MSIT 3250 Assignment 2 - Submission Sheet_Akhil_Banoth

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MSIT 3250 –Assignment #2 Answer Submission Sheet IP Network Identification Exercise 1a For the following IP Address displayed in CIDR notation, determine the Network Identifier in Base-2: 114.35.138.223/20 Answer: 114.35.138.223/20 114.35.138.223 in binary: 01110010 00100011 10001010 11011111 First 20 bits are the network identifier: 01110010 00100011 1000 Network Identifier in Base-2: 01110010001000111000 Exercise 1b For the following IP Address displayed in CIDR notation, determine the Network Identifier in Base-2: 146.24.228.11/13 Answer: 146.24.228.11/13 146.24.228.11 in binary: 10010010 00011000 11100100 00001011 First 13 bits are the network identifier: 1001001 00011000 Network Identifier in Base-2: 1001001000011 Exercise 2a For the following IP Address, identify the Network Address and Broadcast Address: 28.132.93.17/27 Answer: 28.132.93.17/27 28.132.93.17 in binary: 00011100 10000100 01011101 00010001 First 27 bits are the network identifier: 00011100 10000100 01011101 000 Network Address: 28.132.93.0 Broadcast Address: 28.132.93.31

MSIT 3250 –Assignment #2 Answer Submission Sheet IP Network Identification Exercise 2b For the following IP Address, identify the Network Address and Broadcast Address: 00110100 11000010 00101011 01010110 (9 bits for Network Identifier) Answer: 00110100 11000010 00101011 01010110 (9 bits for network identifier) Network Identifier: 00110100 11000010 0010100 Network Address: 00110100 11000010 00101011 Broadcast Address: 00110100 11000010 001010111 Exercise 3a For the following IP Address, identify the Network Address: 177.224.132.227/26 Answer: Network Address: 10110001.11100000.10000100.11000000 In decimal form, this is: Network Address: 177.224.132.192 Exercise 3b For the following IP Address, identify the Network Address: 177.224.132.227/8 Answer: Network Address: 10110001.00000000.00000000.00000000 In decimal form, this is: Network Address: 177.0.0.0 So, the network address for the IP address 177.224.132.227/8 is 177.0.0.0. Exercise 3c For the following IP Address, identify the Network Address: 177.224.132.227/18

MSIT 3250 –Assignment #2 Answer Submission Sheet IP Network Identification Answer: Network Address: 10110001.11100000.10000000.00000000 In decimal form, this is: Network Address: 177.224.128.0 So, the network address for the IP address 177.224.132.227/18 is 177.224.128.0. Exercise 4a Represent the following CIDR mask length in dotted decimal notation: /8 Answer: The CIDR mask length "/8" is represented as the subnet mask 255.0.0.0 in dotted decimal notation. /12 Answer: The CIDR mask length "/12" is represented as the subnet mask 255.240.0.0 in dotted decimal notation. /27 Answer: The CIDR mask length "/27" is represented as the subnet mask 255.255.255.224 in dotted decimal notation Exercise 4b Represent the following dotted decimal subnet masks in CIDR mask length format: 255.224.0.0 Answer: /11 248.0.0.0 Answer: /5

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MSIT 3250 –Assignment #2 Answer Submission Sheet IP Network Identification 255.255.255.240 Answer: /28 Exercise 5a Determine whether the following IP Addresses are on the same network. Be sure to show your work for full credit: 146.16.221.19/28 146.16.221.34/28 Answer: To determine whether the IP addresses are on the same network, we need to compare their network addresses. Here are the steps to calculate the network address: 1. Convert each IP address to binary: 10010010.00010000.11011101.00010011 146.16.221.19 10010010.00010000.11011101.00100010 146.16.221.34 2. Apply the subnet mask (prefix length /28) to each IP address: 10010010.00010000.11011101.00010000 146.16.221.16 10010010.00010000.11011101.00100000 146.16.221.32 3. Compare the network addresses to determine whether the IP addresses are on the same network: The IP addresses 146.16.221.19 and 146.16.221.34 have different network addresses (146.16.221.16 and 146.16.221.32, respectively). Therefore, they are not on the same network. Exercise 5b Some of the below IP Addresses may be on the same network. Determine which, if any, are. Be sure to show your work for full credit: a) 10.129.33.127/11 b) 00001010 10000111 00100001 10000001 (11 bits of mask length) c) 10.135.33.128/11 d) 00001010 10001001 00100010 00001100 (11 bits of mask length) e) 10.126.32.223/11 f) 11.17.32.33/11

