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 12 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2412 subnets
H =
2.IPv4 subnettingHard
The network 192.168.30.0/24 is split with VLSM for these LANs: A: 15 hosts, B: 19 hosts, C: 64 hosts, D: 4 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 B?
A: 15B: 19C: 64D: 4
x =
3.Computer networksMedium
What is the network address of the host 192.168.10.77/26?
- A192.168.10.128
- B192.168.10.0
- C192.168.10.64
- D192.168.10.76
4.Computer networksHard
In a TCP connection, what does the receive window (rwnd) control?
- AThe size of each IP packet header
- BThe retransmission time-out, calculated from the measured round-trip time of each segment
- CHow much unacknowledged data the sender may send, based on the receiver's buffer
- DThe number of open ports
5.Number systemsMedium
Set the bits so they store −2 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −28 bit
U =
6.Number systemsHard
A system must give every one of 2482 items its own binary code of the same length. What is the smallest number of bits b needed?
2482
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 14
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
24 = 16 ≥ 12 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: x = 128
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
C: 64 hosts → 27 = 128 (/25)
B: 19 hosts → 25 = 32 (/27)
A: 15 hosts → 25 = 32 (/27)
D: 4 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .127
B: .128 – .159
A: .160 – .191
D: .192 – .199
Answer
x = 128
LAN B: 192.168.30.128/27.
3.Computer networks
Answer: C · 192.168.10.64
Why this is the answer
/26 subnets start at 0, 64, 128 and 192. 77 lies in the block 64–127, so the network address is 192.168.10.64.
Answer
C · 192.168.10.64
4.Computer networks
Answer: C · How much unacknowledged data the sender may send, based on the receiver's buffer
Why this is the answer
The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.
Answer
C · How much unacknowledged data the sender may send, based on the receiver's buffer
5.Number systems
Answer: U = 254
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.
Invert and add one
2 = 0000 0010
inverted: 1111 1101
+1 → 1111 1110
Read it as unsigned
1111 1110₂ = 256 − 2 = 254
Answer
U = 254
6.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 2482
212 = 4096 ≥ 2482
Answer
b = 12
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