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 −20 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?

N = −208 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1001 0001 XOR 0010 1101. What is the result's value R in decimal?

1001 0001XOR0010 1101
R =
3.Propositional logic & proofsMedium
abcb ∧ ¬(¬c ∧ a)¬(b ∧ ¬(¬c ∧ a))(b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a))111?110?101?100?011?010?001?000?

In how many of the 8 rows of its truth table is (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) true? Call this number N.

(b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a))
N =
4.Propositional logic & proofsHard
¬ca ∧ ¬cc¬a

The premises a → c; a are given. Which conclusion follows from them? Click it.

a → ca
  1. 1¬c
  2. 2a ∧ ¬c
  3. 3c
  4. 4¬a
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010110312140507061120130150140819011110

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C, D) = Σm(0, 1, 2, 4, 9, 10, 12). How many sum terms (factors) N does it have?

F(A, B, C, D) = Σm(0, 1, 2, 4, 9, 10, 12)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
CLKD12345678

A positive-edge-triggered D flip-flop starts with Q = 0 and receives the inputs shown at 8 rising clock edges. How many times N does Q change state?

Q₀ = 0D
N =
7.Number theory & modular arithmeticMedium

Find the inverse of 31 modulo 75: the number x with 0 < x < 75 and 31·x ≡ 1 (mod 75).

a = 31n = 75
31−1 mod 75 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (2021, 11)d = ?

An RSA key uses the primes p = 43 and q = 47 and the public exponent e = 11. Find the private exponent d (0 < d < φ(n)).

p = 43q = 47e = 11
d =
9.Algorithms & complexityMedium

What does the recurrence T(n) = T(n − 1) + O(n) solve to?

  1. AO(2ⁿ)
  2. BO(n)
  3. CO(n²)
  4. DO(n log n)
10.Algorithms & complexityHard

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. AComparisons take logarithmic time each
  2. BBinary search is the only possible algorithm, and it needs log₂ n comparisons
  3. CEvery element must be compared at least once
  4. DA decision tree with n possible outcomes must have depth at least log₂ n
11.Data structuresMedium

Which abstract data type does Dijkstra's algorithm use to choose the next vertex to process?

  1. AA set with no ordering
  2. BA stack
  3. CA FIFO queue
  4. DA min-priority queue
12.Data structuresHard

Which data structure supports an LRU (least recently used) cache with O(1) get and put?

  1. AA single sorted array
  2. BA binary search tree only
  3. CA hash map combined with a doubly linked list
  4. DA stack only
13.Graph theoryMedium

A connected graph has 8 vertices and 10 edges. How many edges k must be removed to leave a spanning tree?

|V| = 8|E| = 10
k =
14.Graph theoryHard
139347517188ABCDEFG

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 7|E| = 9
W =
15.Databases & SQLMedium

Third normal form (3NF) removes:

  1. AAtomic values
  2. BComposite keys
  3. CRedundant indexes on columns that are not part of a key
  4. DTransitive dependencies between non-key attributes
16.Databases & SQLHard

Which window function assigns 1, 2, 3, ... to rows within each partition without ties?

  1. AROW_NUMBER()
  2. BCOUNT()
  3. CDENSE_RANK()
  4. DRANK()
17.Computer networksMedium

What is a cyclic redundancy check (CRC) used for in Ethernet frames?

  1. AChoosing the best route
  2. BDetecting transmission errors
  3. CCorrecting every bit error automatically
  4. DEncrypting the frame
18.Computer networksHard

A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?

  1. AThe default gateway
  2. BThe physical cable
  3. CName resolution (DNS)
  4. DThe network card driver
19.IPv4 subnettingMedium
/24 → /2711111111111111111111111111100000network bitshost bits

The network 192.168.98.0/24 is divided into /27 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 2?

192.168.98.0/24/27subnet 2 (subnet 1 starts at .0)
x =
20.IPv4 subnettingHard

The network 192.168.54.0/24 is split with VLSM for these LANs: A: 3 hosts, B: 43 hosts, C: 102 hosts, D: 9 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 D?

