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FLORIDA INSTITUE OF TECHNOLOGY
ELECTRICAL AND COMPUTER ENGINEERING PROGRAMS
ECE 5535 COMPUTER NETWORKS 2
CSE 5631 ADVANCED COMPUTER NETWORKS
Test #2
26 March 2003
NAME ________________________________Student Number (last 4 digits)_________
1. Congestion Control.
There are two broad classes of type of services that Network Layer provides. When a
connection is established from a source to a destination, packets can be sent via a Connection
Oriented Service (Virtual-Circuit) or Connectionless Oriented Service (Datagram).
A. Which type of connection service allows a simpler Control over Congestion? Explain
why.
(5 points)
B. In Datagram Subnets using Choke Packets is one of the methods that can be applied to
Congestion Control.
i.
Briefly describe this method in the case when destination node D is congested and
the sending node is A. Packets are routed from A,B,C to D as presented in the
figure bellow:
(3 points)
A
ii.
B
C
D
How many hops does it take for the congested node to get relief (i.e., reduced flow)
from the source A.
(2 points)
iii. Is there a way to provide an immediate effect and reduced the flow to the
destination node? If yes explain how.
(5 points)
C. Describe major differences between the WARNING BIT method and the RED (Random
Early Detection) method.
(5 points)
This is a closed book and closed notes exam.
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ANSWER:
A. Connection Oriented Service - Virtual-Circuit. The service can be negotiated during the
setup of the Virtual-Circuit. All necessary resources are reserved then and guaranteed for
the life of that connection. If a catastrophic failure happens to occur (a router failure
along the path) the new Virtual-Circuit can be set up and transmition completed.
B.
i. D sends a Choke Packet to back to the Source A via Router C. Router C in turn
forwards that packet to router B which sends it to A. Source node A reduces the
flow to D.
ii. 6 – Hops (D to C, C to B, B to A, and A to B, B to C and finally C to D).
iii. Yes. The relief can be immediate if the load is distribution across all routers in the
path. First router to share the load is B which is the first one to receive the
CHOKE packet by reducing the flow to D. As CHOKE packet arrives at node C
from B it too reduces the flow to B and finally originating node A does the same
think.
C. Warning BIT method explicitly sets a BIT that indicates congestion notification to the
source; whereas RED implicitly notifies the source by simply dropping one of its packets.
This is a closed book and closed notes exam.
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2. Quality of Service
A. Enumerate four parameters/factors that determine quality of service.
(4 points)
B. Briefly describe each parameter under A, and name at least one application for each
parameter that requires High Control to achieve good Quality of Service?
(8 points)
C. For transmition of Sound data, Jitter Control is very important. Briefly describe one way
to lower the jitter?
(4 points)
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ANSWER:
A.
i. Reliability
ii. Delay
iii. Jitter
iv. Bandwidth
B.
i. Reliability – how often bits gets corrupted.
 e-mail,
 File transfer, etc.
ii. Delay
– how long does it takes to send a packet from source to destination.
 Telephony
 Videoconferencing
iii. Jitter
– how variable the delay is.
 Audio on Demand
 Video on Demand
iv. Bandwidth – is the frequency range of the signal that is being transmitted.
 Videoconferencing
 Video on Demand.
C. One can use Buffering to temporarily store incoming packets arriving at varying time
intervals. The packets can be released at regular time intervals ensuring minimal jitter.
This is a closed book and closed notes exam.
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3. The main protocol for the integrated services architecture is Resource Reservation
Protocol (RSVP).
A. What is this protocol used for?
(2 points)
B. Can congestion occur when using this protocol? Why?
(4 points)
C. The network bellow uses RSVP with multicast trees for host 1 and host 2. Suppose that
host 3 requires a channel of bandwidth 2 MB/sec for a flow from host 1 and another
channel of bandwidth 1 MB/sec for a flow from the host 2. At the same time host 4
requests a channel of bandwidth of 2 MB/sec for a flow from host 1 and host 5 requests a
channel of bandwidth of 1 MB/sec for a flow from host 2. How much total bandwidth
will be reserved for these requests at routers A, B, C, E, H, J, K, and L? Note: Host 1 and
2 each provide only one type of service.
