CS Atlas GATE 2027
Computer Networks course
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Computer Networks/Phase 0 · Foundations

Chapter review L0.2 · Item 8 of 90

Math and units toolkit

Read at your own pace. Try the questions before opening the answers.

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Chapter recap

Units: 1 B = 8 b; rates decimal (k = 10³), sizes binary (K = 2¹⁰) unless the question says otherwise; time to push out = bits ÷ bit/s. Binary/hex: place values 1, 2, 4…; one hex digit = 4 bits; k bits → 2ᵏ values (max 2ᵏ − 1). Bit operations: XOR marks differences and is carry-free addition; 1's-complement addition wraps the carry around; modulo-2 division XORs at each leading 1 and leaves a remainder one bit shorter than the divisor. Logs and probability: bits for N values = ⌈log₂ N⌉; exactly one of n = n·p·(1 − p)ⁿ⁻¹; (1 − 1/n)ⁿ → 1/e.

Merged rule card

From 0.2.1 Bits, bytes and prefixes: powers of 10 for rates, powers of 2 for sizes

  • 1 B = 8 b. Lower-case b = bit, upper-case B = byte.
  • Rates: k = 10³, M = 10⁶, G = 10⁹ (bps).
  • Sizes: K = 2¹⁰, M = 2²⁰, G = 2³⁰ (bytes) — unless the question says 1000.
  • Time to push out = (size in bits) ÷ (rate in bit/s).
  • ms = 10⁻³ s, µs = 10⁻⁶ s, ns = 10⁻⁹ s.

From 0.2.2 Binary, decimal and hex; powers of 2

  • Binary places: 1, 2, 4, 8, 16, 32, 64, 128, …; MSB left, LSB right.
  • Hex: one digit = 4 bits; A–F = 10–15; group from the right.
  • k bits → 2ᵏ values, 0 … 2ᵏ − 1.
  • Bits needed for N values = smallest k with 2ᵏ ≥ N.
  • 2¹⁰ = 1024 · 2¹⁶ = 65,536 · 2²⁰ ≈ 1.05 M · 2³² ≈ 4.29 × 10⁹.

From 0.2.3 XOR, 1's complement addition and modulo-2 division

  • AND (mask), OR, NOT (flip), XOR (1 where bits differ).
  • a ⊕ a = 0, a ⊕ 0 = a, (a ⊕ k) ⊕ k = a.
  • 1's complement = flip all bits; 1's-complement addition wraps the end carry back in.
  • Modulo-2: + = − = XOR, no carries; bits ↔ polynomial.
  • Mod-2 division: leading 1 → XOR divisor; leading 0 → XOR zeros; remainder = (divisor length − 1) bits.

From 0.2.4 Exponents, logarithms and basic probability

  • 2ᵃ·2ᵇ = 2ᵃ⁺ᵇ; 2¹⁰ ≈ 10³ (estimates only).
  • log₂ x = power of 2 giving x; bits for N values = ⌈log₂ N⌉.
  • P(none) = (1−p)ⁿ · P(≥ 1) = 1 − (1−p)ⁿ · P(exactly 1) = n·p·(1−p)ⁿ⁻¹.
  • (1 − 1/n)ⁿ → 1/e ≈ 0.368; 1/(2e) ≈ 0.184.
  • 1 + 2 + … + 2ᵏ⁻¹ = 2ᵏ − 1; expected tries = 1/p.

Practice questions

  1. NAT · 2 marks

    A 3 MB file (MB = 2²⁰ B) is pushed out at 24 Mbps. Time in seconds (3 decimals)?

    Write your answer before opening the key.

  2. MCQ · 1 mark

    Which value is 0x7F?

    1. 127
    2. 128
    3. 255
    4. 112
  3. NAT · 1 mark

    16-bit 1's-complement sum of 0xFFFF and 0x0002, in decimal?

    Write your answer before opening the key.

  4. MCQ · 2 marks

    Remainder of 100100000 ÷ 1101 (modulo-2)?

    1. 001
    2. 101
    3. 011
    4. 000
  5. MSQ · 1 mark

    Which are equal to 2¹⁶?

    1. 65,536
    2. 0x10000
    3. 64 × 1024
    4. 10⁵ approximately within 1 %
  6. NAT · 1 mark

    8 stations each send with p = 0.125. P(none sends)? (4 decimals)

    Write your answer before opening the key.

  7. MCQ · 1 mark

    Minimum bits to number 0 to 1023?

    1. 9
    2. 10
    3. 11
    4. 1023
  8. MCQ · 1 mark

    100 Mbps expressed in bytes per second is:

    1. 12.5 × 10⁶
    2. 12.5 × 2²⁰
    3. 800 × 10⁶
    4. 100 × 2²⁰
  9. NAT · 1 mark

    n = 4, p = 1/4: P(exactly one) to 4 decimals?

    Write your answer before opening the key.

Answer key

Try the questions first. Open when you are ready to check.
  1. 1.04..1.06 — 3 × 2²⁰ × 8 = 25,165,824 bits ÷ 2.4 × 10⁷ = 1.048576 s.

    Review lesson 0.2.1

  2. A — 7 × 16 + 15 = 127.
    • B: 128 = 0x80.
    • C: 255 = 0xFF.
    • D: 112 = 0x70.

    Review lesson 0.2.2

  3. 2 — 65535 + 2 = 65537 = 0x10001 → wrap: 0x0001 + 1 = 0x0002 = 2.

    Review lesson 0.2.3

  4. A — Long division with XOR gives remainder 001.
    • B: Arithmetic slip; recheck each XOR.
    • C: Arithmetic slip; recheck each XOR.
    • D: The divisor does not divide exactly.

    Review lesson 0.2.3

  5. ABC — 2¹⁶ = 65,536 = 0x10000 = 64 × 1024.
    • D: 65,536 is 34 % off 10⁵.

    Review lesson 0.2.2

  6. 0.343..0.344 — 0.875⁸ ≈ 0.3436.

    Review lesson 0.2.4

  7. B — 1024 values = 2¹⁰ → 10 bits.
    • A: 9 bits give only 512 values.
    • C: More than needed.
    • D: That is the largest value, not a bit count.

    Review lesson 0.2.4

  8. A — Divide bits by 8; the rate stays decimal: 12.5 × 10⁶ B/s.
    • B: A rate is never converted with 2²⁰.
    • C: Multiplying by 8 goes the wrong way.
    • D: Neither divided by 8 nor decimal.

    Review lesson 0.2.1

  9. 0.4218..0.4219 — 0.75³ = 0.421875.

    Review lesson 0.2.4

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