A: 3B: 43C: 102D: 9
x =

3Solutions

1.Number systems

Answer: U = 236

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

20 = 0001 0100
inverted: 1110 1011
+1 → 1110 1100

Read it as unsigned

1110 1100₂ = 256 − 20 = 236

Answer

U = 236
2.Number systems

Answer: R = 188

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

XOR gives 1 where the two bits differ.

1001 0001
XOR 0010 1101
= 1011 1100

Value of the result

1011 1100₂ = 188

Answer

R = 188
3.Propositional logic & proofs

Answer: N = 2

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

a = 1, b = 1, c = 1: a ∧ c = 1, b ∧ (a ∧ c) = 1, ¬c = 0, ¬c ∧ a = 0, ¬(¬c ∧ a) = 1, b ∧ ¬(¬c ∧ a) = 1, ¬(b ∧ ¬(¬c ∧ a)) = 0, (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) = 0
a = 1, b = 1, c = 0: a ∧ c = 0, b ∧ (a ∧ c) = 0, ¬c = 1, ¬c ∧ a = 1, ¬(¬c ∧ a) = 0, b ∧ ¬(¬c ∧ a) = 0, ¬(b ∧ ¬(¬c ∧ a)) = 1, (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) = 0
a = 1, b = 0, c = 1: a ∧ c = 1, b ∧ (a ∧ c) = 0, ¬c = 0, ¬c ∧ a = 0, ¬(¬c ∧ a) = 1, b ∧ ¬(¬c ∧ a) = 0, ¬(b ∧ ¬(¬c ∧ a)) = 1, (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) = 0
a = 1, b = 0, c = 0: a ∧ c = 0, b ∧ (a ∧ c) = 0, ¬c = 1, ¬c ∧ a = 1, ¬(¬c ∧ a) = 0, b ∧ ¬(¬c ∧ a) = 0, ¬(b ∧ ¬(¬c ∧ a)) = 1, (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) = 0
a = 0, b = 1, c = 1: a ∧ c = 0, b ∧ (a ∧ c) = 0, ¬c = 0, ¬c ∧ a = 0, ¬(¬c ∧ a) = 1, b ∧ ¬(¬c ∧ a) = 1, ¬(b ∧ ¬(¬c ∧ a)) = 0, (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) = 1
a = 0, b = 1, c = 0: a ∧ c = 0, b ∧ (a ∧ c) = 0, ¬c = 1, ¬c ∧ a = 0, ¬(¬c ∧ a) = 1, b ∧ ¬(¬c ∧ a) = 1, ¬(b ∧ ¬(¬c ∧ a)) = 0, (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) = 1
a = 0, b = 0, c = 1: a ∧ c = 0, b ∧ (a ∧ c) = 0, ¬c = 0, ¬c ∧ a = 0, ¬(¬c ∧ a) = 1, b ∧ ¬(¬c ∧ a) = 0, ¬(b ∧ ¬(¬c ∧ a)) = 1, (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) = 0
a = 0, b = 0, c = 0: a ∧ c = 0, b ∧ (a ∧ c) = 0, ¬c = 1, ¬c ∧ a = 0, ¬(¬c ∧ a) = 1, b ∧ ¬(¬c ∧ a) = 0, ¬(b ∧ ¬(¬c ∧ a)) = 1, (b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)) = 0

Count the rows with value 1

(b ∧ (a ∧ c)) ↔ ¬(b ∧ ¬(¬c ∧ a)): 0, 0, 0, 0, 1, 1, 0, 0
N = 2

Answer

N = 2
4.Propositional logic & proofs

Answer: c

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.

Rows with all premises true

a = 1, c = 1: ¬c = 0, a ∧ ¬c = 0, c = 1, ¬a = 0

Conclusion

c (modus ponens)

Answer

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

Answer: N = 4

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.

ABCD000111100001111010110312140507061120130150140819011110

Group the 0s

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

(C′ + D′)
(B′ + D′)
(B′ + C′)
(A′ + B + C + D)

Minimal POS form

F = (C′ + D′)(B′ + D′)(B′ + C′)(A′ + B + C + D)
N = 4

Answer

N = 4
6.Digital logic: gates, K-maps & flip-flops

Answer: N = 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.

