IPv4 subnetting
- Hosts per subnet
- 2(32 − prefix) − 2 (/31: 2, /32: 1)
- Block size
- 256 − (mask octet)
- Network address
- IP AND mask
- Broadcast address
- IP OR (NOT mask): all host bits set to 1
- Subnets from n borrowed bits
- 2n
Computer Networks: cram pack
Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.
1Formulas and methods
Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- From decimal
- divide by the base, read the remainders bottom-up
- Hex ↔ binary
- one hex digit = 4 bits
- Two’s complement
- invert the bits and add 1
2Practice problems
1.IPv4 subnettingMedium
The network 192.168.10.0/24 must be divided into at least 6 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/246 subnets
H =
2.IPv4 subnettingHard
The network 192.168.130.0/24 is split with VLSM for these LANs: A: 11 hosts, B: 42 hosts, C: 6 hosts, D: 65 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN C?
A: 11B: 42C: 6D: 65
x =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A64
- B30
- C126
- D62
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AThe same MAC address on both hosts
- BA longer cable between the hosts
- CA hub connected to both VLANs
- DA router or layer 3 switch to route between the VLANs
5.Number systemsMedium
The 8-bit pattern 1001 0010 stores a signed integer in two's complement. What is its value N in decimal?
1001 00108 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1000 1000 >> 2. What is the result's value R in decimal?
1000 1000>> 2
R =
3Solutions
1.IPv4 subnetting
Answer: H = 30
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
Bits to borrow
23 = 8 ≥ 6 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 208
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
D: 65 hosts → 27 = 128 (/25)
B: 42 hosts → 26 = 64 (/26)
A: 11 hosts → 24 = 16 (/28)
C: 6 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .127
B: .128 – .191
A: .192 – .207
C: .208 – .215
Answer
x = 208
LAN C: 192.168.130.208/29.
3.Computer networks
Answer: D · 62
Why this is the answer
A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.
Answer
D · 62
4.Computer networks
Answer: D · A router or layer 3 switch to route between the VLANs
Why this is the answer
VLANs are isolated at layer 2, so traffic between them must be routed, for example by a router on a trunk ("router on a stick") or a layer 3 switch.
Answer
D · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: N = −110
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Add the place values
The other bits keep their normal positive values.
−128 + 16 + 2 = −110
Answer
N = −110
6.Number systems
Answer: R = 34
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Bit by bit
Shifting right by 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1000 1000 >> 2
= 0010 0010
Value of the result
0010 0010₂ = 34
Answer
R = 34
ExamProblems · examproblems.com/courses/computer-networks