Lab 10.4.1 IP Addressing OverviewEstimated time: 30 min.Objectives:This Lab will focus on your ability to accomplish the following tasks:Name the five different classes of IP addresses Describe the characteristics and use of the different IP address classes Identify the Class of an IP address based on the network number Determine which part (octets) of an IP address is the network ID and which part is the host ID Identify valid and invalid IP host addresses based on the rules of IP addressing Define the range of addresses and default subnet mask for each class Background:This lab will help you develop an understanding of IP addresses and how TCP/IP networks operate. IP addresses are used to uniquely identify individual TCP/IP networks and hosts (computers and printers) on those network in order for devices to communicate. Workstations and servers on a TCP/IP network are called "HOSTS" and each will have a unique IP address which is referred to as its "HOST" address. TCP/IP is the most widely used protocol in the world. The Internet or World Wide Web uses only IP addressing. In order for a host to access the Internet, it must have an IP address.In its basic form, the IP address has two parts; a Network Address and a Host Address. The network portion of the IP address is assigned to a company or organization by the Internet Network Information Center (InterNIC). Routers use the IP address to move data packets between networks. IP Addresses are 32 bits long (with current version IPv4) and are divided into 4 octets of 8 bits each. They operate at the network layer, Layer 3 of the OSI model, (the Internetwork Layer of the TCP/IP model) and are assigned statically (manually) by a network administrator or dynamically (automatically) by a Dynamic Host Configuration Protocol (DHCP) Server. The IP address of a workstation (host) is a "logical address" meaning it can be changed. The MAC address of the workstation is a 48-bit "physical address" which is burned into the NIC and cannot change unless the NIC is replaced. The combination of the logical IP address and the physical MAC address help route packets to their proper destination.There are 5 different classes of IP addresses and depending on the class, the network and host part of the address will use a different number of bits. In this lab you will work with the different classes of IP addresses and become familiar with the characteristics of each. The understanding of IP addresses is critical to your understanding of TCP/IP and Internetworks in general.Tools / Preparation:This is primarily a written lab exercise but you may want to use Control Panel / Network to review some real network IP addresses. The following resources will be required:PC workstation with Windows operating system (Win 95, 98, NT or 2000) installed on PC and access to the Windows Calculator. Notes:Step 1 Review IP Address classes and Their Characteristics.Explanation: There are 5 classes of IP addresses (A thru E). Only the first 3 classes are used commercially. We will discuss a class A network address in the table to get started. The first column is the class of IP address. The second column is the first octet which must fall within the range shown for a given class of address. The class A address must start with a number between 1 and 126. The first bit of a class "A" address is always a zero meaning the High Order Bit (HOB) or the 128 bit cannot be used. 127 is reserved for loop back testing. The first octet alone defines the network ID for a class A network address. The default subnet mask uses all binary ones (decimal 255) to mask the first 8 bits of the class A address. The default subnet mask helps routers and hosts determine if the destination host is on this network or another one. Since there are only 126 class A networks, the remaining 24 bits (3 octets) can be used for hosts. Each class A network can have 2^24 power (2 to the 24th power) or over 16 million hosts. It is common to subdivide the network into smaller groupings called subnets using a custom subnet mask which will be discussed in the next lab.The network or host or host portion of the address can not be all ones or all zeros. As an example, the class A address of 118.0.0.5 is a valid IP address since the network portion (first eight bits equal to 118) is not all zeros and the host portion (the last 24 bits) is not all zeros or all ones. If the host portion were all zeros it would be the network address itself. If the host portion were all 1's it would be a broadcast for the network address. The value of any octet can never be greater than decimal 255 or binary 11111111.Cls 1st Octet Decimal Range 1st Octet High Order Bits Network / Host ID (N=Network, H=Host) Default Subnet Mask Number of Networks Hosts per Network (usable addresses) A 1 126* 0 N.H.H.H 255.0.0.0 126 (27 2) 16,777,214 (2 24 2) B 128 191 1 0 N.N.H.H 255.255.0.0 16,382 (214 - 2) 65,534 (2 16 2) C 192 223 1 1 0 N.N.N.H 255.255.255.0 2,097,150 (221 2) 254 (2 8 2) D 224 239 1 1 1 0 Reserved for Multicasting E 240 254 1 1 1 1 0 Experimental, used for research * Class A address 127 cannot be used and is reserved for loopback and diagnostic functionsStep 2 Basic IP Addressing.Task: Use the IP address chart and your knowledge of IP address classes to answer the following questions.What is the decimal and binary range of the first octet of all possible class "B" IP addresses?Decimal: From: ________ To: ________Binary: From: ________ To: ________Which octet(s) represent the network portion of a class C IP address? ______________Which octet(s) represent the host portion of a class "A" IP address? ________________ Step 3 Determine the host and network portion of the IP address.Task: With the following IP host addresses, indicate the Class of each address, the Network Address or ID, the Host portion, the Broadcast Address for this network and the default Subnet Mask.Explanation: The host portion will be all zeros for the network ID. Enter just the octets that make up the host. The host portion will be all ones for a broadcast. The network portion of the address will be all ones for the subnet mask.Fill in the following table: Host IP Address Addr. Class Network Address Host Address Network Broadcast Address Default Subnet Mask 216.14.55.137 123.1.1.15 150.127.221.244 194.125.35.199 175.12.239.244 Given an IP address of 142.226.0.15 What is the binary equivalent of the second octet? ___________________ What is the Class of the address? _________________________________ What is the network address of this IP