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Chapter 04
1. Describe three approaches to detecting errors, including how they work, the
probability of detecting an error, and any other benefits or limitations.
Parity Checking
: one additional bit is added to each byte in the message. Probability
of detecting an error: 50%. You will know that an error occurred but not what the
error was.
CheckSum
: a checksum (typically 1 byte) is added to the end of the message.
Calculated by adding the decimal value of each character in the message, divide the
sum by 255, using the remainder as the checksum. If the two values are equal, the
message is presumed to contain no error. 95% error detection rate.
Cyclic Redundancy Check
: one of the most popular methods. Adds 8,16,24, or 32
bits to the message. The message is treated as one long binary number, P. Before
transmission the data link layer divides P by a fixed binary number, G, resulting in a
whole number , Q, and a remainder, R/G. So, P/G = Q + R/G. The remainder, R, is
appended to the message as the error-checking characters before transmission. The
receiving hardware divides the received message by the same G, which generates an
R. The receiving hardware checks to ascertain whether the received R agrees with the locally
generated R. If it doesn't, the message is assumed to be an error. 100% detection rate for all
errors of the same length as the CRC or less. It will detect 99.998% of all bursts longer than 16
bits, whereas CRC-32 will detect about 99.99999998% of all burst errors longer than 32 bits.
2. Compare and contrast stop-and-wait ARQ and continuous ARQ.
Stop-and-wait ARQ, the sender stops and waits for a response from the receiver after
each message or data packet. After receiving a packet, the receiver sends either an
acknowledgement (ACK) if the message was received w/t error, or a negative ack.
(NAK) if the message contained an error. If it's a NAK, the sender resends the
previous message. If it's an ACK, the sender continues w/ the next message. Stop-
and-wait ARQ is by definition, a half-duplex transmission technique.
-With continuous ARQ, the sender doesn't wait for an ACK after sending a message;
it immediately sends the next one. While the messages are being transmitted, the
sender examines the stream of returning acks. If it receives an NAK, the sender
retransmits the needed messages. This is a full duplex transmission technique, b/c
both the sender and receiver are transmitting simultaneously.
3. How efficient would a 6-bit code be in asynchronous transmission if it had 1 parity
bit, 1 start bit, and 2 stop bits. (Some old equipment uses 2 stop bits.)
Total bits = 10 (6 +1+1+2)
= 600/10 = 60%
So, efficiency is greater than 50% in case of 6-bit code
Chapter 5
1.
What does the transport layer do?
The transport layer links application software in the application layer with the network
and is responsible for the end-to-end delivery of the message. The transport layer sits
between the application layer and the network layer. The transport layer accepts messages
from the application layer and packetizes them. Packetizing means to take one outgoing
message from the application layer and break it into a set of smaller packets for
transmission through the network. Conversely, it also means to take the incoming set of
smaller packets form the network layer and reassemble them into one message for the
application layer.
2.
What does the network layer do?
The network layer performs three important functions: addressing, routing, and breaking
long messages into smaller packets for transmission by the data link layer. The network
layer sits between the application layer and the data link layer. The network layer accepts
messages from the application layer and formats and addresses them for transmission by
the data link layer. The network layer also accepts individual messages from the data link
layer and organizes them into coherent messages that it passes to the application layer.
3.
Compare and contrast the three types of addresses used in a network.
When users work with application software, they typically use the application layer
address (e.g., entering an Internet address into a browser, such as www.cba.uga.edu).
When a user types of an Internet address into a Web browser, the request is passed to the
network layer as part of an application layer packet formatted using the HTTP standard.
The network layer software translates this application layer address into a network layer
address. The network layer protocol used on the Internet is TCP/IP, so this Web address
(www.cba.uga.edu) is translated into an TCP/IP address (usually just called an IP address
for short) which is four bytes long when using IPv4 (e.g., 128.192.78.5).
The network layer then determines the best route through the network to the final
destination. Based on this routing, the network layer identifies the data link layer address
of the next computer to which the message should be sent. If the data link layer is running
Ethernet, then the network layer IP address would be translated into an Ethernet address
(e.g., 00-0F-00-81-14-00).
4.
How does TCP establish a session?
TCP sets up a virtual circuit between the sender and the receiver. The transport layer
software sends a special packet (called a SYN, or synchronization characters) to the
receiver requesting that a connection be established. The receiver either accepts or rejects
the connection, and together, they settle on the packet sizes the connection will use. Once
the connection is established, the packets flow between the sender and the receiver,
following the same route through the network.
5.
What is a subnet and why do networks need them?
