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 −76 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?

N = −768 bit
U =
2.Number systemsHard
click the bits to set them012806403201608040201

Set the bits to the 8-bit result of 1110 0100 >> 3. What is the result's value R in decimal?

1110 0100>> 3
R =
3.Propositional logic & proofsMedium
pqr¬r ∨ ¬(q ∧ p)(p ∨ (q ↔ r)) → (¬r ∨ ¬(q ∧ p))¬((p ∨ (q ↔ r)) → (¬r ∨ ¬(q ∧ p)))111?110?101?100?011?010?001?000?

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

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

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

¬∀x (F(x) ∨ G(x))
  1. 1∀x (¬F(x) ∨ ¬G(x))
  2. 2∃x (¬F(x) ∧ ¬G(x))
  3. 3∃x (¬F(x) ∨ ¬G(x))
  4. 4∀x (¬F(x) ∧ ¬G(x))
5.Digital logic: gates, K-maps & flip-flopsMedium
ABC00011110010011130214150716

Use the Karnaugh map to find the minimal product-of-sums (POS) form of F(A, B, C) = Σm(1, 3, 4, 5, 6). How many sum terms (factors) N does it have?

F(A, B, C) = Σm(1, 3, 4, 5, 6)
N =
6.Digital logic: gates, K-maps & flip-flopsHard
FFQ0​Q0​FFQ1​Q1​FFQ2​Q2​FFQ3​Q3​CLKMOD-12

A mod-12 counter has to count through 12 different states. What is the minimum number of flip-flops n it needs?

N = 12
n =
7.Number theory & modular arithmeticMedium

Find the inverse of 16 modulo 19: the number x with 0 < x < 19 and 16·x ≡ 1 (mod 19).

a = 16n = 19
16−1 mod 19 =
8.Number theory & modular arithmeticHard
mc = me​ mod ncm = cd​ mod nm(n, e) = (1763, 17)d = ?

An RSA key uses the primes p = 41 and q = 43 and the public exponent e = 17. Find the private exponent d (0 < d < φ(n)).

p = 41q = 43e = 17
d =
9.Algorithms & complexityMedium

Why do many library sorting routines switch to insertion sort for very small subarrays?

  1. AInsertion sort has little overhead and is fast on small or nearly sorted inputs
  2. BRecursive algorithms cannot handle arrays below 10 elements
  3. CInsertion sort is O(n log n) in the worst case
  4. DInsertion sort uses less memory than any other sorting algorithm on large inputs
10.Algorithms & complexityHard

Topological sorting is possible only for:

  1. AGraphs with negative weights
  2. BUndirected graphs
  3. CDirected acyclic graphs
  4. DComplete graphs
11.Data structuresMedium

What is the purpose of a sentinel (dummy) node in a linked list?

  1. ATo prevent the list from growing beyond a fixed maximum number of nodes
  2. BTo simplify code by removing special cases at the head or tail
  3. CTo make the list sorted
  4. DTo store the number of elements
12.Data structuresHard

In Kruskal's minimum spanning tree algorithm, what is the union-find structure used for?

  1. AStoring each vertex's current distance from the source during the search
  2. BChecking whether an edge's two endpoints are already connected
  3. CChoosing the starting vertex
  4. DSorting the edges by weight
13.Graph theoryMedium

A connected graph has 17 vertices and 18 edges. How many edges k must be removed to leave a spanning tree?

|V| = 17|E| = 18
k =
14.Graph theoryHard
751311211133ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

start: Around 3
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
15.Databases & SQLMedium

A view in SQL is:

  1. AA stored query that behaves like a virtual table
  2. BAn index type
  3. CA copy of a table on disk
  4. DA screen in the database client for browsing tables
16.Databases & SQLHard

When is decomposing R into R1 and R2 guaranteed to be lossless?

  1. AWhen every attribute of R appears in both R1 and R2
  2. BWhen R1 and R2 have no attributes in common
  3. CWhen R1 and R2 have the same number of columns
  4. DWhen the common attributes of R1 and R2 form a superkey of R1 or of R2
17.Computer networksMedium

What is the role of a default gateway on a host?

  1. AIt is the router used for traffic to other networks
  2. BIt is the DNS server
  3. CIt assigns MAC addresses
  4. DIt is the switch port the host is connected to
18.Computer networksHard

TCP congestion control's "slow start" phase:

  1. AKeeps the window fixed at one segment
  2. BIncreases the congestion window exponentially until a threshold or loss
  3. CHalves the congestion window every round trip until a packet loss is detected
  4. DIs used only by UDP
19.IPv4 subnettingMedium
/?11111111111111111111111000000000network bitshost bits

A department needs 373 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?

373 hosts
p =
20.IPv4 subnettingHard
third octet 3 … 10012806403201608041211commonvaries

A router has routes to these networks: 10.117.3.0/24, 10.117.4.0/24, 10.117.5.0/24, 10.117.6.0/24, 10.117.7.0/24, 10.117.8.0/24, 10.117.9.0/24, 10.117.10.0/24. What is the longest prefix length p of a single summary route that covers all of them?

