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
A department needs 654 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
654 hosts
p =
2.IPv4 subnettingHard
The network 192.168.236.0/24 is split with VLSM for these LANs: A: 5 hosts, B: 22 hosts, C: 18 hosts, D: 123 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: 5B: 22C: 18D: 123
x =
3.Computer networksMedium
What is a cyclic redundancy check (CRC) used for in Ethernet frames?
- AEncrypting the frame
- BChoosing the best route
- CDetecting transmission errors
- DCorrecting every bit error automatically
4.Computer networksHard
With sliding-window protocols, how does Go-Back-N differ from Selective Repeat after one frame is lost?
- ANeither ever resends lost frames
- BBoth resend the whole window from the start
- CGo-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
- DGo-Back-N resends only the lost frame; Selective Repeat resends that frame and all later ones
5.Number systemsMedium
Write −35 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −358 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1000 1001 << 3. What is the result's value R in decimal?
1000 1001<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: p = 22
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).
Host bits needed
29 − 2 = 510 < 654
210 − 2 = 1022 ≥ 654 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: x = 160
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: 123 hosts → 27 = 128 (/25)
B: 22 hosts → 25 = 32 (/27)
C: 18 hosts → 25 = 32 (/27)
A: 5 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .127
B: .128 – .159
C: .160 – .191
A: .192 – .199
Answer
x = 160
LAN C: 192.168.236.160/27.
3.Computer networks
Answer: C · Detecting transmission errors
Why this is the answer
The sender computes a CRC over the frame; the receiver recomputes it and discards the frame if they differ. CRC detects errors but does not correct them.
Answer
C · Detecting transmission errors
4.Computer networks
Answer: C · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
Why this is the answer
A Go-Back-N receiver discards out-of-order frames, so the sender must repeat everything after the loss. Selective Repeat buffers them and asks only for the missing one.
Answer
C · Go-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
5.Number systems
Answer: U = 221
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.
Invert and add one
35 = 0010 0011
inverted: 1101 1100
+1 → 1101 1101
Read it as unsigned
1101 1101₂ = 256 − 35 = 221
Answer
U = 221
6.Number systems
Answer: R = 72
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 left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1000 1001 << 3
= 0100 1000
Value of the result
0100 1000₂ = 72
Answer
R = 72
ExamProblems · examproblems.com/courses/computer-networks