Computer Science Fundamentals: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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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

Propositional logic & proofs

Rows of a truth table
2ⁿ
Implication
p → q ≡ ¬p ∨ q ≡ ¬q → ¬p
Negated implication
¬(p → q) ≡ p ∧ ¬q
Biconditional
p ↔ q ≡ (p → q) ∧ (q → p)
De Morgan
¬(p ∧ q) ≡ ¬p ∨ ¬q, ¬(p ∨ q) ≡ ¬p ∧ ¬q
Distributive laws
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r), p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Negated quantifiers
¬∀x P(x) ≡ ∃x ¬P(x), ¬∃x P(x) ≡ ∀x ¬P(x)
Modus ponens, modus tollens
p → q, p ⊢ q; p → q, ¬q ⊢ ¬p

Method

  1. Count the different variables: the truth table has 2ⁿ rows, starting with all of them 1.
  2. Give every subformula its own column and fill the columns from the inside out.
  3. Apply the rule of each connective: → is 0 only for 1 → 0, ↔ is 1 when both sides agree.
  4. For equivalences, compare the final columns row by row; for an argument, look only at the rows in which every premise is 1.
  5. To negate a quantified statement, move ¬ inwards, swap ∀ and ∃, and negate the inner formula.

Digital logic: gates, K-maps & flip-flops

Basic gates
A·B, A + B, A′, (A·B)′, (A + B)′, A ⊕ B
Boolean rules
X + X′ = 1, X·X′ = 0, X + XY = X, XY + XY′ = X
De Morgan
(X·Y)′ = X′ + Y′, (X + Y)′ = X′·Y′
Karnaugh map order
00, 01, 11, 10 (Gray code)
Group of 2ᵏ cells
loses k variables
4-to-1 multiplexer
F = A′B′I0 + A′BI1 + AB′I2 + ABI3
Flip-flops
D: Q⁺ = D, T: Q⁺ = T ⊕ Q, JK: Q⁺ = JQ′ + K′Q
Counters
n-bit: 2ⁿ states; mod-N: smallest n with 2ⁿ ≥ N; fout = fclk/N
Ripple counter
fmax = 1/(n·tpd)
Ring and Johnson counters
n and 2n states

Method

  1. For a circuit, work from the inputs to the output one gate at a time and write the value on every wire, or write its expression and fill in the truth table.
  2. For simplification, put the 1s (and the don’t-cares as X) into a Karnaugh map with Gray-code rows and columns: 00, 01, 11, 10.
  3. Cover the 1s with as few and as large groups of 1, 2, 4, 8 or 16 cells as possible, wrapping round the edges; each group is one product term made of the variables that stay constant. Group the 0s for the POS form.
  4. For flip-flops and counters, read the inputs at each active clock edge, write the state after every edge, and remember that an n-bit counter wraps round after 2ⁿ states.

Number theory & modular arithmetic

Division with remainder
a = q·n + r, 0 ≤ r < n
Euclid
gcd(a, b) = gcd(b, a mod b)
Least common multiple
lcm(a, b) = a·b/gcd(a, b)
Inverse
a·a⁻¹ ≡ 1 (mod n), gcd(a, n) = 1
Linear congruence
a·x ≡ b → x ≡ a⁻¹·b (mod n)
Euler’s totient
φ(n) = n·∏(1 − 1/p), φ(pᵏ) = pᵏ − pᵏ⁻¹
Fermat
aᵖ⁻¹ ≡ 1 (mod p)
RSA
n = pq, φ = (p − 1)(q − 1), d = e⁻¹ mod φ, c = mᵉ mod n, m = cᵈ mod n

Method

  1. Reduce every number modulo n first; remainders lie between 0 and n − 1.
  2. For a gcd, divide repeatedly with Euclid’s algorithm; the last nonzero remainder is the gcd.
  3. For an inverse, run the extended algorithm and keep the coefficient of a; add n if it is negative.
  4. Solve a·x ≡ b by multiplying with a⁻¹; combine two congruences by writing x = r₁ + m₁k.
  5. For powers, write the exponent in binary and square repeatedly; in RSA, d = e⁻¹ mod φ(n) with φ(n) = (p − 1)(q − 1).

