8. C, 255.255.240.0. 150.50.0.0

Subnet 150.50.0.0 to support 7 subnets.

150.50.0.0 is a Class B with a default subnet mask of 255.255.0.0 (11111111.11111111.00000000.00000000 in binary)

To figure out how many bits we need to change from 0 to 1 we can use the formula (2^n)-2=Requirement. When you do the math you come up with the “Binary Truth Line” which is : 2-4-8-16-32-64-128-256-512-1024-2048………….

Count from left to right along the line until you pass seven. You might think 8 would be correct, but you must remember to subtract 2 for the network/wire address and the broadcast address that you cannot use. So, we must go to 16. Now, 16 is the 4 number over so we need to borrow 4 bits from the host portion of the address. Since we are solving for networks/subnets we count those 4 bits from left to right starting at the end of the default Class B mask that we cannot change. In Binary it would look like this: (11111111.11111111.11110000.00000000). Everything to the right stays a 0 indicating the number of host in each subnet. The new subnet mask is 255.255.240.0. You get this number by adding the 1’s together in each octet (128+64+32+16+8+4+2+1=255) and (128+64+32+16=240).

What does this do for us. Now instead of having one big “unmanageable” network we have many smaller “manageable” networks. Each of our networks will have approximately 4096 users each.

9. A, 255.255.255.192

Subnet 209.168.19.0 to support 4 subnets. The “trick here is they did not say “usable”, which means that want “actual” networks. The difference is we do not need to take the (-2) from the formula into consideration.

209.168.19.0 is a Class C with a default subnet mask of 255.255.255.0 (11111111.11111111.11111111.00000000 in binary)

To figure out how many bits we need to change from 0 to 1 we can use the formula (2^n)-2=Requirement. When you do the math you come up with the “Binary Truth Line” which is : 2-4-8-16-32-64-128-256-512-1024-2048………….

Count from left to right along the line until you REACH four. Since we are dealing with the actual available number of networks and not the usable networks that we can assign, we stop here. Now, 4 is the second number over from the left so we need to borrow 2 bits from the host portion of the address. Since we are solving for networks/subnets we count those 2 bits from left to right starting at the end of the default Class C mask that we cannot change. In Binary it would look like this: (11111111.11111111.11111111.11000000). Everything to the right stays a 0 indicating the number of host in each subnet. The new subnet mask is 255.255.255.192. You get this number by adding the 1’s together in each octet (128+64+32+16+8+4+2+1=255) and (128+64=192).

What does this do for us. Now instead of having one big “unmanageable” network we have many smaller “manageable” networks. Each of our networks will have approximately 62 users each. There are 64 addresses in each subnet, but only 62 usable. The other two addresses are the network and broadcast addresses that we cannot use. They are used by the networking devices.

23. A, Use PING from the UNIX workstation

PING is used to verify that you have connectivity to any device on the network that you need to reach. This command can be performed from any operating system because it is a basic DOS command.

If you cannot PING a server, then you know you cannot pull file from it. In this case there is either something wrong with the server, the network between your PC and the server, or something could be wrong with your PC. Further trouble shooting steps would be required. 29. B, 255.255.255.0

Let’s use the same Class B network address from the previous question to show this answers and how solving it is a little different than before.

Subnet 150.50.0.0 to support 175 hosts/devices/PCs

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150.50.0.0 is a Class B with a...