You have been given the 172.30.0.0/16 network. Your companyrequires 100 subnets with at least 500 hosts per subnet. What prefixlength should you use?
Original Network: 172.30.0.0/16
/16 means:
- Network bits: 16
- Host bits: 32 - 16 = 16
- Total addresses: 2^16 = 65,536
Requirements:
✓ At least 100 subnets
✓ At least 500 hosts per subnet
How many bits to create at least 100 subnets?
Bits Borrowed │ Formula │ Subnets Created │ Enough?
──────────────┼─────────┼─────────────────┼─────────
1 │ 2^1 │ 2 │ ❌
2 │ 2^2 │ 4 │ ❌
3 │ 2^3 │ 8 │ ❌
4 │ 2^4 │ 16 │ ❌
5 │ 2^5 │ 32 │ ❌
6 │ 2^6 │ 64 │ ❌
7 │ 2^7 │ 128 │ ✅ (≥100)
Answer: We need to borrow at least 7 bits for subnets
How many bits to have at least 500 hosts?
Host Bits │ Formula │ Total Addresses │ Usable Hosts │ Enough?
──────────┼─────────┼─────────────────┼──────────────┼─────────
7 │ 2^7 │ 128 │ 128-2=126 │ ❌
8 │ 2^8 │ 256 │ 256-2=254 │ ❌
9 │ 2^9 │ 512 │ 512-2=510 │ ✅ (≥500)
Answer: We need at least 9 bits for hosts
We have 16 host bits in /16. Let's split them:
Original host bits: 16
Needed for subnets: 7 bits (gives 128 subnets)
Needed for hosts: 9 bits (gives 510 hosts)
──
Total: 16 bits ✅ Perfect fit!
New prefix calculation:
Original prefix: /16
Bits borrowed: + 7
────
New prefix: /23
Check /23:
┌────────────────────────────────────────────────────────────┐
│ VERIFICATION │
├────────────────────────────────────────────────────────────┤
│ │
│ Prefix: /23 │
│ Network bits: 23 │
│ Host bits: 32 - 23 = 9 │
│ ┌─────────────────────────────────────────────────────┐ │
│ │ SUBNETS: │ │
│ │ Bits borrowed from /16 = 23 - 16 = 7 bits │ │
│ │ Number of subnets = 2^7 = 128 subnets │ │
│ │ Need 100 → Have 128 ✅ │ │
│ └─────────────────────────────────────────────────────┘ │
│ │ HOSTS: │ │
│ │ Host bits = 9 │ │
│ │ Addresses per subnet = 2^9 = 512 │ │
│ │ Usable hosts = 512 - 2 = 510 │ │
│ │ Need 500 → Have 510 ✅ │ │
└────────────────────────────────────────────────────────────┘
What about /22?
/22:
- Host bits = 10 → 2^10 - 2 = 1022 hosts ✅
- Subnet bits = 22 - 16 = 6 → 2^6 = 64 subnets ❌ (need 100)
FAILS: Not enough subnets!
What about /24?
/24:
- Subnet bits = 24 - 16 = 8 → 2^8 = 256 subnets ✅
- Host bits = 8 → 2^8 - 2 = 254 hosts ❌ (need 500)
FAILS: Not enough hosts!
172.30.0.0/16 Original:
172 . 30 . 0 . 0
────────────────────── ─────────────────────
11111111.11111111 00000000.00000000
└───────┬───────┘ └────────┬────────┘
16 Network bits 16 Host bits
172.30.0.0/23 After Subnetting:
────────────────────────────────── ─────────
11111111.11111111.0000000 0.00000000
└────────────┬────────────┘ └────┬────┘
23 Network bits 9 Host bits
(16 original + 7 borrowed)
└──────────────┘└───────┘ └─────────┘
Original 16 7 bits 9 bits
for subnets for hosts
(128 subnets) (510 hosts)
LAN1: 45 hosts
LAN2: 64 hosts
LAN3: 14 hosts
LAN4: 9 hosts
R1 is connect to R2 P2P
Subnet the 192.168.5.0/24 network to provide sufficient addressing for each LAN. (Also, the point-to-point connection between R1 and R2).
Assign the first usable address to the PC in each LAN.
Assign the last usable address to the router's interface in each LAN.
Configure static routes on each router so that all PCs can ping eachother.
Network
Hosts Needed
Hosts + Network + Broadcast
LAN2
64 hosts
64 + 2 = 66 addresses
LAN1
45 hosts
45 + 2 = 47 addresses
LAN3
14 hosts
14 + 2 = 16 addresses
LAN4
9 hosts
9 + 2 = 11 addresses
P2P
2 hosts
2 + 2 = 4 addresses
Why? We must allocate largest subnets first to avoid address overlap.
1. LAN2: 64 hosts (LARGEST)
2. LAN1: 45 hosts
3. LAN3: 14 hosts
4. LAN4: 9 hosts
5. P2P: 2 hosts (SMALLEST)
2⁶ - 2 = 62 ❌ (not enough)
2⁷ - 2 = 126 ✅ (enough!)
Host bits = 7
Network bits = 32 - 7 = 25
Subnet mask = /25 (255.255.255.128)
Block size = 128
2⁵ - 2 = 30 ❌ (not enough)
2⁶ - 2 = 62 ✅ (enough!)
Host bits = 6
Network bits = 32 - 6 = 26
Subnet mask = /26 (255.255.255.192)
Block size = 64
2³ - 2 = 6 ❌ (not enough)
2⁴ - 2 = 14 ✅ (exactly enough!)
Host bits = 4
Network bits = 32 - 4 = 28
Subnet mask = /28 (255.255.255.240)
Block size = 16
2⁴ - 2 = 14 ✅ (enough!)