Each organization must assign the IP addresses it has received to specific computers on
its networks. In general, IP addresses are assigned so that all computers on the same local
area network have a similar address. For example, suppose a university has just received
a set of Class B addresses starting with 128.184.x.x. It is customary to assign all the
computers in the same LAN numbers that start with the same first three digits, so the
Business School LAN might be assigned 128.184.56.x while the Computer Science LAN
might be assigned 128.184.55.x (see Figure 6-8). Likewise, all the other LANs at the
university and the backbone network that connects them would have a different set of
numbers. Each of these LANs are called a TCP/IP subnet because they are logically
grouped together by IP number. Knowing whether a computer is on your subnet or not it
very important for message routing.
6.
How does TCP/IP perform address resolution from URLs into network layer addresses?
It Uses one or more Domain Name Service servers to resolve the address.
7.
How does TCP/IP perform address resolution from IP addresses into data link layer
addresses?
It uses ARP to identify the MAC address of the next node(that packet must be forward)
8.
How does static routing differ from dynamic routing? When would you use static
routing? When would you use dynamic routing?
With static routing, the routing table is developed by the network manager, and changes
only when computers are added to or removed from the network. For example, if the
computer recognizes that a circuit is broken or unusable (e.g., after the data link layer
retry limit has been exceeded without receiving an acknowledgment), the computer will
update the routing table to indicate the failed circuit. If an alternate route is available, it
will be used for all subsequent messages. Otherwise, messages will be stored until the
circuit is repaired. When new computers are added to the network, they announce their
presence to the other computers, who automatically add them into their routing tables.
Static routing is commonly used in networks that have few routing options. For example,
most LANs are connected to the backbone network in only one place. There is only one
route from the LAN to the backbone, so static routing is used.
Dynamic routing (or adaptive routing) is used when there are multiple routes through a
network and it is important to select the best route. Dynamic routing attempts to improve
network performance by routing messages over the fastest possible route, away from
busy circuits and busy computers. An initial routing table is developed by the network
manager, but is continuously updated by the computers themselves to reflect changing
network conditions, such as network traffic. Routers can monitor outgoing messages to
see how long they take to transmit and how long it takes for the receiving computer to
acknowledge them. Based on this monitoring the router can effectuate table updating.
9.
Suppose that a client computer (128.192.98.130) in Building B in Figure 5-17 requests a
large Web page from the Web server 2 in the Data Center (www2.anyorg.com). Assume
that the client computer has just been turned on and does not know any addresses other
than those in its configuration tables. Assume that all gateways and Web servers know all
network layer and data link layer addresses. Explain what messages would be sent and
how they would flow through the network to deliver the Web page request to the server.
When a client computer (128.192.98.130) in Building A requests a large web page from
Web server 2 (www2.anyorg.com) in the Data Center, the process involves several steps.
This process illustrates how network messages flow through the network to deliver a web
page request to the server, with various devices and protocols involved in routing and
communication.
10. Hands-on activity 5D in the book.
Deliverable:
Binary representation of a byte
Decimal
10000000
128
11000000
192
11100000
224
11110000
240
11111000
248
11111100
252
11111110
254
11111111
255
Deliverables:
1.
Number of
Number of
Maximum
Subnet mask in binary and decimal
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Related Questions
True/False
The number 0.5 in decimal notation, when expressed in the IEEE 754 single-precision binary format studied in class, would be expressed as follows:
Sign bit = 0
Exponent = 0111 1110 (+126 in decimal)
Mantissa = 000 0000 0000 0000 0000 0000
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Question 11
Not yet answered
P Flag question
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s
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EV:
V7:
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x bytes
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11:35 AM
46°F Mostly cloudy a @
11/20/2021
hp
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pg u
delete
home
end
num
backspace
lock
}
P
home
4
enter
pause
↑ shift
1
end
ctri
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Solve only part c
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value_a = 179
value_b = 215
sum = value a + value b =
(enter answer in decimal)
value_c = 137
value d = 183
sum = value_c + value_d =|
(enter answer in decimal)
In this exercise, the following two16-bit positive integers are to be
added:
value_a = 46861
value_b
= 51071
If only 16 bits are available to represent the answer, then the result
will be,
sum = value_a + value_b
(enter answer in decimal)
You have designed a number system which uses 19-bit positive
integers
In your system what is the result of,
value a = 393710
value_b = 394509
sum = value_a + value_b
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True/False
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True/False
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Computer Science
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(Note :
The problem should be solved in
separate paper with steps and scanned copy
should be submitted along with other answers
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%3D
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V FIXED (4,1).
X FLOAT(4),
Z FIXED(3):
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BOG
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O True
O False
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