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    Hey I've been studying IPv4 recently and I need help with this question.

    What is the network ID address of the host with an IPv4 address
    192.168.50.100 with a subnet mask of 255.255.255.240?

    Any help is appreciated =)
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    (Original post by HasanKazmi)
    Hey I've been studying IPv4 recently and I need help with this question.

    What is the network ID address of the host with an IPv4 address
    192.168.50.100 with a subnet mask of 255.255.255.240?

    Any help is appreciated =)
    Have you figured out the answer? If not, have you looked into subnetting? If not, try looking into it, if you struggle even more I will take you through it step by step! Just quote me!
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    (Original post by Binary Freak)
    Have you figured out the answer? If not, have you looked into subnetting? If not, try looking into it, if you struggle even more I will take you through it step by step! Just quote me!
    Hii!! Thanks for the reply!

    I got an answer but I'm not sure if it is correct or not. Don't understand much about subnetting right now.

    The answer I got is 192.168.50.64

    I hope I didn't get it wrong. Though I'll be glad if you could take me through subnetting step by step. =)

    I appreciate your help! Thank you!
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    (Original post by HasanKazmi)
    Hii!! Thanks for the reply!

    I got an answer but I'm not sure if it is correct or not. Don't understand much about subnetting right now.

    The answer I got is 192.168.50.64

    I hope I didn't get it wrong. Though I'll be glad if you could take me through subnetting step by step. =)

    I appreciate your help! Thank you!
    Sorry about the delay in my response, I've been revising for upcoming exams.

    I haven't checked through the answer myself, but what you got is wrong - The /28 subnet (255.255.255.240) has 16 hosts per network; with only 14 usable (Not-Reserved).

    As for taking you through it step-by-step, I could do it tomorrow or on Monday definitely, if you want to learn more by then, then I'd suggest taking a read of Mastering Binary Math And Subnetting
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    (Original post by Binary Freak)
    Sorry about the delay in my response, I've been revising for upcoming exams.

    I haven't checked through the answer myself, but what you got is wrong - The /28 subnet (255.255.255.240) has 16 hosts per network; with only 14 usable (Not-Reserved).

    As for taking you through it step-by-step, I could do it tomorrow or on Monday definitely, if you want to learn more by then, then I'd suggest taking a read of Mastering Binary Math And Subnetting
    Hi,
    Thanks for replying, I'll read that few times =) And please do help me by taking me through it step by step so I can understand it better =). I'll really appreciate it.
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    (Original post by HasanKazmi)
    Hi,
    Thanks for replying, I'll read that few times =) And please do help me by taking me through it step by step so I can understand it better =). I'll really appreciate it.
    Yeah I'll go through it with you tomorrow then since I've got a crud load of revision done today.

    May I ask, how you got 192.168.50.64?
    This spoiler is the answer and calculation
    Spoiler:
    Show

    The answer to the question is actually 192.168.50.96/28 -
    ================================ ============================
    We know that subnet mask is 255.255.255.240, or /28 (Using CIDR (CIDR = Routing prefix/subnet bits) notation) = 11111111.11111111.11111111.11110 000 (Subnet in Binary)

    - From CIDR notation/binary conversion, we know that 28 of the 32 bits are used for the subnet mask, leaving 4 remaining for the host portion - 2^4 = 16 total addresses per subnet (Including the reserved Network ID & Broadcast Address), excluding these we have 14 usable hosts per subnet - Meaningless for this question.


    We know that the IP address is 192.168.50.100 = 11000000. 10101000. 00110010. 01100100
    We know the network ID is used to identify the first address of each subnet. So the host bits are set all 0, as shown below (The | identifies the host portion)

    Comparing the two
    11111111.11111111.11111111.1111| 0000 - Subnet Mask
    11000000.10101000.00110010.0110| 0100 - IP Address
    11000000.10101000.00110010.0110| 0000 - Network ID
    11000000.10101000.00110010.0110| 1111 - Broadcast Address

    1. Set the host portion of the address to all 0 (See Network ID)
    2. Convert the Network ID back to decimal -
    11000000. 10101000. 00110010. 01100000 = 192.168.50.96
    From what I've told you in the spoiler answer the following..

    What is the broadcast address for the address 192.168.21.117/26?
    If you need help, just ask (I hope my explaining wasn't/isn't too bad)
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    (Original post by Binary Freak)
    Yeah I'll go through it with you tomorrow then since I've got a crud load of revision done today.

    May I ask, how you got 192.168.50.64?
    This spoiler is the answer and calculation
    Spoiler:
    Show

    The answer to the question is actually 192.168.50.96/28 -
    ================================ ============================
    We know that subnet mask is 255.255.255.240, or /28 (Using CIDR (CIDR = Routing prefix/subnet bits) notation) = 11111111.11111111.11111111.11110 000 (Subnet in Binary)

    - From CIDR notation/binary conversion, we know that 28 of the 32 bits are used for the subnet mask, leaving 4 remaining for the host portion - 2^4 = 16 total addresses per subnet (Including the reserved Network ID & Broadcast Address), excluding these we have 14 usable hosts per subnet - Meaningless for this question.


