MSIT 3250 Assignment 5 - Submission Sheet_Akhil_Banoth

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MSIT 3250 –Assignment #4 Answer Submission Sheet VLSM Exercise 1 Convert the following IPv6 address given in binary to its canonical text representation shown in step 1: 11010011 00000101 11111100 10101010 00000000 11000000 11100111 00111100 01010000 11000001 10000101 00001111 00100100 11011011 10100011 01100110 Answer: Step 1: Split the binary string into 8 groups of 16 bits each: 1101001100000101 1111110010101010 0000000011000000 1110011100111100 0101000011000001 1000010100001111 0010010011011011 1010001101100110 Step 2: Convert each group of 16 bits into its hexadecimal equivalent: D305 FCAA 00C0 E73C 50C1 850F 24DB A366 Step 3: Insert a colon (:) between each pair of hexadecimal values: D305:FCAA:00C0:E73C:50C1:850F:24DB:A366 The canonical text representation of the given binary address is: D305:FCAA:00C0:E73C::50C1:850F:24DB:A366. Exercise 2 Apply leading zero removal to the following address: 00000000 10001111 11000110 00000001 00001011 00111110 11111110 00011000 11101110 01110000 00111001 11111110 01010100 11000001 00000001 11100111 Be sure to convert the address to its canonical text notation repress entation and make sure to eliminate all leadings 0 digits as appropriate. Answer: Step 1: Split the binary string into 8 groups of 16 bits each: 0000000010001111 1100011000000001 0000101100111110 1111111000011000 1110111001110000 0011100111111110 0101010011000001 0000000111100111 Step 2: Convert each group of 16 bits into its hexadecimal equivalent: 008F C601 0B3E FE18 EE70 39FE 54C1 01E7

MSIT 3250 –Assignment #4 Answer Submission Sheet VLSM Step 3: Insert a colon (:) between each pair of hexadecimal values: 008F:C601:0B3E: FE18:EE70:39FE:54C1:01E7 The canonical text representation of the given binary address is: 008F: C601:0B3E: FE18::EE70:39FE:54C1:01E7 Applying leading zero removal to the above canonical text: 8F: C601: B3E: FE18 :: EE70 : 39FE : 54C1:1E7 Exercise 3 Apply zero compression to the following address: 00000000 10001111 11000110 00000001 00001011 00111110 11111110 00011000 11101110 01110000 00111001 11111110 01010100 11000001 00000001 11100111 Be sure to convert the address to its canonical text notation representation and apply zero compression. In this exercise, DO NOT apply leading zero removal. Answer: Step 1: Split the binary string into 8 groups of 16 bits each: 0000000010001111 1100011000000001 0000101100111110 1111111000011000 1110111001110000 0011100111111110 0101010011000001 0000000111100111 Step 2: Convert each group of 16 bits into its hexadecimal equivalent: 008F C601 0B3E FE18 EE70 39FE 54C1 01E7 Step 3: Insert a colon (:) between each pair of hexadecimal values: 008F:C601:0B3E: FE18:EE70:39FE:54C1:01E7 The canonical text representation of the given binary address is: 008F: C601: 0B3E:FE18 :: EE70 : 39FE : 54C1: 01E7 Applying the zero compression to the above canonical text is: 008F:C60:0B3E:FE18::EE70:39FE:54C1:01E7

MSIT 3250 –Assignment #4 Answer Submission Sheet VLSM Exercise 4 Apply zero compression AND leading zero removal to the following address: 00000000 00000000 00000000 00000000 00000000 00000000 00000000 10101011 00001001 00010000 00000000 00000000 11110001 01010110 00001101 01010101 Be sure to convert the address to its canonical text notation representation and apply both zero compression AND leading zero removal. Answer: Step 1: Split the binary string into 8 groups of 16 bits each: 0000000000000000 0000000000000000 0000000000000000 0000000010101011 0000100100010000 0000000000000000 1111000101010110 0000110101010101 Step 2: Convert each group of 16 bits into its hexadecimal equivalent: 0000 0000 0000 00AB 0910 0000 F156 0D55 Step 3: Insert a colon (:) between each pair of hexadecimal values: 0000:0000:0000:00AB 0910:0000:F156:0D55 The canonical text representation of the given binary address is: 0000:0000:0000:00AB::0910:0:F156:0D55 Applying both the zero compression AND leading zero removal we have the result as: AB::910:0:F156:D55 Exercise 5 Imagine that a computer is assigned IPv4 address 245.19.1.99. Create its corresponding IPv4‐ mapped IPv6 address as demonstrated in step 9, making sure to zero compress the result. Answer:  First, we start with 80 0 bits, which corresponds to 5 groups of 4 hexadecimal digits: 0000:0000:0000:0000:0000  Second, we append the 16 1 bits, which corresponds to 1 group of hexadecimal digits: 0000:0000:0000:0000:0000:FFFF  Third, we append the IPv4 address: 0000:0000:0000:0000:0000:FFFF:245.19.1.99.

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MSIT 3250 –Assignment #4 Answer Submission Sheet VLSM  Fourth, we zero compress the result: ::FFFF:245.19.1.99. Binary representation of the IPv4 address 245.19.1.99 to binary: 11110101 00010011 00000001 01100011 The Hexadecimal value to this is : F513:0163 Therefore, the corresponding IPv4-mapped IPv6 address for the IPv4 address 245.19.1.99 is: :: FFFF: F513:163, with the IPv4 address embedded in the last 32 bits of the IPv6 address.

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