CIS 3200 Networking Fundamentals
Solution Exam 4
Friday, May 2nd 2003
Part 1: Security
Q.1. In Denial of Service (DoS) attacks: (Circle all correct answers)
a.
b.
c.
d.
The attacker hacks into multiple clients and plants Zombie programs on them
The attacker intercepts messages and passwords
The attacker sends a stream of messages to the victim
All of the above
/4 .
Q.2. What kind of security systems is usually used to protect a network against
Denial of Service attacks? (Circle all correct answers)
a.
b.
c.
d.
Packet firewalls
Encryption systems
Application firewalls
All of the above
/4 .
Q.3. In Public encryption techniques, when Partner A sends a message to Partner B:
(Circle all correct answers)
a.
b.
c.
d.
Partner B should use Partner A’s private key to decrypt the message
Partner A should encrypt the message with his/her own private key
Partner A should encrypt the message with Partner B’s public key
Partner B should use his/her own private key to decrypt the message
/4 .
Q.4. In Malicious content attacks, the illicit content could be
(Circle all correct answers)
a.
b.
c.
d.
e.
A wrong IP address
A TCP header
A virus
A worm
An unsolicited e-mail
/4 .
Q.5. Application firewall usually examine IP, TCP, and UDP headers to decide allowing or
denying access to incoming messages.
a. True
b. False
/4 .
Q.6. In Symmetric encryption/Decryption, there is no need for secured keys’ exchange.
a. True
b. False
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Q.7. Trojan horses are viruses that propagate across system by themselves
a. True
b. False
/4 .
Part 2: Network Management
Q.8. A networking component has been operating continuously for 20 days. During that time, it
has failed twice, resulting in downtime of 2 hours. Calculate the availability of the component
during that 20-days period using the Approximation method.
20 days = 20 x 24 hours = 480 hours (Total available time)
Downtime = 2 hours
/8
Availability = (480-2)/480 = .9958
Q.9. If a component has a MTBF of 455 hours and a transaction takes 10 seconds, calculate the
reliability of the component using the Reliability equation R(t) = e-bt, where b is equal to
1/MTBF and t the transaction time.
b = 1/455 = .002197
t = 10 seconds = .16666 minutes = .002777 hours
/8
R(.002777) = e-(.002197 x .002777) = .999
Q.10. In centralized network management, the Network Management software
(usually called the Manager) is installed in the managed device.
a. True
b. False
/4 .
Q.11. In centralized network management, there is (only) one Management Information Base,
usually stored on the remote computer.
a. True
b. False
/4 .
Q.12. In Agent-Manager communications, if the Agent detects a condition the Manager should
know about, it can send a message to the Manager. That kind of message is usually called:
CIS 3200 Networking Fundamentals
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a.
b.
c.
e.
a failure message
a routing message
a forwarding message
None of the above
/4 .
Q.13. In Agent-Manager communications: (Circle all correct answers)
a. GET commands tell the Agent to retrieve certain information and return
this information to the Manager
b. The Manager creates SET commands
c. All the above
e. Neither a nor b
/4 .
Part 3: The Internet
Q.14. The host name is the official address for computers connected to the Internet.
a. True
b. False
/4 .
Q.15. The IP address is a string of 32 bits.
a. True
b. False
/4 .
Q.16. Your home PC doesn’t need a host name when your are on the Internet.
a. True
b. False
/4 .
Q.17. The host name is the address that is included in outgoing messages.
a. True
b. False
/4 .
Q.18 Does a server need a host name to be reach on the Internet? Explain.
The answer could be YES or NO. The reason why is that servers can
function and be reached using their IP address (without a host name). But,
since human being cannot easily memorize and use IP address, host names
are used
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Q.19. Using the conversion system illustrated below, convert the following IP address to dotted
decimal notation: 10001011 01000101 00001000 00000011. (Spaces are included to facilitate
reading.)
Position
(N)
7
6
5
4
3
2
1
0
Place Value
(2N)
Bit
Decimal
Your answer:
128
64
32
16
8
4
2
1
10001011 = 139
01000101 = 69
00001000 = 8
/8
00000011 = 3
Q.20. Based on the table shown below, give the class and the network part of each of the
following IP addresses.
Class
Leftmost
bits
Class A 0xxx
Class B 10xx
Class C 110x
IP address
Network Part
Length
Address range
8 bits
16 bits
24 bits
0.x.x.x to 127.x.x.x
128.0.x.x to 191.255.x.x
192.0.0.x to 223.255.255.x
Class ?
Network part (in dotted decimal notation) ?
222.245.217.8
C
222.245.217
191.255.255.255
B
191.255
211.123.231.9
C
211.123.231
110.111.120.228
A
110
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