tag:blogger.com,1999:blog-6240379334240712627.post5686191697014699296..comments2023-08-28T12:42:44.108+01:00Comments on Subnetting Made Easy And Other Cisco Tidbits: Subnetting Made Easy - Critically Acclaimed!Chris Bloomfieldhttp://www.blogger.com/profile/05456433782746614889noreply@blogger.comBlogger78125tag:blogger.com,1999:blog-6240379334240712627.post-61554717271889042312017-11-24T18:04:00.881+00:002017-11-24T18:04:00.881+00:00Boundaries and Block size in the first section swu...Boundaries and Block size in the first section swung it for me. I keep getting these questions correct now, time after time - Thank you so muchAnonymoushttps://www.blogger.com/profile/01875685256856052948noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-37973823373988660282016-01-14T19:14:28.360+00:002016-01-14T19:14:28.360+00:00Hi Kate. They have given you a subnet mask which g...Hi Kate. They have given you a subnet mask which gives you a block size of 32 in the third octet. Therefore the subnet address for that mask will be the following:<br /><br />a.b.0.0<br />a.b.32.0<br />a.b.64.0<br />a.b.96.0<br />a.b.128.0<br />a.b.160.0<br />a.b.192.0<br />a.b.224.0Chris Bloomfieldhttps://www.blogger.com/profile/05456433782746614889noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-50433594597412905022016-01-12T12:27:20.729+00:002016-01-12T12:27:20.729+00:00I have a problem with the following question.
As...I have a problem with the following question. <br /><br />Assuming a subnet mask of 255.255.224.0, which of the following would be valid host addresses?<br /><br />A. 124.78.103.0<br />B. 125.67.32.0<br />C. 125.78.160.0<br />D. 126.78.48.0<br />E. 176.55.96.0 <br />F. 186.211.100.0<br /><br />The answers are A,D and F. I dont know how to get the answers. Plz helpKatehttps://www.blogger.com/profile/04867445903435542989noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-56040612610785828242015-09-12T22:49:55.734+01:002015-09-12T22:49:55.734+01:00Hi Chris,
your prediction was quite right - I did...Hi Chris,<br /><br />your prediction was quite right - I did laugh while doing the subnetting exercises!<br /><br />Thank you, man, you are great! :)<br /><br />Cheers,<br />SvetoslavSvetoslavhttps://www.blogger.com/profile/10656749769967739209noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-30767901241528539682015-09-12T22:48:43.999+01:002015-09-12T22:48:43.999+01:00Hello Chris,
you were quite right in your predict...Hello Chris,<br /><br />you were quite right in your prediction - I did laugh while doing the subnetting exercises.<br /><br />Thank you, man, you are great! :)<br /><br />Cheers,<br />SvetoslavSvetoslavhttps://www.blogger.com/profile/10656749769967739209noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-16426506311472485682015-08-31T12:04:56.924+01:002015-08-31T12:04:56.924+01:00Hi,
Great tutorial! Having been searching for a ex...Hi,<br />Great tutorial! Having been searching for a explanation of sunbathing for days and this by far the best. <br />CheersAnonymoushttps://www.blogger.com/profile/15244227021089556466noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-28966649442186407552015-05-23T05:38:50.744+01:002015-05-23T05:38:50.744+01:00Dear Chris,
Refer Hanson's comment on 23 Sept...Dear Chris,<br /><br />Refer Hanson's comment on 23 September 2013 at 21:24<br /><br />"Enter the maximum number of valid subnets and hosts per subnet that you can get from the network 172.29.0.0 255.255.254.0"<br /><br />The subnets have to be 172.29.0.0, 2.0, 4.0 so on so forth for a total of 2^7 (7 bits borrowed)=128 subnets.<br /><br />Kindly correct if wrong.<br /><br />I loved your subnetting technique. Do you have any other simplified stuff out there to share.<br /><br />Anonymoushttps://www.blogger.com/profile/09472197879052686548noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-33833322432932373132014-08-10T17:21:12.099+01:002014-08-10T17:21:12.099+01:00Joe,
CIDR represents the number of 1s in the mask...Joe,<br /><br />CIDR represents the number of 1s in the mask, or the binary conversion from decimal.<br /><br />255.255.128.0 is 11111111.11111111.10000000.00000000<br /><br />By counting the number of 1s, you have 17, /17.<br /><br />The other way to look at it is from a binary chart.<br /><br />Recalling boundaries, for each class, /8, /16, /24, for A, B, C, respectively. Subtract the given CIDR value from the default mask and remainder is as follows.<br /><br />.128 /1<br />.192 /2<br />.224 /3<br />.240 /4<br />.248 /5<br />.252 /6<br />.254 /7<br />.255 /8<br /><br />/23 is 255.255.254.0, 23-16=7, /7 is .254<br />/12 is 255.240.0.0, 12-8=4, /4 is .240<br /><br />Hope that helps. :)uneekhttps://www.blogger.com/profile/14490666380789262946noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-56044451917024951152014-04-14T15:43:45.467+01:002014-04-14T15:43:45.467+01:00Hi Tom,
