Summer 2024 — Fully Solved

Answer all three. Try each part first, then reveal.

Q1 (CO1) — Foundations & IAS

1aWhat is computer architecture? Show the possible organization of an embedded system.

1bShow the generation-wise advances in IC technology in tabular form.

Generation	Technology
1st	Vacuum tubes
2nd	Transistors
3rd	Small/Medium-Scale ICs (SSI/MSI)
4th	Large-Scale ICs (LSI)
5th	Very/Ultra Large-Scale ICs (VLSI / ULSI)

1cSketch the structure of the IAS computer with its components.

Q2 (CO3) — Performance

2abigA: 250 ps, CPI 2.0. B: 500 ps, CPI 1.2 (same ISA, same program). Which is faster and by how much?

Same ISA + same program ⇒ same IC, so it cancels. $\text{CPU time} = \text{IC} \times \text{CPI} \times T$ :

\begin{aligned} A&: \text{IC} \times 2.0 \times 250 \text{ ps} = 500 \cdot \text{IC ps} \\ B&: \text{IC} \times 1.2 \times 500 \text{ ps} = 600 \cdot \text{IC ps} \end{aligned}

A is faster. Ratio = $\frac{600}{500} = 1.2$ → A is 1.2× faster than B.

2bbig500 MHz, classes A/B/C = 1/2/3 cycles. C1: 5B/1B/1B. C2: 10B/1B/1B. Faster by MIPS? by execution time?

\begin{aligned} \text{Clock rate} &= 500 \text{ MHz} = 5 \times 10^{8} \text{ Hz} \\ \text{Cycles} &= \sum (\text{count} \times \text{CPI}) \\ \text{C1 cycles} &= 5 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 = 10 \times 10^{9} &\qquad \text{IC}_1 &= 7 \times 10^{9} \\ \text{C2 cycles} &= 10 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 = 15 \times 10^{9} &\qquad \text{IC}_2 &= 12 \times 10^{9} \end{aligned}

\begin{aligned} \text{Exec time} &= \frac{\text{cycles}}{\text{clock rate}} \\ T_1 &= \frac{10 \times 10^{9}}{5 \times 10^{8}} = 20 \text{ s} &\qquad T_2 &= \frac{15 \times 10^{9}}{5 \times 10^{8}} = 30 \text{ s} \end{aligned}

\begin{aligned} \text{MIPS} &= \frac{\text{IC}}{\text{time} \times 10^{6}} \\ \text{MIPS}_1 &= \frac{7 \times 10^{9}}{20 \times 10^{6}} = 350 &\qquad \text{MIPS}_2 &= \frac{12 \times 10^{9}}{30 \times 10^{6}} = 400 \end{aligned}

By MIPS: C2 wins (400 > 350). By execution time: C1 wins (20 s < 30 s). Classic pitfall — higher MIPS (C2) is actually the slower program. Trust execution time.

Q3 (CO2) — MIPS

3aDraw all MIPS-32 addressing modes.

3bbigConvert sub $s0,$s1,$s2 to mid-level and low-level. (opcode 000000, funct 100000.)

3cAddresses of the last byte and word in MIPS-32.

$2^{32}$ bytes. Last byte = $2^{32} - 1$ ; last word = $2^{32} - 4$ (words are 4 bytes, aligned).