Graph theory

Handshake lemma
Σ deg(v) = 2|E|
Complete graphs
|E(Kn)| = n(n − 1)/2, |E(Km,n)| = m·n
Regular graph
|E| = n·k/2
Tree and forest
|E| = n − 1, |E| = n − c
Leaves of a tree
L = 2 + Σ (deg(v) − 2) over deg(v) ≥ 3
Euler
circuit: all degrees even; path: exactly 2 odd degrees
Hamilton circuits in Kn
(n − 1)!/2
Dijkstra update
d(v) = min(d(v), d(u) + w(u, v))
Walks of length L
(Mᴸ)ij

Method

  1. Write down the degree of every vertex; the degrees add up to twice the number of edges.
  2. For Euler questions, check that the graph is connected and count the vertices with odd degree.
  3. For trees, use n − 1 edges and the handshake lemma.
  4. For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
  5. For a minimum spanning tree, sort the edges by weight and skip every edge that closes a cycle (Kruskal), or grow one tree from a start vertex (Prim).

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

2Practice problems

1.Number systemsMedium
8-bit two's complement0−12806403201608040201

Set the bits so they store −16 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −168 bit
U =
2.Number systemsHard
32 bit1 px512 px512 pxM = ? KiB

An uncompressed image is 512 × 512 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?

512 × 512 px32 bit/px1 KiB = 1024 B
M = KiB
3.Propositional logic & proofsMedium
pqr¬q ∧ (q ∨ p)((p ∧ r) → r) → (¬q ∧ (q ∨ p))¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p)))111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is ¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))) true? Call this number N.

¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p)))
N =
4.Propositional logic & proofsHard
∃x ∀y ¬R(x, y)∀x ∀y ¬R(x, y)∀x ∃y ¬R(x, y)∃y ∀x ¬R(x, y)

Which statement is equivalent to ¬∀x ∃y R(x, y), with the negation moved all the way inside? Click it.

¬∀x ∃y R(x, y)
  1. 1∃x ∀y ¬R(x, y)
  2. 2∀x ∀y ¬R(x, y)
  3. 3∀x ∃y ¬R(x, y)
  4. 4∃y ∀x ¬R(x, y)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010011302040507160121130151141819111010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 3, 6, 8, 9, 11, 13, 14). How many literals L does it contain in total (a primed variable counts as one literal)?

F(A, B, C, D) = Σm(0, 3, 6, 8, 9, 11, 13, 14)
L =
6.Digital logic: gates, K-maps & flip-flopsHard
TQ0​Q0​TQ1​Q1​TQ2​Q2​CLKUP

A 3-bit binary up counter starts at 5 (decimal, Q = 101). Which count Q (in decimal) does it show after 24 clock pulses?

Q0 = 524n = 3
Q =
7.Number theory & modular arithmeticMedium

Find the inverse of 13 modulo 71: the number x with 0 < x < 71 and 13·x ≡ 1 (mod 71).

a = 13n = 71
13−1 mod 71 =
8.Number theory & modular arithmeticHard

Solve 15x ≡ 5 (mod 25). Give the smallest positive solution x.

15x ≡ 5 (mod 25)
x =
9.Algorithms & complexityMedium

In the activity selection problem, which greedy rule always gives a maximum number of non-overlapping activities?

  1. APick the activity that starts earliest
  2. BPick the activity with the fewest overlaps first, then repeat with the compatible ones
  3. CPick the shortest activity first
  4. DPick the activity that finishes earliest, then repeat with the compatible ones
10.Algorithms & complexityHard

What did the Cook–Levin theorem establish?

  1. AThe halting problem is undecidable
  2. BThe Boolean satisfiability problem (SAT) is NP-complete
  3. CEvery NP problem can be solved in exponential time
  4. DP is not equal to NP
11.Data structuresMedium

In a circular queue stored in an array of capacity n, how is the next position after index i computed?

