assignment 1

halfway done, need to do q 5-9 tonight
excellent assignment, the questions are relevant to the material and the workload is relatively large
hopefully it will cover a lot of what can be asked in the exam

battling with the subnet section of question 2.
anyone else battling?

can anyone pls help met to calculate returned time?
the question states :
A computer sends a timestamp request to another computer. It receives the corresponding timestamp
reply at 3:46:07 A.M. The values of the original timestamp, receive timestamp and transmit timestamp
are 13,560,000, 13,562,000 and 13,564,300 respectively.

i am trying to work out the receiving trip time which is returned – transmit.

Hi jxharding.
With which subquestions do you struggle in question 2? So far I'm winning a little bit

just sorted out 2.7.3 + 2.7.4 thanks Ivan


the return time of the question in the forum posted : convert 3:46:07 A.M. to milliseconds
thanks eugmar!

ivan, did you finish the whole assignment?
im just doing 8.2 and then done at last

Solved 2.7.3 + 2.7.4 just now.
It's just 2.8 is giving me a headache. Sovled about half way I think. Moving on to question 3.

Still have the rest of the assigment to do over the weekend. Whoopie!

But as you said jxharding, it is an excellent assigment, and enjoying it. It's just that we have to cover a lot of work to complete the assignment. Quite hectic.

also battled big time with 2.8
just did the last part on how many addresses were left
2 points down the drain, but i'll have to wait for the answer sheet for that one

Assignment 2 at least looks less work ,but same type of questions, will start way earlier on that one

Any have ideas about question 9.2?

The Autonomous System (AS) number in an organization is 24101. Find the range of multicast addresses that the organization can use in the GLOP block.

I know you can use the AS number as the two middel octet in the range of 256 mulicast addresses. My question is how do one process the AS number to obtain the range of addresses?

Turn it into 16 bit number and slice it in half I guess. Thats what I did.

 
  ,= ,-_-. =.
 ((_/)o o(\_))
  `-'(. .)`-'
      \_/

http://ilanpillemer.com
Entia non sunt multiplicanda praeter necessitatem

Hi Ivan,

You can find some more info on the GLOP block in RFC 3171 (check out http://tools.ietf.org/html/rfc3171)

Section 8 (in that RFC) makes it quite clear what you must do. It is quoted here for you:

8. GLOP Block (233/8)

Addresses in the GLOP block are globally scoped statically assigned
addresses. The assignment is made by mapping a domain's autonomous
system number into the middle two octets of 233.X.Y.0/24. The
mapping and assignment is defined in [RFC2770].

The answer should be obvious now.

Hey Telstart,

Ivan said... " know you can use the AS number as the two middel octet in the range of 256 mulicast addresses. My question is how do one process the AS number to obtain the range of addresses?"

 
  ,= ,-_-. =.
 ((_/)o o(\_))
  `-'(. .)`-'
      \_/

http://ilanpillemer.com
Entia non sunt multiplicanda praeter necessitatem

Hi ilanpillemer.
I can answer my question now thanks to Telstar resources.
Quoted from [tools.ietf.org]:

For purposes of the experiment described here, the IANA has allocated 233/8. The remaining 24 bits will be administered in a manner similar to that described in RFC 1797:

0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| 233 | 16 bits AS | local bits |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

2.1. Example
Consider, for example, AS 5662. Written in binary, left padded with 0s, we get 0001011000011110. Mapping the high order octet to the second octet of the address, and the low order octet to the third octet, we get 233.22.30/24.

Thanks for all your input guys!