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Syllabus

Instructor: Dr. Wesam Ashour

Office: B231

Email: Washour@iugaza.edu

Office Hours: Saturday 10:00 11:00, 13:0014:00 Sunday 11:00 13:00. Wednesday 10:00 11:00.

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Syllabus

Course: ECOM 3310 – Computer Architecture

Contents: 1- Ch.1: Introduction.

2- Ch.2: Instructions: Language of the Computer.

3- Ch.3: Arithmetic for Computers.

4- Ch.4: Assessing and Understanding Performance.

5- Ch.5: The Processor: Datapath and Control.

6- Ch.6: Enhancing Performance with Pipelining.

7- Ch.7: Large and Fast: Exploiting Memory Hierarchy

8- Ch.8: Input/Output Systems

We will cover as much as we can!

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Syllabus 

Grades: 1- 30% Mid Term Exam.

2- 50% Final Exam.

3- Teaching Assistants control %20.

Midterm Exam: Tuesday, 16th November 2010, 11:00am-12:30pmFinal Exam: Saturday, 15th January 2011, 11:00-14:00

Important Notice:

Sometimes (& Suddenly) I will test you by giving Quizzes at the beginning of the lecture! - To solve it easily and to get the highest mark, you should always read the last two lectures before attending the class – Good Luck.

Important Advice:

From first day in the semester you should work very hard and you must activate your mind to think!

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Chapter 2

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lw $t0, 1200($t1)

add $t0, $s2, $t0

sw $t0, 1200($t1)

What is the MIPS machine language code for these three instructions?

1200 (in base 10) = 0000 0100 1011 0000 (binary)

$t0 register is number 8

$t1 register is number 9

$s2 register is number 18

Opcode -> lw is 35 (100011), sw is 43 (101011), R-type (000000) add func is 32 (100000)

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lw $t0, 1200($t1)

add $t0, $s2, $t0

sw $t0, 1200($t1)

What is the MIPS machine language code for these three instructions?

1200 (in base 10) = 0000 0100 1011 0000 (binary)

$t0 register is number 8

$t1 register is number 9

$s2 register is number 18

Opcode -> lw is 35 (100011), sw is 43 (101011), R-type (000000) add func is 32 (100000)

100011 01001 01000 0000010010110000

000000 10010 01000 01000 00000 100000

101011 01001 01000 0000010010110000

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or $t0, $t1, $t2

ori $t0, $t1, 255

sll $t0, $t1, 4

and $t0, $t1, $t2

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if (save[i] >= k)

Code;

Assume save[i] is stored in $t1 and k is stored in $t2

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slt $t0, $t1, $t2

bne $t0, $zero, Exit

Code

Exit:

if (save[i] >= k)

Code;

Assume save[i] is stored in $t1 and k is stored in $t2

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if (save[i] < k)

Code;

Assume save[i] is stored in $t1 and k is stored in $t2

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slt $t0, $t1, $t2

beq $t0, $zero, Exit

Code

Exit:

if (save[i] < k)

Code;

Assume save[i] is stored in $t1 and k is stored in $t2

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if (save[i] > k)

Code;

Assume save[i] is stored in $t1 and k is stored in $t2

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slt $t0, $t1, $t2

bne $t0, $zero, Exit

beq $t1, $t2, Exit

Code

Exit:

if (save[i] > k)

Code;

Assume save[i] is stored in $t1 and k is stored in $t2

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if (save[i] <= k)

Code;

Assume save[i] is stored in $t1 and k is stored in $t2

Try it by yourself

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