  1. A(i + 1) mod n
  2. Bi + 1
  3. C(i − 1) mod n
  4. D2i + 1
12.Data structuresHard

Which property must a red-black tree satisfy?

  1. AEvery red node must have red children, so red nodes always form unbroken chains to the leaves
  2. BEvery path from a node to its descendant leaves has the same number of black nodes
  3. CThe root must be red
  4. DEvery node must be red
13.Graph theoryMedium
ABCDEFEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 6|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
78823474ABCDEF

Use Dijkstra’s algorithm to find the length d(E) of the shortest path from A to E in the weighted graph.

start: Aend: E
d(E) =
15.Databases & SQLMedium

Which statement about HAVING is correct?

  1. AIt cannot be used with GROUP BY
  2. BIt sorts the results
  3. CIt can use aggregate functions to filter groups
  4. DIt must come before WHERE in the SELECT statement
16.Databases & SQLHard

Which SQL feature is best suited to finding every employee below a manager in an organisation tree of unknown depth?

  1. AA single self-join of the employees table on manager_id
  2. BA UNIQUE constraint on manager_id
  3. CA recursive common table expression (WITH RECURSIVE)
  4. DA GROUP BY on manager_id
17.Computer networksMedium

What is the purpose of the TCP three-way handshake?

  1. ATo encrypt the data
  2. BTo find the IP address of a host before data can be sent to it
  3. CTo close a connection quickly
  4. DTo establish a connection and synchronise sequence numbers
18.Computer networksHard

With sliding-window protocols, how does Go-Back-N differ from Selective Repeat after one frame is lost?

  1. AGo-Back-N resends only the lost frame; Selective Repeat resends that frame and all later ones
  2. BBoth resend the whole window from the start
  3. CGo-Back-N resends the lost frame and all later ones; Selective Repeat only the lost one
  4. DNeither ever resends lost frames
19.IPv4 subnettingMedium
/?11111111111111111111111111000000network bitshost bits

A department needs 49 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

49 hosts
p =
20.IPv4 subnettingHard
last octet = 192112816403201608040201networkhost

A host has the address 10.20.234.192/28. What is the fourth octet x of its network address?

10.20.234.192/28
x =

3Solutions

1.Number systems

Answer: U = 240

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

16 = 0001 0000
inverted: 1110 1111
+1 → 1111 0000

Read it as unsigned

1111 0000₂ = 256 − 16 = 240

Answer

U = 240
2.Number systems

Answer: M = 1024 KiB

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.

Bits and bytes

512·512·32 = 8388608 bit = 1048576 B

Convert to KiB

1048576/1024 = 1024 KiB

Answer

M = 1024 KiB
3.Propositional logic & proofs

Answer: N = 6

Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.

Truth table row by row

p = 1, q = 1, r = 1: p ∧ r = 1, (p ∧ r) → r = 1, ¬q = 0, q ∨ p = 1, ¬q ∧ (q ∨ p) = 0, ((p ∧ r) → r) → (¬q ∧ (q ∨ p)) = 0, ¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))) = 1
p = 1, q = 1, r = 0: p ∧ r = 0, (p ∧ r) → r = 1, ¬q = 0, q ∨ p = 1, ¬q ∧ (q ∨ p) = 0, ((p ∧ r) → r) → (¬q ∧ (q ∨ p)) = 0, ¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))) = 1
p = 1, q = 0, r = 1: p ∧ r = 1, (p ∧ r) → r = 1, ¬q = 1, q ∨ p = 1, ¬q ∧ (q ∨ p) = 1, ((p ∧ r) → r) → (¬q ∧ (q ∨ p)) = 1, ¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))) = 0
p = 1, q = 0, r = 0: p ∧ r = 0, (p ∧ r) → r = 1, ¬q = 1, q ∨ p = 1, ¬q ∧ (q ∨ p) = 1, ((p ∧ r) → r) → (¬q ∧ (q ∨ p)) = 1, ¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))) = 0
p = 0, q = 1, r = 1: p ∧ r = 0, (p ∧ r) → r = 1, ¬q = 0, q ∨ p = 1, ¬q ∧ (q ∨ p) = 0, ((p ∧ r) → r) → (¬q ∧ (q ∨ p)) = 0, ¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))) = 1
p = 0, q = 1, r = 0: p ∧ r = 0, (p ∧ r) → r = 1, ¬q = 0, q ∨ p = 1, ¬q ∧ (q ∨ p) = 0, ((p ∧ r) → r) → (¬q ∧ (q ∨ p)) = 0, ¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))) = 1
p = 0, q = 0, r = 1: p ∧ r = 0, (p ∧ r) → r = 1, ¬q = 1, q ∨ p = 0, ¬q ∧ (q ∨ p) = 0, ((p ∧ r) → r) → (¬q ∧ (q ∨ p)) = 0, ¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))) = 1
p = 0, q = 0, r = 0: p ∧ r = 0, (p ∧ r) → r = 1, ¬q = 1, q ∨ p = 0, ¬q ∧ (q ∨ p) = 0, ((p ∧ r) → r) → (¬q ∧ (q ∨ p)) = 0, ¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))) = 1

