Fixed length instructions vs Variable length instructions.
Fixed length instructions:-
All the records in the file are of same size.
Leads to memory wastage.
Access of the records is easier and faster.
Exact location of the records can be determined: location of ith record would be.n*(i-1), where n is the size of every record.
Variable length instructios:-
Different records in the file have different sizes.
Memory efficient.
Access of the records is slower.
Large cache blocks vs small cache blocks.
The “largest” block size case is best for repeated, in-order memory accesses (best for spacial locality) and worst for repeated, random memory accesses (worst for temporal locality).
The “smallest” block size case is worst for repeated, in-order memory accesses (worst for spacial locality) and best for repeated, random memory accesses (best for temporal locality). It is good for out-of-order memory accesses, depending on the size of the working set.
Simple instruction set vs complex instruction set.
Characteristic of Simple instruction set:
The simple instruction set provides commands to the processor, to tell it what it needs to do. The instruction set consists of addressing modes, instructions, native data types, registers, memory architecture, interrupt, and exception handling, and external I/O.
Characteristic of Complex instruction set:
Complex instruction, hence complex instruction decoding.
Instruction are larger than one word size.
Instruction may take more than single clock cycle to get executed.
Less number of general purpose register as operation get performed in memory itself.
Complex Addressing Modes.
More Data types.
Assignment No. 03
Semester: Fall 2019
Advance Computer Architecture – CS501
Total Marks: 20
Due Date: 20-01-2020
Please carefully read the following instructions before attempting assignment
Objective of Assignment
Objective of this assignment is to increase the learning capabilities of the students about:
• Interrupt Driven I/O
• CPU Polling for I/O
• Direct Memory Access
Rules for Assignment
It should be clear that your assignment will get credit only if:
• The assignment is submitted before or on the due date.
• The submitted assignment file is not corrupted or damaged.
• The assignment is not copied (from another student or internet).
Uploading Instructions
Read the following instructions carefully before uploading assignment:
• Upload/Submit assignment in your VULMS assignment interface.
• No assignment will be accepted through email.
• Your assignment must be with .doc extension. (Any other format will not be accepted)
NOTE
No assignment will be accepted after the due date via email in any case (whether it is the case of load shedding or internet malfunctioning etc.). Hence, refrain from uploading assignment in the last hour of deadline. It is recommended to upload solution file at least two days before its closing date.
If you find any mistake or confusion in assignment (Question statement), please consult with your instructor before the deadline. After the deadline, no queries will be entertained in this regard. Moreover, keep checking announcements section.
For any query, feel free to email at:
cs501@vu.edu.pk
Thank You
Question # 01 08 Marks
Suppose a 160GB hard drive is connected to computer system having a 32-bit Pentium-3 processor which is operating at a frequency of 500 MHz. The system employs DMA as I/O interface for data transfer. Before the data can be transferred using DMA, the processor needs the DMA controller to be set-up which requires 3000 clock cycles. Then, for each data transfer, additional 1000 clock cycles are required to handle interrupt. The hard drive transfers data using data blocks of 64 KB each at data transfer rate of 40 MB/s. Considering the given scenario, answer the following questions. What will be the total CPU time required to transfer the file using DMA?
Question # 02 12 Marks
Assume a 64-bit processor operating at 600 MIPS having following devices attached to it:
• A hard drive with a transfer rate of 8 MB/s.
• A mouse which is required to be polled 30 times/seconds.
• An optical drive connected through a 32-bit data bus with a transfer rate of 2MB/sec.
The processor employs polling technique to select the device to perform I/O operations where each polling operation requires 400 instructions. Determine the percentage of CPU time required to poll each of three devices.
Best of Luck
Assignment No. 02
Semester: Fall 2019
Advance Computer Architecture – CS501
Total Marks: 20
Due Date: 28-11-2019
Please carefully read the following instructions before attempting assignment
OBJECTIVE OF ASSIGNMENT
Objective of this assignment is to increase the learning capabilities of the students about:
• Encoding of Assembly Instructions
• Behavioral RTL description of instruction
• ISA of Modified EAGLE
• Address Bus and Data Bus in Modified EAGLE
RULES FOR ASSIGNMENT
It should be clear that your assignment will get credit only if:
• The assignment is submitted before or on the due date.
