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 1669 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1669 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.53.15/28. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.53.15/28
x =
3.Computer networksMedium
A signal travels 2000 km through fibre at about 2 × 10⁸ m/s. What is the propagation delay?
- A1 ms
- B0.1 ms
- C10 ms
- D100 ms
4.Computer networksHard
The count-to-infinity problem occurs in which kind of routing protocol, and what helps against it?
- AStatic routing; adding default routes
- BPath-vector; longer AS paths
- CLink-state; flooding link updates
- DDistance-vector; split horizon and poison reverse
5.Number systemsMedium
The 8-bit pattern 1001 1000 stores a signed integer in two's complement. What is its value N in decimal?
1001 10008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0111 >> 2. What is the result's value R in decimal?
1100 0111>> 2
R =
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1669
211 − 2 = 2046 ≥ 1669 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 15
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
24 = 16
Subnet of this host
network: ⌊15/16⌋·16 = 0
broadcast: 0 + 16 − 1 = 15
usable hosts: 1 … 14
Answer
x = 15
3.Computer networks
Answer: C · 10 ms
Why this is the answer
2 × 10⁶ m / 2 × 10⁸ m/s = 0.01 s = 10 ms. Propagation delay depends on distance and medium, not packet size.
Answer
C · 10 ms
4.Computer networks
Answer: D · Distance-vector; split horizon and poison reverse
Why this is the answer
After a failure, distance-vector routers can keep advertising stale routes to each other, raising the metric slowly. Split horizon, poison reverse and a maximum metric limit this.
Answer
D · Distance-vector; split horizon and poison reverse
5.Number systems
Answer: N = −104
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 + 8 = −104
Answer
N = −104
6.Number systems
Answer: R = 49
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.
1100 0111 >> 2
= 0011 0001
Value of the result
0011 0001₂ = 49
Answer
R = 49
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