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  1. Junior Member
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    #1

    Default How do I find valid host addresses with a subnet mask?

    I have a problem with the following question.

    Assuming a subnet mask of 255.255.224.0, which of the following would be valid host addresses?

    A. 124.78.103.0
    B. 125.67.32.0
    C. 125.78.160.0
    D. 126.78.48.0
    E. 176.55.96.0
    F. 186.211.100.0

    The answers are A,D and F. I dont know how to get the answers. Does anyone have an idea?
    Thanks
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  3. Member
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    #2
    The subnet mask is

    255.255.224.0 which I always write at

    255.255.11100000.00000000

    The first subnet will be the value of the last bit turned on which is 32 (corresponding to the 1 immediately to the left of 00000.00000000)

    So for an ip address of x.x.x.x the first subnet would be

    x.x.32.0, then
    x.x.64.0
    x.x.96.0
    x.x.128.0
    x.x.160.0
    x.x.192.0

    and so on

    A is a valid address in the x.x.96.0 subnet
    B is network ID, x.x.32.0 not a valid host
    C is network ID, x.x.160.0 not a valid host
    D is a valid address in the x.x.32.0 subnet
    B is network ID, x.x.96.0 not a valid host
    F is a valid address in the x.x.96.0 subnet

    So the answer is A,D and F
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  4. Johan Hiemstra Site Admin Webmaster's Avatar
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    #3
    Great explanation spacemancw.
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  5. Senior Member
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    #4
    lots of math in this question, i was wondering if i could get these question types on my 218 exam
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  6. Junior Member
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    #5
    Hi all,

    Heaps of subnet question on the exam!!

    Try this:

    256-224=32

    32+32=64 32+64=96 32+96=128 32+128=160
    32+160=192 32+192=224

    So your subnets are:
    0, 32, 64, 96, 128, 160, 192, 224

    Valid hosts are:
    1-30, 33-62, 65-94, 97-126, 161-190, 193-222

    Broadcast addresses:
    31, 63, 95, 127, 159, 191, 223

    The valid hosts will be 1 more than the sub and 2 less than the next sub
    eg. subnet 32+1=33 first valid host, next subnet 64-2=62 last valid host

    If you can count by 2's, 4's, 8's, 16's, 32's, 64's then good

    If like me you can only count by 1's don't panic just write out the 16X
    table and use it as a refference too easy.

    PS this is how you spell colour.
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  7. Junior Member
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    #6

    Default helpppppppppppp

    ip address 172.18.14.17/20
    subnet 255.255.248.0 not sureeeeeeeee

    i need to find the following

    the address block
    no. no valid networks
    no. of hosts on each network
    1st valid host on each network
    last valid host on each network
    broadcast address on each network
    ip addy above belongs to which network

    got the following but i dont know if right

    256-248=16 addy block
    networks 30
    host 2046

    help pleaseeeeeeeeeee.....
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  8. Senior Member
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    #7
    I'll help you out, just because I've never seen anyone resurrect a 5.5 year old thread before

    You find the network by ANDing the bits of the subnet mask and ip address

    10101100.00010010.00001111.00010001 (172.18.15.17)
    11111111.11111111.11111000.00000000 (255.255.248.0)
    ==============================
    10101100.00010010.00000000.00000000

    So the network address is 172.168.0.0

    You have 12 bits for the hosts, so 2^12-2 = 4094

    You have 4 bits for the network, so 2^4 = 16

    If you want to find the broadcast address, put 1s in for all your host bits:
    10101100.00010010.00001111.11111111 = 172.168.15.255

    For the next subnet, increment your subnet bits by 1.
    10101100.00010010.00010000.00000000 = 172.168.16.0

    And repeat...
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  9. One Man Wolfpac NetAdmin2436's Avatar
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    #8
    Quote Originally Posted by dynamik
    I'll help you out, just because I've never seen anyone resurrect a 5.5 year old thread before

    You find the network by ANDing the bits of the subnet mask and ip address

    10101100.00010010.00001111.00010001 (172.18.15.17)
    11111111.11111111.11111000.00000000 (255.255.248.0)
    ==============================
    10101100.00010010.00000000.00000000

    So the network address is 172.168.0.0

    You have 12 bits for the hosts, so 2^12-2 = 4094

    You have 4 bits for the network, so 2^4 = 16

    If you want to find the broadcast address, put 1s in for all your host bits:
    10101100.00010010.00001111.11111111 = 172.168.15.255

    For the next subnet, increment your subnet bits by 1.
    10101100.00010010.00010000.00000000 = 172.168.16.0

    And repeat...
    Actually, with Sabby's example....wouldn't it be like this?....or am I retarded?