address? ____________________ Is this a valid IP host address (Y/N) ? _____________________________ Why or why not? _______________________________________________________ _______________________________________________________________________________________________________________________________________________________________________________________________________________Which is the maximum number of hosts you can have with a class C network address? ______How many class B networks are there? _____________How many hosts can each class B network have ? ______________How many octets are there in an IP address? ________ How many bits per octet? ________Step 4 Determine which IP host addresses are valid for commercial networks.Task: For the following IP host addresses determine which are valid for commercial networks. Why or why not?. Explanation: Valid means it could be assigned to a workstation, server, printer, router interface etc.Fill in the following table. IP Address Valid Address? (Yes/No) Why or why not? 150.100.255.255 175.100.255.18 195.234.253.0 100.0.0.23 188.258.221.176 127.34.25.189 224.156.217.73 LAB 10.4.1.1 IP ADDRESSING ANSWERSStep 2What is the decimal and binary range of the first octet of all possible class "B" IP addresses?Decimal: From: 128 To: 191Binary: From: 10000000 To: 10111111Which octet(s) represent the network portion of a class "C" IP address? The first three octetsWhich octet(s) represent the host portion of a class "A" IP address? The last three octetsStep 3Fill in the following Host IP Address Addr. Class Network Address Host Address Network Broadcast Address Default Subnet Mask 216.14.55.137 C 216.14.55.0 137 216.14.55.255 255.255.255.0 123.1.1.15 A 123.0.0.0 1.1.15 123.255.255.255 255.0.0.0 150.127.221.244 B 150.127.0.0 221.244 150.127.255.255 255.255.0.0 194.125.35.199 C 194.125.35.0 199 194.125.35.255 255.255.255.0 175.12.239.244 B 175.12.0.0 239.244 175.12.255.255 255.255.0.0 Given an IP address of 142.226.0.15 a. What is the binary equivalent of the second octet? 11100010b. What is the Class of the address? Class Bc. What is the network address of this IP address? 142.226.0.0 (First two octets followed by zeros)d. Is this a valid IP host address? Yese. Why or why not? The network part of the address (the first 2 octets or 16 bits) is 142.226 which is a valid class B address. The host part (last 2 octets or 16 bits) is 0.15 or 00000000.00001111. Since all 16 bits of the host ID are not all zeros or all ones it is a valid host address.Which is the maximum number of hosts you can have with a class C network address? 254 The host portion of the address cannot be all ones (11111111 binary or 255 decimal) or all zeros (00000000 binary or 0 decimal). There are 8 bits or 2^8 power = 265 minus 2 = 254.How many class B networks are there? 16,382 Although there are 16 bits in the first two octets for networks, the first two bits are always 10 and do not change. That leaves 14 bits or 2^14 power minus 2 = 16,382How many hosts can each class B network have ? 65,534. There are 16 bits left in the host portion of the address and you cannot use the all zeros (network address) or all ones (broadcast address) value. That means 16 bits or 2^16 power minus 2 = 65,534How many octets are there in an IP address? 4 Four How many bits per octet? 8 Eight Step 4fill in the following table: IP Address Valid Address? (Yes/No) Why or why not? 150.100.255.255 NO 150.100.0.0.is a Class B network. This is a broadcast address for a class B (host portion 3rd and 4th octets is all ones) and cannot be used for a host address. 175.100.255.18 YES 175.100.0.0 is a class B network. The host portion is the 3rd and 4th octets (16 bits taken together) 11111111.00010010 and is not all zeros or all ones. It is valid even though the 3rd octet is all ones. 195.234.253.0 NO 195.234.253.0 is a Class C network. This is the network address or ID for this network and cannot be used for a host address since all the host bits are zeros. 100.0.0.23 YES 100.0.0.0 is a Class A network. The host portion of the address is the 2nd, 3rd and 4th octets (24 bits taken together) 00000000.00000000.00010111 and is not all zeros or all ones. It is valid even though the 2nd and 3rd octets are all zeros. 188.258.221.176 NO This would be a class B network but is invalid because the 2nd octet is greater than 255. No octet can be greater then 255 (all ones) in any IP address (network or host). 127.34.25.189 NO This would be a Class A network but is invalid because the 127 can't be used in the first octet since it is reserved for diagnostic testing. 224.156.217.73 NO This is a Class D network and class D is reserved for multicasting and can't be used as a commercial IP address. Lab 10.6.6 Subnet Mask 1 Estimated time: 45 min.Objectives:This Lab will focus on Class C subnet masks and your ability to accomplish the following tasks:Cite some reasons why a subnet mask would be needed Distinguish between a Default Subnet Mask and a Custom Subnet Mask Determine the subnets available with a particular IP network address and Subnet mask Given a network address and requirements for how many subnets and hosts, be able to determine what subnet mask should be used Given a network address and a subnet mask, be able to determine the number of subnets and host per subnet that can be created as well as useable subnets and useable number of hosts Use the "ANDing" process to determine if a destination IP address is Local or Remote Identify valid and invalid IP host address based on a given a Network number and subnet mask Background:This lab will help you understand the basics of IP subnet masks and their use with TCP/IP networks. The subnet mask can be used to split up an existing network into "subnetworks" or "subnets". This may be done to 1) reduce the size of the broadcast domains (create smaller networks with less traffic), 2) to allow LANs in different geographical locations to communicate or 3) for security reason to separate one LAN from another. Routers separate subnets and the router determines when a packet can go from one subnet to another. Each router a packet goes through is considered a "hop". Subnet masks help workstations, servers and routers in an IP network determine if the destination host for the packet they the want to send is on their own network or another network. Default subnet masks were discussed in a prior lab. This Lab will review the Default Subnet Mask and then focus on Custom Subnet Masks which will use more bits than the default subnet mask by "borrowing" these