Count the rows with value 1

¬(((p ∧ r) → r) → (¬q ∧ (q ∨ p))): 1, 1, 0, 0, 1, 1, 1, 1
N = 6

Answer

N = 6
4.Propositional logic & proofs

Answer: ∃x ∀y ¬R(x, y)

Quantifiers and arguments: ¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x). Move a negation inwards, swap every quantifier it passes, and negate the inner formula with the propositional laws, such as ¬(a → b) ≡ a ∧ ¬b. An argument is valid when the conclusion is 1 in every row in which all premises are 1. A single row with true premises and a false conclusion is a counterexample and shows the argument is invalid.

Move the negation inwards

¬∀x ∃y R(x, y) ≡ ∃x ∀y ¬R(x, y)

Answer

∃x ∀y ¬R(x, y)

The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.

5.Digital logic: gates, K-maps & flip-flops

Answer: L = 12

Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.

ABCD000111100001111010011302040507160121130151141819111010

Group the 1s

Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.

B′C′D′
B′CD
BCD′
AC′D

Minimal SOP form

F = B′C′D′ + B′CD + BCD′ + AC′D
L = 12

Answer

L = 12
6.Digital logic: gates, K-maps & flip-flops

Answer: Q = 5

Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.

Full cycles

Every 8 pulses bring the counter back to the same state.

24 = 3·8 + 0

Count after the remaining pulses

Q = (5 + 0) mod 8 = 5
Q = 101₂

Answer

Q = 5
7.Number theory & modular arithmetic

Answer: 13−1 mod 71 = 11

Inverses and congruences: a has an inverse modulo n exactly when gcd(a, n) = 1. The extended Euclidean algorithm writes 1 = s·n + t·a, and then a⁻¹ ≡ t (mod n). A congruence a·x ≡ b (mod n) with gcd(a, n) = 1 has the solution x ≡ a⁻¹·b. Two congruences x ≡ r₁ (mod m₁) and x ≡ r₂ (mod m₂) with coprime moduli have one solution modulo m₁·m₂ (Chinese remainder theorem). Euler’s φ(n) counts the numbers from 1 to n that are coprime to n: φ(n) = n·∏(1 − 1/p).

Extended Euclidean algorithm

Each remainder r is written as r ≡ t·13 (mod 71).

r = 71, t = 0
r = 13, q = 5, t = 1
r = 6, q = 2, t = −5
r = 1, q = 6, t = 11

Inverse

1 ≡ 11·13 (mod 71)
x = 11
check: 13·11 = 143 = 2·71 + 1

Answer

13−1 mod 71 = 11
8.Number theory & modular arithmetic

Answer: x = 2

Inverses and congruences: a has an inverse modulo n exactly when gcd(a, n) = 1. The extended Euclidean algorithm writes 1 = s·n + t·a, and then a⁻¹ ≡ t (mod n). A congruence a·x ≡ b (mod n) with gcd(a, n) = 1 has the solution x ≡ a⁻¹·b. Two congruences x ≡ r₁ (mod m₁) and x ≡ r₂ (mod m₂) with coprime moduli have one solution modulo m₁·m₂ (Chinese remainder theorem). Euler’s φ(n) counts the numbers from 1 to n that are coprime to n: φ(n) = n·∏(1 − 1/p).