    10101100.00010010.00001110.00010001 (172.18.14.17)
    11111111.11111111.11111000.00000000 (255.255.248.0)
    ==============================
    10101100.00010010.00000000.00000000

    So the network address is 172.18.0.0

    You have 12 bits for the hosts, so 2^12-2 = 4094

    You have 4 bits for the network, so 2^4 = 16

    If you want to find the broadcast address, put 1s in for all your host bits:
    10101100.00010010.00001111.11111111 = 172.18.15.255

    For the next subnet, increment your subnet bits by 1.
    10101100.00010010.00010000.00000000 = 172.18.16.0

    Sabby,
    This comes in handy to double check your work.
    http://www.techexams.net/ip-subnet-calculator/
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  10. Senior Member
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    #9
    Whoops. I did the binary right, but yea, 18 not 168. That one IP was off too, but fortunately it didn't matter in the grand scheme of things. Remember kids, don't drink and subnet
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  11. Senior Member ConstantlyLearning's Avatar
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    #10
    Quote Originally Posted by dynamik
    Whoops. I did the binary right, but yea, 18 not 168. That one IP was off too, but fortunately it didn't matter in the grand scheme of things. Remember kids, don't drink and subnet
    crazy talk, nothing better than knockin back a few cans and hitting up subnettingquestions.com.
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  12. Member
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    #11

    Default Re: How do I find valid host addresses with a subnet mask?

    Quote Originally Posted by rc240sx
    I have a problem with the following question.

    Assuming a subnet mask of 255.255.224.0, which of the following would be valid host addresses?

    A. 124.78.103.0
    B. 125.67.32.0
    C. 125.78.160.0
    D. 126.78.48.0
    E. 176.55.96.0
    F. 186.211.100.0

    The answers are A,D and F. I dont know how to get the answers. Does anyone have an idea?
    Thanks
    According to me none of the options are correct because Question is to find valid HOST addresses(which is not in the option).A, D and F are valid Network addresses...

    Question should be "which of the following would be valid NETWORK addresses?"


    Well,If I am wrong because of any misconception, Do correct me.
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  13. Senior Member ConstantlyLearning's Avatar
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    #12

    Default Re: How do I find valid host addresses with a subnet mask?

    Quote Originally Posted by *StarFire
    Quote Originally Posted by rc240sx
    I have a problem with the following question.

    Assuming a subnet mask of 255.255.224.0, which of the following would be valid host addresses?

    A. 124.78.103.0
    B. 125.67.32.0
    C. 125.78.160.0
    D. 126.78.48.0
    E. 176.55.96.0
    F. 186.211.100.0

    The answers are A,D and F. I dont know how to get the answers. Does anyone have an idea?
    Thanks
    According to me none of the options are correct because Question is to find valid HOST addresses(which is not in the option).A, D and F are valid Network addresses...

    Question should be "which of the following would be valid NETWORK addresses?"


    Well,If I am wrong because of any misconception, Do correct me.
    You're wrong...and dude, look at the date of that post you've quoted.


    Take option A. as an example. (124.78.103.0)

    The network address is 124.78.96.0
    Broadcast address is 124.78.127.255
    Host range is 124.78.96.1 - 124.78.127.254

    124.78.103.0 falls within the host range.
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  14. Senior Member
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    #13
    Just because there is a ZERO in the last OCTET doesn't automatically make it a Network Address. Especially with Classless and large Network Addressing Class A and B!

    The valid Host range for A D and F are:

    A) 124.78.96.1 - 124.78.127.254 with a Network address of 124.78.96.0 and broadcast of 124.78.127.255.

    D) 126.78.32.1 - 126.78.63.254, 124.78.32.0 Network, 124.78.63.255 broadcast.

    F) 186.211.96.1 - 186.211.127.254, 186.211.96.0 Network, 186.211.127.255 broadcast.
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  15. Member
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    #14
    Thanks Friendss ..........
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  16. Member
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    #15
    I dont care how old this is, its STILL GREAT, and always valid!

    It took me a while to get it, but I get it now. It is a great explanation.
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  17. Junior Member Registered Member
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    #16

    Default I want to re-OPEN this thread. WAY 2 COOL!

    Quote Originally Posted by amtt81 View Post
    I dont care how old this is, its STILL GREAT, and always valid!

    It took me a while to get it, but I get it now. It is a great explanation.
    I want to re-OPEN this thread. WAY 2 COOL!
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  18. Senior Member
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    #17
    Haha awesome first post to open a thread that is 10 years old!
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  19. Junior Member Registered Member
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    #18
    Now i am the one who opened this 2 year old thread! lol
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  20. Junior Member Registered Member
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    #19

    Default Which two addresses are available host addresses on the subnet 192.168.73.46/27?

    Which two addresses are available host addresses on the subnet 192.168.73.46/27?
    • 192.168.73.13 and 192.168.73.37
    • 192.168.73.37 and 192.168.73.55
    • 192.168.73.62 and 192.168.73.13
    • 192.168.73.55 and 192.168.73.37
    • None of the above

    Can someone help me out by solving this?
    The answer is 192.168.73.37 and 192.168.73.55
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  21. Senior Member
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    #20
    It's a poorly-written question. You won't find anything unclear like this on the actual exam.

    First, "192.168.73.37 and 192.168.73.55" and "192.168.73.55 and 192.168.73.37" contain identical host addresses. One should not be asked to choose between them, as they are both equally correct or incorrect.