MSIT 3250 –Assignment #2 Answer Submission Sheet IP Network Identification Answer: To determine whether the IP addresses are on the same network, we need to compare their network addresses. Here are the steps to calculate the network address: 1. Convert each IP address to binary: 00001010.10000001.00100001.01111111 10.129.33.127 00001010.10000111.00100001.10000000 10.135.33.128 00001010.01111110.00100000.11011111 10.126.32.223 00001011.00010001.00100000.00100001 11.17.32.33 2. Apply the subnet mask (prefix length /11) to each IP address: 00001010.00000000.00000000.00000000 10.0.0.0 00001010.00000000.00000000.00000000 10.0.0.0 00001010.00000000.00000000.00000000 10.0.0.0 00001011.00000000.00000000.00000000 11.0.0.0 3. Compare the network addresses to determine whether the IP addresses are on the same network: - The IP addresses 10.129.33.127 and 10.135.33.128 have different network addresses (10.0.0.0 and 10.0.0.0, respectively). - The IP addresses 10.129.33.127 and 10.126.32.223 have the same network address (10.0.0.0). - None of the other IP addresses are on the same network. Therefore, only IP addresses 10.129.33.127 and 10.126.32.223 are on the same network (with a network address 10.0.0.0). Exercise 5c Determine which, if any, of the below IP Addresses are on the same network. Be sure to show your work for full credit: a) 172.16.32.17/29 b) 172.16.33.14/29 c) 10101100 00010000 00100000 00010110 (29 bits of mask length) d) 172.16.32.23/29 e) 172.16.32.14/29 f) 10101100 00010000 00100000 00011000 (29 bits of mask length) Answer: The IP addresses given are: 172.16.32.17/29 172.16.33.14/29 10101100 00010000 00100000 00010110 (29 bits of mask length) 172.16.32.23/29 172.16.32.14/29 10101100 00010000 00100000 00011000 (29 bits of mask length)

MSIT 3250 –Assignment #2 Answer Submission Sheet IP Network Identification 1. Convert each IP address to binary: 172.16.32.17/29 10101100.00010000.00100000.00010001 172.16.33.14/29 10101100.00010000.00100001.00001110 10101100.00010000.00100000.00010110 (29 bits mask) 172.16.32.23/29 10101100.00010000.00100000.00010111 172.16.32.14/29 10101100.00010000.00100000.00001110 10101100.00010000.00100000.00011000 (29 bits mask) 2. Apply the 29-bit subnet mask to get the network identifiers: 172.16.32.17/29 10101100.00010000.00100000.00010000 (172.16.32.16) 172.16.33.14/29 10101100.00010000.00100001.00001000 (172.16.33.8) 10101100.00010000.00100000.00010000 (172.16.32.16) 172.16.32.23/29 10101100.00010000.00100000.00010000 (172.16.32.16) 172.16.32.14/29 10101100.00010000.00100000.00010000 (172.16.32.16) 10101100.00010000.00100000.00010000 (172.16.32.16) 3. Compare the network identifiers:  172.16.32.17/29, 10101100.00010000.00100000.00010110, 172.16.32.23/29, 172.16.32.14/29, and 10101100.00010000.00100000.00011000 all have network address 172.16.32.16  Only 172.16.33.14/29 has a different network address of 172.16.33.8 Therefore, the 5 IP addresses are on the same network, while 172.16.33.14/29 is on a different network. Exercise 6a Complete a bitwise AND operation on the following bit sequence: 10101101 AND 11011001 Answer: 10101101 = 10101101 11011001 = 11011001

