...Jumlah Subnet, Jumlah Host per Subnet, Blok Subnet, dan Alamat Host- Broadcast. Penulisan IP address umumnya adalah dengan 192.168.1.2. Namun adakalanya ditulis dengan 192.168.1.2/24, apa ini artinya? Artinya bahwa IP address 192.168.1.2 dengan subnet mask 255.255.255.0. Lho kok bisa seperti itu? Ya, /24 diambil dari penghitungan bahwa 24 bit subnet mask diselubung dengan binari 1. Atau dengan kata lain, subnet masknya adalah: 11111111.11111111.11111111.00000000 (255.255.255.0). Konsep ini yang disebut dengan CIDR (Classless Inter-Domain Routing) yang diperkenalkan pertama kali tahun 1992 oleh IEFT. Pertanyaan berikutnya adalah Subnet Mask berapa saja yang bisa digunakan untuk melakukan subnetting? Ini terjawab dengan tabel di bawah: Subnet Mask | Nilai CIDR | 255.128.0.0 | /9 | 255.192.0.0 | /10 | 255.224.0.0 | /11 | 255.240.0.0 | /12 | 255.248.0.0 | /13 | 255.252.0.0 | /14 | 255.254.0.0 | /15 | 255.255.0.0 | /16 | 255.255.128.0 | /17 | 255.255.192.0 | /18 | 255.255.224.0 | /19 | | Subnet Mask | Nilai CIDR | 255.255.240.0 | /20 | 255.255.248.0 | /21 | 255.255.252.0 | /22 | 255.255.254.0 | /23 | 255.255.255.0 | /24 | 255.255.255.128 | /25 | 255.255.255.192 | /26 | 255.255.255.224 | /27 | 255.255.255.240 | /28 | 255.255.255.248 | /29 | 255.255.255.252 | /30 | | SUBNETTING PADA IP ADDRESS CLASS C Ok, sekarang mari langsung latihan saja. Subnetting seperti apa yang terjadi dengan sebuah NETWORK ADDRESS 192.168.1.0/26 ? ...
Words: 1498 - Pages: 6
...46 subnets and provide information for subnets #1, #4, #5, and #46 Step 1: Convert the network address to binary 215.251.145.0/24 is a class C address with a subnet mask of 24 bits of ones First Octet 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=0, 16=1, 8=0, 4=1, 2=1, 1=1 215 = 11010111. Second Octet = 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=0, 2=1, 1=1 251 = 11111011 Third Octet = 128.64.32.16.8.4.2.1, 128=1, 64=0, 32=0, 16=1, 8=0, 4=0, 2=0, 1=1 145 = 10010001 Fourth Octet = 0 11010111.11111011.10010001.00000000 Network address Step 2: Convert Subnet mask into binary First Octet 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=1, 2=1, 1=1 255 = 11111111 Second Octet 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=1, 2=1, 1=1 255 = 11111111 Third Octet 128.64.32.16.8.4.2.1, 128=1, 64=1, 32=1, 16=1, 8=1, 4=1, 2=1, 1=1 255 = 11111111 Forth Octet = 0 11111111.11111111.11111111.00000000 Subnet mask Step 3: Calculate the subnets required 46 subnets required per instructions: Use the formula 2 to the power of X to find out how many bits to borrow from the subnet mask. 2^5= 32 and is insufficient to cover the subnets required so, 2^6 = 64 x is 6 we need 6 bits from last octet of the subnet mask to cover our subnets. I need 46 subnets but I have total 64 available so that leaves me with; 64-46 = 18 extra subnets to use at a later date. Original Network address in binary: 11010111.11111011.10010001.00000000 Replace subnet bits...