(8 points)
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ANSWER:
A. It is used to make reservation of resources when requesting for a particular service. Other
protocols are used to transfer data.
B. No. All resources needed are reserved with RSVP. If insufficient resources are available
then the connection is not established.
C. A=2MB/sec, B=0, C=1MB/sec. D=0, E=3MB/sec. F=0, G=0, H=3MB/sec, I=0,
J=3MB/sec, K=2MB/sec, L=1MB/sec.
This is a closed book and closed notes exam.
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4. The IP Protocol.
Bellow is the figure depicting IPv4 header.
A. A router is blasting out IP packets whose total length (data plus header is) is 1024 bytes.
Assuming that packets live for 10 sec what is the maximum line speed that router can
operate at without danger of cycling through the IP datagram Identification number
space?
(5 points)
B. Suppose that host A is connected to a router R1, R1 is connected to another router, R2,
and R2 is connected to host B. Suppose that a TCP message that contains 900 bytes of
data and 20 bytes of TCP header is passed to the IP code at host A for delivery to B.
Show the Total length, Identification, DF (Don’t Fragment field), MF (More
Fragments) and Fragment offset fields of the IP deader in each packet transmitted
over the three links. Assume that link A-R1 can support a maximum frame size of 1024
bytes including a 140 byte frame header, link R1-R2 can support a maximum frame size
of 512 bytes including a 14-byte frame header, link R1-R2 can support a maximum frame
size of 512 bytes, including an 8-byte frame header, and link R2-B can support a
maximum frame size of 512 bytes including a 12-byte header. Note that fragments must
be a multiple of 8 bytes except the last one. 8 bytes is elementary fragment unit.
(12 points)
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ANSWER:
A. If the bit rate of the line is b bits/sec, the number of packets/sec that the router can emit is
b/(8*1024) = b/8192. The number of seconds it takes to emit a packet is 8192/b. To put
216 = 65536 packets takes 23*210*216/b sec. Equating this equation with packet lifetime of
10 sec, we get: 229/b = 10. Then b is about 53687091 bps = 51.2 Mbps.
B. TCP Packet = 900 (data) + 20 (IP header) = 920 Bytes.
IP Packet = 920 (data) + 20 (IP header) = 940 Bytes
Link A-R1: Can support 1024 bytes including 14-byte frame header (e.g., 1010-byte IP
data packet) thus no fragmentation is needed.
Length = 940, ID = x, DF = 0, MF = 0, Offset = 0.
Link R1-R2: Can support 512 bytes including 8-byte header (e.g., 512-8=504 bytes of IP
data packet and 504 – 20 (IP header) = 484). Since IP data in a frame must be multiple of
8 bytes:
8*x ≤ 484 => x ≤ 60.5 bytes. Use x = 60 bytes
IP data 1 = 60*8 = 480 Bytes
IP packet size 1 = 480 + 20 = 500 Bytes.
IP data 2 = 920 – 480 = 440 Bytes
IP packet size 2 = 440 + 20 = 460 Bytes.
[1]
[2]
Length = 500, ID = x, DF = 0, MF = 1, Offset = 0.
Length = 460, ID = x, DF = 0, MF = 0, Offset = 60.
Link R2-B: Can support 512 bytes including 12-byte header (e.g., 512-8=500 bytes of IP
data packet and 500 – 20 (IP header) = 480). This implies that no additional
fragmentation is needed. Thus
[1]
[2]
Length = 500, ID = x, DF = 0, MF = 1, Offset = 0.
Length = 460, ID = x, DF = 0, MF = 0, Offset = 60.
This is a closed book and closed notes exam.
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5. The IP Addresses.
Bellow is the figure depicting IPv4 Address Formats.
A. Suppose that instead of using 16 bits for the network part of class B address
originally, 20 bits had been used. How many class B networks would there have
been?
(2 points)
B. Convert the IP address whose hexadecimal representation is C22F1582 to dotted
decimal notation?
(2 points)
C. A network on the Internet has a subnet mask of 255.255.240.0. What is the maximum
number of hosts it can handle?
