1) Let's convert this into binary format 255.255.255.240 Let's convert all octets to binary separately Converting 255 to binary Divide 255 successively by 2 until the quotient is 0 > 255/2 = 127, remainder is 1 > 127/2 = 63, remainder is 1 > 63/2 = 31, remainder is 1 > 31/2 = 15, remainder is 1 > 15/2 = 7, remainder is 1 > 7/2 = 3, remainder is 1 > 3/2 = 1, remainder is 1 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 11111111 So, 255 of decimal is 11111111 in binary 255 in binary is 11111111 Converting 255 to binary Divide 255 successively by 2 until the quotient is 0 > 255/2 = 127, remainder is 1 > 127/2 = 63, remainder is 1 > 63/2 = 31, remainder is 1 > 31/2 = 15, remainder is 1 > 15/2 = 7, remainder is 1 > 7/2 = 3, remainder is 1 > 3/2 = 1, remainder is 1 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 11111111 So, 255 of decimal is 11111111 in binary 255 in binary is 11111111 Converting 255 to binary Divide 255 successively by 2 until the quotient is 0 > 255/2 = 127, remainder is 1 > 127/2 = 63, remainder is 1 > 63/2 = 31, remainder is 1 > 31/2 = 15, remainder is 1 > 15/2 = 7, remainder is 1 > 7/2 = 3, remainder is 1 > 3/2 = 1, remainder is 1 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 11111111 So, 255 of decimal is 11111111 in binary 255 in binary is 11111111 Converting 240 to binary Divide 240 successively by 2 until the quotient is 0 > 240/2 = 120, remainder is 0 > 120/2 = 60, remainder is 0 > 60/2 = 30, remainder is 0 > 30/2 = 15, remainder is 0 > 15/2 = 7, remainder is 1 > 7/2 = 3, remainder is 1 > 3/2 = 1, remainder is 1 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 11110000 So, 240 of decimal is 11110000 in binary 240 in binary is 11110000 ====================================================================================== || 255.255.255.240 in binary notation is 11111111.11111111.11111111.11110000 || ====================================================================================== Remove all dots to form 11111111111111111111111111110000 Remove all 0's from the right side to form 1111111111111111111111111111 Number of 1's in this is 28 so, Subnet mask is /28 Answer: -------- 203.120.45.23/28 2) IP Address: 172.255.20.30 ---------------------------------------- Let's first convert this into binary format 172.255.20.30 Let's convert all octets to binary separately Converting 172 to binary Divide 172 successively by 2 until the quotient is 0 > 172/2 = 86, remainder is 0 > 86/2 = 43, remainder is 0 > 43/2 = 21, remainder is 1 > 21/2 = 10, remainder is 1 > 10/2 = 5, remainder is 0 > 5/2 = 2, remainder is 1 > 2/2 = 1, remainder is 0 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 10101100 So, 172 of decimal is 10101100 in binary 172 in binary is 10101100 Converting 255 to binary Divide 255 successively by 2 until the quotient is 0 > 255/2 = 127, remainder is 1 > 127/2 = 63, remainder is 1 > 63/2 = 31, remainder is 1 > 31/2 = 15, remainder is 1 > 15/2 = 7, remainder is 1 > 7/2 = 3, remainder is 1 > 3/2 = 1, remainder is 1 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 11111111 So, 255 of decimal is 11111111 in binary 255 in binary is 11111111 Converting 20 to binary Divide 20 successively by 2 until the quotient is 0 > 20/2 = 10, remainder is 0 > 10/2 = 5, remainder is 0 > 5/2 = 2, remainder is 1 > 2/2 = 1, remainder is 0 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 10100 So, 20 of decimal is 10100 in binary 20 in binary is 00010100 Converting 30 to binary Divide 30 successively by 2 until the quotient is 0 > 30/2 = 15, remainder is 0 > 15/2 = 7, remainder is 1 > 7/2 = 3, remainder is 1 > 3/2 = 1, remainder is 1 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 11110 So, 30 of decimal is 11110 in binary 30 in binary is 00011110 ==================================================================================== || 172.255.20.30 in binary notation is 10101100.11111111.00010100.00011110 || ==================================================================================== Subnet mask is /20 For Calculating network ID, keep first 20 bits of 10101100.11111111.00010100.00011110 and set all remaining bits to 0. so, network ID in binary is 10101100.11111111.00010000.00000000 10101100.11111111.00010000.00000000: ---------------------------------------- 10101100.11111111.00010000.00000000 Let's convert all octets to decimal separately Converting 10101100 to decimal => 10101100 => 1x2^7+0x2^6+1x2^5+0x2^4+1x2^3+1x2^2+0x2^1+0x2^0 => 1x128+0x64+1x32+0x16+1x8+1x4+0x2+0x1 => 128+0+32+0+8+4+0+0 => 172 10101100 in decimal is 172 Converting 11111111 to decimal => 11111111 => 1x2^7+1x2^6+1x2^5+1x2^4+1x2^3+1x2^2+1x2^1+1x2^0 => 1x128+1x64+1x32+1x16+1x8+1x4+1x2+1x1 => 128+64+32+16+8+4+2+1 => 255 11111111 in decimal is 255 Converting 00010000 to decimal => 00010000 => 0x2^7+0x2^6+0x2^5+1x2^4+0x2^3+0x2^2+0x2^1+0x2^0 => 0x128+0x64+0x32+1x16+0x8+0x4+0x2+0x1 => 0+0+0+16+0+0+0+0 => 16 00010000 in decimal is 16 Converting 00000000 to decimal => 00000000 => 0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+0x2^0 => 0x128+0x64+0x32+0x16+0x8+0x4+0x2+0x1 => 0+0+0+0+0+0+0+0 => 0 00000000 in decimal is 0 ==================================================================================== || 10101100.11111111.00010000.00000000 in decimal notation is 172.255.16.0 || ==================================================================================== =========================================== || So, Network ID is 172.255.16.0 || =========================================== IP Address: 172.255.14.30 ---------------------------------------- Let's first convert this into binary format 172.255.14.30 Let's convert all octets to binary separately Converting 172 to binary Divide 172 successively by 2 until the quotient is 0 > 172/2 = 86, remainder is 0 > 86/2 = 43, remainder is 0 > 43/2 = 21, remainder is 1 > 21/2 = 10, remainder is 1 > 10/2 = 5, remainder is 0 > 5/2 = 2, remainder is 1 > 2/2 = 1, remainder is 0 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 10101100 So, 172 of decimal is 10101100 in binary 172 in binary is 10101100 Converting 255 to binary Divide 255 successively by 2 until the quotient is 0 > 255/2 = 127, remainder is 1 > 127/2 = 63, remainder is 1 > 63/2 = 31, remainder is 1 > 31/2 = 15, remainder is 1 > 15/2 = 7, remainder is 1 > 7/2 = 3, remainder is 1 > 3/2 = 1, remainder is 1 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 11111111 So, 255 of decimal is 11111111 in binary 255 in binary is 11111111 Converting 14 to binary Divide 14 successively by 2 until the quotient is 0 > 14/2 = 7, remainder is 0 > 7/2 = 3, remainder is 1 > 3/2 = 1, remainder is 1 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 1110 So, 14 of decimal is 1110 in binary 14 in binary is 00001110 Converting 30 to binary Divide 30 successively by 2 until the quotient is 0 > 30/2 = 15, remainder is 0 > 15/2 = 7, remainder is 1 > 7/2 = 3, remainder is 1 > 3/2 = 1, remainder is 1 > 1/2 = 0, remainder is 1 Read remainders from the bottom to top as 11110 So, 30 of decimal is 11110 in binary 30 in binary is 00011110 ==================================================================================== || 172.255.14.30 in binary notation is 10101100.11111111.00001110.00011110 || ==================================================================================== Subnet mask is /20 For Calculating network ID, keep first 20 bits of 10101100.11111111.00001110.00011110 and set all remaining bits to 0. so, network ID in binary is 10101100.11111111.00000000.00000000 10101100.11111111.00000000.00000000: ---------------------------------------- 10101100.11111111.00000000.00000000 Let's convert all octets to decimal separately Converting 10101100 to decimal => 10101100 => 1x2^7+0x2^6+1x2^5+0x2^4+1x2^3+1x2^2+0x2^1+0x2^0 => 1x128+0x64+1x32+0x16+1x8+1x4+0x2+0x1 => 128+0+32+0+8+4+0+0 => 172 10101100 in decimal is 172 Converting 11111111 to decimal => 11111111 => 1x2^7+1x2^6+1x2^5+1x2^4+1x2^3+1x2^2+1x2^1+1x2^0 => 1x128+1x64+1x32+1x16+1x8+1x4+1x2+1x1 => 128+64+32+16+8+4+2+1 => 255 11111111 in decimal is 255 Converting 00000000 to decimal => 00000000 => 0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+0x2^0 => 0x128+0x64+0x32+0x16+0x8+0x4+0x2+0x1 => 0+0+0+0+0+0+0+0 => 0 00000000 in decimal is 0 Converting 00000000 to decimal => 00000000 => 0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+0x2^0 => 0x128+0x64+0x32+0x16+0x8+0x4+0x2+0x1 => 0+0+0+0+0+0+0+0 => 0 00000000 in decimal is 0 =================================================================================== || 10101100.11111111.00000000.00000000 in decimal notation is 172.255.0.0 || =================================================================================== ========================================== || So, Network ID is 172.255.0.0 || ========================================== Answer: ------- Network addresses od addresses are 172.255.16.0 and 172.255.0.0 which are different. so, these devices are not on the same subnet.
IP address is: 203.120.45.23 and subnet mask is: 255.255.255.240. Represent this information in CIDR format. IP...
A/ Given the following IP address from the Class B address range using the default subnet mask: 100.110.0.0. Your network plan requires no more than 64 hosts on a subnet. When you configure the IP address in Cisco IOS software, which value should you use as the subnet mask? 255.255.0.0 255.255.128.0 255.255.255.128 255.255.255.252 B/ Identify how many valid host addresses can you have on 192.168.27.32 network with a subnet mask of 255.255.255.240. (2^4) - 2 (2^3) – 2 (2^2)...
using the information given below . what is the ip address , subnet mask , default gateway? ** corrections # of bits in subnet for A :6 # of bits in submet for B: 3 PC-A R1 G0/0 R GO/1 S1 PC-B ? assign the IP ADDRESS information with given information Given, IP address: 202.44.66.0/24 For Subnet A: It needs 2 valid hosts 2 valid hosts + 2 (network and broadcast) = 22 i.e., It needs 2 bit as host...
14. What is the range of assignable IP addresses for a subnet containing an IP address of 172.16.1.10/192 172.16.0.1-172.16.31254 172.16.0.1-172.16.63.254 172.16.0.0-172.16.31.255 172.16.0.1-172.16.31.255 172.16.0.0-172.16.63.254 15. What subnet mask should be used to subnet the 192.168.10.0 network to support number of subnets and IP addresses per subnet shown in the following topology? 255.255.255.0 255.255.255.128 255.255.255.192 255.255.255.224 255.255.255.240
Given an IP address and mask of 192.168.10.0/24 255.255.255.0 (address / mask) subnet A has 100 hosts subnet b has 50 hosts Specification Subnet A Subnet B Number of bits in the subnet IP mask (binary) New IP mask (decimal) Maximum number of usable subnets (including the 0th subnet) Number of usable hosts per subnet IP Subnet First IP Host address Last IP Host address Description Subnet A Subnet B First IP address Last IP address Maximum number of hosts
(8 points) A company has a network address of 178.145.20.64 with a subnet mask of 255.255.255.192. The company wants to break this address block into 4 subnets of equal size. a. What Subnet Mask should they configure on internal devices to achieve this? b. How many hosts per subnet do they obtain? c. List out the 4 Subnet IDs, the first and last valid host address, and the broadcast address for each of these subnets. 5. a. Subnet Mask: b....
(a) What is the network number of the IP address 140.100.120.02 if the subnet mask is 255.255.224.0? (b) Does it make sense to have a subnet mask equal to 255.255.224.7? why or why not? (c) Suppose an organization needs 8000 IP addresses. How many Class C addresses would be needed? If they were consecutive, describe the CIDR addressing scheme.
In a class B subnet, we know the IP address of one of the hosts and the subnet mask as given below: IP Address: 188.48.82.176 Subnet mask: 255.255.255.248 1) What is the first valid host address in the subnet? (in dotted decimal) 2) What is the last valid address in the subnet? (in dotted decimal) 3) What is the subnet network address? (in dotted decimal) 4) What is the subnet broadcast address? (in dotted decimal) 5) What is the number...
Given the following IP address and subnet mask, what is the network identifier? • IP Address: 18.132.219.175 . Subnet Mask: 255.255.248.0 Network ID: BLANK-1 BLANK-1 Add your answer Question 28 Given the following IP address and subnet mask, what is the network identifier? - IP Address: 18.132.219.175 . Subnet Mask: 255.255.248.0 • Broadcast Address: BLANK-1 BLANK-1 Add your answer
(8 points) A company has a network address of 178.145.20.64 with a subnet mask of 255.255.255.192. The company wants to break this address block into 4 subnets of equal size. a. What Subnet Mask should they configure on internal devices to achieve this? b. How many hosts per subnet do they obtain? c. List out the 4 Subnet IDs, the first and last valid host address, and the broadcast address for each of these subnets. 5. a. Subnet Mask: b....
Task 1: Design a Logical LAN Topology Step 1: Design an IP addressing scheme Given the IP address block of 192.168.7.0 /24, design an IP addressing scheme that satisfies the following requirements Subnet Subnet A Subnet B Number of Hosts 110 54 The 0 subnet is used. No subnet calculators may be used. Create the smallest possible subnets that satisfy the requirements for hosts. Assign the first usable subnet to Subnet A. Subnet A Specification Student Input Number of bits...