, station A, at one end of the network, sends off a frame. Let the propagation
time be  . Just before the frame gets to the other end, the most distant station, B, starts transmitting.
When B detects that it is receiving more power than it is putting out, it knows that a collision has
occurred, so it aborts its transmission and generates a 48-bit noise burst to warn all other stations. At
about time 2, the sender sees the noise burst and aborts its transmission, too. It then waits a
random time before trying again. If a station tries to transmit a very short frame, it is conceivable that
a collision occurs, but the transmission completes before the noise burst gets back at 2. The sender
will then incorrectly conclude that the frame was successfully sent. To prevent this situation from
occurring, all frames must take more than 2 to send. This corresponds to 64 bytes.
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Final Exam, CPSC 558
December 19, 2002
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Cynthia Michele Vincent
8. Given the following TCP trace
a) What congestion control algorithm is used between 0 - 0.5 sec? (2 pt)
b) Why is congestion control window flat between 0.5 - 2.0 sec? (3 pt)
c) What congestion control algorithm is used between 2 and just after 3.5 sec? (2 pt)
d) What triggers the change in the slope of congestion window size at just after 4.0 sec?
(2 pt)
e) How will the trace change if Fast Recovery algorithm is used? (3 pt)
f)
Describe how fast transmission works and explain why it is not effective when there
are multiple packet loss during RTT. (4 pt)
a) Slow Start is the congestion control algorithm used between 0 – 0.5 sec.
b) There are no acknowledgements (ACKs) arriving because several packets were lost that
is why congestion control window is flat between 0.5 – 2.0 sec.
c) Additive Increase is the congestion control algorithm used between 2 and just after 3.5
sec.
d) Packet loss is what triggers the change in the slope of congestion window size at just after
4.0 sec.
e) The slow start phase that happens between when fast retransmit detects a lost packet
and additive increase begins is removed from the trace.
f) Fast Transmission works this way: Everytime a data packet arrives at the receiving side,
the receiver responds with an acknowledgement, even if this sequence number has already been
acknowledged. Thus, when a packet arrives out of order, that is, TCP cannot yet acknowledge the
data the packet contains because earlier data has not yet arrived, TCP resends the same
acknowledgement it sent the last time (duplicate ACK). When the sending side sees a duplicate
ACK, it knows that the other side must have received a packet out of order, which suggests that an
earlier packet might have been lost. Since it is also possible that the earlier packet has only been
delayed rather than lost, the sender waits until it sees some number of duplicate ACKs and tehn
retransmits the missing packet. (page 472 – 473 of textbook).
According to page 474 of textbook, for a small window size, there will not be enough packets
in transit to cause enough duplicate ACKs to be delivered. Given enough lost packets, the sliding
window eventually blocks the sender until a timeout occurs. Given a 64 KB max advertised window
size, TCP’s fast retransmit mechanism is able to detect up to 3 dropped packets per window in
practice. The figure below demonstrates why Fast Transmission is not effective when there are
multiple packet loss during RTT:
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Final Exam, CPSC 558
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Cynthia Michele Vincent
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
9. Suppose a router has built up a routing table as shown below. The router can deliver
packets directly to interface 0 or interface 1, or it can forward packets to R2, R3, or R4.
Describe what router does with a packet addressed to each of the following destinations (10
pt).
Subnet Number
128.96.39.0
128.96.39.128
128.96.40.0
192.4.153.0
Default
Subnet Mask
255.255.255.128
Next Hop
Interface 0
Interface 1
R2
R3
R4
a) 128.96.39.10
b) 128.96.40.151
c) 128.96.40.12
d) 192.4.153.95
e) 192.4.153.11
Given IP number 128.96.39.10 with subnet mask 255.255.255.128, the result of ANDing
them gives 128.96.39.0. Therefore, the subnet number of 128.96.39.10 is 128.96.39.0. The result
128.96.39.0 match the subnet number 128.96.39.0, so the right entry is Interface 0. The router
forwards the packet to next hop which is Interface 0. .
Given IP number 128.96.40.151 with subnet mask 255.255.255.128, the result of ANDing
them gives 128.96.40.128. Therefore, the subnet number of 128.96.40.151 is 128.96.40.128. The
result 128.96.40.128 does not match the subnet of the entry, so the right entry is R4 which is the
default. The router forwards the packet to next hop which is R4. (There was an entry 128.96.39.128
– I hope it’s not a typo or my answer R4 would be wrong.)
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Final Exam, CPSC 558
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Given IP number 128.96.40.12 with subnet mask 255.255.255.128, the result of ANDing
them gives 128.96.40.0. Therefore, the subnet number of 128.96.40.12 is 128.96.40.0. The result
128.96.40.0 match the subnet number 128.96.40.0, so the right entry is R2. The router forwards the
packet to next hop which is R2.
Given IP number 192.4.153.95 with subnet mask 255.255.255.192, the result of ANDing
them gives 192.4.153.64. Therefore, the subnet number of 192.4.153.95 is 192.4.153.64. The result
192.4.153.64 does not match the subnet of the entry so the right entry is R4 which is the default. The
router forwards the packet to next hop which is R4.
