Question

An organization is granted the block 130.34.12.64/26. The organization needs to have four subnets. What are the subnet addresses and the range of address for each subnet?

Answer 1

/26 means subnet is 255.255.255.192

if you make it /28 that is subnet 255.255.255.240 you can get four subnets with 14 use able IPs in each subnet

Subnets

Netmask:   255.255.255.240 = 28  11111111.11111111.11111111.1111 0000

Wildcard:  0.0.0.15              00000000.00000000.00000000.0000 1111

Network:   130.34.12.64/28       10000010.00100010.00001100.0100 0000 (Class B)

Broadcast: 130.34.12.79          10000010.00100010.00001100.0100 1111

HostMin:   130.34.12.65          10000010.00100010.00001100.0100 0001

HostMax:   130.34.12.78          10000010.00100010.00001100.0100 1110

Hosts/Net: 14                    

Network:   130.34.12.80/28       10000010.00100010.00001100.0101 0000 (Class B)

Broadcast: 130.34.12.95          10000010.00100010.00001100.0101 1111

HostMin:   130.34.12.81          10000010.00100010.00001100.0101 0001

HostMax:   130.34.12.94          10000010.00100010.00001100.0101 1110

Hosts/Net: 14                    

Network:   130.34.12.96/28       10000010.00100010.00001100.0110 0000 (Class B)

Broadcast: 130.34.12.111         10000010.00100010.00001100.0110 1111

HostMin:   130.34.12.97          10000010.00100010.00001100.0110 0001

HostMax:   130.34.12.110         10000010.00100010.00001100.0110 1110

Hosts/Net: 14                    

Network:   130.34.12.112/28      10000010.00100010.00001100.0111 0000 (Class B)

Broadcast: 130.34.12.127         10000010.00100010.00001100.0111 1111

HostMin:   130.34.12.113         10000010.00100010.00001100.0111 0001

HostMax:   130.34.12.126         10000010.00100010.00001100.0111 1110

Hosts/Net: 14                    

Subnets:   4  

Hosts:     56

Answer 2

Answer:

Each subnetwork has 16 addresses. 64+32 = 96. This is the starting address of the 3rd block. 130.34.12.96/28

Explanation:

Designing Subnets

- The number of subnets$\mathrm{N}$ must be a power of 2 .

1. Find the Subnet Mask

- 1-Bits: 1's from default mask $+\log _{2} \mathrm{~N}$

- $\quad$ 0-Bits: The remaining bits (of the 32-bit IP address)

2. Number of Addresses in Each Subnet $=2$ #of 0 -Bits.

3. Range of Addresses in ach Subnet.

Slash notation

A.B.C.D/n

Attach the #of 1s in the mask (a.k. a. prefix length) to a classless address.

Slash notation is also called CIDR notation.

The suffix length is 6 . This means the total number of addresses in the block is $64\left(2^{6}\right)$. If we create four subnets, each subnet will have 16 addresses.

The given block 130.34.12.64/26

Here subnet mask will be, $11111111.11111111 .11111111 .11000000(26 1 's)$

$255.255 .255 .192(11111111.11111111 .11111111 .11000000) (MASK)$

$130.34 .12 .64 (IP Address) & (10000010.00100010 .00001100 .10000000)$

$130.34 .12 .64 \quad(10000010.00100010 .00001100 .10000000)$\\

The range of the IP's available to the Organization

$00000000000000000000000000111111=64$

Dividing it into four subnets will give,

Let us first find the subnet prefix (subnet mask). We need four subnets, which means we need to add two more 1 s to the site prefix. The subnet prefix is then $/ 28$.

Subnet 1: $130.34 .12 .64 / 28 \quad \rightarrow 130.34 .12 .79 / 28$

Subnet 2:$130.34 .12 .80 / 28 \longrightarrow 130.34 .12 .95 / 28 \text {. }$

Subnet 3: $130.34 .12 .96 / 28 \longrightarrow 130.34 .12 .111 / 28$

Subnet 4: $130.34 .12 .112 / 28 \longrightarrow 130.34 .12 .127 / 28$

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