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.
Alter 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 form /0 to/32
Answer: 
Explanation:
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
Which of the following would represent the largest possible contagious block of remaining addresses?
I don’t know if your answer to this is correct or not?
Because if its asking about the next network which has the largest number of hosts possible then it should be /29 which would have 32 subnets with 8 hosts per subnet?
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I’m looking at this as marketing /28, IT /28, HR /27, Finance /26, sales /26. The most contiguous block remaining is /26 with 30 hosts unused from the finance dept using only 30 hosts from a /26 subnet leaves 30 usable ip’s left.
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I’m thinking this is the right answer: The directions at the top say, “…Identify the UNUSED portion of the SUBNET by responding to the question on the graphic”. The question on the graphic says, “Which of these gives you the LARGEST contiguous block of remaining addresses?” You put those two statements together and you have, “Given the IP address and the # of hosts required for each department, which department had the most host addresses remaining?” The answer is Finance which required a /26 to allow for 32 hosts. A /26 provides 64 host ID’s, 2 of which are unusable, leaving you with 62 valid host IP addresses. The largest portion (contiguous block) of UNUSED (remaining) IP addresses on a given subnet is 62 – 32 = 30.
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no when you use /26 that means you have 26 network bits, and 6 host bits, the max # of addresses you can get from 6 bits are 64. A bit has only 2 possible outcomes which are 1 or 0, so 2 to the 6th power gives you 64.
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This question came up on my exam.
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It thought they meant the remaining addresses once everything is subnetted. This would be 64-2=62. Subnetting as testing enumerates above.
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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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