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MSIT 3250 –Assignment #2 Answer Submission Sheet IP Network Identification 10101101 AND 11011001 = 10001001 Exercise 6b Complete a Bitwise AND operation on the following Base-10 numbers (this will require conversion to Base-2 first) and convert the result back to Base-10: 196 AND 224 Answer: Converting 196 and 224 to binary: 196 = 11000100 224 = 11100000 Performing a bitwise AND operation on the binary sequences: 11000100 & 11100000 ----------- 11000000 Converting the binary result back to Base-10: 11000000 = 192 Therefore, the result of the Bitwise AND operation on 196 and 224 is 192. Exercise 7a To determine whether the following IP Addresses are on the same network, use the following steps: 10.20.30.39/30 10.20.30.40/30 1) Convert each address from CIDR to Base-2 2) Calculate the subnet mask for each address in Base-2 3) Perform a Bitwise AND operation from each pair of IP Address/Subnet Mask to calculate the Network Address in Base-2 4) (Not needed by computers, but do it anyways) convert the Base-2 Network Address for each back to Dotted Decimal Notation

MSIT 3250 –Assignment #2 Answer Submission Sheet IP Network Identification 5) Compare the two Network Addresses to determine whether they are on the same network. Answer: 1. Convert IP addresses to binary: 10.20.30.39/30 00001010 00010100 00011110 00100111 10.20.30.40/30 00001010 00010100 00011110 00101000 2. Calculate subnet masks: /30 means the subnet mask is 11111111 11111111 11111111 11111100 in binary 3. Apply bitwise AND: 10.20.30.39: 00001010 00010100 00011110 00100111 11111111 11111111 11111111 11111100 (subnet mask) 00001010 00010100 00011110 00100100 (network address) 10.20.30.40: 00001010 00010100 00011110 00101000 11111111 11111111 11111111 11111100 (subnet mask) 00001010 00010100 00011110 00100100 (network address) 4. Convert network addresses to dotted decimal: Both network addresses are 10.20.30.36 5. Since the network addresses match, the IP addresses are on the same network. Exercise 7b Using the steps outlined in Exercise 7a, determine which of the following IP Addresses (if any) are on the same network. Be sure to show your work for full credit: 192.168.200.17/14 192.167.201.124/14 192.168.199.254/14 192.169.203.1/14 192.172.205.13/14 191.168.202.72/14 Answer: To determine if the IP addresses are on the same network, we need to compare their network addresses. Here are the steps to calculate the network address: 1. Convert each IP address to binary:

MSIT 3250 –Assignment #2 Answer Submission Sheet IP Network Identification 11000000.10101000.11001000.00010001 192.168.200.17 11000000.10100111.11001001.01111100 192.167.201.124 11000000.10101000.11000111.11111110 192.168.199.254 11000000.10101001.11001011.00000001 192.169.203.1 11000000.10101100.11001101.00001101 192.172.205.13 10111111.10101000.11001010.01001000 191.168.202.72 2. Apply the subnet mask to each IP address: 11000000.10100000.00000000.00000000 192.168.192.0 11000000.10100000.00000000.00000000 192.167.192.0 11000000.10100000.00000000.00000000 192.168.192.0 11000000.10100000.00000000.00000000 192.169.192.0 11000000.10100000.00000000.00000000 192.172.192.0 10111111.10100000.00000000.00000000 191.168.192.0 3. Compare the network addresses to determine which IP addresses are on the same network: - 192.168.200.17 and 192.168.199.254 are on the same network - None of the other IP addresses are on the same network

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MSIT 3250 Assignment 2 - Submission Sheet_Akhil_Banoth

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