Jumat, 05 April 2013

[Tugas Jarkom - 2 TI B] Tugas 6.7.3.1 Cisco Networking Academy

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Tugas Jaringan Komputer (26 Maret 2013)
CCNA Exploration 4.0 Networks Fundamental


Activity 6.7.3: IPv4 Address Subnetting Part 1
Learning Objectives
Upon completion of this activity, you will be able to determine network information for a given IP address and network mask.

Background
This activity is designed to teach how to compute network IP address information from a given IP
address.

Scenario
When given an IP address and network mask, you willbe able to determine other information about the
IP address such as:
•Network address
•Network broadcast address
•Total number of host bits
•Number of hosts

Task 1: For a given IP address, Determine Network Information.
Given:
Host IP Address   172.25.114.250
Network Mask   255.255.0.0 (/16)
Find:
Network Address, Network Broadcast,  Address total Number of Host Bits, Number of Hosts

Step 1: Translate Host IP address and network mask into binary notation.
Convert the host IP address and network mask to binary:
                                172              25           114           250
IP Address        10101100  11001000  01110010  11111010
Network Mask   11111111  11111111  00000000  00000000
                               255             255             0              0
Step 2: Determine the network address.
1. Draw a line under the mask.
2. Perform a bit-wise AND operation on the IP address and the subnet mask.
Note:1 AND 1 results in a 1; 0 AND anything results in a 0.
3. Express the result in dotted decimal notation.
4. The result is the network address for this host IP address, which is 172.25.0.0.

IP Address   10101100  11001000  01110010  11111010 (172.25.114.250)
Subnet Mask   11111111  11111111  00000000  00000000 (255.255.0.0)
Network Address   10101100  11001000  00000000  00000000 ( 172.25.0.0)

Step 3: Determine the broadcast address for the network address
The network mask separates the network portion of the address from the host portion. The network
address has all 0s in the host portion of the address and the broadcast address has all 1s in the host
portion of the address.

                                  172             25             0                  0
Network Add.     10101100  11001000  00000000  00000000
Mask                    11111111  11111111  00000000  00000000
Broadcast.           10101100  11001000  11111111  11111111
                                 172              25             255            255

By counting the number of host bits, we can determine the total number of usable hosts for this network.
Host bits: 16
Total number of hosts:
2^16 = 65,536
65,536 – 2 = 65,534 (addresses that cannot use the all 0saddress, network address, or the all 1s
address, broadcast address.)

Add this information :
  • Host IP Address   = 172.25.114.250
  • Network Mask     = 255.255.0.0 (/16)
  • Network Address = 172.25.0.0
  • Network Broadcast Address = 172.25.255.255
  • Total Number of Host Bits   = Host bits 16 and total number of host 65.534


Task 2: Challenge
For all problems:
Create a Subnetting Worksheet to show and record all work for each problem.
Problem 1
Host IP Address 172.30.1.33
Network Mask 255.255.0.0 (/16)
Network Address  (?) = 172.30.1.0
Network Broadcast Address (?) = 172.30.255.255
Total Number of Host Bits (?)  and Number of Hosts (?) =  Host bits 16 and total number of host 65.534 



Problem 2
Host IP Address 172.30.1.33
Network Mask 255.255.255.0 (/24)
Network Address  (?) = 172.30.1.0
Network Broadcast Address (?) = 172.30.1.255
Total Number of Host Bits (?) = 8
Number of Hosts (?) : 2^8 - 2 = 256 - 2 = 254 hosts

Problem 3
Host IP Address 192.168.10.234
Network Mask 255.255.255.0 (/24)
Network Address  (?) = 192.168.10.0
Network Broadcast Address (?) = 192.168.1.255
Total Number of Host Bits (?) = 8
Number of Hosts (?) = 2^8 -2 = 256 - 2 = 254 hosts

Problem 4
Host IP Address 172.17.99.71
Network Mask 255.255.0.0 (/16)
Network Address  (?) = 172.17.0.0
Network Broadcast Address (?) = 172.17.255.255
Total Number of Host Bits (?) = 16
Number of Hosts (?) : 2^16-2 = 65,536 – 2 = 65,534 hosts

Problem 5
Host IP Address 192.168.3.219
Network Mask 255.255.0.0 (/16)
Network Address  (?) = 192.168.0.0
Network Broadcast Address (?) = 192.168.255.255
Total Number of Host Bits (?) = 16
Number of Hosts (?): 2^16-2 = 65,536 – 2 = 65,534 hosts

Problem 6
Host IP Address 192.168.3.219
Network Mask 255.255.255.224
Network Address  (?) = 192.168.3.192
Network Broadcast Address (?) = 192.168.3.223
Total Number of Host Bits (?) = 5
Number of Hosts (?): 2^5-2 = 32 -2 = 30 hosts

Task 3: Clean Up
Remove anything that was brought into the lab, and leave the room ready for the next class.

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