Danielle Gardner

Fundamentals of Computer Science

Dr. Kihlstrom

October 3, 2006

 

Chapter 7- Exc #5, 7, 8, 9, 13

 

5. The address field is needed in an Ethernet LAN protocol because it is needed to help specify the destination and the source of each data pocket that is both sent and delivered otherwise a node would not know when to send or receive. If you were trying to broadcast everyone in the network, then it would be helpful to omit the address field or create a special broadcast address that knows to send to everyone and every receiver knows to accept that request. For instance, creating a special address called ÒALLÓ so everyone knows to open it once received and not ignore the message.

 

7. An advantage to the creation of centralized LAN would be to help enforce and control operations so that simple changes can be made by the centralized node that would then be ÔpropagatedÕ throughout the rest of the system. It would be like giving authority to a manager that determines the mobility and frequency of specific messages being sent and received by different nodes. It will also provide a more straightforward route for migrating between other networks, proving to be more cost effective, increasingly error-free and greater security thanks to a more central manager. 

 

8. False, because you cannot guarantee that it would be delivered in a particular amount of time. The ARQ (Automatic Repeat Request) where after an error has been detected then retransmission of a new and unblemished copy of the original message is sent, ensuring that every message sent eventually arrives at the proper destination; however, there are no specific time constraints.

 

9.  A. The smallest number of point-to-point communication links would be N-1.

           

     B. You would use a Ôredundant connection link-structureÓ so therefore you would add more links, some going diagonal and some going horizontal and maybe some going vertical so you donÕt run the risk of having a disconnected network.

 

13.

            a. From node A to G there are seven simple paths.

                        ABCG

                        ABECG

                        ABEDCG

                        ADECG

                        ADCG

                        ADEBCG

                        AFG

           

            b. Shortest path is from node A to Node G would A to D to E to C to

            G. The overall delay would be 10.

 

           

 

c. If node E fails, this does change the shortest path. The new shortest path would be three paths with a delay of eleven:

                        ABCG

                        AFG

                        ADCG