• The submitted assignment file is not corrupted or damaged.
• The assignment is not copied (from another student or internet).
UPLOADING INSTRUCTIONS
Read the following instructions carefully before uploading assignment:
• Upload/Submit assignment in your VULMS assignment interface.
• No assignment will be accepted through email.
• Your assignment must be with .doc extension. (Any other format will not be accepted)
NOTE
No assignment will be accepted after the due date via email in any case (whether it is the case of load shedding or internet malfunctioning etc.). Hence, refrain from uploading assignment in the last hour of deadline. It is recommended to upload solution file at least two days before its closing date.
If you find any mistake or confusion in assignment (Question statement), please consult with your instructor before the deadline. After the deadline, no queries will be entertained in this regard. Moreover, keep checking announcements section.
For any query, feel free to email at:
cs501@vu.edu.pk
Best of Luck
Question # 01 10 Marks
Consider the below given table consisting of assembly instructions belonging to different processors.
Instruction
Processor
Hexadecimal
Code
Behavioral RTL
JPL R5, [26]
Falcon - A
STS R7, R2 (100)
Falcon - E
STACC R4, 36
Modified EAGLE
DIV R2
EAGLE
SHIFTL R5, R2, 7
FALCON - A
Write machine code (in the hexadecimal representation) and behavioral RTL description for each instruction given in table.
Question # 02 10 Marks
Consider the below given tables belonging to Modified EAGLE architecture.
• Table 1 shows the contents of registers R1, R2 and memory labels A, B, C. The contents of memory labels are memory addresses.
Register
Contents
Memory Label
Address
R1
0015h
A
AB10h
R2
25CBh
B
3320h
C
AB0Eh
Table 1: Contents in Registers and Memory labels
• Table 2 represents byte- aligned memory map and shows the value stored at each memory address.
Memory Address
Memory Contents
Memory Address
Memory Contents
AB0Eh
15h
3320h
CEh
AB0Fh
20h
3321h
55h
AB10h
56h
3322h
39h
AB11h
EFh
3323h
20h
Table 2: Contents at different memory addresses
• Table 3 contains the instructions of an assembly program for Modified EAGLE. You are required to complete Table 3 by writing the values of destination operand, 16-bit address bus and 16-bit data bus after each instruction is executed.
Write the complete steps for calculating the values of Destination Operand, Data Bus & Address Bus after the execution of each instruction.
Instruction
Contents stored in Destination Operand
Data Bus
Address Bus
Calculation Steps
LDACC B
SUB R1
LDACC C
ADD R2
STACC A
Table 3: Data Bus and Address Bus Contents for Modified Eagle
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Fixed length instructions vs Variable length instructions.
Fixed length instructions:-
All the records in the file are of same size.
Leads to memory wastage.
Access of the records is easier and faster.
Exact location of the records can be determined: location of ith record would be.n*(i-1), where n is the size of every record.Variable length instructios:-
Different records in the file have different sizes.
Memory efficient.
Access of the records is slower.Large cache blocks vs small cache blocks.
The “largest” block size case is best for repeated, in-order memory accesses (best for spacial locality) and worst for repeated, random memory accesses (worst for temporal locality).
The “smallest” block size case is worst for repeated, in-order memory accesses (worst for spacial locality) and best for repeated, random memory accesses (best for temporal locality). It is good for out-of-order memory accesses, depending on the size of the working set.
Simple instruction set vs complex instruction set.
Characteristic of Simple instruction set:
The simple instruction set provides commands to the processor, to tell it what it needs to do. The instruction set consists of addressing modes, instructions, native data types, registers, memory architecture, interrupt, and exception handling, and external I/O.
Characteristic of Complex instruction set:
Complex instruction, hence complex instruction decoding.
Instruction are larger than one word size.
Instruction may take more than single clock cycle to get executed.
Less number of general purpose register as operation get performed in memory itself.
Complex Addressing Modes.
More Data types.
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