Words: 1170 - Pages: 5
...0.0/8 subnetted to 10808 subnets and provide information for subnets #1, #726, #4213, and #10808 SUMMARY Subnet #: 1 | | Network Address | 64.0.4.0 | Broadcast Address | 64.0.7.255 | Subnet Mask: | 255.255.252.0 | Range of IP Addresses: | 64.0.4.1 to 64.0.7.254 | Subnet #: 726 | | Network Address | 64.11.88.0 | Broadcast Address | 64.11.91.255 | Subnet Mask: | 255.255.252.0 | Range of IP Addresses: | 64.11.88.1 – 64.11.91.254 | Subnet #: 4213 | | Network Address | 64.65.212.0 | Broadcast Address | 64.65.215.255 | Subnet Mask: | 255.255.252.0 | Range of IP Addresses: | 64.65.212.1 – 64.65.215.254 | Subnet #: 10808 | | | Network Address | 64.168.224.0 | | Broadcast Address | 64.168.227.255 | | Subnet Mask: | 255.255.252.0 | | Range of IP Addresses: | 64.168.224.1 – 64.168.227.254 | | Subnet #1: Converting the IP address 64.0.0.0/8 into binary numbers 64.0.0.0 = 01000000.00000000.00000000.00000000 The /8 at the end of the IP address means that the subnet mask will consist of eight 1’s for the first octet and then 24 0s following Network Bits – Host Bits 11111111.00000000.00000000.00000000 255. 0. 0. 0. Both binary numbers are lined up: 01000000.00000000.00000000.00000000 - IP Address 11111111.00000000.00000000.00000000 - Subnet Mask Left most bits w/ the 1’s = Network Bits The rest is Host bits. We will take host bits and add them to the network portion. Since we have 10808 subnets, 2 to the power of 14 is...
Words: 832 - Pages: 4
...Running head: DETERMINING SUBNET CONFIGURATION Determining Subnet Configuration for Variety of Networks Alvin Miles AIU Online 09/20/2012 Determining Subnet Configuration for Variety of Networks In this final week IP assignment we are given some IP addresses and are ask several questions on anything from what is the broadcast address of this IP, what are the range of valid host IP addresses, to subnet ranges. Before we go too deep in trying to answer these questions I must first explain what classful routing protocol is as this will help explain some of my answers. In classful routing protocol all your networks will be using different classes and are separated only by a router (Lewis, 2011). Also classful routing protocol will only look at IP addresses and not the subnet mask. This is a problem as you can’t change the subnet mask in any of your network (Lewis, 2011). Most networks today use the classless routing protocol. This protocol supports VLSM and different sizes network within the same IP address class (Lewis, 2011). The first IP address we are given is 12.2.201.2, from this we are ask the question of what is the network address or network number? The network address would fall within the Class “A” and would give us 12.0.0.0. The second question asks us to find out what the broadcast address of the IP address 211.106.32.0 that has a subnet mask of 255.255.255.224. To get the broadcast address we will need to find out how many subnets and hosts that we have...
Words: 1211 - Pages: 5
...Subnet and Router Configuration Author Note This research is being submitted on August 18th, 2013, for Christine Stagnetto-Zweig’s N235/CET2629 Section 02 Cisco Networking Fundamentals and Routing course. Subnet and Router Configuration Given the IP address 154.78.0.0 calculate the subnet masks that would offer 18 subnets, and have the ability to add 6 per year for the next five years, with each subnet having up to 580 hosts? With This IP address (154.78.0.0) in order to give my client 18 available Subnets also with an option of adding 6 more subnets for the next five years and with each subnet having up to 580 hosts per subnet. I have given this much thought and the subnet mask I would use is 255.255.252.0 this will give me 64 available subnets for growth along with 1,022 hosts per subnet I know that’s a little more than 580 hosts but in my mind having doubled that wouldn’t hurt. This will make my first host IP address be 154.78.0.1 and my last host address be 154.78.3.254 leaving 154.78.3.255 my broadcast address and 154.78.0.0 my network address. Indicate the steps and command that you would use to access and configure the router/switch. Configuring the Fast Ethernet Interface Steps | Command | Purpose | 1 | Router>enable | EXEC Mode, Privileged/pswd | 2 | Router# configure terminal | global configuration mode | 3 | Router# show ip interface brief | brief status of the interfaces that are configured for IP | 4 | Router(config)# interface fastethernet...
