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 complement?−128?64?32?16?8?4?2?1

Write −85 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −858 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1101 1010 << 3. What is the result's value R in decimal?

1101 1010<< 3
R =
3.Propositional logic & proofsMedium
pqr¬(r ∧ ¬p)¬(r ∧ ¬p) ↔ q¬((¬q → r) → p) → (¬(r ∧ ¬p) ↔ q)111?110?101?100?011?010?001?000?

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

¬((¬q → r) → p) → (¬(r ∧ ¬p) ↔ q)
N =
4.Propositional logic & proofsHard
∃x ∀y ¬R(x, y)∀x ∀y ¬R(x, y)∃y ∀x ¬R(x, y)∀x ∃y ¬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∃y ∀x ¬R(x, y)
  4. 4∀x ∃y ¬R(x, y)
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111011111

In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.

F = Σm(1, 8, 12, 13, 15)
  1. 1AB = 00, CD = 00
  2. 2AB = 00, CD = 01
  3. 3AB = 00, CD = 11
  4. 4AB = 00, CD = 10
  5. 5AB = 01, CD = 00
  6. 6AB = 01, CD = 01
  7. 7AB = 01, CD = 11
  8. 8AB = 01, CD = 10
  9. 9AB = 11, CD = 00
  10. 10AB = 11, CD = 01
  11. 11AB = 11, CD = 11
  12. 12AB = 11, CD = 10
  13. 13AB = 10, CD = 00
  14. 14AB = 10, CD = 01
  15. 15AB = 10, CD = 11
  16. 16AB = 10, CD = 10
6.Digital logic: gates, K-maps & flip-flopsHard
DQ0​Q0​DQ1​Q1​DQ2​Q2​DQ3​Q3​DQ4​Q4​DQ5​Q5​CLKRING

A ring counter is made of 6 D flip-flops with the last output fed back to the first input, and starts with a single 1. How many different states N does it go through?

n = 6
N =
7.Number theory & modular arithmeticMedium

Solve 27x ≡ 5 (mod 73). Give the solution x with 0 < x < 73.

27x ≡ 5 (mod 73)
x =
8.Number theory & modular arithmeticHard

71 is prime. Use Fermat’s little theorem to find 1094716 mod 71.

p = 71k = 94716
1094716 mod 71 =
9.Algorithms & complexityMedium

Dynamic programming is useful when a problem has:

  1. AOverlapping subproblems and optimal substructure
  2. BA single loop
  3. COnly random inputs
  4. DNo subproblems at all, so it can be solved in one step
10.Algorithms & complexityHard

What are the time and space costs of computing the n-th Fibonacci number with a loop that keeps only the last two values?

  1. AO(n) time and O(n) space
  2. BO(n) time and O(1) space
  3. CO(log n) time and O(n) space
  4. DO(2ⁿ) time and O(1) space
11.Data structuresMedium

What is the worst-case height of a binary search tree with n nodes if keys are inserted in sorted order?

  1. A1
  2. Bn (it becomes a linked list)
  3. C√n
  4. Dlog₂ n
12.Data structuresHard

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(n)
  2. BO(α(n)), the inverse Ackermann function, which is practically constant
  3. CO(log n) exactly
  4. DO(n²)
13.Graph theoryMedium

A tree has 4 vertices of degree 3, 4 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

4 × deg 34 × deg 53 × deg 2
L =
14.Graph theoryHard
351317647ABCDE

Prim’s algorithm builds a minimum spanning tree starting from vertex B. Click the edge it adds as edge number 2.

start: Bk = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge BE
15.Databases & SQLMedium

Why are indexes created on columns?

  1. ATo make inserts and updates faster on those columns
  2. BTo save disk space
  3. CTo prevent NULL values
  4. DTo speed up searches and joins on those columns
16.Databases & SQLHard

Given the functional dependencies A → B, B → C and C → D on R(A, B, C, D), what is the closure A+?

  1. A{A, B}
  2. B{A}
  3. C{A, B, C, D}
  4. D{A, B, C}
17.Computer networksMedium

Why is UDP often preferred for live video calls?

  1. AUDP is required for all video
  2. BLow delay matters more than retransmitting every lost packet
  3. CUDP guarantees every packet arrives, so the picture never freezes
  4. DUDP encrypts data automatically
18.Computer networksHard

By Shannon's theorem, what is the capacity of a 1 MHz channel with a signal-to-noise ratio of 255?

  1. A2 Mbit/s
  2. B255 Mbit/s
  3. C1 Mbit/s
  4. D8 Mbit/s
19.IPv4 subnettingMedium
/24 → /2611111111111111111111111111000000network bitshost bits

The network 192.168.181.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 4?

192.168.181.0/24/26subnet 4 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard
last octet = 128112806403201608040201networkhost

A host has the address 10.20.236.128/28. What is the last number (fourth octet) x of its subnet's broadcast address?

