FIRST MIDTERM ANSWERS JehanFranois Pris jfparissbcglobal net A

FIRST MIDTERM ANSWERS Jehan-François Pâris jfparis@sbcglobal. net

A Problem with the first problem set • Problem 5 states – Simplify the expression (ABC)’ (A + DE) (A + B’C) (B + E) and convert it into a form that can be represented using a programmable logic array. • This means transforming the expression into sum of products

A Problem with the first problem set • Neglecting null products, we have – (ABC)’ (A + DE) (A + B’C) (B + E) = (A’ + B’ + C’) (A+ DE) (A +B’C) (B + E) = A’B’CDE + AB’E + AB’CE + AB’DE + ABC’ + AC’E + AB’C’DE + AC’E = B’CDE + AB’E + ABC’ + AC’E • This is fine but we can eliminate one of the three products

A Problem with the first problem set • Let us rewrite AC’E = ABC’E + AB’C’E • We would them have – B’CDE + AB’E + ABC’E + AB’C’E = B’CDE + AB’E(1 + C’) + ABC’(1 + E) = B’CDE + AB’E + ABC’ • Solution obtained using Karnaugh maps – Those you have it will have 20/20.

First Question • Simplify the following Boolean expression. (10 points) – w (v x y + y) +w' y + v w x + y

Answer – w (v x y + y) + w' y + v w x + y = v w x y + w’ y + v w x + y = v w x + y

Second Question • Give a simplified implementation for the following expression using only NAND gates. (20 points) – w' (x y) + w x y' + x' y'

Answer w' x' y (w'x'y)' ((w'x'y)'. y)' = w'x'y + y'

Third Question • Consider a huge program that consists of a purely sequential part that takes two hours and another part that takes eight hours. What is the maximum speedup we can achieve by parallelizing the second part of the program? (5 points)

Amdahl’s Law • The total speedup for the whole system will be • The maximum possible speedup when Ti 0

Answer – The maximum speedup we can achieve is a factor of (8 + 2)/2 = 5.

Fourth Question • Server motherboard A has a SPEC CPU 2006 rating of 31. 4 while server B has a rating of 29. 7. Which one of the two motherboards is faster? (5 points)

Answer • Motherboard _ A _ is faster than the other motherboard.

Fifth Question (I) • Assume we have a very basic microprocessor doing 4 -bit arithmetic. (3× 5 points) – How would you represent the decimal value – 8 in signed arithmetic?

Answer • – 8 is represented in 4 -bit arithmetic by 1000

Fifth Question (II) – What would be its result of adding 0100 to 0101 assuming that • The numbers being added were unsigned integers? • The numbers being added were signed integers?

Answer – 0100 + 0101 = 1001 • If the numbers being added are unsigned integers, the answer is +9 • If the numbers being added are signed integers, the answer is INVALID due to an overflow condition

Sixth Question • Implement a two-bit counter going to the cycle 00, 01, 10, 11, 00, … when its input is on. You may use the flipflops and the gates of your choice. (15 points)

Answer • State transition diagram 00 01 11 10

Answer • Four states means two flipflops – One for lesser significant bit: • Triggered by each transition – One for more significant bit: • Triggered by transition occurring when LSB = 1 • Use T flipflops

Answer T X Q’ Q Clock T From chapter II notes Y Q’ Q Clock

Seventh Question • Given the following MIPS instructions, – lw r 1, a(r 2) addi r 1, r 2, #n addiu r 1, r 2, #n andi r 1, r 2, #n ori r 1, r 2, #n – Clear the contents of register r 1? (5 points) – Copy the contents of register r 2 into

Answer – Clear the contents of register r 1? (5 points) • ori r 1, $0, 0 • andi r 1, $0, 0 • addi r 1, $0, 0 (slower? ) – Copy the contents of register r 2 into register r 1? (5 points) • ori r 1, r 2, 0 • addi r 1, r 2, 0 (slower? ) • addr 1, r 2, $0

Eighth Question • Unlike the IBM/360 instruction set, the MIPS instruction set has fixed-size instructions. – What is the main advantage of having fixed-size instructions? (5 points) – Which group of MIPS instructions takes advantage of this property and how? (5 points)

Answer – What is the main advantage of having fixed-size instructions? (5 points) • Can predict the exact location of an instruction before decoding the instructions that precede –Allows prefetching and pipelining

Answer – Which group of MIPS instructions takes advantage of this property and how? (5 points) • I-format jump instructions multiply their offset by 4 because all instructions are 4 -byte wide

Ninth Question • Explain why the very large instruction set of the x 86 family of processors is at the same time a nightmare and a blessing for both Intel and AMD. (2× 5 points)

Answer • A nightmare to implement • A blessing because it discourages the competition