215

Figure 6-20  Borrowing 3 Bits to Create Subnets
				Start with					–	192.168.1.0 (/24)	Address:	11000000.10101000.00000001.00000000
				This Address
										255.255.255.0	Mask:	11111111.11111111.11111111.00000000
				Make Eight				0		192.168.1.0 (/27)	Address:	11000000.10101000.00000001.00000000
					Subnets					255.255.255.224	Mask:	11111111.11111111.11111111.111		00000
								1		192.168.1.32 (/27)	Address:	11000000.10101000.00000001.00100000		00000
										255.255.255.224	Mask:	11111111.11111111.11111111.111
								2		192.168.1.64 (/27)	Address:	11000000.10101000.00000001.01000000		00000
										255.255.255.224	Mask:	11111111.11111111.11111111.111
								3		192.168.1.96 (/27)	Address:	11000000.10101000.00000001.01100000
										255.255.255.224	Mask:	11111111.11111111.11111111.111		00000
			0
									4	192.168.1.128 (/27)	Address:	11000000.10101000.00000001.10000000
		2	1
										255.255.255.224	Mask:	11111111.11111111.11111111.111		00000
			Router A
								5		192.168.1.160 (/27)	Address:	11000000.10101000.00000001.10100000		00000
			5							255.255.255.224	Mask:	11111111.11111111.11111111.111
		4					3			192.168.1.192 (/27)	Address:	11000000.10101000.00000001.11000000
									6
			Router B							255.255.255.224	Mask:	11111111.11111111.11111111.111		00000
								7		192.168.1.224 (/27)	Address:	11000000.10101000.00000001.11100000		00000
										255.255.255.224	Mask:	11111111.11111111.11111111.111

Three bits are borrowed to

provide eight subnets.

To accommodate six networks, subnet 192.168.1.0 /24 into address blocks using this formula:

2³ = 8

To get at least six subnets, borrow 3 host bits. A subnet mask of 255.255.255.224 provides the 3 additional network bits.

To calculate the number of hosts, begin by examining the last octet. Notice these subnets:

0 = 00000000

32 = 00100000

64 = 01000000

96 = 01100000

128 = 10000000

160 = 10100000