Given IP number 192.4.153.11 with subnet mask 255.255.255.192, the result of ANDing
them gives 192.4.153.0. Therefore, the subnet number of 192.4.153.11 is 192.4.153.0. The result
192.4.153.0 match the subnet number 192.153.0, so the right entry is R3. The router forwards the
packet to next hop which is R3.
Note:
The router does the ANDing I calculated above.
The router ANDs the packet’s
destination address with the subnet mask for each entry in turn; if the result matches the subnet
number of the entry, then this is the right entry to use, and it forwards the packet to the next hop
router indicated. If the result does not match the subnet of the entry, then the router takes the packet
to the default route. (page 314 of textbook).
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
10. Why does IPv4 have fragmentation reassembly done at the endpoint rather than at the
next router? (3 pt) Why does IPv6 abandon fragmentation entirely? (3 pt) Describe how the
size of the Path MTU is determined in IPv6. (4 pt)
Fragmentation produces smaller, valid IP datagrams that can be readily reassembled into the
original datagram upon receipt, independent of the order of their arrival. Only the destination node
that needs the packets in order (not the router).
IPv6 abandons fragmentation to make a simpler header; the routers can process packets
more quickly improving throughput.
The size of the Path MTU is determined in IPv6 by using a Path MTU Discovery - process by
which a node learns the PMTU (Path MTU) of a path. Source determines Path MTU; and source
sends probe message of various sizes dynamically, until destination is reached.
The basic idea of Path MTU Discovery is that a source node initially assumes that the PMTU
of a path is the (known) MTU of the first hop in the path. If any of the packets sent on that path are
too large to be forwarded by some node along the path, that node will discard them and return
ICMPv6 Packet Too Big messages [ICMPv6]. Upon receipt of such a message, the source node
reduces its assumed PMTU for the path based on the MTU of the constricting hop as reported in the
Packet Too Big message.
The Path MTU Discovery process ends when the node's estimate of the PMTU is less than or
equal to the actual PMTU. Note that several iterations of the packet-sent/Packet-Too-Big-messagereceived cycle may occur before the Path MTU Discovery process ends, as there may be links with
smaller MTUs further along the path.
Alternatively, the node may elect to end the discovery process by ceasing to send packets
larger than the IPv6 minimum link MTU. (Source: http://www.cis.ohio-state.edu/cgibin/rfc/rfc1981.html)
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Final Exam, CPSC 558
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11. The RED gateway mechanism maintains two queue thresholds: MIN and MAX. Explain
how a RED gateway process an arriving packet in terms of these two thresholds using RED
mechanism. Explain why it is possible for the queue in RED gateway to contain more than
MAX packets (10 pt).
A RED gateway computes an average queue length using weighted running average.
AvgLen = (1 - Weight) * AvgLen + Weight * SampleLen
where 0 < Weight < 1 (usually 0.002) and
SampleLen is queue length each time a packet arrives
RED has 2 queue length thresholds that trigger certain activity:
MinThreshold and
MaxThreshold. When a packet arrives at the gateway, RED computes the AvgLen with these
two thresholds according to the following rules:
if(AvgLen <= MinThreshold) then
queue the packet
if(MinThreshold < AvgLen < MaxThreshold) then
calculate probability P
drop arriving packet with probability P
if(MaxThreshold <= AvgLen) then
drop arriving packet
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Final Exam, CPSC 558
December 19, 2002
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Cynthia Michele Vincent
12. List real-time application QoS metrics (2 pt). Describe how Integrated Service works and
Integrated Service’s main drawback (5 pt.) Describe how Differentiated Service is designed to
address this issue (5 pt).
The real-time application QoS metrics are (based on slides from C.S. Wu):
a) Data loss (packet loss)
b) Transmission rate (bandwidth)
c) End-to-end delay (latency)
d) Jitter (variation in latency, based on slides -- student presentation)
How Integrated Service works:
According to CPSC 558’s student presentation, router
reserves resources at each hop or router per flow. These resources are pre-allocated for a flow and
are dynamically allocated to flows. Each flow can be accepted or rejected by the router. According to
IEEE Report, Integrated Service performs resource reservation using a dynamic signaling protocol
and employs admission control, packet classification, and intelligent scheduling to achieve desired
QoS. This model is relatively complex and has difficulties in scaling to large backbones. This
architecture is based on the Resource Reservation Protocol (RSVP). RSVP is an IETF Internet
Standard (RFC 2205) protocol for allowing an application to dynamically reserve network bandwidth.
It enables applications to request a specific QoS for a data flow. Hosts and routers use RSVP to
deliver QoS requests to the routers along the path, and to maintain router and host state to provide
the requested service, usually bandwidth and latency. Bandwidth reservation is based on mean data
rate, the largest amount of data the router will keep in its queue, and the minimum QoS required.
The figure below shows how Integrated Service works. The source and destination hosts
exchange RSVP signaling messages to establish packet classification and forwarding state at each
node along the path between them. (Source: http://www.juniper.net/techcenter/techpapers/20001903.html#figure11)

