Exam Pattern & Cheat Sheet

CSE 214 midterm · Full marks 30 · 1 hr 30 min · answer all 3 questions. Four past papers analysed below — the skeleton barely changes.

The recurring skeleton

Q	CLO	Topic	What it asks (every semester)
Q1	CO1	Ch.1 + IAS	Moore’s law · IAS structure/memory/formats/size · IAS assembly + machine code · abstraction · embedded/IC generations
Q2	CO3	Performance	CPU-time & CPI comparison · MIPS vs execution time · Amdahl’s Law (derive + numeric) · BIPS
Q3	CO2	MIPS	addressing modes · R/I/J formats · C → assembly → machine code · last byte/word · max offset

Master one worked example of each sub-type and you can answer the entire paper. Each solved past paper below is one click away.

If the question is about…	Go to
Moore’s law, translation hierarchy, RISC/CISC, abstraction	Ch.1 Concepts
IAS structure, memory size, formats, programs	IAS chapter
CPI, MIPS pitfall, Amdahl, BIPS	Performance chapter
MIPS formats, addressing modes, encoding	MIPS chapter

Performance

\text{CPU time} = \text{IC} \times \text{CPI} \times T = \frac{\text{IC} \times \text{CPI}}{\text{clock rate}} \qquad \text{total cycles} = \text{IC} \times \text{CPI}

\text{MIPS} = \frac{\text{IC}}{\text{exec time} \times 10^{6}} \qquad \text{BIPS} = \frac{\text{IC}}{\text{exec time} \times 10^{9}} = \frac{\text{clock rate}}{\text{CPI} \times 10^{9}}

\text{Amdahl: new time} = (1 - F) + \frac{F}{S} \qquad \text{Speedup} = \frac{1}{(1 - F) + \dfrac{F}{S}} \qquad \text{max speedup} = \frac{1}{1 - F}

IAS

4096 \text{ words} \times 40 \text{ bits}

· address = 12 bits ·

\text{size} = 40 \times 2^{12} = 163{,}840 \text{ bits} = 20{,}480 \text{ bytes} = 4096 \text{ words}

\text{rows} = 2^{\text{addr bits}}

\text{addr bits} = \log_2(\text{rows})

· number word = 1 sign + 39 · instr word =

2 \times (\text{op}_{8} + \text{addr}_{12})

LOAD=00000001 · ADD=00000101 · SUB=00000110 · STOR=00100001 · LSH =

\times 2

MIPS
word = 32 bits = 4 bytes · byte-addressed · last byte =

2^{32} - 1

· last word =

2^{32} - 4

R-format

I-format

J-format

branch dest =

(\text{PC} + 4) + 4 \times \text{offset}

· max offset =

2^{15} - 1

· $s0..=16.. · $t0..=8..

Instr	Type	op	funct
add	R	000000	100000
sub	R	000000	100010
slt	R	000000	101010
lw	I	100011	—
sw	I	101011	—
beq / bne	I	000100 / 000101	—
addi	I	001000	—
j / jal	J	000010 / 000011	—

Register numbers: $zero=0 · $v0–1=2–3 · $a0–3=4–7 · $t0–7=8–15 · $s0–7=16–23 · $t8–9=24–25 · $gp=28 · $sp=29 · $fp=30 · $ra=31.