Figure 16 Binary Place Values Question 1 Answer

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173*160.1-

172,15,255,254 172.ifl.25s.255

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Answer: 172.16.0.0 is 1 subnet (broadcast domain) with 254 hosts (28 - 2). The first host for this 172.16.0.1. The last host for this subnet is 172.16.0.254, and the broadcast address is 172.16.0.2, representation of this is critical to understanding bits and boundaries. Often it is helpful to write ou ranges, such as th e fo llowing:

• 172 .16.0.1 through 172.16.0.254 (hosts on subnet 172.16.0.0/24)

• 172.16.0.255 (broadcast on subnet 172.16.0.0/24)

This is a Clas s B addres s with a default subnet mask of /16. The given mask is /24, which means tl were borrowed to provide more networks (subnets). This is subnetting. There are 28 = 256 availab this scenario wit h 254- hosts ( 28 - 2) on each one. These subnets increment by 1 because the lowes bit is in the 1 or 20 binary position. The next two subnets are 172.16.1.0/24 and 172.16.2.0/24.

UseFigure 1-7 ta veri fy your ca lculations and to relate the following general rules:

The rightm ost avai I able host bit is turned on (1) for the first host. All other host bits are off (C

The rightmost available h ost bit is turned of! (0) for t he last Vost. All other host bits are on (1

All host bits are on (1) for the b roadcast address. The broadcast for a subnet is one less than subnet.

Subnets iecrement by the l owest 1 bit (nig Comost bit) in dhe mas Tt The Tubnet inclement and aubn ets are ciocle d id Figune 1-n.

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