[Tugas 4 - Jaringan Komputer] Subnetting Workshop

Exercises:

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 ID = 201.114.168.0

Subnetmask = 255.255.255.240 | bit = 11111111.11111111.11111111.11110000 | /28

Bit subnet (N) = 4

Sum of subnet = 2^N - 2  |  2⁴ – 2 = 16 – 2 = 14 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 ID = 222.72.157.0

Subnetmask = 255.255.255.248 | bit = 11111111.11111111.11111111.11111000 | /29

Bit subnet (N) = 5

Bit host (n) = 3

Sum of subnet = 2^N - 2  |  2⁵ – 2 = 32 – 2 = 30 subnets.

Sum of host each subnets = 2ⁿ - 2  |  2³ – 2 = 8 – 2 = 6 host.

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?

Provide at least six usable subnets = 2^N – 2

6 < 2^N – 2  |  N = 3

192.162.10.	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 ID on last useable subnet : 192.162.10.192, Beginning IP address = 192.168.10.193 /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?

Network ID = 199.215.210.0

Subnetmask = 255.255.255.252  |  bit = 11111111.11111111.11111111.11111100  |  /30

Bit host (n) = 2

Sum of host each subnets = 2ⁿ - 2  |  2² – 2 = 4 – 2 = 2 host.

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?

Network ID : 190.90.x.x  |  Class B

Refer to first page on modul, high-order bits for class B is bit 10

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 ID : 162.160.x.x  |  Class B

At least 60 useable subnets = 2^N – 2

60 < 2^N – 2  |  N = 6

Subnet mask bit = 11111111.11111111.11111100.00000000  |  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 ID = 150.150.x.x

Subnet mask = 255.255.192.0  |  bit = 11111111.11111111.11000000.00000000  |  /18

Bit host (n) = 14

Sum of host each subnets = 2ⁿ - 2  |  2¹⁴ – 2 = 16.384 – 2 = 16.382 host.

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 ID = 145.19.x.x = Class B

At least 100 useable subnets < 2^N – 2

100 < 2^N – 2  |  N = 7

Subnet mask bit = 11111111.11111111.11111110.00000000 = 255.255.254.0

First Octet	Second Octet	Third Octet		Fourth Octet	Decimal of 3rd Octet
145.	19.	0000000	0	00000000	0
		0000001	0		2
		0000010	0		4
		0000011	0		6
		0000100	0		8
		0000101	0		10
		0000110	0		12
		0000111	0		14
		0001000	0		16
		0001001	0		18
		0001010	0		20
		0001011	0		22
		0001100	0		24
		0001101	0		26
		.etc	.etc		.etc

Ending IP address : 145.19.17.254

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 ID : 33.x.x.x = Class A

At least 125 subnet < 2^N – 2

125 < 2^N – 2  |  N = 7

Subnet mask bit = 11111111.11111110.00000000.00000000 = 255.254.0.0 = /15

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 ID : 55.0.0.0 à Class A

Subnet mask : 255.240.0.0  |  bit = 11111111.11110000.00000000.00000000  |  /12

Bit subnet (N) = 4

Bit host (n) = 20

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

Sum of host each subnets = 2ⁿ - 2  |  2²⁰ – 2 = 1.048.576 – 2 = 1.048.574 host.