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 436 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
436 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.187.224/29. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.187.224/29
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ATDMA
- BCSMA/CD
- CCSMA/CA
- DToken passing
4.Computer networksHard
Why does a browser warn when a site presents a self-signed certificate?
- AThe certificate is always expired
- BThe site must be using HTTP instead of HTTPS
- CNo trusted certificate authority vouches that the key belongs to that site
- DSelf-signed certificates use no encryption at all, so the traffic is sent in clear text
5.Number systemsMedium
Write −113 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1138 bit
U =
6.Number systemsHard
A system must give every one of 4452 items its own binary code of the same length. What is the smallest number of bits b needed?
4452
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 23
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
28 − 2 = 254 < 436
29 − 2 = 510 ≥ 436 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
2.IPv4 subnetting
Answer: x = 231
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).
Block size
23 = 8
Subnet of this host
network: ⌊224/8⌋·8 = 224
broadcast: 224 + 8 − 1 = 231
usable hosts: 225 … 230
Answer
x = 231
3.Computer networks
Answer: B · CSMA/CD
Why this is the answer
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer
B · CSMA/CD
4.Computer networks
Answer: C · No trusted certificate authority vouches that the key belongs to that site
Why this is the answer
Browsers trust certificates signed, directly or through a chain, by CAs in their trust store. A self-signed certificate could be made by anyone, such as a man-in-the-middle.
Answer
C · No trusted certificate authority vouches that the key belongs to that site
5.Number systems
Answer: U = 143
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
113 = 0111 0001
inverted: 1000 1110
+1 → 1000 1111
Read it as unsigned
1000 1111₂ = 256 − 113 = 143
Answer
U = 143
6.Number systems
Answer: b = 13 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
212 = 4096 < 4452
213 = 8192 ≥ 4452
Answer
b = 13
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