Re: difficult ccna question
George, you will find answers here http://www.3com.com/nsc/501302s.html cheers Larry --- George <[EMAIL PROTECTED]> wrote: > if you have a class B network with a 10-bit subnet > mask, how many subnet and > how many hosts do you have? > > a. 1022 subnets, 62 hosts > b. 62 subnets, 8190 hosts > c. 8190 subnets, 254 hosts > d. 254 subnets , 126 hosts > > > _ > FAQ, list archives, and subscription info: > http://www.groupstudy.com/list/cisco.html > Report misconduct and Nondisclosure violations to > [EMAIL PROTECTED] > > > = "Wear a smile and have friends; wear a scowl and have wrinkles." - George Eliot "the greatest glory is not in never falling, but rising up each time we fall." "The greatest man is not he who does not fall but he who falls and rises again to win" __ Do You Yahoo!? Get email at your own domain with Yahoo! Mail. http://personal.mail.yahoo.com/ _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
RE: difficult CCNA question
Class be is a 255.255.0.0 with 10 subnet bit ( bits over and above the default mask) is a 255.255.255.192 that is host portions of 62 ip addresses ( 64 available IPs less broadcast and networks address) Therefore answer is A -Original Message- From: Larry Osei-Kwaku [mailto:[EMAIL PROTECTED]] Sent: 22 March 2001 11:56 To: George; [EMAIL PROTECTED] Subject: Re: difficult ccna question George, you will find answers here http://www.3com.com/nsc/501302s.html cheers Larry --- George <[EMAIL PROTECTED]> wrote: > if you have a class B network with a 10-bit subnet > mask, how many subnet and > how many hosts do you have? > > a. 1022 subnets, 62 hosts > b. 62 subnets, 8190 hosts > c. 8190 subnets, 254 hosts > d. 254 subnets , 126 hosts > > > _ > FAQ, list archives, and subscription info: > http://www.groupstudy.com/list/cisco.html > Report misconduct and Nondisclosure violations to > [EMAIL PROTECTED] > > > = "Wear a smile and have friends; wear a scowl and have wrinkles." - George Eliot "the greatest glory is not in never falling, but rising up each time we fall." "The greatest man is not he who does not fall but he who falls and rises again to win" __ Do You Yahoo!? Get email at your own domain with Yahoo! Mail. http://personal.mail.yahoo.com/ _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
RE: difficult ccna question
Just remember the following fomulas: (2^# of masked bits) - 2 = Total # of subnets (2^# of unmasked bits - 2 = Total # of hosts Based on this the correct answer is A. -Original Message- From: George [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 12:16 AM To: [EMAIL PROTECTED] Subject: difficult ccna question if you have a class B network with a 10-bit subnet mask, how many subnet and how many hosts do you have? a. 1022 subnets, 62 hosts b. 62 subnets, 8190 hosts c. 8190 subnets, 254 hosts d. 254 subnets , 126 hosts _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
RE: difficult ccna question
Based on the subnet mask (255.255.192.0) subtract 256 from 192 which is 64. Therefore 64 is your first subnet. To get the second subnet, add 64 to the first subnet. To get the third add 64 to the second subnet and so on. Continue in this fashion until you reach 192. Remember that you cannot use the ranges 172.16.0.1 - 172.16.0.62 and 172.16.255.193 - 172.16.255.254 (network and broadcast respectively) unless your router is configured to do so. Subnet Host Range 1 172.16.0.65 - 172.16.0.126 2 172.16.0.129 - 172.16.0.190 3 172.16.0.193 - 172.16.0.254 4 172.16.1.1 - 172.16.1.62 5 172.16.1.65 - 172.16.1.126 6 172.16.1.129 - 172.16.1.190 7 172.16.1.193 - 172.16.1.254 8 172.16.2.1 - 172.16.2.62 9 172.16.2.65 - 172.16.2.126 10 172.16.2.129 - 172.16.2.190 . . . . . . 1015 172.16.253.193 - 172.16.253.254 1016 172.16.254.1 - 172.16.254.62 1017 172.16.254.65 - 172.16.254.126 1018 172.16.254.129 - 172.16.254.190 1019 172.16.254.193 - 172.16.254.254 1020 172.16.255.1 - 172.16.255.62 1021 172.16.255.65 - 172.16.255.126 1022 172.16.255.129 - 172.16.255.190 -Joshua -Original Message- From: Lowell Sharrah [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 8:14 AM To: [EMAIL PROTECTED] Subject: RE: difficult ccna question how do you know where the first subnet begins? >>> Joshua Beining <[EMAIL PROTECTED]> 03/22/01 10:34AM >>> Just remember the following fomulas: (2^# of masked bits) - 2 = Total # of subnets (2^# of unmasked bits - 2 = Total # of hosts Based on this the correct answer is A. -Original Message- From: George [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 12:16 AM To: [EMAIL PROTECTED] Subject: difficult ccna question if you have a class B network with a 10-bit subnet mask, how many subnet and how many hosts do you have? a. 1022 subnets, 62 hosts b. 62 subnets, 8190 hosts c. 8190 subnets, 254 hosts d. 254 subnets , 126 hosts _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