The resource reservation setup protocol (RSVP) allows individual applications to request
resources from routers and then install per-flow state along the path of the packet flow.

Flow specifications provide a syntax that allows applications to specify their specific resource
requirements.

A packet classification process examines incoming packets and decides which of the various
classes of service should be applied to each packet.

An admission control process determines whether a requested reservation can be supported,
based on the availability of both local and network resources.
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Final Exam, CPSC 558
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
A policing and shaping process monitors each flow to ensure that it conforms to its traffic
profile.

A packet scheduling process distributes network resources (buffers and bandwidth) among
the different flows.
The main drawback of Integrated Service is scalability. According to student presentation,
Integrated Service is limited to the number of flows that can be supported, and that every router in
path must support Integrated Service.
According to IEEE report, Integrated Service model is
relatively complex and has difficulties in scaling to large backbones.
Based on CPSC 558’s Student Presentation, Differentiated Services addresses the drawback
of Integrated Service by making Differentiated Services extendable (by incremental deployment) and
by making it reduce router processing and resources and eliminates per flow signaling. Based on
IEEE report, Differentiated Service is based on traffic aggregation rather than individual perapplication instance flows. (Traffic aggregation reduces router processing.) The Differentiated
Services model largely focuses on the use of the ToS field in IPv4 header or the IPv6 Traffic Class
octet as a QoS mechanism. These bits are used to mark a packet to receive a particular forwarding
treatment, or per-hop behavior, at each network node. Classification, marking and policing are done
at the network edges and only packet handling requirements need to be provided in the core of the
network. (Source: http://www.ewh.ieee.org/reg/7/canrev/canrev36/agarwal.pdf.)
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Final Exam, CPSC 558
December 19, 2002
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Cynthia Michele Vincent
13. In WLAN, explain what is hidden node problem and what is exposed node problem?
Explain how CDMA/CA addresses these two problems (10 pt).
Stations in wired networks and WLANs must coordinate their access to and use of the shared
communication media.
The media access control (MAC) protocol for wired networks called the
Carrier Sense Multiple Access Protocol with Collision Detection (CSMA/CD) didn’t work for WLANs
because collision detection is difficult to achieve in wireless networks. Therefore, a new protocol
called the Carrier Sense Multiple Access Protocol with Collision Avoidance (CSMA/CA) had to
be developed. In wired networks, CSMA/CD protocol requires a station to sense the channel to
determine if the channel is idle. If it is, then the station is free to transmit without causing a collision.
CSMA/CD protocol introduces two kinds of problems if used in WLANs because it cannot assume
that stations hear one another. When a station senses that the channel is free in the station’s local
area, it doesn’t necessarily mean it’s actually free in the receiver’s area. It also doesn’t necessarily
mean that the channel is busy when the station senses an activity within the receiver’s area. The
problem is called the Hidden Node Problem. Some stations are hidden to others who want to
transmit so that they cannot hear one another. When they try to transmit at the same time, a collision
will result. The other problem is called the Exposed Node Problem. When a station is exposed to
another station that is transmitting, it thinks the channel is busy and it concludes that it cannot
transmit. This, however, is a false conclusion because it can actually transmit to another station
belonging to another region without causing a collision. The two problems demonstrated above are
handled and solved by the CSMA/CA protocol for WLANs (Figure 1). This protocol requires a sender
to send a Request To Send (RTS) frame to its receiver. The receiver must respond with a Clear To
Send (CTS) frame to reserve access to the channel. Then, the sender sends its intended data, and
the receiver will send an Acknowledgement (ACK). The use of the RTS and CTS frames helps to
avoid collisions because the receiver's transmitted CTS frame will be heard by all stations within the
receiver's region. Any station that hears the CTS frame from one station is not allowed to transmit to
