Step 1:

Each common serial connection obliges one subnet. Subsequently, you require two subnets for the serial connections between Router A and Routers B and C. You should likewise have one subnet per Ethernet interface on every switch. Since you have three Ethernet networks, you require three subnets. Utilizing this extremely straightforward tallying strategy, you find that you require a sum of five subnets. Shockingly, you have been doled out a Class C address. The network address 211.212.10.0 takes into account a solitary network of 254 hosts. You must borrow host ID bits to make this address work for you

Step 2:

Determining Number of Bits You Can Borrow

In Step 2, you must focus the number of bits that you can borrow. This number changes relying upon the kind of network address you begin with. For Class An addresses, you have 24 host ID bits, yet you can just borrow up to 22. For Class B addresses, you have 16 host ID bits, yet you must have at least two host bits; in this manner, you can borrow 14 bits. Your Class C address (211.212.10.0) has eight aggregate host ID bits, yet you can just borrow a most extreme of six. The most effortless approach to focus the number of bits you can borrow is to compose the number of octets that contain host ID bits in double. In the Class C sample network 211.212.10.0, you have the accompanying bits to “play” with: 00000000

Step 3:

Determining Number of Bits You Must Borrow to Get Needed Number of Subnets

After you focus the number of subnets you need and the number of bits you can borrow, you must figure the number of host ID bits you must borrow to get the required number of subnets. The equation for deciding the number if bits you must borrow is 2n-2= # of subnets. The n speaks to the number of bits you borrow. At the end of the day, raise two to the number’s force of bits you borrow and subtract two from that number. The outcome is the number of useable subnets made when you borrow that number of bits. For the case network, you require five subnets. On the off chance that you borrow three bits, the equation’s outcome is six usable subnets 2^3 = 8-2 = 6.

Step 4:

Turning On Borrowed Bits and Determining Decimal Value

In Step 4, utilizing the bits you decided were accessible in Step 2, you turn on (set to 1) the number of bits decided you must borrow in Step 3. You should dependably start with the high-arrange bits (the bits beginning on the left of a twofold number). Utilizing the number of bits you can work with and the number of bits you must borrow (from Step 3), your outcome is the accompanying: 11100000. As it were, from the eight aggregate bits from Step 2 (six of which you could borrow), you borrow three host ID bits. In Step 4, you likewise need to focus the decimal estimation of the octets from which you borrow host ID bits.

Step 5:

Determining New Subnet Mask

Step 5 figures the new subnet mask after you borrow the host ID bits in Step 4. You must include the decimal quality from Step 4 to the default subnet mask for the class of address you are sub-netting. The case is a Class C address so the default mask is 255.255.255.0. The new mask in the wake of borrowing three bits gets to be 255.255.255.224.

 

Step 6:

Finding Host/Subnet Variable

In Step 6, you must locate the high’s least request (bits beginning from the left) turned “on.” Step 6 takes all of you the path back to prior in the section to the qualities found in every bit position inside of the octet. Our case characterizes the octets from which we borrow as 11100000. The most astounding request bit turned on speaks to 25, which approaches 32.

Step 7:

Determining Range of Addresses

The last step permits you to take the Host/Subnet variable from Step 6 (32) and make your subnet ranges. Utilizing the Class C network over, the scope of subnets when you borrow three bits are:

211.212.10.0 to 211.212.10.31

211.212.10.32 to 211.212.10.63

211.212.10.64 to 211.212.10.95

211.212.10.96 to 211.212.10.127

211.212.10.128 to 211.212.10.159

211.212.10.160 to 211.212.10.191

211.212.10.192 to 211.212.10.223

211.212.10.224 to 211.212.10.255

IP addresses can’t be all ones or all zeros; in this manner, by and large the first scope of addresses and the last scope of addresses are unusable. In some you can utilize the first scope of addresses, or subnet 0. Just sure makers’ hardware, for example, Cisco Systems, completely bolsters the utilization of subnet zero. In each subnet, the first IP address is unusable on the grounds that it speaks to the subnet ID. The last address is additionally unusable in light of the fact that it is the telecast address for the subnet. Because of these two limitations, in subnet one, 211.212.10.33 is the first useable host ID and 211.212.10.62 is the last useable host ID.

 

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