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identify the unused portion of the subnet by responding to the question on the graphic.

HOTSPOT
Corporate headquarters provided your office a portion of their class B subnet to use at a new office location. Allocate the minimum number of addresses (using
CIDR notation) needed to accommodate each department.

After accommodating each department, identify the unused portion of the subnet by responding to the question on the graphic. All drop downs must be filled.
Instructions: When the simulation is complete, please select the Done button to submit.
All Networks have the range from /0 to/32
Hot Area:

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Answer:

Explanation:
An IPv4 address consists of 32 bits. The first x number of bits in the address is the network address and the remaining bits are used for the host addresses. The
subnet mask defines how many bits form the network address and from that, we can calculate how many bits are used for the host addresses.
The formula to calculate the number of hosts in a subnet is 2n – 2. The “n” in the host’s formula represents the number of bits used for host addressing. If we apply
the formula (22 – 2), we can determine that the following subnets should be configured:
Sales network – /26 – This will provide up to 62 usable IP addresses (64-2 for subnet and broadcast IP)
HR network – /27 – This will provide for up to 30 usable IP’s (32-2)
IT – /28 – This will provide for up to 14 usable IP’s (16-2)
Finance – /26 – Note that a /27 is 32 IP addresses but 2 of those are reserved for the network and broadcast IP’s and can’t be used for hosts.
Marketing – /28
If we add up how many IP blocks are used that is 64+32+16+64+16=192.
A /24 contains 256 IP addresses, so 256-192=64.
So the last unused box should be a /26, which equates to 64 addresses

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3 Comments on “identify the unused portion of the subnet by responding to the question on the graphic.

  1. Christopher Schaffner says:

    The question at no time indicates if you can or cannot use the same / designation twice. In regards to Marketing I knew it had to be a /28 but since it was already used I didn’t think I could use it again. I get that they want us to think around corners on this stuff but this isn’t one of those questions where its needed to be vague and have us just figure it out. Something like this I should think would need clearer verbiage to clarify the point where we are able to use a designation more than once.




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    1. christy says:

      They honestly have given more than enough information to answer this
      correctly, You need to understand Variable Length Subnet Masking to
      do this.

      You start with a 172.30.232.0/24

      So you only have 256 host bits (8 bits in the last octet; 2^8=256)
      since its a /24 they give you to subnet

      From the Question you need

      1 subnet with 57 hosts
      1 subnet with 32 hosts
      1 subnet with 23 hosts
      1 subnet with 12 hosts
      1 subnet with 9 hosts

      To fit 57 hosts we need a block size of at least 64
      To fit 32 hosts we need a block size of at least 64
      To fit 23 hosts we need a block size of at least 32
      To fit 12 hosts we need a block size of at least 16
      To fit 9 hosts we need a block size of at least 16

      Since we are given 172.30.232.0/24 our new mask for the first subnet will be a /26 to fit 64 hosts (minus 1 for the network and minus another 1 for the broadcast. In total 62 hosts. So we will start with the address we are given with just a /26 mask

      So our first subnet will be 172.30.232.0/26 for Sales
      Our second subnet will be 172.30.232.64/26 for Finance
      Our third subnet will be 172.30.232.128/27 for HR
      Our fourth subnet will be 172.30.232.160/28 for IT
      Our fifth subnet will be 172.30.232.176/28 for Marketing

      So our first subnet will have host space from 172.30.232.1 to
      172.30.232.63 (note 63 will be the broadcast so 172.30.232.62 will
      be the last usable host in that subnet)for the 172.30.232.0 Network

      And our second subnet will have host space from 172.30.232.65 to
      172.30.232.127 (note 127 will be the broadcast so 172.30.232.126 will
      be the last usable host in that subnet) for the 172.30.232.64 Network

      This is why our 3rd subnet has to start at .128 as if you had any
      less you wouldn’t have the 64 hosts spaces for the second subnet

      So our third subnet will have host space from 172.30.232.129 to
      172.30.232.159 (note 159 will be the broadcast so 172.30.232.158 will
      be the last usable host in that subnet) for the 172.30.232.128 Network

      Since our 3rd subnet needs only 32 addresses the next subnet will
      be .128 + 32 = 160 so it will start at .160

      Our fourth subnet will have host space from 172.30.232.161 to
      172.30.232.175 (note 175 will be the broadcast so 172.30.232.174 will
      be the last usable host in that subnet) for the 172.30.232.160 Network

      Since our fourth subnet only needs 16 addresses the next subnet will
      be .160 + 16 = 176 so it will start at .176

      So finally our fifth subnet will have host space from 172.30.232.177 to
      172.30.232.191 (note 191 will be the broadcast so 172.30.232.190 will
      be the last usable host in that subnet)for the 172.30.232.176 Network

      Since our fifth subnet only needs 16 addresses the next subnet would
      start at .176 + 16 = 192 so it would start at .192 but we dont need any more subnets, so since we had 256 to work with; 256 minus the 192 that
      we used would leave us with 64 address spaces left (256 – 192 = 64)
      And a 64 block size would require a /26 mask just like our first 2 subnets

      Its really helpful to right out the mask values with their block
      sizes and how many bits you have used

      Bits used 1 2 3 4 5 6 7 8
      Mask Values 128 192 224 240 248 252 254 255
      Block Sizes 128 64 32 16 8 4 2 1

      for a /24 you deal with the last octet so have from /25 to /32
      for practical purposes you wont use /31 and /32 as they only leave
      1 host bit for /31 and 0 host bits for /32

      Mask Values 128 192 224 240 248 252 254 255
      block sizes 128 64 32 16 8 4 2 1
      CIDR /25 /26 /27 /28 /29 /30 /31 /32

      It is easy to remember this as from right to left your block sizes
      will be multiples of each other: 2,4,8,16,32,64,128

      And the mask values from left to right will start at 128 and then you
      just add the block size for the next one

      128+64=192
      192+32=224
      224+16=240
      240+8=248
      248+4=252
      252+2=254
      254+1=255

      for a /16 you deal with the 3rd octet so you have from /17 to /24

      Mask Values 128 192 224 240 248 252 254 255
      block sizes 32768 16384 8192 4096 2048 1024 512 256
      CIDR /17 /18 /19 /20 /21 /22 /23 /24

      Note the block sizes are multiples of 2 so from the block size of 128
      its 128×2= 256, 256×2=512, 512×2=1024 and so on. I wont write out anymore but to go beyond the /17 to a /16 the new block size would obviously be 32768 x 2 = 65536. this is why a /16 has 65536 address spaces (65534 usable hosts; remember 1 for the network itself and another for the broadcast address) and so on




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  2. christy says:

    The empty space formatting didn’t work right after I posted that so below is what it should look like in columns.

    Bits used___ 1___2___3___4___5___6___7___8
    Mask Values 128 192 224 240 248 252 254 255
    block sizes 128__64__32__16__8___4___2___1

    for a /24 you deal with the last octet so you have from /25 to /32
    for practical purposes you wont use /31 and /32 as they only leave
    1 host bit for /31 and 0 host bits for /32

    Mask Values_ 128 192 224 240 248 252 254 255
    block sizes 128 64 _32 __16 __8 __4 __2 __1
    CIDR ______ /25 /26 /27 /28 /29 /30 /31 /32

    It is easy to remember this as from right to left your block sizes
    will be multiples of 2: 1,2,4,8,16,32,64,128

    And the mask values from left to right will start at 128 and then you
    just add the block size for the next one

    128+64=192
    192+32=224
    224+16=240
    240+8=248
    248+4=252
    252+2=254
    254+1=255

    for a /16 you deal with the 3rd octet so you have from /17 to /24

    Mask Values__ 128 __192 __224 _240 _248 _252 254 255
    block sizes 32768 16384 8192 4096 2048 1024 512 256
    CIDR________ /17__ /18 _/19 __/20 /21 _/22 _/23 /24

    Note the block sizes are multiples of 2 so from the block size of 128
    its 128×2= 256, 256×2=512, 512×2=1024 and so on. I wont write out anymore but to go beyond the /17 to a /16 the new block size would obviously be 32768 x 2 = 65536. this is why a /16 has 65536 address spaces (65534 usable hosts; remember 1 for the network itself and another for the broadcast address) and so on




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