that station. However, if a station only hears an RTS and not the CTS, it is allowed to transmit. Thus,
it correctly concludes that the channel is free.
Figure 1. Carrier Sense Media Access with Collision Avoidance
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Final Exam, CPSC 558
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To demonstrate the Hidden Node Problem, consider Figure 2 where A, B and B, C are in
their own regions. Let’s assume that A and C are both sending to B. A is hidden to C and C is
hidden to A because they do not fall in the same region. If they simply listen if the line is idle, they will
transmit. If they transmit at the same time, a collision will result.
Figure 2. Hidden Node Problem
To demonstrate the Exposed Node Problem, consider Figure 3 where A, B and B, C and C, D
are in their own regions. Let’s assume that B is sending to A and C wants to send to D. B is exposed
to C because they fall in the same region. If C simply listens if the line is idle, C will notice that the
line is busy since B is transmitting to A and C can hear that transmission. Therefore, C will not
transmit. But, this is unwarranted since it can actually transmit to D without causing a collision; C and
D are in a different region as A and B.
Figure 3. Exposed Node Problem
To demonstrate the solutions of these problems by using CSMA/CA, let us consider Figure 2
again assuming that A and C want to send to B. If A is sending to B, A will send an RTS to B. B will
respond with CTS to A. C can hear this CTS because B is in its region and therefore it will not
transmit at that time. The result is no collision. Now, let us consider Figure 3 assuming that B is
sending to A. B will send an RTS to A, and A will respond with a CTS. Since C cannot hear this
CTS, C is allowed to transmit to D.
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
14. Describe the differences between the IP data network and telecommunication network in
terms of packet transmission, routing table setup, and QoS provided. (10 pt)
Telecommunications Network: For the most part, is still based on circuit switched technology
and the use of conventional telephone numbers (E164).
IP Data Network: is based on a packet switch technology and uses the Internet domain
name system for addressing global customers.
Telecommunication network QoS: end to end delay, packet loss and delay jitter, bandwidth
affect QoS. These are in general difficult to control particularly in multiple networks, but can be
mitigated to some extent by technology developments such as buffering, reducing packet size and
codec improvements.
IP network QoS:
In order to achieve guaranteed grades of service quality for packets
traversing networks within the IP transport plane, it will be necessary to control the well known
technical parameters such as end to end delay, packet loss, delay jitter and bandwidth. The main
architectures and techniques defined for IP QoS are: The Integrated Services and The Differentiated
Services.
The IP routing table is set up by mapping network numbers to next hops. It may also contain
how this information is learned.
Telecommunications routing involves the translation of a number or name to a routing
address. The telecommunications routing tables consist of a route and rules to select one path from
the route for a given connection or bandwidth-allocation request.
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
15. Explain how a long-distance call is routed between end office switches and long distance
switch (5 pt). What are the functions of SS7 signaling? (2 pt).
A tandem switch is a switch that has connections to other switches (end office, toll, cellular)
A toll tandem switch is a switch that is capable of switching long distance traffic. It is owned
by local telephone company to switch calls between local and long distance networks. It interfaces
with long distance switches.
The following is a process describing how a long distance call is routed between an end
office switch and the long distance switch: (Slide 18, C.S.Wu)
1) User dials a long distance number.
2) Central Office (CO) receives the digits and looks up customer profile in the
customer information record to determine which long distance telephone
company to use.
3) CO switch attaches access code to beginning of dialed number and ships the
number out on an intra-office trunk that connects central office switch to the local
toll/tandem switch.
4) Tandem receives digits, examines the access code and sends the call to the long
distance carrier point of presence (POP). Call is switched onto the trunk that
connects the local telephone company’s tandem to the long distance switch.
Switch may be in the same city as the local toll/tandem switch. If not in the same
