Need help on computer organization and architecture course.

Need help on computer organization and architecture course.

You will perform calculations comparing the performance of two computer systems, Machine A and Machine B, against two separate benchmark programs, I and II.  Table

2

.

1

gives the clock rates for Machines A and B, along with the number of clock cycles required to execute each type of instruction.  Table 2.2 gives the number of instructions (in millions) required to run each benchmark program for each of the two machines.  The reference run times (Tref) for benchmarks I and II are

5

0 ms and

7

0 ms, respectively.

Your task is to evaluate and compare Machine A vs. Machine B for the two benchmarks by completing Table 2.

3

.  As you click on cells in Table 2.3, a large green box will appear on the right to help guide you through the problem.

Table 2.1: Clock rate and Cycles per Instruction

Machine

A

B

Clock Rate

1.

8

GHz

3.0 GHz

Arithmetic and Logic

1

1

Load/Store

3 5

Branch

4

3

Other

2

3

Machine

A

B

I

II

Arithmetic and Logic

5

Load/Store

5

7

5

Branch

3

5

4

4

Other

4

4

2

3

Table 2.2: Instruction Counts for benchmarks I and II (millions)

Benchmark

I

II

8

6

7

10

Top of Form

Machine

A

B

Benchmark

I

II

I

II

B

B

Table 2.3: Benchmark Comparison

Clock period (ps)

Instruction Count (millions)

Average Cycles per Instruction

Execution time (ms)

MIPS

Speed ratio

SPEC speed metric

Faster Clock Frequency

Faster Machine

Bottom of Form

Question #1 – CPU Components

Which of the following are components of the CPU?

a.

system bus

b.

I/O module

c.

main memory

d.

I/O buffers

e.

instruction register

f.

registers for accessing I/O devices

g.

registers for accessing memory

h.

program counter

i.

execution unit

Question #2 – IAS Instruction Format

[Modified from Stallings Problem 2.2]

For the Institute for Advanced Studies (IAS) computer from Princeton (see photo at right), use Fig. 2.2 and Table 2.1 from your textbook to determine the following machine code:

Left instruction – take the next instruction from the left half of M(224)
Right instruction – transfer the negative of the absolute value of M(282) to AC

M(X) refers to the data at memory location X, where X is expressed as a 3 digit hexadecimal value.  Express your answer in binary.

Op Code

Address

Table 2.1 – IAS Instruction

LEFT

RIGHT

Op Code

Address

Question #3 – IAS Program

[Modified from Stallings Problem 2.4] Consider the IAS code fragment shown below in Table 3.1.  Six instructions are located at 3 consecutive memory addresses (See Stallings Fig. 2.2b for the instruction format).  Using Table 2.1 from your textbook, translate the 40-bit hexadecimal values in Table 3.1 to their corresponding assembly language instructions.  Notice that the value ‘X’ needs to replaced with a 3-digit address for those instructions in which the ‘X’ appears.  A few examples of legal syntax assembly language instructions are shown in Table 3.2 below.

Notes:

1. All of the data and addresses are in hexadecimal.

2. The IAS computer stores negative numbers in signed-magnitude format (see Fig. 2.2a in your textbook).  The MSB (Most Significant Bit) is 0 for positive numbers and 1 for negative numbers.  The remaining 39 bits give the magnitude.  Notice how this format differs from 2’s complement format.

3. Unless a JUMP instruction is encounted, the IAS first executes the LEFT instruction, then the RIGHT instruction, then proceeds to the next address.

Address

LEFT

RIGHT

Table 3.1 – IAS Code Fragment

Contents (hex)

047

0113E2113F

048

090501004A

049

0213E2113F

LEFT

RIGHT

Table 3.2 – Sample IAS instruction Syntax

LOAD M(A39) 

JUMP+M(0F8,0:19)

ADD |M(2A3)| 

LSH

MUL M(8E0) 

STOR M(49C,28:39)

Now assume the program counter begins at 047:LEFT, and that the value stored at memory location 13E is 90C959CE09.  Complete the following statements about what happens when the code fragment executes:

a. The instruction at location 049:RIGHT be executed. Will or will not?

b. The value stored at memory location 13E after program completion will be

.

c. The value stored at memory location 13F after program completion will be

.

d. The value stored in the accumulator after program completion will be

.

e. After execution, the program counter is at

: and is it left or right?

Again assuming the program counter begins at 047:LEFT, but now that the value stored at memory location 13E is 10C959CE09.  Complete the following statements about what happens when the code fragment executes:

a. The instruction at location 049:LEFT be executed. Will or will not?

b. The value stored at memory location 13E after program completion will be

.
c. The value stored at memory location 13F after program completion will be

.
d. The value stored in the accumulator after program completion will be

.
e. After execution, the program counter is at

: and is it left or right?

Question #4 – Data transfer

A. Describe two causes for performance degradation of a single-bus design as the number of devices connected to the bus increases.

1.

2.

B. How does point-to-point interconnect differ from a shared bus?

Question #5 – Transfer Rate

[Modified from Stallings Problem 3.5] Consider a 32-bit microprocessor with a 64-bit external data bus that is driven by an input clock running at 132 MHz.  Assume this microprocessor has a bus cycle whose duration equals five input clock cycles.

A. What is the bus cycle rate?

bus cycles/sec

B. How many bytes are transferred each bus cycle?

bytes/bus cycle

C. What is the maximum data transfer rate?

bytes/sec

_1425895909.unknown

_1425895917.unknown

_1425895921.unknown

_1425895923.unknown

_1425895924.unknown

_1425895922.unknown

_1425895919.unknown

_1425895920.unknown

_1425895918.unknown

_1425895913.unknown

_1425895915.unknown

_1425895916.unknown

_1425895914.unknown

_1425895911.unknown

_1425895912.unknown

_1425895910.unknown

_1425894028.unknown

_1425895901.unknown

_1425895905.unknown

_1425895907.unknown

_1425895908.unknown

_1425895906.unknown

_1425895903.unknown

_1425895904.unknown

_1425895902.unknown

_1425895897.unknown

_1425895899.unknown

_1425895900.unknown

_1425895898.unknown

_1425895895.unknown

_1425895896.unknown

_1425894029.unknown

_1425894020.unknown

_1425894024.unknown

_1425894026.unknown

_1425894027.unknown

_1425894025.unknown

_1425894022.unknown

_1425894023.unknown

_1425894021.unknown

_1425894016.unknown

_1425894018.unknown

_1425894019.unknown

_1425894017.unknown

_1425894011.unknown

_1425894013.unknown

_1425894015.unknown

_1425894012.unknown

_1425894009.unknown

_1425894010.unknown

_1425894007.unknown

_1425894008.unknown

_1425894006.unknown

_1425894005.unknown

Still stressed with your coursework?
Get quality coursework help from an expert!