Words: 423 - Pages: 2
...How to Calculate Subnets Subnets and Hosts Borrow 2 bits S S H H H H H H # of subnets = 22 = 4 Subnet mask = 2 bits = 128 + 64 = 192 Range of hosts = 26 = 64 TT Range Useable Range Network ID 0 – 63 64 – 127 65 - 126 128 – 191 129 - 190 Broadcast 192 – 255 Address Borrow 3 bits S S S H H H H H # of subnets = 23 = 8 Subnet mask = 3 bits = 128 + 64 + 32 = 224 Range of hosts = 25 = 32 Range 0 – 31 32 – 63 64 – 95 96 – 127 128 – 159 160 – 191 192 – 223 224 – 255 Useable Range 33 - 62 65 - 94 97 -126 129 -158 161 -190 193 -222 Network ID Broadcast Address ©1999 Dan Foss How to Calculate Subnets Decimal/Binary Subnet Ranges Borrow 2 bits S S H H H H H H # of subnets = 22 = 4 = 00000100 Subnet mask = 2 bits = 128 + 64 = 192 = 11000000 Range of hosts = 26 = 64 = 01000000 [Range ……………………………] [Useable Range …………………...] Network ID 0 – 63 00 000000 – 00 111111 64 – 127 01 000000 – 01 111111 65 - 126 01 000001 – 01 111110 128 – 191 10 000000 – 10 111111 129 – 190 10 000001 – 10 111110 Broadcast 192 – 255 11 000000 – 11 111111 Address Borrow 3 bits S S S H H H H H # of subnets = 23 = 8 Subnet mask = 3 bits = 128 + 64 + 32 = 224 = 11100000 Range of hosts = 25 = 32 = 00100000 [Range ……………………………] Network ID 0 – 31 000 00000 – 000 11111 32 – 63 001 00000 – 001 11111 64 – 95 010 00000 – 010 11111 96 – 127 011 00000 – 011 11111 128 – 159 100 00000 – 100 11111 160 – 191 101 00000 – 101 11111 192 – 223 110 00000 – 110 11111 Broadcast 224 – 255 111 00000 – 111 11111 Address...
Words: 942 - Pages: 4
...How to Subnet a Network How to use this paper ♦ Absolute Beginner: Read all Sections 1-4 ♦ N eed a q uick rev iew : Read Sections 2 -4 ♦ J ust need a little h elp : Read Section 4 P a r t I : F o r th e I P B e g in n e r IP N e tw o r k A d d r e ss e s To u nd e r st and ne t w o r k IP ad d r e ssi ng , l e t ’ s t ak e a l o o k at p o st al ad d r e sse s. E v e r y m u st h av e i t s o w n u ni q u e ad d r e ss i n o r d e r f o r m ai l t o b e d e l i v e r e d . A n ad d r e ss co d i f f e r e nt p ar t s su ch as t h e st r e e t , nu m b e r , and ci t y . In a ne t w o r k , e v e r y d e v i ce m u i t s o w n u ni q u e IP ad d r e ss. Th at i s, e v e r y ne t w o r k d e v i ce (p r i nt e r , se r v e r , r o u t e r , e m u st b e i d e nt i f i e d w i t h a se p ar at e IP ad d r e ss. M a n y D U si ng t h ar r i v e at (ap ar t m e e v ic e s o r e p o st al e th e rig h t nt nu m b e b u i l d i ng nsi st s o f st h av e t c.) H o s ts x am p l e , t h i nk ab o u t an ap ar t m e nt b u i l d i ng . In o r d e r f o r t h e m ai l t o ap ar t m e nt , e ach ap ar t m e nt m u st h av e i t s o w n u ni q u e i d e nt i f i e r r ) i n ad d i t i o n t o t h e st r e e t nam e and ad d r e ss. W i t h IP ad d r e b u t a si ng l e IP assu m e t h at A and t h at A B C assi g ne d as f o D e D e D e D e D e v i ce v i ce v i ce v i ce v i ce sse s, an o r g ani z at i o n i s assi g ne d a u ni q u e IP ne t w o r k , su ch as 192.168.1.0...
Words: 11640 - Pages: 47
...CTI BEDFORDVIEW ITNT121 Assignment Networking Bradley Barker RJMMLYX21 9/12/2017 Networking Assignment Contents Question 1 1 Question 2 5 Question 3 42 marks 8 Bibliography 11 Question 1 Scenario Read the scenario below and answer the questions that follow: 1.1 You have been hired as a network expert to design and configure this network. Design an appropriate IP addressing scheme for Big Data College. For each subnet provide the following: (8 marks) a. Subnet address b. First IP address c. Last IP address d. Broadcast address Subnet Address First host Last Host Broadcast Lab 1 218.35.50.0 218.35.50.1 218.35.50.62 218.35.50.63 Lab 2 218.35.50.64 218.35.50.65 218.35.50.126 218.35.50.127 Lab 3 218.35.50.128 218.35.50.129 218.35.50.190 218.35.50.191...