city, long-haul circuits are used to connect the 2 switches.
The functions of SS7 signaling are to set up and manage the connection for a call; to
teardown the connection when call is completed; to manage call forwarding, calling party name and
number display, three-way calling and other intelligent network services. Billing is another function.
(Slide 39, C.S. Wu)
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
16. Suppose an RSVP router suddenly loses its reservation state, but otherwise remains
running.
a) What will happen to the existing reserved flows if the router handles reserved and nonreserved flows via a single FIFO queue (5 pt).
b) What might happen to the existing reserved flows if the router used weighted fair
queuing to segregate reserved and non-reserved traffic? (5 pt)
When the routers implement multiple FIFO queues, one for each priority class (i.e. reserved
class and non-reserved class), the router always transmits packets out of the highest-priority queue
(i.e. reserved class) if that queue is nonempty before moving on to the next priority queue. Within
each priority, packets are still managed in a FIFO manner. Using multiple FIFO queues just allows
high-priority packets to cut to the front of the line. Now, if these multiple FIFO queues become a
single queue, the existing reserved flows (queues) will not be prioritized.
They will not be
discriminated against other traffic sources.
What might happen to the existing reserved flows if the router used weighted fair queuing is
that the existing reserved flows will get a certain share of link that is higher than the non-reserved
traffic. The outcome will be identical to application of Round Robin scheduling with variable size of
quantum. Although the reserved flows will be favored against non-reserved flows, non-reserved flows
will sometimes be transmitted even when reserved flows are already waiting.
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
17. Describe the role of MIB in SNMP (2 pt). Explain how SNMP messages are exchanged to
a) report a network device failure and
b) to collect measurements from a network device. (8 pt)
The role of MIB in SNMP is to specify the variables (resources) that agents in SNMP
maintain.
a) When an agent notices that a significant event has occurred, it immediately reports the
event to all management stations in its configuration list. This report is called the SNMP trap. The
report usually just states that some even has occurred. It is up to the management station to then
issue queries to find all details. Because communication form managed nodes to the management
station is not reliable (is not acknowledged), it is wise for the management station to poll each
managed node occasionally anyway, checking for unusual events, just in case the model of polling at
long intervals with acceleration on receipt of a trap is called trap directed polling.
This model
assumes that each managed node is capable of running an SNMP agent internally. Older devices
are handled by the use of proxy agent, an agent that watches over one or more nonSNMP devices
and communicates with the management station on their behalf, possibly communicating with the
devices themselves using nonstandard protocol. (pg 632 -- Tanenbaum)
b)
SNMP messages are exchanged to collect measurements from a network device.
Monitors acquire counter values by sending an SNMP GET request periodically to routers with
identifer of the counters to collect. Special local counters can be “created” by using local agents
which measure the characteristics of traffic flows. These agents send measurement results to special
devices collectors, and requires some processing in routers (http://www.eurescom.de/~pubdeliverables/p1000-series/p1006/TI_3/p1006ti3.pdf)
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
18. In SIP VoIP model, explain the functions of the following protocols.
a) SIP (2 pt)
b) RTP (2 pt)
c) RTCP (2 pt)
d) SDP (2 pt)
SIP (Session Description Protocol) is a protocol that carries session description for user
capabilities negotiation. It is a protocol to set up calls between gateways; to determine whether party
can be reached via H.323; and to find H.245 gateway/user address. (Slide 22, 23, C.S. Wu). SIP
supports session descriptions that allow participants to agree on a set of compatible media types; and
supports user mobility by proxying and redirecting requests to the user’s current location.
The
services that SIP provide include [RFC2543]: User Location, Call Setup, User Availability, User
Capabilities and Call Handling.
User Location determines the end system to be used for
communication. Call Setup rings and establishes call parameters at both called and calling party.