Divide by the gcd

gcd(15, 25) = 5
3x ≡ 1 (mod 5)

Solve the reduced congruence

3−1 ≡ 2 (mod 5) → x ≡ 2 (mod 5)

Solutions modulo 25

x ∈ {2, 7, 12, 17, 22}
x = 2

Answer

x = 2
9.Algorithms & complexity

Answer: D · Pick the activity that finishes earliest, then repeat with the compatible ones

Why this is the answer

Choosing the earliest finishing time can be proven optimal by an exchange argument; the other rules have counterexamples.

Answer

D · Pick the activity that finishes earliest, then repeat with the compatible ones
10.Algorithms & complexity

Answer: B · The Boolean satisfiability problem (SAT) is NP-complete

Why this is the answer

Cook and Levin showed independently that every problem in NP reduces to SAT in polynomial time, so SAT was the first known NP-complete problem.

Answer

B · The Boolean satisfiability problem (SAT) is NP-complete
11.Data structures

Answer: A · (i + 1) mod n

Why this is the answer

The modulo operation wraps index n − 1 back to 0, so the array is reused without shifting elements.

Answer

A · (i + 1) mod n
12.Data structures

Answer: B · Every path from a node to its descendant leaves has the same number of black nodes

Why this is the answer

Red-black rules: nodes are red or black, root is black, red nodes have black children, and all paths have equal black-height, keeping height ≤ 2 log₂(n + 1).

Answer

B · Every path from a node to its descendant leaves has the same number of black nodes
13.Graph theory

Answer: Euler path, no circuit

Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.

Degrees

deg(A) = 2, deg(B) = 2, deg(C) = 2, deg(D) = 1, deg(E) = 4, deg(F) = 1

Odd vertices

D, F → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

14.Graph theory

Answer: d(E) = 9

Shortest paths and spanning trees: Dijkstra’s algorithm settles one vertex per round: the unsettled vertex with the smallest tentative distance. Its distance is then final, and every neighbour v gets d(v) = min(d(v), d(u) + w(u, v)). Kruskal’s algorithm builds a minimum spanning tree by taking the edges from light to heavy and skipping any edge that would close a cycle; Prim’s algorithm grows one tree from a start vertex by always adding the lightest edge that leaves it. A spanning tree of n vertices has n − 1 edges.

Dijkstra round by round

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): C: 7
C (7): B: 15, E: 9
E (9): D: 12, F: 16, B: 13
D (12)
B (13)
F (16)

Shortest path

A → C → E
d(E) = 9

Answer

d(E) = 9
15.Databases & SQL

Answer: C · It can use aggregate functions to filter groups

Why this is the answer

HAVING filters groups after aggregation and may reference aggregates; WHERE cannot.

Answer

C · It can use aggregate functions to filter groups
16.Databases & SQL

Answer: C · A recursive common table expression (WITH RECURSIVE)

Why this is the answer

A recursive CTE starts with an anchor query and repeatedly joins the previous result to the table, so it handles hierarchies of any depth.

Answer

C · A recursive common table expression (WITH RECURSIVE)
17.Computer networks

Answer: D · To establish a connection and synchronise sequence numbers

Why this is the answer

The handshake sets up a connection and agrees initial sequence numbers on both sides before data is sent.

Answer

D · To establish a connection and synchronise sequence numbers
18.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
19.IPv4 subnetting

Answer: p = 26

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

25 − 2 = 30 < 49
26 − 2 = 62 ≥ 49 → h = 6

Prefix length

p = 32 − 6 = 26

Answer

p = /26
20.IPv4 subnetting

Answer: x = 192

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: ⌊192/16⌋·16 = 192
broadcast: 192 + 16 − 1 = 207
usable hosts: 193 … 206

Answer

x = 192

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