Help answer this question.
rphann
Member Posts: 76 ■■□□□□□□□□
This question is from the Microsoft Training Kit.
For each address block and internally configured subnet mask, determine the number of available subnets and hosts per subnet.
Original Address Block: 208.147.66.0 /25
Internal Configured Subnet Mask: 255.255.255.248
Number of Available Subnets:
Number of Available Hosts per Subnet:
My answers are 16 for Number of Available Subnets and 6 for Number of Available Hosts per Subnet. Although the answers shown in the Training Kit, shows 8 for Number of Available Subnets and 6 for Number of Available Hosts per Subnet. Please let me know which one is the correct answer.
Thanks,
rphann
For each address block and internally configured subnet mask, determine the number of available subnets and hosts per subnet.
Original Address Block: 208.147.66.0 /25
Internal Configured Subnet Mask: 255.255.255.248
Number of Available Subnets:
Number of Available Hosts per Subnet:
My answers are 16 for Number of Available Subnets and 6 for Number of Available Hosts per Subnet. Although the answers shown in the Training Kit, shows 8 for Number of Available Subnets and 6 for Number of Available Hosts per Subnet. Please let me know which one is the correct answer.
Thanks,
rphann
Comments
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mikedisd2 Member Posts: 1,096 ■■■■■□□□□□If it's reserved 5x bits for the subnet, I would have thought that gave 32x subnets with 2^8-2 hosts.
Subnetting isn't my strong point though. -
breaker Member Posts: 21 ■□□□□□□□□□Edit
http://support.microsoft.com/kb/917983
Page 2-66: Number of available subnets incorrect
On page 2-66, in Lesson 3, Practice, Exercise 2 answer block, the last line reads:
"208.147.66.0/25 255.255.255.248 8 6"
It should read:
"208.147.66.0/25 255.255.255.248 16 6" -
breaker Member Posts: 21 ■□□□□□□□□□If it's reserved 5x bits for the subnet, I would have thought that gave 32x subnets with 2^8-2 hosts.
Subnetting isn't my strong point though.
If the mask is /25 then it leaves you with 7 bits
the new internal mask of 248 is /29
so your are using 4 for network and 3 for host
2^4 = 16
and
2^3-2 = 6 -
rphann Member Posts: 76 ■■□□□□□□□□Thanks, Breaker.
So my answer is correct. Thanks for clarifying this for me.
rphann