Using Subnet Shortcuts Using Scenario 2 above that has 14 subnets, perform the following shortcut steps. Compare the results to the Scenario 2 results. They should match exactly The first three subnets have been completed for you 1. Identify the number of bits beyond the default mask bits that will be used for the subnet mask. This scenario includes a Class B subnet mask using 4 bits in the third octet. 11111111111111111000000000000 2. We borrowed 4 bits. Take the lowest valued bit of the ones borrowed and covert it to decimal. So 16 is the value that we are interested in. The 3 octet of the host IP number ranges will increment by 16. 3. To determine the number of possible subnets, reverse the bits so that they are in what is called “low order” as shown below. Convert the bits to decimal format and then subtract 1. 00001111 =15 (# of valid subnets) Starting with increment by the value calculated in Step 2 (16) the same number of times as the possible number of bit combinations(subnets) calculated in Step 3 (15) of Increment by Subersumber found step in 2 Valid range of host Paddresses for Class with subnet of 255.255.240.0 Sargo ncrement by 1 Range 14 octet han nest starting Ending Range 14 och 1616 NN 16.1 NN 31.254 16 16 32 NN 32.1 116 15 31 or hos p115 – 47 -1) 03.1963 o NN 47254 NN 63254 321648 NN46.1 T ANN 141 54. 1579 NN 19.254 ne you reach 14 betwee NNE network octets) Define the ranges of host IDs that would be available for 2 subnets that use the following information (the subnet numbers are not known): Network ID 75.0.0.0, subnet mask 255.255.0.0. Show transcribed image text Using Subnet Shortcuts Using Scenario 2 above that has 14 subnets, perform the following shortcut steps. Compare the results to the Scenario 2 results. They should match exactly The first three subnets have been completed for you 1. Identify the number of bits beyond the default mask bits that will be used for the subnet mask. This scenario includes a Class B subnet mask using 4 bits in the third octet. 11111111111111111000000000000 2. We borrowed 4 bits. Take the lowest valued bit of the ones borrowed and covert it to decimal. So 16 is the value that we are interested in. The 3 octet of the host IP number ranges will increment by 16. 3. To determine the number of possible subnets, reverse the bits so that they are in what is called “low order” as shown below. Convert the bits to decimal format and then subtract 1. 00001111 =15 (# of valid subnets) Starting with increment by the value calculated in Step 2 (16) the same number of times as the possible number of bit combinations(subnets) calculated in Step 3 (15) of Increment by Subersumber found step in 2 Valid range of host Paddresses for Class with subnet of 255.255.240.0 Sargo ncrement by 1 Range 14 octet han nest starting Ending Range 14 och 1616 NN 16.1 NN 31.254 16 16 32 NN 32.1 116 15 31 or hos p115 – 47 -1) 03.1963 o NN 47254 NN 63254 321648 NN46.1 T ANN 141 54. 1579 NN 19.254 ne you reach 14 betwee NNE network octets) Define the ranges of host IDs that would be available for 2 subnets that use the following information (the subnet numbers are not known): Network ID 75.0.0.0, subnet mask 255.255.0.0.

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Answer to Using Subnet Shortcuts Using Scenario 2 above that has 14 subnets, perform the following shortcut steps. Compare the res…