bits from the host portion of the IP address. This creates a three-part address; 1) The original network address assigned, 2) The subnet address made up of the bits borrowed and 3) the host address made up of the bits left after borrowing some for subnets.Tools / Preparation:This is primarily a written lab exercise but you may want to use Control Panel / Network to review some real network IP addresses. The following resources will be required.PC workstation with Windows operating system (Win 95, 98, NT or 2000) installed on the PC and access to the Windows Calculator. Notes:Step 1 IP Address Basics.Explanation: IP network addresses are assigned by the Internet Network Information Center (InterNIC). If your organization has a class "A" IP network address, the first octet (8 bits) is assigned be InterNIC and your organization can use the remaining 24 bits to define up to 16,777,214 hosts on your network. This is a lot of hosts! It is not possible to put all of these hosts on one physical network without separating them with routers and subnets. A workstation may be on one network or subnet and a server may be on another network or subnet. When the workstation needs to retrieve a file from the server it will need to use its subnet mask to determine the network or subnet that the server is on. The purpose of a subnet mask is to help hosts and routers determine the network location where a destination host can be found. Refer to the following table to review IP address classes, default subnet masks and the number of networks and hosts that can be created with each class of network address.Cls 1st Octet Decimal Range 1st Octet High Order Bits Network / Host ID (N=Network, H=Host) Default Subnet Mask Number of Networks Hosts per Network (usable addresses) A 1 126* 0 N.H.H.H 255.0.0.0 126 (27 2) 16,777,214 (2 24 2) B 128 191 1 0 N.N.H.H 255.255.0.0 16,382 (214 - 2) 65,534 (2 16 2) C 192 223 1 1 0 N.N.N.H 255.255.255.0 2,097,150 (221 2) 254 (2 8 2) D 224 239 1 1 1 0 Reserved for Multicasting E 240 254 1 1 1 1 0 Experimental, used for research Step 2 The "ANDing" process.Explanation: Hosts and routers use the "ANDing" process to determine if a destination host is on the same network or not. The ANDing process is done each time a host wants to send a packet to another host on an IP network. If you want to connect to a server, you may know the IP address of the server you want to connect to or you may just enter the host name (e.g. www.cisco.com) and a Domain Name Server (DNS) will convert the host name to an IP address. First the source host will compare (AND) its own IP address to its own subnet mask. The result of the ANDing is to identify the network where the source host resides. It will then compare the destination IP address to its own subnet mask. The result of the 2nd ANDing will be the network that the destination host is on. If the source network address and the destination network address are the same they can communicate directly. If the results are different then they are on different networks or subnets and will need to communicate through routers or may not be able to communicate at all.ANDing depends on the subnet mask. A default subnet mask for a Class C network is 255.255.255.0 or 11111111.111111111.111111111.00000000. This is compared to the source IP address bit for bit. The first bit of the IP address is compared to the first bit of the subnet mask and the second bit to the second etc. If the two bits are both ones, then the ANDing result is a ONE. If the two bits are a zero and a one or two zeros then the ANDing result is a ZERO. Basically this means that a combination of 2 ones results in a ONE, anything else is a zero. The result of the ANDing process is the network or subnet number that the source or destination address is on. Step 3 Two Class C networks using the default subnet mask.Explanation: This example will show how a Class C default subnet mask can be used to determine which network a host is on. A default subnet mask does not break an address into subnets. If the default subnet mask is used then the network is not being "subnetted". Host X (source) on network 200.1.1.0 has an IP address of 200.1.1.5 and wants to send a packet to host Z (destination) on network 200.1.2.0 and has an IP address of 200.1.2.8. All hosts on each network are connected to hubs or switches and then to a router. Remember that with a Class C network address ARIN assigns the first 3 octets (24 bits) as the network address so these are two different class C networks. This leaves one octet (8 bits) for hosts so each class C network could have up to 254 hosts (2^8 power = 256 2 = 254).The ANDing process will help the packet get from host 200.1.1.5 on network 200.1.1.0 to host 200.1.2.8 on network 200.1.2.0 using the following steps.Host X compares its own IP address to its own subnet mask using the ANDing process Host X IP address 200.1.1.5 11001000.00000001.00000001.00000101 Subnet Mask 255.255.255.0 11111111.11111111.11111111.00000000 ANDing Result (200.1.1.0) 11001000.00000001.00000001.00000000 NOTE: The result of step 3a of the ANDing process is the network address of host X which is 200.1.1.0 Next host X compares the IP address of the Host Z destination to its own subnet mask using the ANDing process.Host Z IP address 200.1.2.8 11001000.00000001.00000010.00001000 Subnet Mask 255.255.255.0 11111111.11111111.11111111.00000000 ANDing Result (200.1.2.0) 11001000.00000001.00000010.00000000 NOTE: The result of step3b ANDing process is the network address of host Z which is 200.1.2.0.Host X compares the ANDing results from step A and the ANDing result from step B and they are different. Host X now knows that host Z is not in its Local Area Network (LAN) and it must send the packet to its "Default Gateway" which is the IP address of the router interface of 200.1.1.1 on network 200.1.1.0. The router will then repeat the ANDing process to determine which router interface to send the packet out.Step 4 One Class C network using a Custom subnet mask.Explanation: This example uses a single Class C network address (200.1.1.0) and will show how a class C custom subnet mask can be used to determine which subnetwork (or subnet) a host is on and to route packets from one subnetwork to another. Remember that with a class C network address ARIN assigns the first 3 octets (24 bits) as the network address. This leaves 8 bits (one octet) for hosts so each class C network could have up to 254 hosts (2^8 power = 256 2 = 254).Perhaps you want less than 254 host (workstations and