Question

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12. What is the broadcast address of the network? 10.117.49.179 255.248.0.0
  
13. What is the broadcast address of the network 10.155.219.81/13

14. What is the subnet mask for a /17?

15. What is the subnet mask for a /9?

16. What is the subnet mask for a /29?

17. What is the subnet mask for a /30?

Answer 1

12)
IP Address: 10.117.49.179
----------------------------------------
Let's first convert this into binary format
10.117.49.179
Let's convert all octets to binary separately
Converting 10 to binary
Divide 10 successively by 2 until the quotient 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 1010
So, 10 of decimal is 1010 in binary
10 in binary is 00001010

Converting 117 to binary
Divide 117 successively by 2 until the quotient is 0
   > 117/2 = 58, remainder is 1
   > 58/2 = 29, remainder is 0
   > 29/2 = 14, remainder is 1
   > 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 1110101
So, 117 of decimal is 1110101 in binary
117 in binary is 01110101

Converting 49 to binary
Divide 49 successively by 2 until the quotient is 0
   > 49/2 = 24, remainder is 1
   > 24/2 = 12, remainder is 0
   > 12/2 = 6, remainder is 0
   > 6/2 = 3, remainder is 0
   > 3/2 = 1, remainder is 1
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 110001
So, 49 of decimal is 110001 in binary
49 in binary is 00110001

Converting 179 to binary
Divide 179 successively by 2 until the quotient is 0
   > 179/2 = 89, remainder is 1
   > 89/2 = 44, remainder is 1
   > 44/2 = 22, remainder is 0
   > 22/2 = 11, remainder is 0
   > 11/2 = 5, remainder is 1
   > 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 10110011
So, 179 of decimal is 10110011 in binary
179 in binary is 10110011

====================================================================================
||    10.117.49.179 in binary notation is 00001010.01110101.00110001.10110011    ||
====================================================================================

Subnet address: 255.248.0.0
----------------------------------------
Let's convert this into binary format
255.248.0.0
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 248 to binary
Divide 248 successively by 2 until the quotient is 0
   > 248/2 = 124, remainder is 0
   > 124/2 = 62, remainder is 0
   > 62/2 = 31, remainder is 0
   > 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 11111000
So, 248 of decimal is 11111000 in binary
248 in binary is 11111000

Converting 0 to binary
Divide 0 successively by 2 until the quotient is 0
Read remainders from the bottom to top as
So, 0 of decimal is  in binary
0 in binary is 00000000

Converting 0 to binary
Divide 0 successively by 2 until the quotient is 0
Read remainders from the bottom to top as
So, 0 of decimal is  in binary
0 in binary is 00000000

==================================================================================
||    255.248.0.0 in binary notation is 11111111.11111000.00000000.00000000    ||
==================================================================================

Remove all dots to form 11111111111110000000000000000000
Remove all 0's from the right side to form 1111111111111
Number of 1's in this is 13
so, Subnet mask is /13

For Calculating broadcast ID, keep first 13 bits of 00001010.01110101.00110001.10110011 and set all remaining bits to 1.
so, broadcast ID in binary is 00001010.01110111.11111111.11111111
00001010.01110111.11111111.11111111:
----------------------------------------
00001010.01110111.11111111.11111111
Let's convert all octets to decimal separately
Converting 00001010 to decimal
=> 00001010
=> 0x2^7+0x2^6+0x2^5+0x2^4+1x2^3+0x2^2+1x2^1+0x2^0
=> 0x128+0x64+0x32+0x16+1x8+0x4+1x2+0x1
=> 0+0+0+0+8+0+2+0
=> 10
00001010 in decimal is 10

Converting 01110111 to decimal
=> 01110111
=> 0x2^7+1x2^6+1x2^5+1x2^4+0x2^3+1x2^2+1x2^1+1x2^0
=> 0x128+1x64+1x32+1x16+0x8+1x4+1x2+1x1
=> 0+64+32+16+0+4+2+1
=> 119
01110111 in decimal is 119

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 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

======================================================================================
||    00001010.01110111.11111111.11111111 in decimal notation is 10.119.255.255    ||
======================================================================================
===============================================
||    So, broadcast ID is 10.119.255.255    ||
===============================================