    Second, "192.168.73.46/27" is not a subnet ID.

    I leave the math to you. See above, or another popular thread on subnetting technique--

    Subnetting Made Easy
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  22. BMG FTW!!
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    #21
    range of network: 192.168.73.32 - 192.168.73.63

    and the options are wrong. 2nd & 4th options are same, and I never saw any options in that format.
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  23. Junior Member
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    #22
    Hi everybody!

    Can't believe it has passed 10 years already since somebody started this post. It made me wonder what I was doing back than. Anyways need some help/explanation to a newbie with next question, please.

    What is he last valid host on the subnetwork 172.16.242.0 255.255.254.0?According to the subnettingquestions.com answer should be: 172.16.243.254

    Here is what I know for now.

    172.16.242.0 255.255.254.0
    1 bit left in the 3rd octet which equals 1 when converted from binary. So 242 + 1 = 243 for the 3rd octet.
    and we have 8 bits left in the 4th octet which is 255 - 1 for the last host.

    Just want to make sure I'm doing it right and maybe there is some formula or anything else I should know?

    Last edited by Elias38; 07-12-2013 at 06:22 AM.
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    #23
    Quote Originally Posted by Elias38 View Post
    It made me wonder what I was doing back than.
    I was in college (Summer 2003) and taking "Introduction to C++" and "Computer Imaging I" that semester.
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    #24
    Quote Originally Posted by Elias38 View Post
    Hi everybody!

    Can't believe it has passed 10 years already since somebody started this post. It made me wonder what I was doing back than. Anyways need some help/explanation to a newbie with next question, please.

    What is he last valid host on the subnetwork 172.16.242.0 255.255.254.0?According to the subnettingquestions.com answer should be: 172.16.243.254

    Here is what I know for now.

    172.16.242.0 255.255.254.0
    1 bit left in the 3rd octet which equals 1 when converted from binary. So 242 + 1 = 243 for the 3rd octet.
    and we have 8 bits left in the 4th octet which is 255 - 1 for the last host.

    Just want to make sure I'm doing it right and maybe there is some formula or anything else I should know?

    Quote Originally Posted by Elias38 View Post
    Hi everybody!

    Can't believe it has passed 10 years already since somebody started this post. It made me wonder what I was doing back than. Anyways need some help/explanation to a newbie with next question, please.

    What is he last valid host on the subnetwork 172.16.242.0 255.255.254.0?According to the subnettingquestions.com answer should be: 172.16.243.254

    Here is what I know for now.

    172.16.242.0 255.255.254.0
    1 bit left in the 3rd octet which equals 1 when converted from binary. So 242 + 1 = 243 for the 3rd octet.
    and we have 8 bits left in the 4th octet which is 255 - 1 for the last host.

    Just want to make sure I'm doing it right and maybe there is some formula or anything else I should know?


    That's not a good way to do it. If for instance, if it was a /20, then are you going to add 16 to 242? It would be wrong. The way you are thinking makes the number of the octet you are working on, the network base every time. But it isn't always the base.

    Use the magic bit technique.

    in each octet there are 8 bits. Each 1 reps a number. If 0 it doesn't equal anything.

    1 1 1 1 1 1 1 1
    128 64 32 16 8 4 2 1

    In your case the 24th place is in the OFF position. It's a Zero. The last active bit is the 23rd place, which is a 2.

    11111110 = 254 (11111111.11111111.11111110.00000000 is 255.255.254.0)

    Right? Because 128+64+32+16+8+4+2 = 254 . The last "1" is missing, since it's a 0

    So, you look at the last active bit and see what it's value is. Since the value is 2, you then know you can have your network by 2's.

    0,2,4,6,8,10....all the way up to 242. SO, your answer is right. It is 172.16.242.0 - 172.16.243.255.

    But, if you thought for instance that the Network base was 172.16.241.0, it would be wrong since counting by twos means the network base would be even. In fact, just a tip..the base is always even and the broadcast always odd. Usable hosts can be either though.
    You can use the MAGIC bit and count from zero up until you run into your network address you are trying to figure out.

    Another example.

    172.16.240.0 /19
    That is a submask of:
    255.255.224.0

    ok. Count the bits on the subnet mask..there are 19 right?

    11111111.11111111.11100000.00000000

    Find the last one which is the magic bit and see it's value. It is 32.

    Count 32 from 0 and find your address in the range.

    0-32-64-96-128-160-192-224-256

    So you can see your address (240) is in between 224 and 256.

    So, network base is 172.16.224.0 - 172.16.255.255

    The magic bit allows you to count the networks inside of the octet.

    One more.

    172.16.224.0 /17

    mask= 255.255.128.0

    11111111.11111111.10000000.0000000

    What is the value of the last 1? 128

    So count from zero. 0 - 128 -256 So Network base is

    172.16.0.0 - 172.16.127.255

    And

    172.16.128.0 - 172.16.255.255 (the 240 would be inside of this range)


    Hope this helps someone.
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