RE: difficult ccna question
Wouldn't the first subnet be 172.16.0.0? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of Joshua Beining Sent: Thursday, March 22, 2001 11:43 AM To: 'Lowell Sharrah' Cc: '[EMAIL PROTECTED]' Subject: RE: difficult ccna question Based on the subnet mask (255.255.192.0) subtract 256 from 192 which is 64. Therefore 64 is your first subnet. To get the second subnet, add 64 to the first subnet. To get the third add 64 to the second subnet and so on. Continue in this fashion until you reach 192. Remember that you cannot use the ranges 172.16.0.1 - 172.16.0.62 and 172.16.255.193 - 172.16.255.254 (network and broadcast respectively) unless your router is configured to do so. Subnet Host Range 1 172.16.0.65 - 172.16.0.126 2 172.16.0.129 - 172.16.0.190 3 172.16.0.193 - 172.16.0.254 4 172.16.1.1 - 172.16.1.62 5 172.16.1.65 - 172.16.1.126 6 172.16.1.129 - 172.16.1.190 7 172.16.1.193 - 172.16.1.254 8 172.16.2.1 - 172.16.2.62 9 172.16.2.65 - 172.16.2.126 10 172.16.2.129 - 172.16.2.190 . . . . . . 1015 172.16.253.193 - 172.16.253.254 1016 172.16.254.1 - 172.16.254.62 1017 172.16.254.65 - 172.16.254.126 1018 172.16.254.129 - 172.16.254.190 1019 172.16.254.193 - 172.16.254.254 1020 172.16.255.1 - 172.16.255.62 1021 172.16.255.65 - 172.16.255.126 1022 172.16.255.129 - 172.16.255.190 -Joshua -Original Message- From: Lowell Sharrah [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 8:14 AM To: [EMAIL PROTECTED] Subject: RE: difficult ccna question how do you know where the first subnet begins? >>> Joshua Beining <[EMAIL PROTECTED]> 03/22/01 10:34AM >>> Just remember the following fomulas: (2^# of masked bits) - 2 = Total # of subnets (2^# of unmasked bits - 2 = Total # of hosts Based on this the correct answer is A. -Original Message- From: George [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 12:16 AM To: [EMAIL PROTECTED] Subject: difficult ccna question if you have a class B network with a 10-bit subnet mask, how many subnet and how many hosts do you have? a. 1022 subnets, 62 hosts b. 62 subnets, 8190 hosts c. 8190 subnets, 254 hosts d. 254 subnets , 126 hosts _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
RE: difficult ccna question
Technically yes. But like most things you cannot get something for free. There is a price for subnetting which is that you loose the first subnet (network - all zeros) and the last subnet (broadcast - all ones). Note that with some routers you can configure them to use these subnets. But be careful, because some OS's and devices do not react well to using then. HTH. -Joshua -Original Message- From: David A. Lauer [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 11:14 AM To: [EMAIL PROTECTED] Subject: RE: difficult ccna question Wouldn't the first subnet be 172.16.0.0? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of Joshua Beining Sent: Thursday, March 22, 2001 11:43 AM To: 'Lowell Sharrah' Cc: '[EMAIL PROTECTED]' Subject: RE: difficult ccna question Based on the subnet mask (255.255.192.0) subtract 256 from 192 which is 64. Therefore 64 is your first subnet. To get the second subnet, add 64 to the first subnet. To get the third add 64 to the second subnet and so on. Continue in this fashion until you reach 192. Remember that you cannot use the ranges 172.16.0.1 - 172.16.0.62 and 172.16.255.193 - 172.16.255.254 (network and broadcast respectively) unless your router is configured to do so. Subnet Host Range 1 172.16.0.65 - 172.16.0.126 2 172.16.0.129 - 172.16.0.190 3 172.16.0.193 - 172.16.0.254 4 172.16.1.1 - 172.16.1.62 5 172.16.1.65 - 172.16.1.126 6 172.16.1.129 - 172.16.1.190 7 172.16.1.193 - 172.16.1.254 8 172.16.2.1 - 172.16.2.62 9 172.16.2.65 - 172.16.2.126 10 172.16.2.129 - 172.16.2.190 . . . . . . 