User Availability determines the willingness of the called party to engage in communications. User
Capabilities determines the media and media parameters to be used. Finally, Call Handling transfers
and
terminates
calls.
(Source:
ftp://ftp.netlab.ohio-state.edu/pub/jain/courses/cis788-
99/voip_protocols/index.html#3. Session Initiation Protocol(SIP))
RTP is a protocol for carrying real time audio/video data (Slide 13, C.S. Wu). RTP supports
the transfer of real-time media (audio and video) over packet switched networks. The RTP allows the
receiver to detect any losses in packets and also provide timing information so that the receiver can
correctly compensate for delay jitter. The functions provided by RTP are sequencing, payload
identification, frame indication, source identification and intramedia synchronization. Sequence
numbers are used in the RTP packet is used for detecting lost packets. Payload Identification is
included in each RTP packet to describe the encoding of the media. Frame Indication (by frame
marker bit) is used indicate the beginning and end of the frame. Source Identification is included to
determine originator of a frame. Intramedia Synchronization is used to compensate for the different
delay jitter for packets within the same stream by providing timestamps which are needed by the playout buffers. (Source: ftp://ftp.netlab.ohio-state.edu/pub/jain/courses/cis78899/voip_protocols/index.html#3.4 Sample SIP Operation)
RTCP is a protocol to monitor or to provide feedback on the transmission and reception
quality of data carried by RTP. (Slide 13,C.S. Wu). In a RTP session, participants periodically send
RTCP packets to obtain useful information about QoS etc. The additional services that RTCP
provides to the participants are QoS feedback, Session Control, Identification, and Intermedia
Synchronization. The QoS feedback reports the quality of service with info about number of lost
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
packets, Round Trip Time, jitter . This info is used by the sources to adjust their data rate. The
Session Control allows participants to indicate that they are leaving a session.
Identification:
Information are included in the RTCP packets so that all the users can know the identities of the other
users for that session. Finally, Intermedia Synchronization synchronize the video and audio sent over
different streams at the receiver so that they play together.
(Source:
ftp://ftp.netlab.ohio-
state.edu/pub/jain/courses/cis788-99/voip_protocols/index.html#3.4 Sample SIP Operation)
SDP is a protocol that describe multimedia session for the purpose of session
announcement, session invitation, etc. It conveys information about media streams in multimedia
sessions to allow the recipients of a session description to participate in the session. SDP includes
the following information: [RFC2327]: Session name and purpose, Address and port number, Start
and stop times, Information to receive those media, Information about the bandwidth to be used by
the conference and Contact information for the person responsible for the session
The above information is conveyed in a simple textual format. When a call is set up using
SIP, the INVITE message contains an SDP body describing the session parameters acceptable to the
calling party. The response from the callee includes a SDP body describing the capabilities of the
callee. In general, SDP must convey enough information to be able to join a session and to announce
the resources to be used to non-participants that may need to know. (Source: ftp://ftp.netlab.ohiostate.edu/pub/jain/courses/cis788-99/voip_protocols/index.html#3.4 Sample SIP Operation).
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Final Exam, CPSC 558
December 19, 2002
Submitted by
Cynthia Michele Vincent
19. In H.323 VoIP model, explain the functions of the following components:
a) Gateway (2 pt)
b) Gatekeeper (2 pt)
Gateway is an interface that connects H.323 to other networks. Gateway converts data
formats, signaling and control messages, audio/video codec between different networks. (Source:
Slide 16, C.S. Wu).
Gatekeeper has the following functions:

Monitors devices (from Lecture, Dec. 10, 2002)

Handle control and management functions to registered terminals within its H.323
zone.

Registration and Admission Control (using H.225 RAS protocol)

Manages bandwidth; limits number of calls

Translates addresses (LAN-IP)

Handles the routing request form H.245 or Q.931 for connecting two users; to help a
terminal find a gateway

Responsible for performing the appropriate translation of both signaling info and the
media stream that need to be carried over the telephone network.
This is the last page. Merry Christmas!
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