Q after every edge

A D flip-flop copies D at each rising edge: Q⁺ = D.

1: D = 1 → Q = 1
2: D = 1 → Q = 1
3: D = 1 → Q = 1
4: D = 0 → Q = 0
5: D = 1 → Q = 1
6: D = 1 → Q = 1
7: D = 0 → Q = 0
8: D = 1 → Q = 1

Count the changes

0 → 1 → 1 → 1 → 0 → 1 → 1 → 0 → 1
N = 5

Answer

N = 5
7.Number theory & modular arithmetic

Answer: 31−1 mod 75 = 46

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·31 (mod 75).

r = 75, t = 0
r = 31, q = 2, t = 1
r = 13, q = 2, t = −2
r = 5, q = 2, t = 5
r = 3, q = 1, t = −12
r = 2, q = 1, t = 17
r = 1, q = 2, t = −29

Inverse

1 ≡ −29·31 (mod 75)
x = 46
check: 31·46 = 1426 = 19·75 + 1

Answer

31−1 mod 75 = 46
8.Number theory & modular arithmetic

Answer: d = 527

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.

Modulus and totient

n = 43·47 = 2021
φ(n) = 42·46 = 1932

Extended Euclidean algorithm

r = 1932, t = 0
r = 11, q = 175, t = 1
r = 7, q = 1, t = −175
r = 4, q = 1, t = 176
r = 3, q = 1, t = −351
r = 1, q = 3, t = 527

Private exponent

d = 527
check: 11·527 = 5797 = 3·1932 + 1

Answer

d = 527
9.Algorithms & complexity

Answer: C · O(n²)

Why this is the answer

The work sums to about n² / 2, which is the recurrence of selection sort or of quicksort with the worst pivot.

Answer

C · O(n²)
10.Algorithms & complexity

Answer: D · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

D · A decision tree with n possible outcomes must have depth at least log₂ n
11.Data structures

Answer: D · A min-priority queue

Why this is the answer

A min-priority queue, usually a binary heap, returns the unvisited vertex with the smallest distance efficiently.

Answer

D · A min-priority queue
12.Data structures

Answer: C · A hash map combined with a doubly linked list

Why this is the answer

The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).

Answer

C · A hash map combined with a doubly linked list
13.Graph theory

Answer: k = 3

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.

Edges of a spanning tree

8 − 1 = 7

Edges to remove

k = 10 − 7 = 3

Answer

k = 3
14.Graph theory

Answer: W = 40

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.

Kruskal’s algorithm

BD (3): take
CE (4): take
DF (5): take
DE (7): take
GE (8): take
BC (9): skip, closes a cycle
AC (13): take

Total weight

W = 3 + 4 + 5 + 7 + 8 + 13 = 40

Answer

W = 40
15.Databases & SQL

Answer: D · Transitive dependencies between non-key attributes

Why this is the answer

In 3NF no non-key attribute depends on another non-key attribute, such as storing city and postcode where city depends on postcode.

Answer

D · Transitive dependencies between non-key attributes
16.Databases & SQL

Answer: A · ROW_NUMBER()

Why this is the answer

ROW_NUMBER gives unique sequential numbers; RANK leaves gaps after ties; DENSE_RANK does not.

Answer

A · ROW_NUMBER()
17.Computer networks

Answer: B · 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

B · Detecting transmission errors
18.Computer networks

Answer: C · Name resolution (DNS)

Why this is the answer

A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.

Answer

C · Name resolution (DNS)
19.IPv4 subnetting

Answer: x = 32

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

25 = 32

Subnet number 2

network: (2 − 1)·32 = 32
broadcast: 32 + 32 − 1 = 63

Answer

x = 32
20.IPv4 subnetting

Answer: x = 192

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

C: 102 hosts → 27 = 128 (/25)
B: 43 hosts → 26 = 64 (/26)
D: 9 hosts → 24 = 16 (/28)
A: 3 hosts → 23 = 8 (/29)

Allocation, largest first

C: .0 – .127
B: .128 – .191
D: .192 – .207
A: .208 – .215

Answer

x = 192

LAN D: 192.168.54.192/28.

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