1015 172.16.253.193 - 172.16.253.254 1016 172.16.254.1 - 172.16.254.62 1017 172.16.254.65 - 172.16.254.126 1018 172.16.254.129 - 172.16.254.190 1019 172.16.254.193 - 172.16.254.254 1020 172.16.255.1 - 172.16.255.62 1021 172.16.255.65 - 172.16.255.126 1022 172.16.255.129 - 172.16.255.190 -Joshua -Original Message- From: Lowell Sharrah [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 8:14 AM To: [EMAIL PROTECTED] Subject: RE: difficult ccna question how do you know where the first subnet begins? >>> Joshua Beining <[EMAIL PROTECTED]> 03/22/01 10:34AM >>> Just remember the following fomulas: (2^# of masked bits) - 2 = Total # of subnets (2^# of unmasked bits - 2 = Total # of hosts Based on this the correct answer is A. -Original Message- From: George [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 12:16 AM To: [EMAIL PROTECTED] Subject: difficult ccna question if you have a class B network with a 10-bit subnet mask, how many subnet and how many hosts do you have? a. 1022 subnets, 62 hosts b. 62 subnets, 8190 hosts c. 8190 subnets, 254 hosts d. 254 subnets , 126 hosts _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
RE: difficult ccna question
Can you give examples of OSs that don't support all ones? -Original Message- From: Joshua Beining [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 2:44 PM To: 'David A. Lauer' Cc: '[EMAIL PROTECTED]' Subject: RE: difficult ccna question Technically yes. But like most things you cannot get something for free. There is a price for subnetting which is that you loose the first subnet (network - all zeros) and the last subnet (broadcast - all ones). Note that with some routers you can configure them to use these subnets. But be careful, because some OS's and devices do not react well to using then. HTH. -Joshua -Original Message- From: David A. Lauer [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 11:14 AM To: [EMAIL PROTECTED] Subject: RE: difficult ccna question Wouldn't the first subnet be 172.16.0.0? -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of Joshua Beining Sent: Thursday, March 22, 2001 11:43 AM To: 'Lowell Sharrah' Cc: '[EMAIL PROTECTED]' Subject: RE: difficult ccna question Based on the subnet mask (255.255.192.0) subtract 256 from 192 which is 64. Therefore 64 is your first subnet. To get the second subnet, add 64 to the first subnet. To get the third add 64 to the second subnet and so on. Continue in this fashion until you reach 192. Remember that you cannot use the ranges 172.16.0.1 - 172.16.0.62 and 172.16.255.193 - 172.16.255.254 (network and broadcast respectively) unless your router is configured to do so. Subnet Host Range 1 172.16.0.65 - 172.16.0.126 2 172.16.0.129 - 172.16.0.190 3 172.16.0.193 - 172.16.0.254 4 172.16.1.1 - 172.16.1.62 5 172.16.1.65 - 172.16.1.126 6 172.16.1.129 - 172.16.1.190 7 172.16.1.193 - 172.16.1.254 8 172.16.2.1 - 172.16.2.62 9 172.16.2.65 - 172.16.2.126 10 172.16.2.129 - 172.16.2.190 . . . . . . 1015 172.16.253.193 - 172.16.253.254 1016 172.16.254.1 - 172.16.254.62 1017 172.16.254.65 - 172.16.254.126 1018 172.16.254.129 - 172.16.254.190 1019 172.16.254.193 - 172.16.254.254 1020 172.16.255.1 - 172.16.255.62 1021 172.16.255.65 - 172.16.255.126 1022 172.16.255.129 - 172.16.255.190 -Joshua -Original Message- From: Lowell Sharrah [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 8:14 AM To: [EMAIL PROTECTED] Subject: RE: difficult ccna question how do you know where the first subnet begins? >>> Joshua Beining <[EMAIL PROTECTED]> 03/22/01 10:34AM >>> Just remember the following fomulas: (2^# of masked bits) - 2 = Total # of subnets (2^# of unmasked bits - 2 = Total # of hosts Based on this the correct answer is A. -Original Message- From: George [mailto:[EMAIL PROTECTED]] Sent: Thursday, March 22, 2001 12:16 AM To: [EMAIL PROTECTED] Subject: difficult ccna question if you have a class B network with a 10-bit subnet mask, how many subnet and how many hosts do you have? a. 1022 subnets, 62 hosts b. 62 subnets, 8190 hosts c. 8190 subnets, 254 hosts d. 254 subnets , 126 hosts _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
Re: difficult ccna question
