Question

Using a 23 bit network prefix, what network would the IPv4 address 123.154.89.114 be forwarded to?  Use dotted decimal format and do not include the mask in your answer.

Answer 1

123.154.88.0

Explanation:
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IP Address: 123.154.89.114
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Let's first convert this into binary format
123.154.89.114
Let's convert all octets to binary separately
Converting 123 to binary
Divide 123 successively by 2 until the quotient is 0
   > 123/2 = 61, remainder is 1
   > 61/2 = 30, remainder is 1
   > 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 1111011
So, 123 of decimal is 1111011 in binary
123 in binary is 01111011

Converting 154 to binary
Divide 154 successively by 2 until the quotient is 0
   > 154/2 = 77, remainder is 0
   > 77/2 = 38, remainder is 1
   > 38/2 = 19, remainder is 0
   > 19/2 = 9, remainder is 1
   > 9/2 = 4, remainder is 1
   > 4/2 = 2, remainder is 0
   > 2/2 = 1, remainder is 0
   > 1/2 = 0, remainder is 1
Read remainders from the bottom to top as 10011010
So, 154 of decimal is 10011010 in binary
154 in binary is 10011010

Converting 89 to binary
Divide 89 successively by 2 until the quotient is 0
   > 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 1011001
So, 89 of decimal is 1011001 in binary
89 in binary is 01011001

Converting 114 to binary
Divide 114 successively by 2 until the quotient is 0
   > 114/2 = 57, remainder is 0
   > 57/2 = 28, remainder is 1
   > 28/2 = 14, remainder 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 1110010
So, 114 of decimal is 1110010 in binary
114 in binary is 01110010

=====================================================================================
||    123.154.89.114 in binary notation is 01111011.10011010.01011001.01110010    ||
=====================================================================================

Subnet mask is /23

For Calculating network ID, keep first 23 bits of 01111011.10011010.01011001.01110010 and set all remaining bits to 0.
so, network ID in binary is 01111011.10011010.01011000.00000000
01111011.10011010.01011000.00000000:
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01111011.10011010.01011000.00000000
Let's convert all octets to decimal separately
Converting 01111011 to decimal
Converting 01111011 to decimal
01111011
=> 0x2^7+1x2^6+1x2^5+1x2^4+1x2^3+0x2^2+1x2^1+1x2^0
=> 0x128+1x64+1x32+1x16+1x8+0x4+1x2+1x1
=> 0+64+32+16+8+0+2+1
=> 123
01111011 in decimal is 123

Converting 10011010 to decimal
Converting 10011010 to decimal
10011010
=> 1x2^7+0x2^6+0x2^5+1x2^4+1x2^3+0x2^2+1x2^1+0x2^0
=> 1x128+0x64+0x32+1x16+1x8+0x4+1x2+0x1
=> 128+0+0+16+8+0+2+0
=> 154
10011010 in decimal is 154

Converting 01011000 to decimal
Converting 01011000 to decimal
01011000
=> 0x2^7+1x2^6+0x2^5+1x2^4+1x2^3+0x2^2+0x2^1+0x2^0
=> 0x128+1x64+0x32+1x16+1x8+0x4+0x2+0x1
=> 0+64+0+16+8+0+0+0
=> 88
01011000 in decimal is 88

Converting 00000000 to decimal
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

====================================================================================
||    01111011.10011010.01011000.00000000 in decimal notation is 123.154.88.0    ||
====================================================================================
===========================================
||    So, Network ID is 123.154.88.0    ||
===========================================

so, IP address 123.154.89.114 is forwarded to network 123.154.88.0