LoDoGg Forums - CCNA - Subnetting and more Subnetting

Refer to Exhibit

A Network Administrator is adding two new hosts to Switch A. Which three values could be used for the configuration of these hosts? (Choose Three)

A. Host 1 IP Address 192.168.1.79
B. Host 1 IP Address 192.168.1.64
C. Host 1 Default Gateway 192.168.1.78
D. Host 2 IP Address 192.168.1.128
E. Host 2 Default Gateway 192.168.1.129
F. Host 2 IP Address 192.168.1.190

~~~~~~~~~
My Logic is Workstation A and B are using Vlan's 10 & 20 so that's where will focus on.

Vlan10
fa0/0.10
ip address 192.168.1.78 255.255.255.224

Vlan20
ip address 192.168.1.130 255.255.255.192

fa0/0.10 Subnet break down

IP Address - 192.168.1.78
Subnet    - 255.255.255.224

To figure out the / value you know 255.255.255.0 is a /24.
I would then figure out the last octet.

128 + 64 + 32 + 16 + 8 + 4 2 1
|       |
128 + 64 + 32 =224

255.255.255.224 would then be a /27 and you know there is 32 hosts because this is the last octet in your math so you then can do the following chart:

Remember you need to remove a number for the Broadcast address:
1   - 31
32 - 63
64 - 95     192.168.1.78 (78 - 95 is your range) Answer A, C
96 - 127
128 - 159
160 - 191
192 - 223
224 - 255

Now for the fa0/0.20 Subnet break down

IP Address - 192.168.1.130
Subnet       - 255.255.255.192

To figure out the / value you know 255.255.255.0 is a /24.
I would then figure out the last octet.

128 + 64 + 32 + 16 + 8 + 4 2 1
|        |
128 + 64 = 192

255.255.255.192 would then be a /26 and you know there is 64 hosts because this is the last octet in your math so you then can do the following chart:

Remember you need to remove a number for the Broadcast address:
1      - 63
64    - 127
128 - 191     192.168.1.130 (128 - 191 is your range) Answer F
192 - 255

A. Host 1 IP Address 192.168.1.79 (In Range)
B. Host 1 IP Address 192.168.1.64 (Not in Range)
C. Host 1 Default Gateway 192.168.1.78 (Vlan10's IP is the Gateway)
D. Host 2 IP Address 192.168.1.128 (Not in Range)
E. Host 2 Default Gateway 192.168.1.129 (Not Vlan 20's IP)
F. Host 2 IP Address 192.168.1.190 (In Range)

Edited by lodogg - 06 December 2007 at 9:02pm

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Hoagie914 Members Profile Send Private Message Find Members Posts Add to Buddy List Guest	Quote Reply Topic: CCNA - Subnetting and more Subnetting Posted: 06 December 2007 at 10:59pm
	Nice shortcut! However, remember when calculating # of hosts to subtract 2 from the final number. 1 for the loopback address & 1 for the broadcast address. The correct # of hosts is 30 & 62.

lodogg Members Profile Send Private Message Find Members Posts Visit Members Homepage Add to Buddy List Admin Group Joined: 25 January 2003 Online Status: Offline Posts: 1831	Quote Reply Posted: 06 December 2007 at 8:53pm
	Refer to Exhibit A Network Administrator is adding two new hosts to Switch A. Which three values could be used for the configuration of these hosts? (Choose Three) A. Host 1 IP Address 192.168.1.79 B. Host 1 IP Address 192.168.1.64 C. Host 1 Default Gateway 192.168.1.78 D. Host 2 IP Address 192.168.1.128 E. Host 2 Default Gateway 192.168.1.129 F. Host 2 IP Address 192.168.1.190 ~~~~~~~~~ My Logic is Workstation A and B are using Vlan's 10 & 20 so that's where will focus on. Vlan10 fa0/0.10 ip address 192.168.1.78 255.255.255.224 Vlan20 ip address 192.168.1.130 255.255.255.192 fa0/0.10 Subnet break down IP Address - 192.168.1.78 Subnet - 255.255.255.224 To figure out the / value you know 255.255.255.0 is a /24. I would then figure out the last octet. 128 + 64 + 32 + 16 + 8 + 4 2 1 \| \| 128 + 64 + 32 =224 255.255.255.224 would then be a /27 and you know there is 32 hosts because this is the last octet in your math so you then can do the following chart: Remember you need to remove a number for the Broadcast address: 1 - 31 32 - 63 64 - 95 192.168.1.78 (78 - 95 is your range) Answer A, C 96 - 127 128 - 159 160 - 191 192 - 223 224 - 255 Now for the fa0/0.20 Subnet break down IP Address - 192.168.1.130 Subnet - 255.255.255.192 To figure out the / value you know 255.255.255.0 is a /24. I would then figure out the last octet. 128 + 64 + 32 + 16 + 8 + 4 2 1 \| \| 128 + 64 = 192 255.255.255.192 would then be a /26 and you know there is 64 hosts because this is the last octet in your math so you then can do the following chart: Remember you need to remove a number for the Broadcast address: 1 - 63 64 - 127 128 - 191 192.168.1.130 (128 - 191 is your range) Answer F 192 - 255 A. Host 1 IP Address 192.168.1.79 (In Range) B. Host 1 IP Address 192.168.1.64 (Not in Range) C. Host 1 Default Gateway 192.168.1.78 (Vlan10's IP is the Gateway) D. Host 2 IP Address 192.168.1.128 (Not in Range) E. Host 2 Default Gateway 192.168.1.129 (Not Vlan 20's IP) F. Host 2 IP Address 192.168.1.190 (In Range) Edited by lodogg - 06 December 2007 at 9:02pm