servers) all on one network and you want to create 2 sub-networks and separate them with a router for security reason or to reduce traffic. This will create smaller independent broadcast domains and can improve network performance and increase security since these subnetworks will be separated by a router. Assume you will need at least 2 subnetworks and at least 50 hosts per subnetwork. Since you only have one Class C network address you have only 8 bits in the fourth octet available for a total of 254 possible hosts, you must create a Custom Subnet mask. You will use the custom subnet mask to "BORROW" bits from the host portion of the address. The following steps will help accomplish this:The first step to "subnetting" is to determine how many subnets are needed. In this case you will need 2 subnetworks. To see how many bits you should borrow from the host portion of the network address, add the bit values from right to left until the total is equal to or greater than the number of subnets you will need. Since we need 2 subnets, add the one bit and the two bit which equals three. This is over the number of subnets we need, so we will need to borrow at least two bits from the host address starting from the left side of the octet that contains the host address. Network address: 200.1.1.0 4th octet Host address bits: 1 1 1 1 1 1 1 1 Host address bit values (from right) 128 64 32 16 8 4 2 1 (Add bits starting from the right side (the 1 and the 2) until you get more than the number of subnets needed)Once we know how many bits to borrow we take them from the left side of the first octet of the host address. Every bit we borrow from the host leaves fewer bits for the hosts. Even though we increase the number of subnets, we decrease the number of hosts per subnet. Since we need to borrow 2 bits from the left side, we must show that new value in our subnet mask. Our existing default subnet mask was 255.255.255.0 and our new "Custom" subnet mask is 255.255.255.192. The 192 comes from the value of the first two bits from the left (128 + 64 = 192). These bits now become 1s and are part of the overall subnet mask. This leaves 6 bits for host IP addresses or 2^6 = 64 hosts per subnet.4th Octet borrowed bits for subnet: 1 1 1 1 1 1 1 1 Subnet bit values: (from left side) 128 64 32 16 8 4 2 1 With this information you can build the following table. The first two bits are the Subnet binary value. The last 6 bits are the host bits. By borrowing 2 bits from the 8 bits of the host address you can create 4 subnets with 64 hosts each. The 4 networks created are the "0" net, the "64" net, the "128" net and the "192" net. The "0" net and the "192" net are considered unusable. This is because the "0" net has all zeros in the subnet portion of the address and the 192 net has all ones in the subnet portion of the address.Subnet No. Subnet bits borrowed Binary value Subnet bits Decimal Value Host bits possible binary values (range) (6 bits) Subnet / Host Decimal range Useable? Subnet #0 00 0 000000 111111 0 63 NO Subnet #1 01 64 000000 111111 64 127 YES Subnet #2 10 128 000000 111111 128 191 YES Subnet #3 11 192 000000 111111 192 254 NO Notice that the first subnet always starts at 0 and, in this case, increases by 64 which is the number of hosts on each subnet. One way to determine the number of hosts on each subnet or the start of each subnet is to take the remaining host bits to the power of 2. Since we borrowed two of the 8 bits for subnets and have six bits left, the number of hosts per subnet is 2^6 or 64. Another way to figure the number of host per subnet or the "increment" from one subnet to the next is to subtract the subnet mask value in decimal (192 in the fourth octet) from 256 (which is maximum number of possible combinations of 8 bits) which equals 64. This means you start at 0 for the first network and add 64 for each additional subnetwork. If we take the second subnet (the 64 net) as an example the IP address of 200.1.1.64 cannot be used for a host ID because it is the "network ID" of the "64" subnet (host portion is all zeros) and the IP address of 200.1.1.127 cannot be used because it is the broadcast address for the 64 net (host portion is all ones).Step 5 One Class C network using a Custom Subnet Mask.Task: Use the following information and the previous examples to answer the following subnet related questions.Explanation: Your company has applied for and received a Class C network address of 197.15.22.0. You want to subdivide your physical network into 4 subnets, which will be interconnected by routers. You will need at least 25 hosts per subnet. You will need to use a Class C custom subnet mask and will have a router between the subnets to route packet from one subnet to another. Determine the number of bits you will need to borrow from the host portion of the network address and then the number of bits left for host addresses. (Hint: There will be 8 subnets)Fill in the table below and answer the following questions: Subnet No. Subnet bits borrowed Binary value Subnet bits Decimal & Subnet No. Host bits possible binary values (range) (6 bits) Subnet / Host Decimal range Use? Subnet #0 Subnet #1 Subnet #2 Subnet #3 Subnet #4 Subnet #5 Subnet #6 Subnet #7 Notes:QUESTIONS: Use the table you just developed above to help answer the following questions:Which octet(s) represent the network portion of a Class C IP address? _______________ Which octet(s) represent the host portion of a Class C IP address? _________________ What is the binary equivalent of the Class C network address in the scenario (197.15.22.0)? Decimal Network address: __________ . __________ . __________ . __________ Binary Network address: __________ . __________ . __________ . __________ How many high-order bits were borrowed from the host bits in the fourth octet? _____ What subnet mask must you use (show the subnet mask in decimal and binary)? Decimal Subnet mask: __________ . __________ . __________ . __________ Binary subnet mask: __________ . __________ . __________ . __________ What is the maximum number of subnets that can be created with this subnet mask? _____ What is the maximum number of useable subnets that can be created with this mask? ____ How many bits were left in the 4th octet for host IDs ? ______ How many hosts per subnet can be defined with this subnet mask? _____________ What is the maximum number of hosts that can be defined for all subnets with this scenario (assuming you cannot use the lowest and highest subnet numbers and cannot use the lowest and highest host ID on each subnet) ? _____________________________________________ Is 197.15.22.63 a valid host IP address with this scenario? ____________Why or why not ? _____________________________________________________________Is 197.15.22.160 a valid host IP address with this scenario? ______Why or why not ? _____________________________________________________________Host "A" has an IP address of 197.15.22.126. Host "B" has an IP address of 197.15.22.129. Are these hosts on the same subnet? ______ Why? ________________________________ ___________________________________________________________________________ LAB 10.6.6 SUBNET MASK 1 ANSWERSStep 5Fill in the table below and answer the following questions: Subnet No. Subnet bits borrowed Binary value Subnet bits Decimal & Subnet No. Host bits possible binary values (range) (6 bits) Subnet / Host Decimal range Use? Subnet #0 000 0 (197.15.22.0) 00000 11111 0 31 NO Subnet #1 001 32 (197.15.22.32 00000 11111 32 63 YES Subnet #2 010 64 (197.15.22.64) 00000 11111 64 95 YES Subnet #3 011 96 (197.15.22.96) 00000 11111 96 127 YES Subnet #4 100 128 (197.15.22.128) 00000 11111 128 159 YES Subnet #5 101 160 (127.15.22.160) 00000 11111 160 191 YES Subnet #6 110 192 (127.15.22.192) 00000 11111 192 223 YES Subnet #7 111 224 (127.15.22.224) 00000 11111 224 255 NO Notes: Lab 10.7.5 Subnetmask Estimated time: 45 min.Objectives:This Lab will focus on your ability to accomplish the following tasks:Work with Class B network addresses and subnets Determine the subnets available with a particular IP Network address and Subnet mask Given a network address and requirements for how many subnets and hosts, be able to determine what subnet mask should be used Given a network address and a subnet mask, be able to determine the number of subnets and hosts per subnet that can be created as well as useable subnets and useable number of hosts Assign IP addresses and subnet masks to hosts and router interfaces Use the "ANDing" process to track an IP packet from a local host to a remote host through a router Background:This lab will focus on a Class B network with three subnets and using a subnet mask. This lab will focus on a Class B network with three subnets and using a subnet mask.Tools / Preparation:This is primarily a written lab exercise but you will want to use Control Panel / Network to review some real network IP addresses and the basics covered in Labs 1.10 and 1.11. The following resources will be required:PC workstation with Windows operating system (Win 95, 98, NT or 2000) installed on PC to the Windows Calculator. Notes:Step 1 IP Address BasicsExplanation: For reference, the IP addressing table is included here. IP network addresses are assigned by ARIN. You will work with a Class B.Adr Cls 1st Octet Decimal Range 1st Octet High Order Bits Network / Host ID (N=Network, H=Host) Default Subnet Mask Number of Networks Hosts per Network (usable addresses) A 1 126* 0 N.H.H.H 255.0.0.0 126 (27 2) 16,777,214 (2 24 2) B 128 191 1 0 N.N.H.H 255.255.0.0 16,382 (214 - 2) 65,534 (2 16 2) C 192 223 1 1 0 N.N.N.H 255.255.255.0 2,097,150 (221 2) 254 (2 8 2) D 224 239 1 1 1 0 Reserved for Multicasting E 240 254 1 1 1 1 0 Experimental, used for research Step 2 Class B network address with 3 subnets.Task: Use the information below and from prior labs to help determine your valid subnets and host IP addresses. Answer the following questions.Explanation: Your institution has a Class B network address of 150.193.0.0. This class B network address will be subdivided to accommodate your physical network and you will need at least 50 subnets interconnected with routers. Each subnet needs to be able to handle at least 750 hosts per subnet (workstations, servers and router interfaces). As the network manager for your local campus at the institution, you were given the first 10 of these subnets for use with your local campus. You will be using 6 of these subnets now and will keep the others for future growth. Do NOT use the first or last subnet.What is the binary equivalent of the Class B network address 150.193.0.0 in the exercise? ___________ . _____________ . _____________ . ______________Which octet(s) and how many bits are used to represent the network portion of this network address? _______________________________Which octet(s) and how many bits represent the host portion of this Class B network address? _______________________________How many original Class B networks are there? ____________________What is the total number of hosts that can be created with a Class B network address if it has not been subdivided? _____________________________How many bits must you borrow from the host portion of the network address in order to provide at least 50 subnets and at least 750 hosts per subnet? _______________________What will the Subnet Mask be (using dotted decimal notation) based on the number of bits borrowed in step 6? ____________________________What is the binary equivalent of the subnet mask above: ___________ . ____________ . ____________ . ____________ Step 3 Class B network address with 3 subnets.Task: Use the diagram at the end of the worksheet to fill in the blanks and answer the following questions. Explanation: Be sure to specify all four octets for subnet address and subnet mask. The same subnet mask should be used for all hosts, router interfaces and all subnets. Having a common subnet mask will allow hosts and routers to determine which subnet the IP packet is intended for. Router interfaces will usually be numbered first when assigning IP addresses and hosts will receive higher numbers. Fill in the following table for each of the possible subnets that can be created by borrowing 6 bits for subnets from the third octet (1st host octet). Identify the Network Address, the Subnet Mask, the Subnetwork Address, the range of possible host IP addresses for each subnet, the broadcast address of each subnet and also indicate whether the subnet is useable or not. You will only use 3 of these subnets for the exercise. SN# Network Address Subnet mask Subnetwork Address Range of possible Host IP Addresses Broadcast Address Use? 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th Assign an IP Address and Subnet Mask to router interface A and write it down here. __________________ / ___________________ Assign an IP Address and Subnet Mask of router interface B and write it down here. __________________ / ___________________Assign an IP Address and Subnet Mask of router interface C and write it down here. __________________ / ___________________Assign a host IP Address to Host X on Subnet A and assign an IP address to Host Z on Subnet C (answers may vary). Describe the steps (using ANDing) for the process of sending an IP packet from Host X to Host Z through the router. Remember, when ANDing two 1s together the result is a 1, ANDing any other combination (1 and 0, 0 and 1 or 0 and 0) results in a zero (0). Also, when ANDing two network IP addresses together the result of the ANDing process will be the network (or subnetwork) address of the destination IP address in the packet. Use the information from the diagram above to help assign IP addresses and subnet masks. What is the result of the ANDing process for Host X?Decimal Host X IP addr: _________ . _________ . ___________ . ____________Binary Host X IP addr: _________ . _________ . ___________ . ____________Binary Subnet Mask: _________ . _________ . ___________ . ____________Binary ANDing Result: _________ . _________ . ___________ . ____________Decimal ANDing Result: _________ . _________ . ___________ . ____________What is the result of the ANDing process for Host Z?Decimal Host Z IP addr: _________ . _________ . ___________ . ____________Binary Host Z IP addr: _________ . _________ . ___________ . ____________Binary Subnet Mask: _________ . _________ . ___________ . ____________Binary ANDing Result: _________ . _________ . ___________ . ____________Decimal ANDing Result: _________ . _________ . ___________ . ____________The Decimal ANDing result from questions 7 is the network/subnet that Host X is on. The result from question 8 is the network/subnet that Host Z is on. Are Host X and Host Z on the same network/subnet? ______________What will Host X now do with the packet ? __________________________________________ _________________________________________________________________________________Fill in the blanks in the following diagram with the correct Network and IP addresses. LAB 10.7.6.2 SUBNET MASK 3 ANSWERSStep 21. What is the binary equivalent of the class B network address 150.193.0.0 in the exercise?10010110 . 11000001 . 00000000 . 00000000Which octet(s) and how many bits are used to represent the network portion of this network address?The first and second octets or 16 bitsWhich octet(s) and how many bits represent the host portion of this class B network address? The third and fourth octets or 16 bitsHow many original Class B networks are there? 6,382 (2^14 2)What is the total number of hosts that can be created with a class B network address if it has not been subdivided? 65,534 (2^16 2)How many bits must you borrow from the host portion of the network address in order to provide at least 50 subnets and at least 750 hosts per subnet? 6 bits (2^6 = 64 2 = 62 possible subnetworks. This leaves 10 bits or 2^10 2 = 1022 hosts per subnet)What will the Subnet Mask be (using dotted decimal notation) based on the number of bits borrowed in step 6? 255 . 255 . 252 . 0 (The first 6 bit are borrowed from the left side of the host address. 128 + 64 + 32 + 16 + 8 + 4 = 252)What is the binary equivalent of the subnet mask above:11111111 . 11111111 . 11111100 . 00000000 (The first two octets (16 bits of all 1s) mask the original network address. The first 6 bits of the host portion 3rd octet are all 1s and mask the subnet number. The last 10 bits of 0s are reserved for host Ids.) Step 3Fill in the following table for each of the possible subnets that can be created by borrowing 6 bits for subnets from the third octet (1st host octet). SN# Network Address Subnet mask Subnetwork Address Range of possible Host IP Addresses Broadcast Address Use? 1st 150.193.0.0 255.255.252.0 150.193.0.0 150.193.0.1 3.254 150.193.3.255 No 2nd 150.193.0.0 255.255.252.0 150.193.4.0 150.193.4.1 7.254 150.193.3.255 Y 3rd 150.193.0.0 255.255.252.0 150.193.8.0 150.193.8.1 11.254 150.193.3.255 Y 4th 150.193.0.0 255.255.252.0 150.193.12.0 150.193.12.1 15.254 150.193.3.255 Y 5th 150.193.0.0 255.255.252.0 150.193.16.0 150.193.16.1 19.254 150.193.3.255 Y 6th 150.193.0.0 255.255.252.0 150.193.20.0 150.193.20.1 23.254 150.193.3.255 Y 7th 150.193.0.0 255.255.252.0 150.193.24.0 150.193.24.1 27.254 150.193.3.255 Y 8th 150.193.0.0 255.255.252.0 150.193.28.0 150.193.28.1 31.254 150.193.3.255 Y 9th 150.193.0.0 255.255.252.0 150.193.32.0 150.193.32.1 35.254 150.193.3.255 Y 10th 150.193.0.0 255.255.252.0 150.193.36.0 150.193.36.1 39.254 150.193.3.255 Y Assign an IP Address and Subnet Mask to router interface A and write it down here.150.193.4.1 / 255.255.252.0Assign an IP Address and Subnet Mask of router interface B and write it down here.150.193.8.1 / 255.255.252.0Assign an IP Address and Subnet Mask of router interface C and write it down here.150.193.12.1 / 255.255.252.0Assign a host IP Address to Host X on Subnet A and assign an IP address to Host Z on Subnet C (answers may vary). Describe the steps (using ANDing) for the process of sending an IP packet from Host X to host Z through the router. Use the information from the diagram above and in Lab 1.10 to help assign IP addresses and subnet masks. Host X = 150.193.4.2 (4.1 was used for the router interface on subnet A), Host Z = 150.193.12.2 (12.1 was used for the router interface on subnet C). Host X compares (ANDs) the subnet mask to its own IP address and comes up with its own network/subnet address of 150.193.4.0. It then compares the subnet mask to the IP address of the destination host (150.193.12.2) and comes up with the network/subnet address of the target network (150.193.12.0). Since the two do not match, host X must assume that the destination host is not on its network and it send the packet to its "Default Gateway" or the nearside port of the router (interface A). The router goes through the same process on its incoming interface A and determines that network 150.193.12.2 is on it's C interface. The router forwards the packet to interface C and since the Router also knows the MAC address of hosts directly attached to its interfaces such as host Z, it forwards the packet to the 150.193.12.0 network/subnet LAN and host Z picks it up.What is the result of the ANDing process for host X?Decimal Host X IP addr: 150 . 193 . 4 . 2Binary Host X IP addr: 10010110 . 11000001 . 00000100 . 00000010Binary Subnet Mask: 11111111 . 11111111 . 11111100 . 00000000Binary ANDing Result: 10010110. 11000001 . 00000100 . 00000000Decimal ANDing Result: 150 . 193 . 4 . 0What is the result of the ANDing process for host Z?Decimal Host Z IP addr: 150 . 193 . 12 . 2Binary Host Z IP addr: 10010110 . 11000001 . 00001100 . 00000010Binary Subnet Mask: 11111111 . 