12. Consider the network in figure 1

 

To send a frame to C, host A puts 88-B2-2F-54-1A-0F in the destination MAC address field of the frame

To send a frame to C, host A puts FF-FF-FF-FF-FF-FF in the destination MAC address field of the frame

To send a frame to C, host A puts switch S’s MAC address in the destination MAC address field of the frame d) None of these

13. A 16-bit CRC will

detect all error bursts of length 16 or less

detect most error bursts of length greater than 16

require data to be at least 16 bits

all of these

14. About the start frame delimiter in Ethernet

It is used to synchronize the receiver and sender clock rates

It is based on the physical layer coding violation principle

It is used to synchronize the receiver and sender at the byte level

It is used to synchronize the receiver and sender at the bit level

 

Fill in the blank questions 

(1 pt) Consider 4 nodes with MAC addresses A, B, C, and D connected to interfaces 1, 2, 3, and 4 respectively of a switch, according to a star topology. Assume prior to event 1, the switch table is empty (all entries are blank). 

Suppose event 1 happens: B sends a frame to D.

List all the interface(s) to which the switch forwards the frame. List only the interface numbers, separated by commas, by order of increasing number. Do not add any comment or text, such as “Interfaces are “. If you don’t follow the instructions, the automated grader won’t be able to recognize your answer as correct. A hypothetical example of answer in the correct format is 2, 3, 4. Spaces do not matter.

 

(1 pt) After event 1 has happened, what is/are the non-blank entry/entries of the switch table? List the content of each non-blank entry as a tuple . Write only the MAC address, and the interface number, separated by a comma. If there is more than one non-blank entry, list the tuples by order of increasing MAC address, and separate by a comma.  

As a hypothetical example, if there are two non-blank entries: 

MAC address: C, mapped to interface 1 

MAC address: A, mapped to interface 3 

Your answer should be

 

A S 111.111.111.111 222.222.222.222 74-29-9C-E8-FF-55 49-BD-D2-C7-56-2A 222.222.222.220 1A-23-F9-CD-06-9B 222.222.222.221 88-B2-2F-54-1A-0F 111.111.111.110 111.111.111.112 E6-E9-00-17-BB-4B CC-49-DE-D0-AB-7D