10.20.236.128/28
x =

3Solutions

1.Number systems

Answer: U = 171

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

85 = 0101 0101
inverted: 1010 1010
+1 → 1010 1011

Read it as unsigned

1010 1011₂ = 256 − 85 = 171

Answer

U = 171
2.Number systems

Answer: R = 208

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.

1101 1010 << 3
= 1101 0000

Value of the result

1101 0000₂ = 208

Answer

R = 208
3.Propositional logic & proofs

Answer: N = 7

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

Count the rows with value 1

¬((¬q → r) → p) → (¬(r ∧ ¬p) ↔ q): 1, 1, 1, 1, 0, 1, 1, 1
N = 7

Answer

N = 7
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: AB = 00, CD = 01

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.

Neighbours of each 1

A 1 with a neighbouring 1 can join a group of two or more.

m1: m0, m3, m5, m9 → 0
ABCD000111100001111011111

The isolated 1

m1 = A′B′C′D → AB = 00, CD = 01

Answer

AB = 00, CD = 01
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 6

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.

Sequence for 6 flip-flops

100000 → 010000 → 001000 → 000100 → 000010 → 000001

Number of states

N = 6

Answer

N = 6
7.Number theory & modular arithmetic

Answer: x = 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).

Inverse of 27

r = 73, t = 0
r = 27, q = 2, t = 1
r = 19, q = 1, t = −2
r = 8, q = 2, t = 3
r = 3, q = 2, t = −8
r = 2, q = 1, t = 19
r = 1, q = 2, t = −27
27−1 ≡ 46 (mod 73)

Solve

x ≡ 46·5 = 230 ≡ 11 (mod 73)
check: 27·11 = 297 ≡ 5 (mod 73)

Answer

x = 11
8.Number theory & modular arithmetic

Answer: 1094716 mod 71 = 36

Modular powers and RSA: compute aᵏ mod n by repeated squaring: write k in binary, square, reduce after every step and multiply the powers whose bit is 1. Fermat: aᵖ⁻¹ ≡ 1 (mod p) for a prime p not dividing a, so the exponent can be reduced modulo p − 1. RSA: choose primes p and q, n = p·q and φ(n) = (p − 1)(q − 1); the public exponent e is coprime to φ(n), and the private exponent is d = e⁻¹ mod φ(n). Encrypt with c = mᵉ mod n and decrypt with m = cᵈ mod n.

Reduce the exponent

94716 mod 70 = 6
1094716 ≡ 106 (mod 71)

Small power

106 mod 71 = 36

Answer

1094716 mod 71 = 36
9.Algorithms & complexity

Answer: A · Overlapping subproblems and optimal substructure

Why this is the answer

Storing results of overlapping subproblems (memoisation or tabulation) avoids repeated work, turning exponential algorithms into polynomial ones.

Answer

A · Overlapping subproblems and optimal substructure
10.Algorithms & complexity

Answer: B · O(n) time and O(1) space

Why this is the answer

Each step needs only the two previous values, so the table of the bottom-up method can be replaced by two variables.

Answer

B · O(n) time and O(1) space
11.Data structures

Answer: B · n (it becomes a linked list)

Why this is the answer

Sorted insertions create a degenerate tree with height n, making searches O(n). Balanced trees prevent this.

Answer

B · n (it becomes a linked list)
12.Data structures

Answer: B · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

B · O(α(n)), the inverse Ackermann function, which is practically constant
13.Graph theory

Answer: L = 18

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.

Handshake lemma for a tree

L + 38 = 2(L + 11) − 2
L = 18

Answer

L = 18

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

14.Graph theory

Answer: edge AC

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.

Prim’s algorithm

1. AB (3)
2. AC (5)
3. CE (6)
4. DE (4)

edge number 2

AC

Answer

edge AC
15.Databases & SQL

Answer: D · To speed up searches and joins on those columns

Why this is the answer

Indexes (often B-trees) speed up lookups, but take space and slow down inserts and updates that must maintain them.

Answer

D · To speed up searches and joins on those columns
16.Databases & SQL

Answer: C · {A, B, C, D}

Why this is the answer

Start with A, add B (A → B), then C (B → C), then D (C → D); since A+ contains all attributes, A is a key.

Answer

C · {A, B, C, D}
17.Computer networks

Answer: B · Low delay matters more than retransmitting every lost packet

Why this is the answer

Retransmitting lost packets would add delay; real-time media tolerates some loss, so UDP-based protocols like RTP are used.

Answer

B · Low delay matters more than retransmitting every lost packet
18.Computer networks

Answer: D · 8 Mbit/s

Why this is the answer

C = 10⁶ × log₂(256) = 10⁶ × 8 = 8 Mbit/s. This is the upper limit for error-free data rate on a noisy channel.

Answer

D · 8 Mbit/s
19.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

26 = 64

Subnet number 4

network: (4 − 1)·64 = 192
broadcast: 192 + 64 − 1 = 255

Answer

x = 192
20.IPv4 subnetting

Answer: x = 143

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: ⌊128/16⌋·16 = 128
broadcast: 128 + 16 − 1 = 143
usable hosts: 129 … 142

Answer

x = 143

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