Words: 824 - Pages: 4
...156.192.162.56/255.255.255.0 Address Class : __ ___ Bits Borrowed for Subnet: __________ Bits for Host: __________ Number of Usable Subnets: __________ Number of Usable Hosts/Subnet: _______ Network Address: _____ Usable Address Range: __________ Broadcast Address: ____________________ 205.173.45.9/255.255.255.0 Address Class : __ ___ Bits Borrowed for Subnet: __________ Bits for Host: __________ Number of Usable Subnets: __________ Number of Usable Hosts/Subnet: __________ Network Address: _____________ Usable Address Range: __________ Broadcast Address: ________ 204.167.98.45 (0010 1101)/255.255.255.240 (1111 0000) Address Class : _ ____ Bits Borrowed for Subnet: __________ Bits for Host: __________ Number of Usable Subnets: __________ Number of Usable Hosts/Subnet: __________ Network Address: _204.167.98.32 (0010 0000)/ Usable Address Range: __ Broadcast Address: ____ 201.1.97.33(0010 0001)/255.255.255.248(11111000) Address Class : _____ Bits Borrowed for Subnet: __________ Bits for Host: __________ Number of Usable Subnets: __________ Number of Usable Hosts/Subnet: __________ Network Address: ___201.1.97.32(0010 0000)/ Usable Address Range: _______ Broadcast Address: 13.156.59.109 (0110 1101)/255.255.255.248 Address Class : ___ __ Bits Borrowed for Subnet: __________ Bits for Host: __________ Number of Usable Subnets: _____ Number of Usable Hosts/Subnet: __________ Network Address: __13.156.59.104 (0110 1000)_______________ Usable...
Words: 490 - Pages: 2
...REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST SECTION I: Computing Usable Subnets and Hosts SECTION II: Calculating Subnet Masks Lab NETW 202 WEEK 6 LAB REPORT LATEST ...
Words: 491 - Pages: 2
...Advanced Networks Lab Book 2015/2016 Module Leader: Nauman Israr Office Location: IT 1.06 Email: N.Israr@tees.ac.uk Telephone no: 2693 Course Number and Name: Advanced Networks(COM3038-N-BJ1-2015) Year: 2015/2016 Working Time: Timetable Tutorial Time Only in Lab Name of Student: Name of Lab Instructor: Submission Date: TBA Grade: Submission Method: Introduction The purpose of this lab book is to document your solution for a given set of exercises. The relevant concepts used in each exercise will be covered during the lecture. You must complete these exercises within your timetabled tutorial session. The lab book will count towards your final grade. You will be required to submit completed lab book before the submission deadline. Please check with your tutor about the deadline date, time and submission method. This lab book is 10% of your overall mark. 1 Exercise 1: Duration- One Hour You are a network engineer who has been asked to attend an initial meeting with the management team of ABC, LLC. ABC manufactures electric cars. Its new electric car was just picked up by a Government promotional scheme. ABC is upgrading its manufacturing capacity and hiring new employees. Recently, ABC employees have started saying, “The network is too slow.” They are also experiencing problems sending email, accessing web-based applications, and printing. In the past, when the company was small, it didn’t have these...
Words: 2719 - Pages: 11
...Subnetting Workbook Version 1.1 Instructor’s Edition 11111110 10010101 00011011 10000110 IP Address Classes Class A Class B Class C Class D Class E 1 – 127 128 – 191 192 – 223 224 – 239 240 – 255 (Network 127 is reserved for loopback and internal testing) 00000000.00000000.00000000.00000000 Leading bit pattern 0 Network . Host . Host . Host Leading bit pattern Leading bit pattern 10 110 10000000.00000000.00000000.00000000 Network . Network . Host . Host 11000000.00000000.00000000.00000000 Network . Network . Network . Host (Reserved for multicast) (Reserved for experimental, used for research) Private Address Space Class A Class B Class C 10.0.0.0 to 10.255.255.255 172.16.0.0 to 172.31.255.255 192.168.0.0 to 192.168.255.255 Default Subnet Masks Class A Class B Class C 255.0.0.0 255.255.0.0 255.255.255.0 Produced by: Robb Jones jonesr@careertech.net Frederick County Career & Technology Center Cisco Networking Academy Frederick County Public Schools Frederick, Maryland, USA Special Thanks to Melvin Baker and Jim Dorsch for taking the time to check this workbook for errors. Inside Cover Binary To Decimal Conversion 128 64 1 0 1 1 1 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 1 0 0 32 0 1 1 0 1 0 0 1 1 1 1 0 16 1 1 1 0 1 1 0 1 1 1 1 0 8 0 0 1 0 0 0 0 0 1 0 1 0 4 0 1 1 1 1 0 0 0 0 0 0 1 2 1 1 1 0 1 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 1 Answers 146 119 255 197 246 19 129 49 120 240 59 7 27 170 111 248 32 85 62 3 237 192 128 16 2 146 Scratch Area 64 32 16 4 2 1 119 00011011 10101010...