10.117.3.0/24 … 10.117.10.0/24
p =

3Solutions

1.Number systems

Answer: U = 180

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

76 = 0100 1100
inverted: 1011 0011
+1 → 1011 0100

Read it as unsigned

1011 0100₂ = 256 − 76 = 180

Answer

U = 180
2.Number systems

Answer: R = 28

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 3 moves every bit 3 places to the right; bits that fall off the right end are lost and zeros come in on the left.

1110 0100 >> 3
= 0001 1100

Value of the result

0001 1100₂ = 28

Answer

R = 28
3.Propositional logic & proofs

Answer: N = 1

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

Count the rows with value 1

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

Answer

N = 1
4.Propositional logic & proofs

Answer: ∃x (¬F(x) ∧ ¬G(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 (F(x) ∨ G(x)) ≡ ∃x (¬F(x) ∧ ¬G(x))

Answer

∃x (¬F(x) ∧ ¬G(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 = 2

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.

ABC00011110010011130214150716

Group the 0s

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

(A + C)
(A′ + B′ + C′)

Minimal POS form

F = (A + C)(A′ + B′ + C′)
N = 2

Answer

N = 2
6.Digital logic: gates, K-maps & flip-flops

Answer: n = 4

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.

Number of flip-flops

2ⁿ ≥ 12 → n = 4 (2⁴ = 16)

Answer

n = 4
7.Number theory & modular arithmetic

Answer: 16−1 mod 19 = 6

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·16 (mod 19).

r = 19, t = 0
r = 16, q = 1, t = 1
r = 3, q = 5, t = −1
r = 1, q = 3, t = 6

Inverse

1 ≡ 6·16 (mod 19)
x = 6
check: 16·6 = 96 = 5·19 + 1

Answer

16−1 mod 19 = 6
8.Number theory & modular arithmetic

Answer: d = 593

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 = 41·43 = 1763
φ(n) = 40·42 = 1680

Extended Euclidean algorithm

r = 1680, t = 0
r = 17, q = 98, t = 1
r = 14, q = 1, t = −98
r = 3, q = 4, t = 99
r = 2, q = 1, t = −494
r = 1, q = 2, t = 593

Private exponent

d = 593
check: 17·593 = 10081 = 6·1680 + 1

Answer

d = 593
9.Algorithms & complexity

Answer: A · Insertion sort has little overhead and is fast on small or nearly sorted inputs

Why this is the answer

For small n the simple loop of insertion sort beats the recursion overhead of quicksort or merge sort, even though it is O(n²).

Answer

A · Insertion sort has little overhead and is fast on small or nearly sorted inputs
10.Algorithms & complexity

Answer: C · Directed acyclic graphs

Why this is the answer

A topological order exists if and only if the directed graph has no cycles; it can be found in O(V + E) with DFS or Kahn's algorithm.

Answer

C · Directed acyclic graphs
11.Data structures

Answer: B · To simplify code by removing special cases at the head or tail

Why this is the answer

With a sentinel, every real node has a predecessor, so inserting or deleting at the head needs no special handling.

Answer

B · To simplify code by removing special cases at the head or tail
12.Data structures

Answer: B · Checking whether an edge's two endpoints are already connected

Why this is the answer

Find tells whether both endpoints are in the same component; if not, the edge is added and union merges the components.

Answer

B · Checking whether an edge's two endpoints are already connected
13.Graph theory

Answer: k = 2

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

17 − 1 = 16

Edges to remove

k = 18 − 16 = 2

Answer

k = 2
14.Graph theory

Answer: B

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

1. A (0): C: 7
2. C (7): B: 12, E: 18
3. B (12): D: 25, E: 15
4. E (15): D: 17, F: 28
5. D (17)
6. F (28)

round 3

B

Answer

B
15.Databases & SQL

Answer: A · A stored query that behaves like a virtual table

Why this is the answer

A view stores a SELECT statement; querying it runs the underlying query (materialised views store the result).

Answer

A · A stored query that behaves like a virtual table
16.Databases & SQL

Answer: D · When the common attributes of R1 and R2 form a superkey of R1 or of R2

Why this is the answer

If the shared attributes determine all attributes of R1 or of R2, joining them back produces exactly the original rows and no spurious tuples.

Answer

D · When the common attributes of R1 and R2 form a superkey of R1 or of R2
17.Computer networks

Answer: A · It is the router used for traffic to other networks

Why this is the answer

The host sends packets for non-local destinations to the default gateway, usually its local router.

Answer

A · It is the router used for traffic to other networks
18.Computer networks

Answer: B · Increases the congestion window exponentially until a threshold or loss

Why this is the answer

Slow start grows the congestion window exponentially; after the threshold, congestion avoidance grows it linearly, and loss reduces it.

Answer

B · Increases the congestion window exponentially until a threshold or loss
19.IPv4 subnetting

Answer: p = 23

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

28 − 2 = 254 < 373
29 − 2 = 510 ≥ 373 → h = 9

Prefix length

p = 32 − 9 = 23

Answer

p = /23
20.IPv4 subnetting

Answer: p = 20

Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.

Compare the bits

3 = 00000011
10 = 00001010
common leading bits: 4

Prefix length

p = 16 + 4 = 20

Answer

p = /20

Summary route: 10.117.0.0/20 (it also covers 8 network(s) that are not in the list).

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