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 −120 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −1208 bit
U =
2.Number systemsHard
12 bits, unsigned, N = ?????????????

What is the largest unsigned integer N that fits in 12 bits?

12 bit
N =
3.Propositional logic & proofsMedium
pqr¬r¬r ∨ p¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p)111?110?101?100?011?010?001?000?

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

¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p)
N =
4.Propositional logic & proofsHard
∀x (¬P(x) ∧ ¬Q(x))∃x (¬P(x) ∨ ¬Q(x))∃x (¬P(x) ∧ ¬Q(x))∀x (¬P(x) ∨ ¬Q(x))

Which statement is equivalent to ¬∀x (P(x) ∨ Q(x)), with the negation moved all the way inside? Click it.

¬∀x (P(x) ∨ Q(x))
  1. 1∀x (¬P(x) ∧ ¬Q(x))
  2. 2∃x (¬P(x) ∨ ¬Q(x))
  3. 3∃x (¬P(x) ∧ ¬Q(x))
  4. 4∀x (¬P(x) ∨ ¬Q(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABCD000111100001111010011312040507060121131150141809111010

Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(0, 2, 3, 8, 11, 13, 15). How many product terms N does it have?

F(A, B, C, D) = Σm(0, 2, 3, 8, 11, 13, 15)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
ABCD0001111000011110X001130214X51716X12013X150140809011110

Find the minimal SOP form of F(A, B, C, D) = Σm(3, 4, 6, 7, 10) + d(0, 5, 12, 15), where d marks the don't-care conditions. How many product terms N does it have?

F(A, B, C, D) = Σm(3, 4, 6, 7, 10) + d(0, 5, 12, 15)
N =
7.Number theory & modular arithmeticMedium

Find the smallest positive integer x with x ≡ 2 (mod 5) and x ≡ 3 (mod 7).

x ≡ 2 (mod 5)x ≡ 3 (mod 7)
x =
8.Number theory & modular arithmeticHard
mc = me​ mod n347m = cd​ mod n?(n, e) = (629, 7)d

An RSA key uses p = 17, q = 37 and e = 7. Decrypt the ciphertext c = 347: find the message m.

p = 17q = 37e = 7c = 347
m =
9.Algorithms & complexityMedium

Breadth-first search on an unweighted graph finds:

  1. AThe minimum spanning tree
  2. BA topological order in all graphs
  3. CShortest paths in number of edges from the start node
  4. DShortest paths in graphs that contain negative edge weights
10.Algorithms & complexityHard

What is the difference between an undecidable problem and an intractable one?

  1. AIntractable problems cannot be solved by any algorithm, while undecidable ones only need exponential time
  2. BUndecidable problems are in NP, while intractable problems are in P
  3. CBoth mean the problem has no algorithm at all
  4. DAn undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
11.Data structuresMedium

You have a reference to a node in the middle of a list. In which list can you remove it in O(1)?

  1. AA doubly linked list
  2. BA dynamic array
  3. CA singly linked list
  4. DA sorted array
12.Data structuresHard

The time complexity of decrease-key in a binary heap (given the element's position) is:

  1. AO(n)
  2. BO(log n)
  3. CO(1)
  4. DO(n log n)
13.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
14.Graph theoryHard
5114111466ABCDEF

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

What does the functional dependency A → B mean?

  1. AA is a foreign key referencing B
  2. BEach value of A is associated with exactly one value of B
  3. CA and B are both primary keys
  4. DEach value of B is associated with exactly one value of A
16.Databases & SQLHard

What does the write-ahead logging (WAL) rule require?

  1. AData pages must be written before the log
  2. BA change's log record must reach stable storage before the changed data page does
  3. CEvery query must be logged before it is parsed
  4. DThe log is written only after a crash
17.Computer networksMedium

Which medium access method did classic shared (half-duplex) Ethernet use?

  1. ACSMA/CD
  2. BToken passing
  3. CTDMA
  4. DCSMA/CA
18.Computer networksHard

Which mechanism prevents broadcast loops in a network with redundant switch links?

  1. ADHCP snooping
  2. BSpanning Tree Protocol
  3. CARP caching
  4. DNAT
19.IPv4 subnettingMedium
/24 → /2811111111111111111111111111110000network bitshost bits

The network 192.168.10.0/24 must be divided into at least 12 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?

192.168.10.0/2412 subnets
H =
20.IPv4 subnettingHard
third octet = 89012816403211618040211networkhost

A host has the address 172.20.89.210/21. What is the third octet x of its subnet's broadcast address?

172.20.89.210/21
x =

3Solutions

1.Number systems

Answer: U = 136

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

120 = 0111 1000
inverted: 1000 0111
+1 → 1000 1000

Read it as unsigned

1000 1000₂ = 256 − 120 = 136

Answer

U = 136
2.Number systems

Answer: N = 4095

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.

Largest value

212 − 1 = 4095

Answer

N = 4095
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: ¬p = 0, ¬p ↔ q = 0, ¬(¬p ↔ q) = 1, q → ¬(¬p ↔ q) = 1, ¬(q → ¬(¬p ↔ q)) = 0, ¬r = 0, ¬r ∨ p = 1, ¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p) = 1
p = 1, q = 1, r = 0: ¬p = 0, ¬p ↔ q = 0, ¬(¬p ↔ q) = 1, q → ¬(¬p ↔ q) = 1, ¬(q → ¬(¬p ↔ q)) = 0, ¬r = 1, ¬r ∨ p = 1, ¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p) = 1
p = 1, q = 0, r = 1: ¬p = 0, ¬p ↔ q = 1, ¬(¬p ↔ q) = 0, q → ¬(¬p ↔ q) = 1, ¬(q → ¬(¬p ↔ q)) = 0, ¬r = 0, ¬r ∨ p = 1, ¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p) = 1
p = 1, q = 0, r = 0: ¬p = 0, ¬p ↔ q = 1, ¬(¬p ↔ q) = 0, q → ¬(¬p ↔ q) = 1, ¬(q → ¬(¬p ↔ q)) = 0, ¬r = 1, ¬r ∨ p = 1, ¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p) = 1
p = 0, q = 1, r = 1: ¬p = 1, ¬p ↔ q = 1, ¬(¬p ↔ q) = 0, q → ¬(¬p ↔ q) = 0, ¬(q → ¬(¬p ↔ q)) = 1, ¬r = 0, ¬r ∨ p = 0, ¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p) = 1
p = 0, q = 1, r = 0: ¬p = 1, ¬p ↔ q = 1, ¬(¬p ↔ q) = 0, q → ¬(¬p ↔ q) = 0, ¬(q → ¬(¬p ↔ q)) = 1, ¬r = 1, ¬r ∨ p = 1, ¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p) = 1
p = 0, q = 0, r = 1: ¬p = 1, ¬p ↔ q = 0, ¬(¬p ↔ q) = 1, q → ¬(¬p ↔ q) = 1, ¬(q → ¬(¬p ↔ q)) = 0, ¬r = 0, ¬r ∨ p = 0, ¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p) = 0
p = 0, q = 0, r = 0: ¬p = 1, ¬p ↔ q = 0, ¬(¬p ↔ q) = 1, q → ¬(¬p ↔ q) = 1, ¬(q → ¬(¬p ↔ q)) = 0, ¬r = 1, ¬r ∨ p = 1, ¬(q → ¬(¬p ↔ q)) ∨ (¬r ∨ p) = 1

Count the rows with value 1

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

Answer

N = 7
4.Propositional logic & proofs

Answer: ∃x (¬P(x) ∧ ¬Q(x))

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 (P(x) ∨ Q(x)) ≡ ∃x (¬P(x) ∧ ¬Q(x))

Answer

∃x (¬P(x) ∧ ¬Q(x))

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: 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.

ABCD000111100001111010011312040507060121131150141809111010

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
A′B′D′
ABD

Minimal SOP form

F = B′C′D′ + B′CD + A′B′D′ + ABD
N = 4

Answer

N = 4

Another grouping of the same size exists; it has the same number of terms and literals.

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

Answer: N = 3

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.

Prime implicants

The largest groups that can be formed from 1s and Xs.

BC′D′, BCD, A′C′D′, A′CD, A′B, AB′CD′
ABCD0001111000011110X001130214X51716X12013X150140809011110

Minimal cover

F = A′CD + A′B + AB′CD′
N = 3

Answer

N = 3
7.Number theory & modular arithmetic

Answer: x = 17

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).

First congruence

x = 2 + 5k

Second congruence

2 + 5k ≡ 3 (mod 7) → 5k ≡ 1 (mod 7)
k ≡ 3·1 ≡ 3 (mod 7)

Result

x = 2 + 5·3 = 17
x = 17
check: 17 mod 5 = 2, 17 mod 7 = 3

Answer

x = 17
8.Number theory & modular arithmetic

Answer: m = 97

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.

Private exponent

n = 629, φ(n) = 576
d = 7−1 mod 576 = 247

Decrypt by repeated squaring

247 = 128 + 64 + 32 + 16 + 4 + 2 + 1
3471 ≡ 347 (mod 629)
3472 ≡ 270 (mod 629)
3474 ≡ 565 (mod 629)
3478 ≡ 322 (mod 629)
34716 ≡ 528 (mod 629)
34732 ≡ 137 (mod 629)
34764 ≡ 528 (mod 629)
347128 ≡ 137 (mod 629)
347247 ≡ 347·270·565·528·137·528·137 ≡ 97 (mod 629)

Message

m = 97
check: 977 mod 629 = 347

Answer

m = 97
9.Algorithms & complexity

Answer: C · Shortest paths in number of edges from the start node

Why this is the answer

BFS visits nodes in order of distance (edges) from the source, so it finds shortest unweighted paths in O(V + E).

Answer

C · Shortest paths in number of edges from the start node
10.Algorithms & complexity

Answer: D · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow

Why this is the answer

The halting problem has no algorithm at all; an NP-hard problem such as TSP can be solved by brute force, just not in known polynomial time.

Answer

D · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
11.Data structures

Answer: A · A doubly linked list

Why this is the answer

A doubly linked node knows its predecessor, so you can relink prev and next directly; a singly linked list must search for the predecessor.

Answer

A · A doubly linked list
12.Data structures

Answer: B · O(log n)

Why this is the answer

After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.

Answer

B · O(log n)
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) = 1, deg(B) = 4, deg(C) = 2, deg(D) = 1, deg(E) = 2

Odd vertices

A, D → 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) = 12

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): B: 5, C: 1
C (1): B: 2, E: 12
B (2): D: 6
D (6): F: 12
E (12)
F (12)

Shortest path

A → C → E
d(E) = 12

Answer

d(E) = 12
15.Databases & SQL

Answer: B · Each value of A is associated with exactly one value of B

Why this is the answer

A → B means any two rows with the same A value must have the same B value; A determines B.

Answer

B · Each value of A is associated with exactly one value of B
16.Databases & SQL

Answer: B · A change's log record must reach stable storage before the changed data page does

Why this is the answer

Writing the log first lets the DBMS redo committed work and undo uncommitted work after a crash, supporting atomicity and durability.

Answer

B · A change's log record must reach stable storage before the changed data page does
17.Computer networks

Answer: A · 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

A · CSMA/CD
18.Computer networks

Answer: B · Spanning Tree Protocol

Why this is the answer

STP elects a root bridge and blocks redundant paths, unblocking them if an active link fails.

Answer

B · Spanning Tree Protocol
19.IPv4 subnetting

Answer: H = 14

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).

Bits to borrow

24 = 16 ≥ 12 → b = 4, /28

Hosts per subnet

24 − 2 = 14

Answer

H = 14
20.IPv4 subnetting

Answer: x = 95

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 in the third octet

23 = 8

Subnet of this host

network: 172.20.88.0
broadcast: 172.20.95.255

Answer

x = 95

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