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70-291 ip addressesAccording to the microsoft testing software, the following ip addresses using
VLSM are invalid. Im still trying to understand why they would be invalid. Can anyone offer a quick explanation, or point me to a web site explaining this. invalid class c ip addresses 199.205.15.8/27 199.205.15.64/25 199.205.15.130/26 |

Re: 70-291 ip addresses"Larry" <Larry@discussions.microsoft.com> wrote in message news:C95BE911-BA7F-44CF-B4BF-7C4111B71062@microsoft.com... > According to the microsoft testing software, the following ip addresses > using > VLSM are > invalid. > Im still trying to understand why they would be invalid. Can anyone offer > a > quick explanation, or point me to a web site explaining this. > > invalid class c ip addresses > 199.205.15.8/27 > 199.205.15.64/25 > 199.205.15.130/26 199.205.15.8/27 - This is a valid IP address, it is NOT a valid network address. Think of it this way: every time you add a bit (1) to the subnet mask, you cut it in half. So a /24 (mask 255.255.255.0) has 256 addresses the networks are 192.168.0.0/24, 192.168.1.0/24, 192.168.2.0/24, etc. If you add a bit, making the mask /25, your networks are on boundaries of 128 so your network addresses are 192.168.0.0/25, 192.168.0.128/25, 192.168.1.0/25, 192.168.1.128/25, etc. a /26 would be at boundaries of 64 and a /27 at 32. So valid network addresses for a /26 are 192.168.0.0/26, 192.168.0.64/26, 192.168.0.128/26, etc. Valid networks with a /27 mask are 192.168.0.0/27, 192.168.0.32/17, 192.168.0.64/27, etc. A lot of times, it's not the subnetting that gets you, it's understanding binary arithmetic. Just as in base 10, when we get to the last digit (9) in a column and add one more to it, we zero the column and carry to the next column to the left. So the opposite is shifting to the right. In base 10, when you shift everything one column to the right, it's the same as dividing by 10. In binary, shifting one column to the right is the same as dividing by 2 (0100 binary = 4 decimal, if you shift right one - 0010 binary = 2 decimal or 4 divided by 2). Since the subnet mask defines the number of host bits, every time you add another "1" to the right, you cut the number of hosts in half, and each time you reach the number of hosts and add one more, the hosts bits are all zero'd and the network is incremented. IP address 10.10.10.255/24 - in binary 00001010 00001010 00001010 11111111 S/N mask 255.555.255.0 11111111 11111111 11111111 00000000 so if you add one to the IP address you get 00001010 00001010 00001010 11111111 + 00000000 00000000 00000000 00000001 ------------------------------------------------- = 00001010 00001010 00001011 00000000 in decimal this is 10.10.11.0, the next network. ....kurt |

Re: 70-291 ip addressesThanks Kurt,
Now i understand, the 4 octet of the network address ( i was confusing the term ip address and network address) with a /25 mask can only be 00000000 or 10000000 ie 0 or 128 "Kurt" wrote: > > "Larry" <Larry@discussions.microsoft.com> wrote in message > news:C95BE911-BA7F-44CF-B4BF-7C4111B71062@microsoft.com... > > According to the microsoft testing software, the following ip addresses > > using > > VLSM are > > invalid. > > Im still trying to understand why they would be invalid. Can anyone offer > > a > > quick explanation, or point me to a web site explaining this. > > > > invalid class c ip addresses > > 199.205.15.8/27 > > 199.205.15.64/25 > > 199.205.15.130/26 > > 199.205.15.8/27 - This is a valid IP address, it is NOT a valid network > address. Think of it this way: every time you add a bit (1) to the subnet > mask, you cut it in half. So a /24 (mask 255.255.255.0) has 256 addresses > the networks are 192.168.0.0/24, 192.168.1.0/24, 192.168.2.0/24, etc. > > If you add a bit, making the mask /25, your networks are on boundaries of > 128 so your network addresses are 192.168.0.0/25, 192.168.0.128/25, > 192.168.1.0/25, 192.168.1.128/25, etc. > > a /26 would be at boundaries of 64 and a /27 at 32. > > So valid network addresses for a /26 are 192.168.0.0/26, 192.168.0.64/26, > 192.168.0.128/26, etc. > > Valid networks with a /27 mask are 192.168.0.0/27, 192.168.0.32/17, > 192.168.0.64/27, etc. > > A lot of times, it's not the subnetting that gets you, it's understanding > binary arithmetic. Just as in base 10, when we get to the last digit (9) in > a column and add one more to it, we zero the column and carry to the next > column to the left. So the opposite is shifting to the right. In base 10, > when you shift everything one column to the right, it's the same as dividing > by 10. In binary, shifting one column to the right is the same as dividing > by 2 (0100 binary = 4 decimal, if you shift right one - 0010 binary = 2 > decimal or 4 divided by 2). Since the subnet mask defines the number of host > bits, every time you add another "1" to the right, you cut the number of > hosts in half, and each time you reach the number of hosts and add one more, > the hosts bits are all zero'd and the network is incremented. > > IP address 10.10.10.255/24 - in binary 00001010 00001010 00001010 11111111 > S/N mask 255.555.255.0 11111111 11111111 11111111 > 00000000 > > so if you add one to the IP address you get > 00001010 00001010 00001010 11111111 > + 00000000 00000000 00000000 00000001 > ------------------------------------------------- > = 00001010 00001010 00001011 00000000 > > in decimal this is 10.10.11.0, the next network. > > ....kurt > > > |

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