Subnet Questions -solution

Q4. You are given a major network – 192.168.99.0/255.255.255.0. You have been told to create enough subnets to accommodate the following network topology. Fill in the table below. Three subnets required, thus n=2, Y=4, m=p=6, ∆=64, z=62 Subnet # Subnet ID (NID) First usable host IP Last usable host IP Broadcast IP Subnet mask slash format Subnet mask expanded format 1 192.168.99.0 192.168.99.1 192.168.99.62 192.168.99.63 /26 255.255.255.192 2 192.168.99.64 192.168.99.65 192.168.99.126 192.168.99.127 /26 3 192.168.99.128 192.168.99.129 192.168.99.190 192.168.99.191 /26 4 192.168.99.192 192.168.99.193 192.168.99.254 192.168.99.255 /26 Q5. You are given a major network – 192.168.100.0/255.255.255.0. You have been told to create enough subnets to accommodate the following network topology. Fill in the table below. Our formula: # of SN, Y=2^n, n~# of bits borrowed # of host/SN, Z=2^m -2, m~ all unborrowed bits ∆ = 2^p, p~ remaining bits in the borrowed octet

Q7. You have a Class C network 192.168.0.0/24. You have just created 32 new subnets. Which statement below applies to the resulting subnets? A . One of the subnet being 192.168.0.32 with subnet mask 255.255.255.224 B . One of the subnet being 192.168.0.32 with subnet mask 255.255.255.248 C . One of the subnet being 192.168.0.32 with subnet mask 255.255.248.0 D . One of the subnet being 192.168.0.32 with subnet mask 255.255.255.252 Q8. You are given an IP network address and subnet mask as 155.23.16.0 / 255.255.252.0. How many bits have been borrowed from the host octet? A . 3 bits B . 5 bits C . 6 bits D . 2 bits Q9. You have a network 192.168.99.160/27. Create 8 subnets and fill in the table below: N=3, Y=8, m=p=5, ∆=4, z=30 Subnet # Subnet ID (NID) First usable host IP Last usable host IP Broadcast IP Subnet mask slash format Subnet mask expanded format 1 192.168.99.160 192.168.99.161 192.168.99.162 192.168.99.163 /30 255.255.255.252 2 192.168.99.164 192.168.99.165 192.168.99.166 192.168.99.167 3 192.168.99.168 192.168.99.169 192.168.99.170 192.168.99.171 4 192.168.99.172 192.168.99.173 192.168.99.174 192.168.99.175 5 192.168.99.176 192.168.99.177 192.168.99.178 192.168.99.179 6 192.168.99.180 192.168.99.181 192.168.99.182 192.168.99.183 7 192.168.99.184 192.168.99.185 192.168.99.186 192.168.99.187 8 192.168.99.188 192.168.99.188 192.168.99.190 192.168.99.191

The figure above shows 5 different subnets, each with different host requirements. The given IP address from our ISP is 192.168.1.0/24 . The host requirements are: Network A - 14 hosts Network B - 28 hosts Network C - 2 hosts Network D - 7 hosts Network E - 28 hosts As recommended, we begin the process by subnetting for the largest host requirement first. As it seems, the largest requirements are for NetworkB and NetworkE , each with 28 hosts. Don’t forget the cram table! Let’s apply the formula: usable hosts = 2^n - 2. For networks B and E, 5 bits are borrowed from the host portion and the calculation is 2^5 = 32 - 2. Only 30 usable host addresses are available in this case due to the 2 reserved addresses. Borrowing 5 bits meets the requirement but leaves little room for future growth. So we revert to borrowing 3 bits for subnets leaving 5 bits for the hosts. This allows 8 subnets with 30 hosts each. We have created and will allocate addresses for networks B and E first: