Can any one describe this ?

Question

An ISP is granted a block of addresses starting with 188.50.0.0/16. The ISP wants to distribute these blocks to 100 customers as follows. The first group has 30 medium-size businesses; each needs 128 addresses. The second group has 50 small businesses; each needs 64 addresses. The third group has 20 households; each needs 32 addresses Design the subblocks and give the slash notation for each subblock. Find out how many addresses are still available after these allocations.

Answer 1

ISP granted block of addresses starting with 188.50.0.0/16. So Total no of addresses available now is $2^{16}$=$65536$.

---

Group 1 has 30 medium size businesses each need 128 addresses.So we need $\log_{2}128=7$ bit to define each host.The prefix length for group 1 is then $32-7=25$.

The addresses are :-

1st business :- $188.50.0.0/25 -------188.50.0.127/25$

2nd business :-$188.50.0.128/25 -------188.50.0.255/25$

3rd  business :-$188.50.1.0/25 -------188.50.1.127/25$

4th business :- $188.50.1.128/25 -------188.50.1.255/25$

so on 

29th business:-$188.50.14.0/25 -------188.50.14.127/25$

30th business:-$188.50.14.128/25 -------188.50.14.255/25$

So total address used in group 1  is $30*128=3840$.

---

Group 2 has 50 small size businesses each need 64 addresses.So we need $\log_{2}64=6$ bit to define each host.The prefix length for group 2 is then $32-6=26$.

The addresses are :-

1st business :-$188.50.15.0/26 -------188.50.15.63/26$

2nd business :-$188.50.15.64/26 -------188.50.15.127/26$

3rd business :-$188.50.15.128/26 -------188.50.15.191/26$

4th business :- $188.50.15.192/26 -------188.50.15.255/26$

so on 

49th business :- $188.50.27.0/26 -------188.50.27.63/26$

50th business :- $188.50.27.64/26 -------188.50.27.127/26$

So total address used in group 2  is $50*64=3200$.

---

Group 3 has 20 small household each need 32 addresses.So we need $\log_{2}32=5$ bit to define each host.The prefix length for group 3 is then $32-5=27$.

The addresses are :-

1st Household :- $188.50.27.128/27 -------188.50.27.159/27$

2nd Household :- $188.50.27.160/27 -------188.50.27.191/27$

3rd Household :- $188.50.27.192/27 -------188.50.27.223/27$

4th Household :- $188.50.27.224/27 -------188.50.27.255/27$

so on 

19th Household :- $188.50.29.192/27 -------188.50.29.223/27$

20th Household:- $188.50.29.224/27 -------188.50.29.255/27$

So total address used in group 3  is $20*32=640$.

---

So total number of addresses still available =$65536-(3840+3200+640)$=$57856$.

Similar problem for Practice :- https://uomustansiriyah.edu.iq/media/lectures/5/5_2019_02_02!12_07_44_AM.pdf

Comments

Nice bro.👌

---

Tank you bro