CCNA Exploration 1, Network Fundamentals, Chapter 6 : problem and solution of IP4 Subnetting

The Problem::

The Problem::
Let an ISP is assigned an address block, and consider one of the address from that block is 20.69.0.0 /10

•Subnet the network so that each subnet has at least 3600 hosts ( Host IP loss should be as minimum as possible)
1. How many subnets?
2. Consider subnet # 373, and Find
    a.Network Address of the subnet
    b.First, Last host address
    c.Broadcast address
    d.Unnecessary Host address in the subnet
3. How many Host address losses for subnetting?


Solution::

Given Address: 20.69.0.0 = 20.01000101.0.0
Given Mask : 255.192.0.0 = 255.11000000.0.0
______________________________________
Given Network : 20.64.0.0 = 20.01000000.0.0
Net Bits in given Address = 10
Required 3600 Hosts
 Required Host Bits = 12(why?2¹¹ = 2048, 2¹²= 4096 )
Bits remain for subnetting = 32 –10 –12 = 10  

ANS::

1. Total subnets = 2^{Subnet Bits}= 210= 1024

Binary for 373
•256 128 64 32 16 8 4 2 1
•1 0 1 1 1 0 1 0 1 ( 9 bits only) 

Now consider Subnet Bits, and Host Bits just as place holder

20 . 01_ _ _ _ _ _ . _ _ _ _ _ __ _ ._ _ _ _ _ _ _ _ 

10 places for subnet bits

      •But the binary of 373 has only 9bits
      •Put One additional 0 before the binary number
      •And place it in the place holder of Subnet bits 

20. 01010111.0101_ _ _ _ ._ _ _ _ _ _ _ _

New Network Bits ( Total 20 bits)

20. 01 010111.0101_ _ _ _ ._ _ _ _ _ _ _ _ 

 Now for Subnet address place 0 on host bits

For Broadcast address place 1on host bits

For First Address just add 1with Net address

                                                                    20 . 87 . 80 . 0
                                                                                      + 1
                                                                   ____________
                                                                    20 . 87 . 80 . 1

For Last Address just subtract 1 from broadcast address

                                                                   20 . 87 . 95 . 255
                                                                                           -1
                                                                   ______________
                                                                   20 . 87 . 95 . 254

2.    d) Host IP in each subnet = (2¹²–2) = 4094

              Unnecessary Host IPs = 4094 –3600 = 494

3.     Host IP losses for subnetting

              Total Host IPs in Original Net = 2^{Host Bits}= 2²² –2 = A

               Host IPs in subnets = Total subnet X Host IP in each subnet

               = 2¹⁰x (2¹²–2) = B

               Losses = A –B = (2²²–2) –2¹⁰x (2¹²–2)

               = 2²² –2²²+ 2¹¹–2 = 2046