Exercises - CS 61C Course Notes

1Conceptual Review¶

Question

Solution

2Short Exercises¶

True/False: If a piece of data is both in the cache and in memory, reading it from cache is faster than reading from memory

Solution

True. The cache is smaller and faster than memory.

Conversions: Convert the following numbers into the quantity of bytes each term represents (you may leave your answer in terms of powers of 2). (See precheck section on IEC Prefixes for assistance)

a) 4 KiB

Solution

One KiB is 2¹⁰ and $4 = 2^2$ , so $4 \text{ KiB} = 2^2 \times 2^{10} = 2^{12}$ bytes.

b) 2 MiB

Solution

One MiB is 2²⁰ and $2 = 2^1$ , so $2 \text{ MiB} = 2^1 \times 2^{20} = 2^{21}$ bytes.

c) 8 Kib

Solution

Notice how the unit is Kib (Kibibits) and not KiB (Kibibytes). One Kib is 2¹⁰ bits, so $8 \text{ Kib} = 2^3 \times 2^{10} = 2^{13}$ bits. Because there are $8 = 2^3$ bits in one byte, we divide our answer to get $8 \text{ Kib} = 2^{10}$ bytes.
(Note that $8 \text{ Kib} = 1 \text{ KiB}$ )

d) 24 GiB

Solution

We can factor $24 = 4 \times 6 = 2^2 \times 2 \times 3 = 2^3 \times 3$ . One GiB = 2³⁰, so we can write $24 \text{ GiB} = 2^3 \times 3 \times 2^{30} = 3 \times 2^{33}$ bytes (alternatively, just $24 \times 2^{30}$ bytes).

e) 19 TiB

Solution

Note that 19 cannot be factored or easily representable in powers of 2. Following the same process as above, we can simplify to $19 \text{ TiB} = 19 \times 2^{40}$ bytes $\approx 19$ trillion bytes.