2² - 2 = 2 ✅ (exactly enough!)
Host bits = 2
Network bits = 32 - 2 = 30
Subnet mask = /30 (255.255.255.252)
Block size = 4
Starting from 192.168.5.0:
┌─────────────────────────────────────────────────────────────┐
│ 192.168.5.0/24 │
│ (256 addresses) │
├─────────────────────────────────────────────────────────────┤
│ .0────────────────────────────────────────────.127 │ LAN2 │
│ .128─────────────────────────────.191 │ LAN1 │
│ .192────────────.207 │ LAN3 │
│ .208────────────.223 │ LAN4 │
│ .224────.227 │ P2P │
│ .228────.255 │ FREE │
└─────────────────────────────────────────────────────────────┘
Network Address: 192.168.5.0/25
Subnet Mask: 255.255.255.128
First Usable: 192.168.5.1 ← PC2
Last Usable: 192.168.5.126 ← Router Interface
Broadcast: 192.168.5.127
Usable Hosts: 126
Calculation:
Network: 192.168.5.0 = 192.168.5.00000000
Broadcast: 192.168.5.0 + 127 = 192.168.5.01111111 = 192.168.5.127
First: Network + 1 = 192.168.5.1
Last: Broadcast - 1 = 192.168.5.126
Network Address: 192.168.5.128/26
Subnet Mask: 255.255.255.192
First Usable: 192.168.5.129 ← PC1
Last Usable: 192.168.5.190 ← Router Interface
Broadcast: 192.168.5.191
Usable Hosts: 62
Start after LAN2: 192.168.5.128
Block size: 64
Network: 192.168.5.128
Broadcast: 192.168.5.128 + 63 = 192.168.5.191
First: 192.168.5.129
Last: 192.168.5.190
Network Address: 192.168.5.192/28
Subnet Mask: 255.255.255.240
First Usable: 192.168.5.193 ← PC3
Last Usable: 192.168.5.206 ← Router Interface
Broadcast: 192.168.5.207
Usable Hosts: 14
Start after LAN1: 192.168.5.192
Block size: 16
Network: 192.168.5.192
Broadcast: 192.168.5.192 + 15 = 192.168.5.207
First: 192.168.5.193
Last: 192.168.5.206
Network Address: 192.168.5.208/28
First Usable: 192.168.5.209 ← PC4
Last Usable: 192.168.5.222 ← Router Interface
Broadcast: 192.168.5.223
Start after LAN3: 192.168.5.208
Network: 192.168.5.208
Broadcast: 192.168.5.208 + 15 = 192.168.5.223
First: 192.168.5.209
Last: 192.168.5.222
Network Address: 192.168.5.224/30
Subnet Mask: 255.255.255.252
First Usable: 192.168.5.225 ← R1 (Serial/G0/0/0)
Last Usable: 192.168.5.226 ← R2 (Serial/G0/0/0)
Broadcast: 192.168.5.227
Usable Hosts: 2
Start after LAN4: 192.168.5.224
Block size: 4
Network: 192.168.5.224
Broadcast: 192.168.5.224 + 3 = 192.168.5.227
First: 192.168.5.225
Last: 192.168.5.226
NETWORK DIAGRAM
┌─────────┐ ┌─────────┐
│ PC1 │ │ PC3 │
│ .129 │ │ .193 │
└────┬────┘ └────┬────┘
│ LAN1 │ LAN3
│ 192.168.5.128/26 │ 192.168.5.192/28
┌────┴────┐ P2P Link ┌────┴────┐
│ R1 │ 192.168.5.224/30 │ R2 │
│ │.225 ══════════════════ .226 │ │
│ .190 │ │ .206 │
│ .126 │ │ .222 │
│ LAN2 │ LAN4
│ 192.168.5.0/25 │ 192.168.5.208/28
┌────┴────┐ ┌────┴────┐
│ PC2 │ │ PC4 │
│ .1 │ │ .209 │
└─────────┘ └─────────┘
Subnet
Mask
PC (First)
Router (Last)
Broadcast
192.168.5.0/25
255.255.255.128
.1
.126
.127
192.168.5.128/26
255.255.255.192
.129
.190
.191
192.168.5.192/28
255.255.255.240
.193
.206
.207
192.168.5.208/28
.209
.222
.223
192.168.5.224/30
255.255.255.252
.225 (R1)
.226 (R2)
.227
enable
configure terminal
hostname R1
! LAN1 Interface (45 hosts)
interface GigabitEthernet0/0/0
description LAN1
ip address 192.168.5.190 255.255.255.192
no shutdown
exit
! LAN2 Interface (64 hosts)
interface GigabitEthernet0/0/1
description LAN2
ip address 192.168.5.126 255.255.255.128
! Point-to-Point to R2
interface Serial0/1/0
description P2P to R2
ip address 192.168.5.225 255.255.255.252
! Static Routes to reach R2's networks
ip route 192.168.5.192 255.255.255.240 192.168.5.226
ip route 192.168.5.208 255.255.255.240 192.168.5.226
end
write memory
hostname R2
! LAN3 Interface (14 hosts)
description LAN3
ip address 192.168.5.206 255.255.255.240
! LAN4 Interface (9 hosts)
description LAN4
ip address 192.168.5.222 255.255.255.240
! Point-to-Point to R1
description P2P to R1
ip address 192.168.5.226 255.255.255.252
! Static Routes to reach R1's networks
ip route 192.168.5.0 255.255.255.128 192.168.5.225
ip route 192.168.5.128 255.255.255.192 192.168.5.225
Last changed24 days ago