The Best Networking and Encryption Quiz Questions (Part 2)


Q1.1) What does Address Resolution Protocol do?

Address Resolution Protocol (ARP) is a protocol for mapping an Internet Protocol address (IP address) to a physical machine address that is perceived in the nearby network. It is utilized particularly IPv4, to guide IP network addresses to the equipment addresses utilized by an information link protocol. The protocol works underneath the network layer as a part of the interface between the OSI network and OSI link layer. It is utilized when IPv4 is utilized over Ethernet. A table, as a rule called the ARP reserve, is utilized to keep up a relationship between every MAC address and its comparing IP address. ARP gives the protocol guidelines to making this connection and giving address change in both addresses.

Q1.2) Network adaptors are manufactured by many companies in many countries. How can these companies ensure no two adaptors will have the same MAC address? Identify at least five network interface card makers, and find out how you can tell whether a card was made by any of these companies.

Every producer is relegated a block of MAC addresses by IEEE. Inside of that block, it’s up to the producer to ensure there are no crashes. All manufacturers and/or applications do this. Like, for instance, numerous hypervisors dole out the virtual NIC MAC address by a variation of the VM name, prompting crashes now and again.

Vendors are given a scope of MAC Addresses that can be assigned to their items by the IEEE (Institute of Electrical and Electronics Engineers). Macintosh Address are relegated to Vendors in different estimated blocks as proper. The IEEE offers Registration Authority programs or registries which keep up arrangements of unique identifiers under guidelines and issue unique identifiers to those wishing to enroll them. The IEEE Registration Authority doles out unambiguous names to protests in a way which makes the task accessible to invested individuals.

Devices are not uniquely identified by their MAC addresses. In the past vendors have purposefully or by slip-up allocated the same MAC Address to numerous devices. Manufacturers re-use MAC Addresses and they send cards with copy addresses to various parts of the United States or the world so that there is just a little risk two PCs with network cards with the same MAC Address will wind up on the same network. The following are the biggest manufacturers of NIC:

  • Asus
  • Atheros
  • Cisco
  • D-Link
  • Gigabyte Technology
  • IBM
  • Linksys



You can get some answers accessing so as to concern your NIC creator System Information from Windows, Another strategy for deciding the network card in the PC is by physically taking a gander at the network card. Commonly the network card will list the maker and part number on the genuine card. On the off chance that you can’t find a producer or model number of the network card, yet can find a FCC recognizable proof number, it is prescribed that you perform a FCC look utilizing that number. Extra data about FCC numbers and how to scan for data around a FCC number can be found on our FCC word reference definition page.

Q1.3) What is CSMA/CD? How does it work? Explain why RTT on a LAN is an important parameter for CSMA/CD to work.

Carrier Sense Multiple Access/Collision Detect (CSMA/CD) is the protocol for carrier transmission access in Ethernet networks. On Ethernet, any device can attempt to send an edge whenever. Every device senses whether the line is unmoving and along these lines accessible to be utilized. On the off chance that it is, the device starts to transmit its first edge. In the event that another device has attempted to send in the meantime, a collision is said to happen and the casings are disposed of. Every device then holds up an arbitrary measure of time and retries until fruitful in getting its transmission sent. Standard Ethernet networks use CSMA/CD to physically screen the movement hanging in the balance at taking an interest stations. In the event that no transmission is occurring at the time, the specific station can transmit. On the off chance that two stations endeavor to transmit all the while, this causes a collision, which is detected by every taking an interest station. After an arbitrary time interim, the stations that impacted endeavor to transmit once more. The jam signal or sticking signal is a signal that conveys a 32-bit twofold example sent by an information station to educate alternate stations of the collision and that they should not transmit. The most extreme jam-time is figured as takes after: The greatest permitted distance across of an Ethernet establishment is constrained to 232 bits. This makes a round-outing time of 464 bits. As the opening time in Ethernet is 512 bits, the contrast between space time and round-trek time is 48 bits (6 bytes), which is the most extreme “jam-time”.
Q1.4) What techniques can be used for error-detection in the data link layer?
Error Detection

Errors in the got casings are detected by method for Parity Check and Cyclic Redundancy Check (CRC). In both cases, couple of additional bits are sent alongside real data to affirm that bits got at flip side are same as they were sent. On the off chance that the counter-check at collector’ end comes up short, the bits are viewed as corrupted.

  1. Parity Check

One additional bit is sent alongside the first bits to make number of 1s either even if there should be an occurrence of even parity, or odd if there should arise an occurrence of odd parity.

The sender while making an edge counts the quantity of 1s in it. For instance, if even parity is utilized and number of 1s is even then one bit with quality 0 is included. Thusly number of 1s remains even. If the quantity of 1s is odd, to make it even a bit with quality 1 is included.


The receiver counts the quantity of 1s in an edge. In the event that the number of 1s is even and even parity is utilized, the edge is thought to be not-debased and is acknowledged. On the off chance that the number of 1s is odd and odd parity is utilized, the casing is still not ruined.

On the off chance that a solitary bit flips in travel, the recipient can detect it by tallying the quantity of 1s. Be that as it may, when more than one bits are flawed, then it is hard for the recipient to detect the error.

