Tugas komjar 4

1. 1 | G a l u h M a g h f i r a – 2 1 1 0 1 6 5 0 1 3 – J a r i n g a n K o m p u t e r
Nama : Galuh Maghfira
NRP : 2110165013
Kelas : 1 D4 LJ IT
JARINGAN KOMPUTER
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?
NET id = 201.114.168  Merupakan Kelas C
Netmask = 255.255.255.240 digeser 4 bit = 28
Jaringan yang dipinjam = 32 – 28 = 4 Host
Jumlah Subnet yang dapat digunakan = 24
– 2 = 16-2 = 14 Subnet
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?
NET id = 222.72.157  Merupakan Kelas C
Netmask = 255.255.255.248 digeser 5 bit = 29
Jaringan yang dipinjam = 32-29=3;
Jumlah host dalam sebuah subnet = 23
– 2 = 8-2 = 6 Host
Jumlah Subnet yang dapat digunakan = 25
– 2 = 32 – 2 = 30 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?
NET id = 198.162.10  Merupakan Kelas C
Jumlah host dalam sebuah subnet = 25 Host
Jumlah Subnet yang dapat digunakan = 6 Subnet
Jumlah bit yang di pinjam => 6 = 2N
- 2
6+2 = 2N
N = 3
192.168.10 000 0
001 32
010 64
011 96
100 128
101 160
110 192 [LAST]
111 224

2. 2 | G a l u h M a g h f i r a – 2 1 1 0 1 6 5 0 1 3 – J a r i n g a n K o m p u t e r
IP pertama untuk Subnet Terakhir adalah : 192.168.10.193
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?
NET id = 199.115.210  Merupakan Kelas C
Netmask = 255.255.255.252 digeser 6 bit = 30
Jaringan yang dipinjam = 32-30 = 2
Jumlah host dalam sebuah subnet = 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?
NET id = 190.90  Merupakan Kelas B
Netmask = 255.255.192.0  /18
High Order Bits pada Kelas B adalah = 10 Binary
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?
NET id = 162.160.0.0  Merupakan Kelas B
Jumlah Subnet yang dapat digunakan = 60 <= 2N
– 2
60 <= 64
N = 6  Geser 6 bit = /22
Maka subnetmask nya adalah 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?
NET id = 150.150  Merupakan Kelas B
Netmask = 255.255.192.0 digeser 2 bit = 18
Jaringan yang dipinjam = 32-18 = 14
Jumlah Subnet yang dapat digunakan = 2n
– 2 = 214
– 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?
NET id = 145.19  Merupakan Kelas B
Netmask = 255.255.0.0 digeser 2 bit = 18
Jumlah Subnet yang dapat digunakan => 2n
– 2 = 27
– 2
N = 7
Jaringan yang dipinjam = 7
Jarak tiap Subnet = 2
Subnet ke – 8 => IP awal = 145.19.17.1
IP Akhir = 145.19.17.254

3. 3 | G a l u h M a g h f i r a – 2 1 1 0 1 6 5 0 1 3 – J a r i n g a n K o m p u t e r
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?
NET id = 33.0.0.0  Merupakan Kelas A
Netmask = 255.0.0.
Jumlah Subnet yang dapat digunakan => 2n
– 2 = 125
2n
= 125 + 2
N = 7  Geser 7 bit
Maka Netmasknya adalah = 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?
NET id = 55.0.0.0  Merupakan Kelas A
Netmask = 255.240.0.0 digeser 4 bit = 12
Jaringan yang dipinjam = 32-12 = 20
Jumlah Subnet yang dapat digunakan = 24
– 2 = 14 Subnet
Jumlah host dalam sebuah subnet = 220
– 2 = 1.048 576 - 2 = 1.048.574 Host