Impossible to say. The default mask would...Hi Tom,<br /><br />Impossible to say. The default mask would be /8 or 255.0.0.0.<br /><br />HTH,<br /><br />ChrisChris Bloomfieldhttps://www.blogger.com/profile/05456433782746614889noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-8700105355849442332014-04-13T19:33:58.672+01:002014-04-13T19:33:58.672+01:00hello Chris
If a host has address 10.251.128.10,
w...hello Chris<br />If a host has address 10.251.128.10,<br />what is correct subnetmask? I cannot find a procedure in your blog.<br /><br />Tomconghttps://www.blogger.com/profile/08636604245585072524noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-41215838157386366722014-02-02T16:53:07.001+00:002014-02-02T16:53:07.001+00:00how do I find the nth subnet? For example I have t...how do I find the nth subnet? For example I have the class B address 150.150.0.0/26 and want to find the 500th subnet?<br /><br />Brilliant explanations by the way!Anonymoushttps://www.blogger.com/profile/06211776821302593167noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-70489725084102490952014-02-01T15:39:05.796+00:002014-02-01T15:39:05.796+00:00I have been reading TCP/IP Illustrated and trying ...I have been reading TCP/IP Illustrated and trying the examples on a test network. I got stuck on subnet/supernet addressing. Your examples filled in the gaps. I was back on track in 5 minutes!Anonymoushttps://www.blogger.com/profile/13110220559104423120noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-34641489967463435282014-01-09T20:15:13.412+00:002014-01-09T20:15:13.412+00:00This is by far the best way to do any subnetting q...This is by far the best way to do any subnetting questions!! <br /><br />The only part I need help with now is how to find the CIDR prefix from a subnet mask and visa versa.<br /><br />E.g. 255.255.128.0 is /17 but how do I work this out? Is there a formula?<br /><br />Can anyone help please???<br /><br />Anonymoushttps://www.blogger.com/profile/08856916486709879927noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-23317809377853637382013-11-11T17:02:37.404+00:002013-11-11T17:02:37.404+00:00Hi Chris,
I ran across this problem and was confu...Hi Chris,<br /><br />I ran across this problem and was confused on how i could figure it out. If it gives me the address in CIDR notation I can do it no problem.<br /><br />Here is the problem:<br />How many subnets and hosts per subnet can you get from the network 192.168.130.0 255.255.255.128?<br /><br />Thanks!Anonymoushttps://www.blogger.com/profile/02648199497534731362noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-86804920657456660362013-11-10T17:04:07.199+00:002013-11-10T17:04:07.199+00:00I was looking for help on understanding subnetting...I was looking for help on understanding subnetting when I stumbled onto your site. This is much easier to grasp than any other method I have seen. I am sure I'll have some questions as I study this but your approach to explaining it is far more direct and to the point. Thanks for your commitment to helping those of us who are trying to learn. Aaron B.J.A.B.https://www.blogger.com/profile/15462315367675842824noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-55436854419068250912013-10-24T10:11:37.567+01:002013-10-24T10:11:37.567+01:00Hi
Chris, almost after 10 year in this field i gav...Hi<br />Chris, almost after 10 year in this field i gave up on learning subnetting insted started using online calculators, nothing ever made sense to me <br />man just reading your artical only once made me feel like every thing falls int place <br /><br />thanks ever so much <br /><br />regards<br />AnsurAnonymoushttps://www.blogger.com/profile/16201764107579034802noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-43999394728209494562013-10-24T10:10:31.212+01:002013-10-24T10:10:31.212+01:00Hi
Chris, almost after 10 year in this field i gav...Hi<br />Chris, almost after 10 year in this field i gave up on learning subnetting insted started using online calculators, nothing ever made sense to me <br />man just reading your artical only once made me feel like every thing falls int place <br /><br />thanks ever so much <br /><br />regards<br />AnsurAnonymoushttps://www.blogger.com/profile/16201764107579034802noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-75661435799063753162013-10-13T02:53:32.861+01:002013-10-13T02:53:32.861+01:00Hi Chris,
