I for one am flattered by this message. I am also reminded of the reasons I have enjoyed this group so much. Nothing like a bunch of folks learning together. Finding out different ways of doing things. Sharing those discoveries. Some highlights of my time here include Cthulu's K-mart firewall, and his stories of the tank range, both of which helped me immensely as I prepared for the ACRC test, the global DLCI conundrum, and the EIGRP routing process that refused to summarize, which resulted in the discovery of a reported bug in certain IOS versions that rendered this feature inoperative. Now that the routers appear to be acting as they should, I hope to share them out for a lab or two. Gotta keep up this tradition. The joy of discovery. This voyage we all share. Chuck -----Original Message----- From: Raymond Everson (Rainman) [mailto:[EMAIL PROTECTED]] Sent: Monday, June 26, 2000 1:20 PM To: Michael L. Williams; Daniel Ma; Cthulu, CCIE Candidate; [EMAIL PROTECTED]; Bal Sandhu; Sanjay.Padmanabhan; Chuck Larrieu; Lou Nelson; Aaron K. Dixon; [EMAIL PROTECTED]; Oz; study group Subject: Re: Sample Question Question Hi Michael, Daniel, Charles, Duarte, Bal, Sanjay, Chuck, Lou, Aaron, Sam, Oz... I've learned two things from this (seriies of) question(s). #1: People would MUCH rather talk about meat/potatoes of tech questions than some of the off-subject crap that gets thrown into this discussion group; and #2: There are some genuinely nice and interested people out there. This question raises the paradox which we face when taking the CCIE written, and then evaluating how to handle a lab scenario: I don't have enough information to answer completely. Please let me explain.... Everyone agrees that the answer is "e. none of the above". Review the provided answer all the way at the bottom. Under normal circumstances (with no consideration for zero subnet-enable) .... the statement is correct, this would be an illegal address/mask pair. But, considering the possibilites of the world, I don't see a thing wrong with 193.243.12.43/25. ... and I would have been totally messed up by that statement. ... which is why Sanjay and Michael and, of course Chuck are getting close to being ready for the lab..... attention to detail.... the "Spot the Issue" gotcha in the config with this scenario is a single line of code: zero subnet-enable btw: Chuck Larrieu wins hands down for his completely awesome explanation of how to figure out what evens have in common, and what odds have in common. After that, it's a simple matter of understanding inverse masking.... DOH! Thanks a million, you're a fine bunch. Rainman "Michael L. Williams" wrote: > I hope I'm right =) > > I would say the answer is E. Here's why. > > 193.x.y.z is a Class C address. So the first three octects of the subnet > mask are the "network" portion, and the last octect is the "hosts" portion. > > What they have done here by making the last octect 128 (1000 0000), instead > of 0 (0000 0000), is to split that 256 hosts into two subnets, by taking > that first bit away from the "hosts" and using it for "network". Each subnet > now has up to 128 hosts (of course you have to subtract 2; 1 for the network > and 1 for broadcast, so each subnet technically can only have up to 126 > hosts). > > So, the valid host addressess in the two subnets would be 193.243.12.1 - > 193.243.12.126 (193.243.12.0 is the network address, and 193.243.12.127 is > the broadcast for the 1st subnet), and 193.243.12.129 - 193.243.12.254 > (where 193.243.12.128 is the network address for subnet 2, and > 193.243.12.255 is the broadcast address for subnet 2) > > So, is they ask us the "network address" for the IP address 193.243.12.43 > with a subnet mask of 255.255.255.128, the answer is 193.243.12.0. If there > were to ask the "network address" for 193.243.12.181 with a subnet mask of > 255.255.255.128, the answer is 193.243.12.128. > > Hope this helps some........ > > Mike W. > > ----- Original Message ----- > From: "Raymond Everson (Rainman)" <[EMAIL PROTECTED]> > Newsgroups: groupstudy.cisco > Sent: Saturday, June 24, 2000 6:10 PM > Subject: Sample Question Question > > > I ran across this question some time ago and kept the darned thing > > because (I *guess*) I still have something to learn about IP ... . just > > when I was *sure* I had it down cold. Could you lend a fella a hand? > > > > Given the IP address of 193.243.12.43 and a subnet mask of > > 255.255.255.128, what is the subnet address? > > > > a.194.243.12.32 > > b.193.243.0.0 > > c.194.243.12.43 > > d.193.243.12.128 > > e.None of the above. > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > . > > e. None of the above > > > > The Subnet Address is found by setting all of the host bits in an > > address to 0. > > According to the mask, there are 7 host bits, so applying this rule > > yields a subnet > > address of 193.243.12.0. However, closer inspection yields that there is > > only one > > subnet bit in this Class C address - therefore, this is an illegal > > address/mask pair. > > > > ___________________________________ > > UPDATED Posting Guidelines: http://www.groupstudy.com/list/guide.html > > FAQ, list archives, and subscription info: http://www.groupstudy.com > > Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] > > --- ___________________________________ UPDATED Posting Guidelines: http://www.groupstudy.com/list/guide.html FAQ, list archives, and subscription info: http://www.groupstudy.com Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
