Study Questions

Question 1:

The three key public network components are____________, ______________, and _______________.

Transmitting device, Receiving Device, Channel of Communication

Question 2:

The ability of FDDI and other high-availability fiber networks to wrap connections on either side of a break is due to the dual ______________ rings.

Counter rotating

Question 3:

SNAP stands for ___________________.

Sub-network Access Protocol

Question 4:

Given a small local area network of 12 systems, what is the best device, where best is best performance without additional unneeded functionality, to connect all 12 systems to each other. Assume that all of the systems need to transfer large files frequently between each other, but that there is no need for external access/communication outside of the LAN.

Switch

Question 5:

Given an analog signal with the highest frequencies at slightly below 4KHz, what would be the best sampling frequency, where best means the lowest sampling frequency that will still accurately allow the original signal to be regenerated.

2 KHz

Question 6:

Name a cell-based protocol:

ATM

Question 7:

What mechanism is used by Ethernet for addressing when multiple stations attempt to transmit at the same time? What is the process that it uses, and what is this called?

The Ethernet protocol uses the CSMA/CD or Carrier Sense Multiple Access with Collision Detection mechanism for network media access. When using CSMA/CD, Ethernet network hosts first check to see if any other hosts are transmitting on the network. If no other hosts are transmitting then the network host will attempt to send data. If another network host also attempts to send data at the same time, both network host transmissions fail (this is a collision). Then each host will refrain from attempting to resend the data for a random period of time, and then retry the transmission. Note that Ethernet collisions only occur when shared media, such as a bus topology or hub is used by hosts on the network.

Question 8:

Describe 5 network topologies (briefly) and give an advantage and a disadvantage for each. Rank the technologies in order of least resilient to most resilient to faults.

In a bus topology all hosts are connected to the network over a single cable or network backbone.  The bus topology is relatively simple to install however, it also creates a single point of failure such that if any point in the cable fails, the entire network fails.

The star topology is configured such that each network host uses an individual cable connection to connect to the network. The other end of the cable terminates at a network device such as a hub or switch. The advantage a star topology has over bus topology is that if a network cable is damaged, only one host loses network connection, rather than the entire network failing. However, the network device to which all network cables terminate is a single point of failure for the entire network.

The ring is configured such that all hosts are connected to the network on a single loop where each host is connected to a single cable with a network host to the left and right and data that travels in one direction. One disadvantage of ring topologies is that if there is cable damage or if a device fails, the entire network will fail as well. However, certain ring topologies such as FDDI have overcome this limitation by adding a redundant ring to the configuration in which data travels in the opposite direction. Using the redundant ring configuration is an advantage, which provides fault tolerance that is not supported in bus topologies.

The tree topology is a combination of both the bus and star topologies in which multiple star configured networks are connected to a main bus topology backbone. This configuration is quite scalable and allows for future expansion. However, the bus portion of the network (the ‘backbone’) is a single point of failure that can cause a disconnection between all the star topology networks if the bus portion fails.

The mesh topology is the most fault tolerant of all the topologies mentioned. The two kinds of mesh topologies are full and partial mesh. In full mesh topologies every device has multiple routes to every other device. So if one route fails, network communication can be rerouted over a different connection, providing built in fault tolerance to minimize potential down time. In a partial mesh, only some of the network devices have multiple routes to other devices on the network. While mesh topologies have the advantage of being fault tolerant, the disadvantage is that they are much more expensive and also more complex to configure and maintain.

Question 9:

In what cases would you use an amplifier instead of a repeater? What is it about the behavior of each that makes one better than the other, and what specifically would be an issue for a WAN deployment cross country?

Network transmissions have distance limitations over single network cables due to the signal attenuation that occurs within the cable. Repeaters can be introduced in order to lengthen the distance of a cable run by regenerating a weakened transmission signal (weakened due to attenuation) down a second cable. A repeater differs from an amplifier in that a repeater regenerates only the message transmission signal whereas an amplifier increases the signal strength of both the transmitted signal and any noise on the cable line (amplifiers do not discriminate). Repeaters are signal/protocol specific and reshape the signal, then amplify and then re-time if needed. This means that repeaters do not and cannot boost the entire signal but rather, only the signal for which they were intended. The advantage then of repeaters is that they do not boost noise, but only the desired data signal. In contrast, amplifiers boost the entire signal, including noise. However, amplifiers are best when the need to boost signal is not signal/protocol specific because they are essentially signal/protocol agnostic. An issue with using amplifiers cross country would be that amplifiers would increase the noise to the point where it would drown out the actual data signal so that it is unrecognizable.