(2 points)
D. A large number of consecutive IP addresses are available starting at 198.16.0.0.
Suppose that four organizations, A, B, C, and D, request 4000, 2000, 4000 and 8000
addresses, respectively, and in that order. For each of these, give the first IP address
assigned, the last IP address assigned, and the mask in the w.x.y.z/s notation.
(8 points)
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ANSWER:
A. 2(20-2) = 218. Since all 0’s and 1’s are special there are 218-2 addresses available
(262142).
B. C2.2F.15.82 = 12*16+2.2*16+15.1*16+5.8*16+2=194.47.21.130
C. The mask is 20 bits long, so the network part is 20 bits. The remaining 12 bits are for
the host, so 212 = 4096 host addresses exist.
D. All requests are rounded up to a power of 2.
Start Address
End Address
Mask
A: 198.16.0.0 - 198.16.15.255
198.16.0.0/20
B: 198.16.16.0 - 198.16.23.255
198.16.16.0/21
C: 198.16.32.0 - 198.16.47.255
198.16.32.0/20
D: 198.16.64.0 - 198.16.95.255
198.16.64.0/19
This is a closed book and closed notes exam.
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6. Network Address Translation (NAT).
Within a LAN NAT is used to convert internal IP addresses to true network addresses as
illustrated in the figure bellow:
A. What is the major motivation of IP address conversion using NAT and what kind of
problem does it solve?
(4 points)
B. NAT violates a number of fundamental assumptions of Layered Network
Architecture by using TCP or UDP fields of transport layer header. Enumerate and
very briefly discuss those violations.
(8 points)
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ANSWER:
A. To enable creation of new LANs in spite of lack of IP addresses. Due to rapid growth
of Internet and limitations imposed by the size of address field in IP header there are
very few IP addresses available.
B.
[1] Based on Network Architectural model, IP is supposed to uniquely identify
each machine worldwide. NAT violates this model on top of which the whole
software structure of the Internet is build.
[2] Internet is changed from connectionless-network to a kind of connection
oriented network. NAT box must maintain information for each connection
passing through it. Having the network maintain connection state is a property
of connection-oriented networks. If a NAT box crashes and its mapping table
is lost, all its TCP connections are destroyed. In the absence of NAT router
crashes have no effect on TCP.
[3] NAT violates fundamental rule of protocol layering: layer k may not make
any assumptions about what layer k+1 has put into the payload field – keeping
the layers independent. NAT uses TCP fields (which are supposed to be
known only to Transport layer). If TCP gets upgraded to a new TCP2 protocol
NAT will fail.
[4] Processes on the Internet are not required to use TCP or UDP. If a process
decides to use a protocol other than TCP or UDP, NAT will fail.
This is a closed book and closed notes exam.
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7. IPv6
IPv6 is supposed to alleviate problems of current IPv4. The basic header of the IPv6 is
illustrated in the figure bellow:
A. If a block of 1 million addresses is allocated every picosecond (10-12 second), how long
will the addresses last?
(5 points)
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ANSWER:
A. With 16 bytes there are 2(16*8) = 2128 ≈ 3.4*1038 addresses. If the rate of allocation is
106/10-12 = 1018 per second they will last 3.4*1038/1018 seconds = 3.4*1020 seconds =
3.4*1020/3.15*107 ≈ 1013 years.
This is a closed book and closed notes exam.
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8. Transport Layer
A. Define the task(s) of transport layer.
(2 points)
B. To allow users to access transport service, the transport layer must provide some
operations to application programs (transport service interface offering a number of
primitives). Enumerate a minimal set of such primitives that would allow establishment
of a connection, data transfer, and release of the connection. Briefly describe function of
each primitive.
(8 points)
C. In order to set up and establish a connection, transfer data, and release connection (that is
provide a complete transport service) transport layer uses handshakes. To establish and
release a connection transport protocol uses a three-way handshake. Briefly describe
three-way handshake concept used for setting up a connection. Also describe reasons
why three-way handshake is required.
(5 points)
This is a closed book and closed notes exam.
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ANSWER:
A. To provide efficient, reliable, and const-effective data transport from source machine to
the destination machine. Transport protocols must be able to perform these tasks over
unreliable networks.
B.
C.

To prevent effects of lost packets as well as packets appearing at inopportune
moments.