Thanks for the wonderful explanation. I ...Hi Chris,<br />Thanks for the wonderful explanation. I now understand everything you mentioned except the following part:<br /><br />What subnet mask should you use if you wanted 60 hosts per subnet? The formula is 60 = 2(32 - n) - 2 so you must find n which is 26. This is easy to find as you know that 26 - 2 = 62 so simply subtract 6 from 32 to get the 26. Therefore your mask is /26. <br /><br />Is there another way to explain this, especially finding the value of n, I simply just dont get it.<br /><br />HPHPhttps://www.blogger.com/profile/11435513541548121712noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-75913823715288140922013-10-03T19:31:42.480+01:002013-10-03T19:31:42.480+01:00wow after three classes on this i had become COMPL...wow after three classes on this i had become COMPLETELY overwhelmed by all the different techniques that we were being shown for subnetting but this is by far the best way of doing things (for me at least)<br /><br />thanks a lot for this man you`ve really saved the daymagichttps://www.blogger.com/profile/05785922145500303865noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-338401825442869182013-10-03T19:31:18.872+01:002013-10-03T19:31:18.872+01:00wow after three classes on this i had become COMPL...wow after three classes on this i had become COMPLETELY overwhelmed by all the different techniques that we were being shown for subnetting but this is by far the best way of doing things (for me at least)<br /><br />thanks a lot for this man you`ve really saved the day magichttps://www.blogger.com/profile/05785922145500303865noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-73521587294995401932013-09-25T16:33:59.432+01:002013-09-25T16:33:59.432+01:00Hi Hanson,
So for both of your questions you have...Hi Hanson,<br /><br />So for both of your questions you have a subnet mask of 255.255.254.0 which is equivalent to a /23. That means your block size is 2 to the power of (24 - 23) which equals 2 subnets. The number of hosts is therefore (2 to the power of (32 - 23)) - 2 = 510 hosts.<br /><br />Your two subnets are effectively 172.17.0.0/23 and 172.17.128.0/23 <br /><br />HTHChris Bloomfieldhttps://www.blogger.com/profile/05456433782746614889noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-45582600700411351622013-09-25T16:08:37.273+01:002013-09-25T16:08:37.273+01:00Hi Chris,
Would you help me solve this problem bel...Hi Chris,<br />Would you help me solve this problem below using your method. <br /><br />Enter the maximum number of valid subnets and hosts per subnet that you can get from the network 172.17.0.0 255.255.254.0<br /><br />Thanks,<br />HansonAnonymoushttps://www.blogger.com/profile/02398906618017944474noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-59421172656095752642013-09-23T21:24:50.669+01:002013-09-23T21:24:50.669+01:00Hi Chris,
This is best subnetting method I've...Hi Chris,<br /> This is best subnetting method I've everseen. <br />There is one type of question I keep having wrong answer. Would you please step by step show me with your method for below problem.<br /><br />"Enter the maximum number of valid subnets and hosts per subnet that you can get from the network 172.29.0.0 255.255.254.0"<br />Thanks a lot!<br /><br />HansonAnonymoushttps://www.blogger.com/profile/02398906618017944474noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-36353627077142879722013-09-04T18:49:41.048+01:002013-09-04T18:49:41.048+01:00what to do if the no of subnets is greater than 32...what to do if the no of subnets is greater than 32.i have a question <br />An organization is granted the block 130.56.0.0 in class B. The administrator wants to create 1024 subnets.<br />a. Find the subnet mask.<br />b. Find the number of addresses in each subnet.<br />Anonymoushttps://www.blogger.com/profile/06481075326706721754noreply@blogger.comtag:blogger.com,1999:blog-6240379334240712627.post-34218253719054509232013-08-22T08:17:22.980+01:002013-08-22T08:17:22.980+01:00Hello Chris, the way you explained this subnetting...Hello Chris, the way you explained this subnetting concept made a great help to my studies.<br />i have a doubt.<br /><br />Could you tell me what will be the subnet and host from<br /><i><b>172.29.0.0 255.255.252.0 </b></i><br /><br />thank you..!!Anonymoushttps://www.blogger.com/profile/18249215423422522771noreply@blogger.com