    We know that the IP address is 192.168.50.100 = 11000000. 10101000. 00110010. 01100100
    We know the network ID is used to identify the first address of each subnet. So the host bits are set all 0, as shown below (The | identifies the host portion)

    Comparing the two
    11111111.11111111.11111111.1111| 0000 - Subnet Mask
    11000000.10101000.00110010.0110| 0100 - IP Address
    11000000.10101000.00110010.0110| 0000 - Network ID
    11000000.10101000.00110010.0110| 1111 - Broadcast Address

    1. Set the host portion of the address to all 0 (See Network ID)
    2. Convert the Network ID back to decimal -
    11000000. 10101000. 00110010. 01100000 = 192.168.50.96
    From what I've told you in the spoiler answer the following..

    What is the broadcast address for the address 192.168.21.117/26?
    If you need help, just ask (I hope my explaining wasn't/isn't too bad)
    I set the host portion wrong, from what I understood from the book I'm learning it,
    11|110000 - Subnet Mask
    01|100100 - IP Address
    01|100000 - Network ID

    Did exact as you did then, set the host post to zero's. And ended up getting 192.168.50.64.

    Broadcast address of 192.168.21.117/26?
    I wish I knew how to thank you lol, I found where I completely understood how to do this by attempting this question. I just did not understand how to identify the host portion. In the book it gave one example, very vague, just states to draw a vertical line.

    So, 26 bits of 32 are used for subnet mask, leaves 6. 2^6 = 64, from which 62 is usable.

    IP address 192.168.21.117 = 1100000. 10101000. 00010101. 01110101

    11111111.
    11111111. 11111111. 11|000000 = Subnet Mask
    11000000. 10101000. 00010101. 01|110101 = IP Address
    11000000. 10101000. 00010101. 01|000000 = Network ID
    11000000. 10101000. 00010101. 01|111111 = Broadcast ID

    Network ID = 192.168.21.64
    There are 62 usable address, so Last usable address will be 192.168.21.126
    Broadcast ID = 192.168.21.127

    I'm hoping I haven't made any mistake =) Let me know if I misunderstood something. Otherwise your explaining made so much sense to me! Good luck with your upcoming exams! Thank you so much for sparing some time to explain this to me. I appreciate it =)

    Sorry for my English explanation.
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    (Original post by HasanKazmi)
    I set the host portion wrong, from what I understood from the book I'm learning it,
    11|110000 - Subnet Mask
    01|100100 - IP Address
    01|100000 - Network ID

    Did exact as you did then, set the host post to zero's. And ended up getting 192.168.50.64.

    Broadcast address of 192.168.21.117/26?
    I wish I knew how to thank you lol, I found where I completely understood how to do this by attempting this question. I just did not understand how to identify the host portion. In the book it gave one example, very vague, just states to draw a vertical line.

    So, 26 bits of 32 are used for subnet mask, leaves 6. 2^6 = 64, from which 62 is usable.

    IP address 192.168.21.117 = 1100000. 10101000. 00010101. 01110101

    11111111.
    11111111. 11111111. 11|000000 = Subnet Mask
    11000000. 10101000. 00010101. 01|110101 = IP Address
    11000000. 10101000. 00010101. 01|000000 = Network ID
    11000000. 10101000. 00010101. 01|111111 = Broadcast ID

    Network ID = 192.168.21.64
    There are 62 usable address, so Last usable address will be 192.168.21.126
    Broadcast ID = 192.168.21.127

    I'm hoping I haven't made any mistake =) Let me know if I misunderstood something. Otherwise your explaining made so much sense to me! Thank you loads!!!

    Sorry for my English explanation.
    Ehh, that's a common mistake, I thought it was what you done wrong. Most books don't really make it obvious, some of them presume that the reader has a solid understanding of subnet masks which in some cases ends up screwing a few people over, I myself included. Just remember the vertical line stops when the subnet mask starts displaying 0.. 1 = Network/Subnet portion, and 0 = Host portion.

    As for the question I gave you.. You got the correct answer for that, the network ID is correct, as is the b'cast, and the usable hosts per subnet!

    Also, if you do any additional looking into this, be very careful, some books like to be very tricky, like sometimes a .255 host address can be used, that is when you've got an excess of 255 hosts per subnet..

    For example you can have 172.16.0.1/17, with that you'd have 32766 hosts per subnet, so you could have 172.16.28.255 as a usable host.
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    (Original post by Binary Freak)
    Ehh, that's a common mistake, I thought it was what you done wrong. Most books don't really make it obvious, some of them presume that the reader has a solid understanding of subnet masks which in some cases ends up screwing a few people over, I myself included. Just remember the vertical line stops when the subnet mask starts displaying 0.. 1 = Network/Subnet portion, and 0 = Host portion.

    As for the question I gave you.. You got the correct answer for that, the network ID is correct, as is the b'cast, and the usable hosts per subnet!

    Also, if you do any additional looking into this, be very careful, some books like to be very tricky, like sometimes a .255 host address can be used, that is when you've got an excess of 255 hosts per subnet..

    For example you can have 172.16.0.1/17, with that you'd have 32766 hosts per subnet, so you could have 172.16.28.255 as a usable host.
    Yep, The book didn't make clear at all, I was banging my head on it for some time lol.

    I'm not sure if I'll be looking more into this for the time, maybe in IPv6 as it is more complicated. Thanks for your help =)
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    (Original post by HasanKazmi)
    Yep, The book didn't make clear at all, I was banging my head on it for some time lol.

    I'm not sure if I'll be looking more into this for the time, maybe in IPv6 as it is more complicated. Thanks for your help =)
    Okay

    I'm not sure whether they've officially published IPv6 subnetting, but IPv6 does truncating

    Well if you need help in the future, then just quote me
 
 
 
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