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It240 Week 4 Answer Key

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Answer Key for TCP/IP LAN Plan CheckPoint, Due in Week Four 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 …show more content…

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 177. Since we have to remember about the 2 address that we subtract I changed the number to 177 to account for them so we make sure we have enough addresses. 128 is not enough so we must go to 256. Now, 256 is the 8th number over so we need to RESERVE 8 bits for the host portion of the address. Since we are solving for host we count those 8 bits from right to left starting at the end of the address with 0’s. In Binary it would look like this: (11111111.11111111.11111111.00000000). Everything to the right stays a 0 indicating the number of host in each subnet. Everything to the left becomes a 1. The new subnet mask is 255.255.255.0. You get this number by adding the 1’s together in each octet (128+64+32+16+8+4+2+1=255). Your networks would look something like this: 150.50.0.0 Gateway (150.50.0.1); Range for Host (150.50.0.2 – 150.50.0.254) Broadcast (150.50.0.255) 150.50.1.0 Gateway (150.50.1.1); Range for Host (150.50.1.2 – 150.50.1.254) Broadcast (150.50.1.255) 150.50.2.0 Gateway (150.50.2.1); Range for Host (150.50.2.2 – 150.50.2.254) Broadcast (150.50.2.255) The networks “increment” by 1 in the third octet. Why? If you look at the subnet mask in binary the last 1 bit is in the 1 position in the

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