Chapter 15 – Analyzing Subnet Masks

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Subnet mask conversion: 

  • Subnet masks can be written as 32-bit binary numbers, but not just any binary number.  
  • The value must not interleave 1s and 0s. 
  • If 1s exist, they are on the left.  
  • If 0s exist, they are on the right 

Converting between binary and prefix masks: 

Table 15-3 Example Conversions: Prefix to Binary Prefix Mask Logic Binary Mask / 18 / 28 / 13 Write 18 Is, then 14 ()s, total 32 Write 28 Is, then 4 Os, total 32 Write 13 Is, then 19 ()s, total 32 11000000 00000000 00000000 00000000
Table 15-4 Nine Possible Values in One Octet of a Subnet Mask Binary Mask Octet 00000000 10000000 11000000 11100000 Decimal Equivalent o 128 192 224 240 248 Number of Binary Is o 1 2 3 4 5
11111100 11111110 11111111 252 254 255
172.16.1.0/24 Figure 15-3 172.16.4.0/24 172.16.5.0/24 Subnet 172.16.2.0/24 172.16.2.101 172.16.2.102 Subnet 172.16.3.0/24 172.16.3.101 172.16.3.102 Simple Subnet Design, with Mask /24

SUBNETS: 

Key Topic Defines the size of the prefix (combined network and subnet) part of the addresses in a subnet Defines the size of the host part of the addresses in the subnet Can be used to calculate the number of hosts in the subnet Provides a means for the network designer to communicate the design details—the num- ber of subnet and host bits—to the devices in the network Under certain assumptions, can be used to calculate the number of subnets in the entire classful network Can be used in binary calculations of both the subnet ID and the subnet broadcast address

EXAMPLE: 

consider the case of IP address 8.1.4.5 with mask 255.255.0.0 

1) 255.255.0.0 = /16, so P=16 

2) 8.1.4.5 is class A network so network bits are N=8 

3) S = P – N = 16 – 8 = 8 

4) H = 32 – P = 32 – 16 = 16 

5) 2 to the power of 16 – 2 = 65534 hosts/subnet 

6) 2 to the power of 8 = 256 subnets 

Figure 15-9 shows a visual analysis of the same problem. 16 1 s 16 Os 1-1=16 Figure 15-9 Visual Representation of Problem: 8.1.4.5, 255.255.0.0

EXAMPLE 2: 

consider address 200.1.1.1, mask 255.255.255.252 

1) 255.255.255.252 = 24 + 6 = /30 

2) 200.1.1.1 = class C network, so N = 24 

3) S = P – N = 30 – 24 = 6 

4) H = 32 – 30 = 2 

5) 2 to the power of 2 – 2 = 2 hosts 

6) 2 to the power of 6 = 64 subnets 

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