11111111 . 11111100 . 00000000Binary ANDing Result: 10010110. 11000001 . 00001100 . 00000000Decimal ANDing Result: 150 . 193 . 12 . 0The Decimal ANDing result from questions 6 is the network/subnet that Host X is on. The result from question 7 is the network/subnet that Host Z is on. Are Host X and Host Z on the same network/subnet?NOWhat will host X now do with the packet ? Since the destination host Z is not on the Host X Local Area Network, Host X will send the packet to the "Default Gateway" which is the IP address of Interface A on the routerFill in the blanks in the following diagram with the correct Network and IP addresses. Lab 10.7.5 Subnetmask Estimated time: 45 min.Objectives:This Lab will focus on your ability to accomplish the following tasks:Work with Class B network addresses and subnets Determine the subnets available with a particular IP Network address and Subnet mask Given a network address and requirements for how many subnets and hosts, be able to determine what subnet mask should be used Given a network address and a subnet mask, be able to determine the number of subnets and hosts per subnet that can be created as well as useable subnets and useable number of hosts Assign IP addresses and subnet masks to hosts and router interfaces Use the "ANDing" process to track an IP packet from a local host to a remote host through a router Background:This lab will focus on a Class B network with three subnets and using a subnet mask. This lab will focus on a Class B network with three subnets and using a subnet mask.Tools / Preparation:This is primarily a written lab exercise but you will want to use Control Panel / Network to review some real network IP addresses and the basics covered in Labs 1.10 and 1.11. The following resources will be required:PC workstation with Windows operating system (Win 95, 98, NT or 2000) installed on PC to the Windows Calculator. Notes:Step 1 IP Address BasicsExplanation: For reference, the IP addressing table is included here. IP network addresses are assigned by ARIN. You will work with a Class B.Adr Cls 1st Octet Decimal Range 1st Octet High Order Bits Network / Host ID (N=Network, H=Host) Default Subnet Mask Number of Networks Hosts per Network (usable addresses) A 1 126* 0 N.H.H.H 255.0.0.0 126 (27 2) 16,777,214 (2 24 2) B 128 191 1 0 N.N.H.H 255.255.0.0 16,382 (214 - 2) 65,534 (2 16 2) C 192 223 1 1 0 N.N.N.H 255.255.255.0 2,097,150 (221 2) 254 (2 8 2) D 224 239 1 1 1 0 Reserved for Multicasting E 240 254 1 1 1 1 0 Experimental, used for research Step 2 Class B network address with 3 subnets.Task: Use the information below and from prior labs to help determine your valid subnets and host IP addresses. Answer the following questions.Explanation: Your institution has a Class B network address of 150.193.0.0. This class B network address will be subdivided to accommodate your physical network and you will need at least 50 subnets interconnected with routers. Each subnet needs to be able to handle at least 750 hosts per subnet (workstations, servers and router interfaces). As the network manager for your local campus at the institution, you were given the first 10 of these subnets for use with your local campus. You will be using 6 of these subnets now and will keep the others for future growth. Do NOT use the first or last subnet.What is the binary equivalent of the Class B network address 150.193.0.0 in the exercise? ___________ . _____________ . _____________ . ______________Which octet(s) and how many bits are used to represent the network portion of this network address? _______________________________Which octet(s) and how many bits represent the host portion of this Class B network address? _______________________________How many original Class B networks are there? ____________________What is the total number of hosts that can be created with a Class B network address if it has not been subdivided? _____________________________How many bits must you borrow from the host portion of the network address in order to provide at least 50 subnets and at least 750 hosts per subnet? _______________________What will the Subnet Mask be (using dotted decimal notation) based on the number of bits borrowed in step 6? ____________________________What is the binary equivalent of the subnet mask above: ___________ . ____________ . ____________ . ____________ Step 3 Class B network address with 3 subnets.Task: Use the diagram at the end of the worksheet to fill in the blanks and answer the following questions. Explanation: Be sure to specify all four octets for subnet address and subnet mask. The same subnet mask should be used for all hosts, router interfaces and all subnets. Having a common subnet mask will allow hosts and routers to determine which subnet the IP packet is intended for. Router interfaces will usually be numbered first when assigning IP addresses and hosts will receive higher numbers. Fill in the following table for each of the possible subnets that can be created by borrowing 6 bits for subnets from the third octet (1st host octet). Identify the Network Address, the Subnet Mask, the Subnetwork Address, the range of possible host IP addresses for each subnet, the broadcast address of each subnet and also indicate whether the subnet is useable or not. You will only use 3 of these subnets for the exercise. SN# Network Address Subnet mask Subnetwork Address Range of possible Host IP Addresses Broadcast Address Use? 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th Assign an IP Address and Subnet Mask to router interface A and write it down here. __________________ / ___________________ Assign an IP Address and Subnet Mask of router interface B and write it down here. __________________ / ___________________Assign an IP Address and Subnet Mask of router interface C and write it down here. __________________ / ___________________Assign a host IP Address to Host X on Subnet A and assign an IP address to Host Z on Subnet C (answers may vary). Describe the steps (using ANDing) for the process of sending an IP packet from Host X to Host Z through the router. Remember, when ANDing two 1s together the result is a 1, ANDing any other combination (1 and 0, 0 and 1 or 0 and 0) results in a zero (0). Also, when ANDing two network IP addresses together the result of the ANDing process will be the network (or subnetwork) address of the destination IP address in the packet. Use the information from the diagram above to help assign IP addresses and subnet masks. What is the result of the ANDing