Subnetting/Masking Question
tmharkins
Member Posts: 2 ■□□□□□□□□□
in CCNA & CCENT
I had these two questions on a practice test, but I don't understand how they came to the answers. Any explanations would be greatly appreciated.
1. Which mask, when used on a Class C address will yield 5 subnets w/ 18 hosts on each?
A. 255.255.240.0
B. 255.255.255.248
C. 255.255.255.224 This was listed as correct answer.
D. 255.255.224.0
E. 255.255.255.240
2. What are the first and last usable host IP addresses on network 192.168.1.24/29?
A. 192.168.1.25 and 192.168.1.31
B. 192.168.1.17 and 192.168.1.30 This was listed as correct answer.
C. 192.168.1.24 and 192.168.1.254
D. 192.168.1.1 and 192.168.1.254
E. 192.168.1.17 and 192.168.1.32
1. Which mask, when used on a Class C address will yield 5 subnets w/ 18 hosts on each?
A. 255.255.240.0
B. 255.255.255.248
C. 255.255.255.224 This was listed as correct answer.
D. 255.255.224.0
E. 255.255.255.240
2. What are the first and last usable host IP addresses on network 192.168.1.24/29?
A. 192.168.1.25 and 192.168.1.31
B. 192.168.1.17 and 192.168.1.30 This was listed as correct answer.
C. 192.168.1.24 and 192.168.1.254
D. 192.168.1.1 and 192.168.1.254
E. 192.168.1.17 and 192.168.1.32
Comments

dynamik Banned Posts: 12,312 ■■■■■■■■■□224 = 3 bits for the subnet mask. 2^3 = 8 subnets. You have five bits remaining for hosts, so 2^52 = 30 hosts. No other combination will satisfy the requirements.
/29 will leave you three bits for the hosts, so your networks are going to increment by 8 (2^3) each time. Therefore your networks are:
192.168.1.07
192.168.1.815
192.168.1.1623
192.168.1.2431
192.168.1.3239
etc.
So, um, either it's too late for me to be subnetting, or the question is messed up. I'd say the answer is 25 and 30. Maybe the info should have been 192.168.1.16/28; that would have made B the correct answer. Check the errata. 
Jason0352 Member Posts: 59 ■■□□□□□□□□A CIDR of /29 = a 240 mask with the smallest bit of 16 being the increment.
So start at 0 then increment by 16 each time to get your network IDs
192.168.1.0
192.168.1.16
192.168.1.32
You can stop now because your node address of 192.168.1.24 has to be between 192.168.1.16 and .32. You know that network IDs cant be used for hosts so 192.168.1.17 is the 1st host and 192.168.1.30 is the last host. Broadcast address is always the address before the next subnet in the increment. IE: 192.168.1.31 = broadcast 
captobvious Member Posts: 648A CIDR of /29 = a 240 mask with the smallest bit of 16 being the increment.
So start at 0 then increment by 16 each time to get your network IDs
192.168.1.0
192.168.1.16
192.168.1.32
You can stop now because your node address of 192.168.1.24 has to be between 192.168.1.16 and .32. You know that network IDs cant be used for hosts so 192.168.1.17 is the 1st host and 192.168.1.30 is the last host. Broadcast address is always the address before the next subnet in the increment. IE: 192.168.1.31 = broadcast 
tmharkins Member Posts: 2 ■□□□□□□□□□Thanks for the help guys. I thought there was a problem somewhere!

WillTech105 Member Posts: 216Since we're on the topic of subnetting i have a very BASIC question.
I understand how to subnet, find hosts, subnets, ranges, ect. Theres one SMALL easy math concept that I simply cant get (BTW  math was my worst subject in HS!
When finding the # of networks or hosts, usually you'll take the 128 64 32, ect. and then find the bit necessary to let you gain how many networks/hosts you'd like.
Example
170.50.0.0/16 client wants 1000 hosts.
The answer was listed as 10 bits (1024) but why cant it be 9 or 8 bits?
1000
1024
24
Thats a negative number so wouldn't the correct answer be 11 bits?
Another example is
199.9.10.0 client wants 12 hosts
This is simple...4 bits since
12
8

4 (not a negative number)
So i can see the math there works out but not for the other answer above. And no I didnt conclude 4 bits for the above because it 128 happens to be 4. I am saying 10001024 would result in a negative number so wouldnt it be 11 bits? Did the Dec > Bin calculator and it did give me 10 bits but I dont see the math logic if 10001024 gives you a negative number.
EDIT: I think I solved my problem  i was going about it the wrong way. Its not a matter of subtraction but a matter of "1024 is higher than 1000 and 512 isnt so the answer is 10 bits". Anyone please let me know if this is the "right" way of going about this  thanks!In Progress: CCNP ROUTE 
dynamik Banned Posts: 12,312 ■■■■■■■■■□WillTech105 wrote: »Since we're on the topic of subnetting i have a very BASIC question.
I understand how to subnet, find hosts, subnets, ranges, ect. Theres one SMALL easy math concept that I simply cant get (BTW  math was my worst subject in HS!
When finding the # of networks or hosts, usually you'll take the 128 64 32, ect. and then find the bit necessary to let you gain how many networks/hosts you'd like.
Example
170.50.0.0/16 client wants 1000 hosts.
The answer was listed as 10 bits (1024) but why cant it be 9 or 8 bits?
1000
1024
24
Thats a negative number so wouldn't the correct answer be 11 bits?
You'd have 24 (22 if they're asking about usable addresses), not 24. You need to subtract 1000 from 1024; you have it backwards.
Also, 9 bits will only give you 512 (510), and 8 will only give you 256 (254). I'm not sure why you brought those up since it seems like you have a good enough handle on things to know that.WillTech105 wrote: »Another example is
199.9.10.0 client wants 12 hosts
This is simple...4 bits since
12
8

4 (not a negative number)
So i can see the math there works out but not for the other answer above. I know it is very basic but if anyone could shed some light on this I'd appreciate it!
EDIT: And no I didnt conclude 4 bits for the above because it 128 happens to be 4. I am saying 10001024 would result in a negative number so wouldnt it be 11 bits? Did the Dec > Bin calculator and it did give me 10 bits but I dont see the math logic if 10001024 gives you a negative number.
Where did you get 8 from? I'm not sure what this subtraction strategy is for.
3 bits will give you 8 (6) hosts, and 4 will give you 16 (14), so 4 is the obvious choice. There's really no need for any other math. 
WillTech105 Member Posts: 216^
Yeah I think I was going about it the wrong way  I dont need to subtract to find usable hosts/subnets. I was probably bringing in Dec > Bin conversion which threw me off.
ThanksIn Progress: CCNP ROUTE 
miller811 Member Posts: 897WillTech105 wrote: »^
Yeah I think I was going about it the wrong way  I dont need to subtract to find usable hosts/subnets. I was probably bringing in Dec > Bin conversion which threw me off.
Thanks
Please take a look at this subnetting tutorial, if you have not already.
http://www.techexams.net/forums/ccnaccent/38772subnettingmadeeasy.htmlI don't claim to be an expert, but I sure would like to become one someday.
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