Keamanan Jaringan 4 - Subnetting

N = sum of borrowed host bit .

n = sum of host bit .

1.    How many usable subnets (not theoretical) do you have when using a subnet mask of 255.255.255.240 on Network ID 201.114.168?

Network 201.144.168.0/255.255.255.240 = 201.144.168.0/28

N = 32 – 28 = 4;

Usable Subnet = 2_­^N – 2 = 16-2=14;

14 Usabale subnets.

2.    You are given Network ID 222.72.157, with a subnet mask of 255.255.255.248 to setup. How many subnets and hosts will you have?

Network = 222.72.157.0/255.255.255.248 => 222.72.157.0/29

N = 32 – 29 = 3

Sum of host in a Subnet = 2³ – 2  = 8 - 2 = 6

n = 8 – N => Class C

Sum of subnet = 2⁵ – 2 = 32 – 2 = 30

30 useable subnets and 6 useable hosts on each subnet

3.    You are assigned a Network ID of 198.162.10 and asked to configure the network to provide at least six useable subnets with at least 25 hosts on each subnet. What is the BEGINNING IP address of the LAST useable subnet in the network?

Sum of host in a subnet = 25 => /27

Sum of subnet = 6

Search borrowed network bit => 6 = 2^N – 2, N = 3

IP Address	B.Host	Bit Host	Subnet
192.168.10	000	00000	0
	001	00000	32
	010	00000	64
	011	00000	96
	100	00000	128
	101	00000	160
	110	00000	192
	111	00000	224

Network => 198.162.10.192/27

First IP for last subnet = 198.162.10.193/27

4.    How many useable hosts are on each subnet when the Network ID is 199.215.210 and the subnet mask is 255.255.255.252?

Netmask 255.255.255.252 = /30 => N = 32-30 = 2

IP Class = Class C => First Oktet is 199

Sum of host in each subnet= 2² – 2 = 4 -2 = 2

5.    You are given Network ID 190.90, with a subnet mask of 255.255.192.0 to setup. What are the high-order bits (Leading Bit Values) for this network?

Binary 10,

Class B ,high order bits = Binary “10”,network bit = 14 ,host bit = 16.

6.    You are assigned a Network ID of 162.160 and asked to configure the network to provide at least 60 useable subnets? What would be the subnet mask for this network?

Network = 162.160.0.0

2⁶ – 2 = 64 -2 = 62 , so the sum of borrowed host bit is 6

So the subnetmask is 255.255.252.0  

7.    How many useable hosts are on each subnet when the Network ID is 150.150 and the subnet mask is 255.255.192.0?

Network = 150.150.0.0

Borrowed host bit = 18

Sum of host bit = 32 – 18 = 14.

So the sum of host is

2ⁿ – 2  = 2¹⁴ – 2 = 16.382

8.    You are assigned a Network ID of 145.19 and asked to configure the network to provide at least 100 useable subnets with at 500 hosts on each subnet. What is the ENDING IP address of the EIGHTH useable subnet in the network?

Network = 145.19.0.0

Class = B

Normal mask = 255.255.0.0

Sum of subnet = 2^N – 2 =2⁷ – 2

                                2^N = 102

                               N = 7

So sum of borrowed network bit is 7 .

Range from each subnet is plus 2

8th Subnet

IP first = 145.19.17.1

IP last = 145.19.17.254

IP broadcast = 145.19.17.255

9.    You are a private contractor hired by the large company to setup the network for their enterprise. The Network ID is 33 and you need at least 125 subnets in their large network with at least 125,000 hosts on each of the subnets. What would be the subnet mask for this network?

Network = 33.0.0.0

Class = A

Normal subnet = 255.0.0.0

Sum of subnet = 125 ,

Sum of host  = 125.000

Sum of subnet = 2^N – 2

                                2^N =  125 +2

                                2^N  = 127

                               N = 7.

                                                        

Subnetmask = 255.254.0.0

10.          You are given Network ID 55.0.0.0, with a subnet mask of 255.240.0.0 to setup. How many subnets and hosts will you have?

Network = 55.0.0.0

Subnetmask = 255.240.0.0

Sum of subnet = 2^N – 2 = 2⁴ – 2 = 16 - 2 = 14

Sum of host = 2ⁿ – 2 = 2²⁰ – 2 = 1.048.576 - 2 =  1.04.,574  

14 useable subnets and 1,048,574 useable hosts on each subnet