process for Host X?Decimal Host X IP addr: _________ . _________ . ___________ . ____________Binary Host X IP addr: _________ . _________ . ___________ . ____________Binary Subnet Mask: _________ . _________ . ___________ . ____________Binary ANDing Result: _________ . _________ . ___________ . ____________Decimal ANDing Result: _________ . _________ . ___________ . ____________What is the result of the ANDing process for Host Z?Decimal Host Z IP addr: _________ . _________ . ___________ . ____________Binary Host Z IP addr: _________ . _________ . ___________ . ____________Binary Subnet Mask: _________ . _________ . ___________ . ____________Binary ANDing Result: _________ . _________ . ___________ . ____________Decimal ANDing Result: _________ . _________ . ___________ . ____________The Decimal ANDing result from questions 7 is the network/subnet that Host X is on. The result from question 8 is the network/subnet that Host Z is on. Are Host X and Host Z on the same network/subnet? ______________What will Host X now do with the packet ? __________________________________________ _________________________________________________________________________________Fill in the blanks in the following diagram with the correct Network and IP addresses. LAB 10.7.6.2 SUBNET MASK 3 ANSWERSStep 21. What is the binary equivalent of the class B network address 150.193.0.0 in the exercise?10010110 . 11000001 . 00000000 . 00000000Which octet(s) and how many bits are used to represent the network portion of this network address?The first and second octets or 16 bitsWhich octet(s) and how many bits represent the host portion of this class B network address? The third and fourth octets or 16 bitsHow many original Class B networks are there? 6,382 (2^14 2)What is the total number of hosts that can be created with a class B network address if it has not been subdivided? 65,534 (2^16 2)How many bits must you borrow from the host portion of the network address in order to provide at least 50 subnets and at least 750 hosts per subnet? 6 bits (2^6 = 64 2 = 62 possible subnetworks. This leaves 10 bits or 2^10 2 = 1022 hosts per subnet)What will the Subnet Mask be (using dotted decimal notation) based on the number of bits borrowed in step 6? 255 . 255 . 252 . 0 (The first 6 bit are borrowed from the left side of the host address. 128 + 64 + 32 + 16 + 8 + 4 = 252)What is the binary equivalent of the subnet mask above:11111111 . 11111111 . 11111100 . 00000000 (The first two octets (16 bits of all 1s) mask the original network address. The first 6 bits of the host portion 3rd octet are all 1s and mask the subnet number. The last 10 bits of 0s are reserved for host Ids.) Step 3Fill in the following table for each of the possible subnets that can be created by borrowing 6 bits for subnets from the third octet (1st host octet). SN# Network Address Subnet mask Subnetwork Address Range of possible Host IP Addresses Broadcast Address Use? 1st 150.193.0.0 255.255.252.0 150.193.0.0 150.193.0.1 3.254 150.193.3.255 No 2nd 150.193.0.0 255.255.252.0 150.193.4.0 150.193.4.1 7.254 150.193.3.255 Y 3rd 150.193.0.0 255.255.252.0 150.193.8.0 150.193.8.1 11.254 150.193.3.255 Y 4th 150.193.0.0 255.255.252.0 150.193.12.0 150.193.12.1 15.254 150.193.3.255 Y 5th 150.193.0.0 255.255.252.0 150.193.16.0 150.193.16.1 19.254 150.193.3.255 Y 6th 150.193.0.0 255.255.252.0 150.193.20.0 150.193.20.1 23.254 150.193.3.255 Y 7th 150.193.0.0 255.255.252.0 150.193.24.0 150.193.24.1 27.254 150.193.3.255 Y 8th 150.193.0.0 255.255.252.0 150.193.28.0 150.193.28.1 31.254 150.193.3.255 Y 9th 150.193.0.0 255.255.252.0 150.193.32.0 150.193.32.1 35.254 150.193.3.255 Y 10th 150.193.0.0 255.255.252.0 150.193.36.0 150.193.36.1 39.254 150.193.3.255 Y Assign an IP Address and Subnet Mask to router interface A and write it down here.150.193.4.1 / 255.255.252.0Assign an IP Address and Subnet Mask of router interface B and write it down here.150.193.8.1 / 255.255.252.0Assign an IP Address and Subnet Mask of router interface C and write it down here.150.193.12.1 / 255.255.252.0Assign a host IP Address to Host X on Subnet A and assign an IP address to Host Z on Subnet C (answers may vary). Describe the steps (using ANDing) for the process of sending an IP packet from Host X to host Z through the router. Use the information from the diagram above and in Lab 1.10 to help assign IP addresses and subnet masks. Host X = 150.193.4.2 (4.1 was used for the router interface on subnet A), Host Z = 150.193.12.2 (12.1 was used for the router interface on subnet C). Host X compares (ANDs) the subnet mask to its own IP address and comes up with its own network/subnet address of 150.193.4.0. It then compares the subnet mask to the IP address of the destination host (150.193.12.2) and comes up with the network/subnet address of the target network (150.193.12.0). Since the two do not match, host X must assume that the destination host is not on its network and it send the packet to its "Default Gateway" or the nearside port of the router (interface A). The router goes through the same process on its incoming interface A and determines that network 150.193.12.2 is on it's C interface. The router forwards the packet to interface C and since the Router also knows the MAC address of hosts directly attached to its interfaces such as host Z, it forwards the packet to the 150.193.12.0 network/subnet LAN and host Z picks it up.What is the result of the ANDing process for host X?Decimal Host X IP addr: 150 . 193 . 4 . 2Binary Host X IP addr: 10010110 . 11000001 . 00000100 . 00000010Binary Subnet Mask: 11111111 . 11111111 . 11111100 . 00000000Binary ANDing Result: 10010110. 11000001 . 00000100 . 00000000Decimal ANDing Result: 150 . 193 . 4 . 0What is the result of the ANDing process for host Z?Decimal Host Z IP addr: 150 . 193 . 12 . 2Binary Host Z IP addr: 10010110 . 11000001 . 00001100 . 00000010Binary Subnet Mask: 11111111 . 11111111 . 11111100 . 00000000Binary ANDing Result: 10010110. 11000001 . 00001100 . 00000000Decimal ANDing Result: 150 . 193 . 12 . 0The Decimal ANDing result from questions 6 is the network/subnet that Host X is on. The result from question 7 is the network/subnet that Host Z is on. Are Host X and Host Z on the same network/subnet?NOWhat will host X now do with the packet ? Since the destination host Z is not on the Host X Local Area Network, Host X will send the packet to the "Default Gateway" which is the IP address of Interface A on the routerFill in the blanks in the following diagram with the correct Network and IP addresses.
