Transcript 例外(exception）とは何か

Answers of Exercise 7
1. Explain what are the connection-oriented communication and the connectionless communication.
Give some examples for each of the above two types of communications.
Answer:
Connection-oriented communication operates analogous to a telephone system. When two
computers communicate they must establish a connection through network. After both
computers agree to communicate, the network system forms a data path called a connection.
The connection stays in place until the computers decide it is no longer needed, I.e., they
have no data to exchange. Either computer can choose to terminate the connection.
Examples of connection-oriented communication are Frame Relay, ATM, TCP, FTP, etc.
Connectionless communication operates analogous to the postal mail system. No connection
will be established before exchange data between two computers or end stations. Whenever it
has data to send, a computer must place the data in the appropriate frames format, attach
the address of the computer to which the data should be delivered, and then pass the packet
to the network. The connectionless network system transports the packet to the prescribed
destination and delivers it.
Examples of connectionless communication are Ethernet, Token Ring, FDDI, IP, email, etc.
2. Describe concepts of delay, throughput, bit corruption (error), disordered packet,
duplicate packet, lost packet, flow control and network congestion, respectively.
Answer:
Delay: It specifies how long it takes for a bit or packet of data to travel across the network
from one computer to another.
Answers of Exercise 7
Throughput: It is measure of the rate at which data can be transmitted in a network. The term
bandwidth is sometimes used as a synonym for throughput. The effective throughput is less
than the hardware bandwidth.
Bit corruption (error): Any change of bits introduced as data passes across a network. Bit
corruptions or errors can be caused by electrical interference or hardware malfunctions.
Disordered packet: When two computers communicate over a connectionless network and
exchange data consisted of a sequence of packets. Two connective packets may go different
routes to arrive a same destination. The later packet may arrive a destination early. This results
in the order of packet received is different from that sent.
Duplicated packet: Malfunctioning hardware can cause packets to be duplicated. Duplication
often arise in WANs caused, for example, by retransmission, but can also occur in LANs when
a bridge cycle exists.
Lost packet: When a computer or some network device detects bit errors in a packet, the whole
data of the packet will be discarded. When a switch or a router has no enough memory to store
all received packets, it also discard packets. Thus, the receiver can not receive some packets,
call lost packet.
Flow control: Computers do not all operate at the same speed. Data overrun occurs when a
computer sends data across a network faster than the destination can be absorb it. Consequently
data is lost.
Network congestion: Network congestion arises in network systems that include multiple links.
When more packets arrive than can be sent, the queue grows and effective delay increase.
This situation is known as congestion. If congestion persists, a packet switch will run out of
memory and begin discarding packets. Furthermore, if the situation continues, the entire
network can become unusable. This is known as congestion collapse.
Answers
3. Professionals sometimes refer to a “knee” in the delay curve.
To understand what they mean, plot the effective delay for
D
values of utilization between 0 and 0.95. Explain why
utilization of a network should not approach to 1.
20D0
Answer: The delay curve is given on the right.
The figure shows that the delay increases slowly
when U<0.8, and increases very fast when U>0.9.
That is, network utilization should be less than 0.9,
otherwise, long delay will occur and throughput will
decrease abruptly.
4D0
D0
of Exercise 7
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4. Match the following to one of the ISO 7 layers:
.5 .8 .95
0
a. Route determination.  Network layer
b. Flow control.  Data link layer
c. Provides reliable end-to-end transmission of entire message.  Transportation Layer
d. Reassembly of data packets into a message.  Transportation layer
e. Provides format and code conversion services.  Presentation layer
f. Provides user services such as electronic mail and file transfer.  Application layer
g. Provides verification of login and logout.  Session Layer
h. Mechanical, electrical, and functional interface.  Physical layer
U
5. Why does a stop-and-go protocol have especially low throughput over a satellite channel
that operates at two megabits per second. How to improve throughput using different
flow control method?
Answer: Because propagation delay over a satellite channel is very large, much time will
spend on waiting ACK from a receiver. Thus low throughput will be expected when using
stop-and-go protocol. The sliding window flow control can improve the throughput.