You are a network engineer assigned the responsibility of subnetting a network T
ID: 3720346 • Letter: Y
Question
You are a network engineer assigned the responsibility of subnetting a network The organization has been assigned a/24 network address of 192.36.15.0 with a default subnet mask of 255.255.255.0. Since you learned about subnet masks in your MIS 320 class at GMU, you know that 192.36.15 is the network portion of the IP address and that you have the last eight (8) bits for hosts Based on your organization's LAN design, you are assigned the responsibility of subnetting the IP address to accommodate eight (8) subnets Using all the resources available to you, including your textbook, the Internet, etc. calculate the new subnet mask that will permit the last octet of 192.36.15.0 to be divided among at least eight (8) subnetworks Your new subnet mask is: 255.255.255 This new subnet mask gives you eight (8) subnets each of which can have 32 hosts What are the IP address ranges for the eight (8) new networks? 1: 192.36.15.00- 192.36.15. 192.36.15__-192.36.15 192.36.15.__-192.36.15 192.36.15__- 192.36.15 192.36.15___-192.36.15 192.36.15._- 4 6 .192.36.15 7: .192.36.15 8: 192.36.15 - 192.36.15.255Explanation / Answer
For 8 subnets we need to fix 3 bits so the new subnet mask will be 11111111.11111111.11111111.11100000(in binary)
Which is 255.255.255.224
The IP address ranges for the 8 sunets are:-
1) 192.36.15.00 to 192.36.15.31
2) 192.36.15.32 to 192.36.15.63
3) 192.36.15.64 to 192.36.15.95
4) 192.36.15.96 to 192.36.15.127
5) 192.36.15.128 to 192.36.15.159
6) 192.36.15.160 to 192.36.15.191
7) 192.36.15.192 to 192.36.15.223
8) 192.36.15.224 to 192.36.15.225
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