Words: 11837 - Pages: 48
...up collision and broadcast domains. Subnets can reflect organizational structure and help support security policies. WAN links typically join different subnets. Subnets can define administrative units and hence support the structuring and delegation of administrative tasks. Unfortunately, mastering subnetting can pose difficulties for both professionals and students because of the binary mathematics that underlies the technology. While it is imperative to present subnetting concepts in terms of the underlying binary representation, most texts also present subnetting procedures in binary terms. Such an approach can make it difficult for students to learn how to actually carry out subnetting without tables or other reference materials, even when they understand the basic concepts. This paper presents a simple, alternative method for understanding and implementing subnetting without software, calculators, tables, or other aids. The only knowledge of binary arithmetic required is familiarity with the powers of 2 from 0 to 8 (2x for x = 0, 1, …, 8). With a little decimal arithmetic thrown in, the whole process is simple enough to be carried out mentally. This paper assumes the reader is already somewhat familiar with IP addressing, the role of subnet masks, and the uses for subnetting. It proceeds quickly from a brief introduction to a thorough discussion of simple techniques for determining the number of subnets and hosts, calculating the subnet mask, determining (sub)network id’s...
Words: 7956 - Pages: 32
...Exercise Create 7 subnets from the following Class C IP address: 194.27.56.0 subnet mask: 255.255.255.0 Answer the following questions and upload your answers to blackboard. See the next page for an example of how to perform the exercise. 1. How many subnets? 2. How many hosts per subnet? 3. What is the new subnet mask? 4. What is the block size? 5. Create the subnet table. 6. What are the useable subnet address ranges? Example Create 2 subnets from the following Class C IP address: IP Address: 192.168.1.0 Subnet mask: 255.255.255.0 How many subnets? Steal enough bits from the fourth octet to acquire the desired number of subnet. 2n where n equals 2 - 11000000 22 = 4 subnets (Note useable subnets are 22 - 2 = 2 usable subnets), so stealing 2 bits is enough. How many hosts per subnet? 2y – 2 = number of hosts per subnet. Where y is the number of unmasked bits, or the zeros bits in the fourth octet. - 11000000 26 -2 = 62 hosts per subnet What is the new subnet mask? 11111111.11111111.11111111.11000000 - 255.255.255.192 What is the block size? 256 – subnet mask = block size 256 - 192 = 64 What are the blocks? Start at zero and add the block size to the number to get the next number. Stop when you reach the block size. 0, 64, 128, 192 Create the subnet table. Subnet 0 64 128 192 First Host 1 65 129 193 Last Host 62 126 190 194 Broadcast 63 127 191 255 What are the useable subnet address ranges? ...
Words: 289 - Pages: 2
...8. Your company is assigned the network address 150.50.0.0. You need to create seven subnets on the network. A router on one of the subnets will connect the network to the Internet. All computers on the network will need access to the Internet. What is the correct subnet mask for the network? The answer would be C, 255.255.240.0. It would require seven subnets in order for every computer with an IP address. To reach the seven subnets an individual should mask the initial four or the last 16 bits to allow a maximum of 14 subnets. 9. A company with the network ID 209.168.19.0 occupies four floors of a building. You create a subnet for each floor. You want to allow for the largest possible number of host IDs on each subnet. Which subnet mask should you choose? The correct answer for this scenario is 255.255.255.224, for a total of six subnets and the minimum that is required is four. The host ID’s will utilize a subnet mask that allow a few subnets. This allows the biggest quantity of host. 23. Several users are complaining that they cannot access one of your Windows NT file servers, which has an IP address that is accessible from the Internet. When you get pages, you are not in the server room, Instead you are at another company in a friend’s office that only has a UNIX workstation available, which also has internet access. What can you do to see id the file server is still functioning on the network? To see is the file server is still functioning on the network I can...
Words: 439 - Pages: 2