Well, your numbers are a little off, but the general idea is sound. With the way the question is worded, who knows exactly what the correct mask is. Other than the obvious, maybe we can assume that the questioner is adding 10 additional bits to the native classful mask of 255.255.0.0 for a class B network. This would produce a 255.255.255.192 mask. Now for the math. With 192 in the last octet, this leaves 6 bits for hosts 2^6 = 64 - 2 (don't forget this part!) = 62 usable host addresses You also now have 10 bits for subnetting (additional to the native classful mask) So, 2^10 = 1024 - 2 (don't forget this part here either!) = 1022 usable subnets Notice we can use the same simple formula ( 2^n-2) to find both the number of usable host addresses and subnets. One other thing to be aware of: each subnet ID, which is the first address in every subnet, always has an even number in the subnetted octet. In other words, the first "usable" subnet in the previously mentioned example should be 172.16.0.64. This is just a chracteristic of the process of subnettingsomething about always starting to count with a 0 instead of a 1 ;-} 172.16.0.0 - 172.16.0.63 172.16.0.64 - 172.16.0.127 172.16.0.128 - 172.16.0.191 172.16.0.192 - 172.16.0.255 etc., etc., etc you get the idea The website I think is a great resource for those just getting into TCP/IP is: http://www.learntosubnet.com/ Check it out! BTW: it is true that many legacy device/software won't recognize the first or last subnet. The first subnet, also called the "zero subnet", can be configured for use on Cisco devices with the "ip subnet-zero" command. Although not every vendor supports doing this, I would guess that a vast majority of them do. It was only a short time ago when this was not the case. I can remember when NT 4.0 would not support this. It probably does now, with service packs and all, but it didn't always. Rik "Joshua Beining" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > Based on the subnet mask (255.255.192.0) subtract 256 from 192 which is 64. > Therefore 64 is your first subnet. To get the second subnet, add 64 to the > first subnet. To get the third add 64 to the second subnet and so on. > Continue in this fashion until you reach 192. Remember that you cannot use > the ranges 172.16.0.1 - 172.16.0.62 and 172.16.255.193 - 172.16.255.254 > (network and broadcast respectively) unless your router is configured to do > so. > > Subnet Host Range > 1 172.16.0.65 - 172.16.0.126 > 2 172.16.0.129 - 172.16.0.190 > 3 172.16.0.193 - 172.16.0.254 > 4 172.16.1.1 - 172.16.1.62 > 5 172.16.1.65 - 172.16.1.126 > 6 172.16.1.129 - 172.16.1.190 > 7 172.16.1.193 - 172.16.1.254 > 8 172.16.2.1 - 172.16.2.62 > 9 172.16.2.65 - 172.16.2.126 >10 172.16.2.129 - 172.16.2.190 > . > . > . > . > . > . > > 1015 172.16.253.193 - 172.16.253.254 > 1016 172.16.254.1 - 172.16.254.62 > 1017 172.16.254.65 - 172.16.254.126 > 1018 172.16.254.129 - 172.16.254.190 > 1019 172.16.254.193 - 172.16.254.254 > 1020 172.16.255.1 - 172.16.255.62 > 1021 172.16.255.65 - 172.16.255.126 > 1022 172.16.255.129 - 172.16.255.190 > > -Joshua > -Original Message- > From: Lowell Sharrah [mailto:[EMAIL PROTECTED]] > Sent: Thursday, March 22, 2001 8:14 AM > To: [EMAIL PROTECTED] > Subject: RE: difficult ccna question > > > how do you know where the first subnet begins? > > >>> Joshua Beining <[EMAIL PROTECTED]> 03/22/01 10:34AM >>> > Just remember the following fomulas: > > (2^# of masked bits) - 2 = Total # of subnets > (2^# of unmasked bits - 2 = Total # of hosts > > Based on this the correct answer is A. > > -Original Message- > From: George [mailto:[EMAIL PROTECTED]] > Sent: Thursday, March 22, 2001 12:16 AM > To: [EMAIL PROTECTED] > Subject: difficult ccna question > > > if you have a class B network with a 10-bit subnet mask, how many subnet and > how many hosts do you have? > > a. 1022 subnets, 62 hosts > b. 62 subnets, 8190 hosts > c. 8190 subnets, 254 hosts > d. 254 subnets , 126 hosts > > > _ > FAQ, list archives, and subscription info: > http://www.groupstudy.com/list/cisco.html > Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] > > _ > FAQ, list archives, and subscription info: > http://www.groupstudy.com/list/cisco.html > Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] > > _ > FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html > Report misconduct and Nondisclosure violations to [EMAIL PROTECTED] > _ FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