  1. Cyclic Redundancy Check (CRC)

CRC is an alternate way to deal with detect if the got outline contains substantial data. This system includes paired division of the data bits being sent. The divisor is created utilizing polynomials. The sender performs a division operation on the bits being sent and ascertains the rest of. Before sending the genuine bits, the sender includes the rest of the end of the real bits. Genuine data bits in addition to the rest of called a codeword. The sender transmits data bits as codewords.


At the flip side, the collector performs division operation on codewords utilizing the same CRC divisor. On the off chance that the rest of all zeros the data bits are acknowledged, else it is considered as there some data debasement happened in travel.



Q1.5) What techniques can be used for error-correction in the data link layer?

In the advanced world, error rectification should be possible in two ways:

  • Backward Error Correction: When the beneficiary detects an error in the data got, it demands back the sender to retransmit the data unit.
  • Forward Error Correction: When the beneficiary detects some error in the data got, it executes error-rectifying code, which offers it to auto-some assistance with recovering and to right a few sorts of errors.

The first, Backward Error Correction, is straightforward and must be productively utilized where retransmitting is not costly. For instance, fiber optics. Be that as it may, if there should be an occurrence of remote transmission retransmitting might cost excessively. In the last case, Forward Error Correction is utilized.

To amend the error in data outline, the beneficiary must know precisely which bit in the casing is adulterated. To find the bit in error, excess bits are utilized as parity bits for error detection.

For m data bits, r excess bits are utilized. “r” bits can give 2r blends of data. In m+r bit codeword, there is plausibility that the r bits themselves might get adulterated. So the quantity of r bits utilized must advise about m+r bit areas in addition to no-error data, i.e. m+r+1.


Part 2

Q2.1) A 3000-km-long T1 trunk is used to transmit 64-byte frames using Go-Back-N protocol. If the propagation speed is 6 microsecond/km how many bits should the sequence numbers be?

Propagation time = 6 sec/km × 3000 km = 18 ms
Speed (T1) = 1.536 Mbps
Transmission time for a frame = 64byte × 8 / 1.536Mbps = 0.3 ms
Total transmission time for a frame and its ack
= 2 × propagation time + transmission time for a frame
= 2 × 18ms + 0.3ms = 36.3ms
Window size = 36.3ms / 0.3ms = 121
As 27 = 128 > 121, so the sequence numbers should be 7 bits.






Q2.2 A simple telephone system consists of two end offices and a single toll office to which each end office is connected by a 1-MHzfull-duplex trunk. The average telephone is used to make five calls per 8-hour workday. The mean call duration is 6 min. Ten percent of the calls are long distance (i.e., pass through the toll office). What is the maximum number of telephones an end office can support?

Each user makes 4 calls, 6 minutes each (24 minutes). 10% calls (2.4 minutes) are long distance. Each user has a 4 KHz connection. This means each 1 MHz/4KHz = 256 simultaneous calls can be made. The maximum number of telephones are supported if the long distance calls are distributed uniformly during the 8-hour workdays.

Maximum number of supported telephones is: 256 * 8 * 60 / 2.4

Q2.3 Suppose the information content of a packet is the bit pattern 1011 0110 1110 1011, and an even parity scheme is being used. What would the value of the checksum field be for the case of a two-dimensional parity scheme? Your answer should be such that a minimum-length checksum field issued.

The minimum length checksum field should be 4*4 matrix. For our data, two dimensional (even) parity:
1 0 1 1 1
0 1 1 0 0
1 1 1 0 1
1 0 1 1 1
1 0 0 0 1


Two‐dim Parity = Generalization of the simple (one‐dim) parity scheme:

  • Form an MxN matrix of bits, then
  • Add a (even or odd) parity bit to each row and to each column


Q2.4 Suppose two nodes, A and B, are attached to opposite ends of an 800m cable, and that they each have one frame of 1,500 bits (including all headers and preambles) to send to each other. Both nodes attempt to transmit at time t = 0. Suppose there are four repeaters between A and B, each inserting a 20-bit delay. Assume the transmission rate is 100 Mbps, and CSMA/CD with backoff intervals of multiples of 512 bits is used. After the first collision, A draws K = 0 and B draws K = 1 in the exponential backoff protocol. Ignore the jam signal and the 96-bit time delay.



a. What is the one-way propagation delay (including repeater delays) between A and B in seconds? Assume that the signal propagation speed is 2 * 108 m/sec.

b. At what time (in seconds) is A’s packet completely delivered at B?

At time t

c. Now suppose that only A has a packet to send, and that the repeaters are replaced with switches. Suppose that each switch has a 16-bit processing delay in addition to a store-and-forward delay. At what time, in seconds, is A’s packet delivered at B?

First note, the transmission time of a single frame is given by 1500/ (100Mbps) = 15 micro sec, longer than the propagation delay of a bit.

At time t = 0, both A and B transmit.

At time t = 4.8µsec, both A and B detect a collision, and then abort.

At time t = 9.6µsec last bit of B’s aborted transmission arrives at A.

At time t =14.4µsec first bit of A’s retransmission frame arrives at B.

Q2.5. Two CSMA/CD stations are each trying to transmit long (multiframe) files. After each frame is sent, they contend for the channel, using the binary exponential backoff algorithm. What is the probability that the contention ends on round k, and what is the mean number of rounds per contention period?

Binary operational backoff:


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