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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. -
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 AccessRules 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:
[email protected]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
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Assignment No. 02
Semester: Fall 2019
Advance Computer Architecture – CS501
Total Marks: 20Due 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 EAGLERULES 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:
[email protected]Best of Luck
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
Question # 01 10 Marks
Consider the below given table consisting of assembly instructions belonging to different processors.Write machine code (in the hexadecimal representation) and behavioral RTL description for each instruction given in table.
Register Contents Memory Label Address R1 0015h A AB10h R2 25CBh B 3320h C AB0Eh
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.Table 1: Contents in Registers and Memory labels
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 represents byte- aligned memory map and shows the value stored at each memory address.Table 2: Contents at different memory addresses
Instruction Contents stored in Destination Operand Data Bus Address Bus Calculation Steps LDACC B SUB R1 LDACC C ADD R2 STACC A
• 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.Table 3: Data Bus and Address Bus Contents for Modified Eagle
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Advance Computer Architecture (CS501)
Assignment # 01
Fall 2019
Total marks = 20Deadline Date
12th Nov 2019Please carefully read the following instructions before attempting assignment.
RULES FOR MARKING
It should be clear that your assignment would not get any credit if:
The assignment is submitted after the due date.
The submitted assignment does not open or file is corrupt.
Strict action will be taken if submitted solution is copied from any other student or from the internet.You should consult the recommended books to clarify your concepts as handouts are not enough.
You are supposed to submit your assignment in .doc or docx format.
Any other formats like scan images, PDF, zip, rar, ppt and bmp etc will not be accepted.Objective:
Objective of this assignment is to increase the learning capabilities of the students about
• Performance Measurement of a processor
• Performance Comparison of processors
• Classification of Instruction Set Architecture for different machinesNOTE
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.
For any query, feel free to email at:
[email protected]Questions No 01 10 marks
Calculate the total execution time required by the processor to execute the program. If CPI for ALU is decreased by 20% and CPI for Load/Store is increased by 10%, then calculate the execution time.
Suppose we have a program which contains 200 instructions of different types. We want to execute this program on a 500 MHz processor. The ratio of each type of instruction in the program as well as clocks per instruction for each type of instruction is given below:
Questions No 02 10 marks
Write assembly language program for 0-address and 1-address machines to evaluate the following expression.
D = A(B+C) – 2AC/B + C2
Note: A, B, C and D are memory labels.Good Luck!
SOLVED CS501 Assignment 1 Solution and Discussion
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Advance Computer Architecture (CS501)
Assignment # 01
Fall 2019
Total marks = 20Deadline Date
12th Nov 2019Please carefully read the following instructions before attempting assignment.
RULES FOR MARKING
It should be clear that your assignment would not get any credit if:
The assignment is submitted after the due date.
The submitted assignment does not open or file is corrupt.
Strict action will be taken if submitted solution is copied from any other student or from the internet.You should consult the recommended books to clarify your concepts as handouts are not enough.
You are supposed to submit your assignment in .doc or docx format.
Any other formats like scan images, PDF, zip, rar, ppt and bmp etc will not be accepted.Objective:
Objective of this assignment is to increase the learning capabilities of the students about
• Performance Measurement of a processor
• Performance Comparison of processors
• Classification of Instruction Set Architecture for different machinesNOTE
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.
For any query, feel free to email at:
[email protected]Questions No 01 10 marks
Suppose we have a program which contains 200 instructions of different types. We want to execute this program on a 500 MHz processor. The ratio of each type of instruction in the program as well as clocks per instruction for each type of instruction is given below:
- Calculate the total execution time required by the processor to execute the program.
- If CPI for ALU is decreased by 20% and CPI for Load/Store is increased by 10%, then calculate the execution time.
Questions No 02 10 marks
Write assembly language program for 0-address and 1-address machines to evaluate the following expression.
D = A(B+C) – 2AC/B + C2
Note: A, B, C and D are memory labels.
Good Luck!
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Solution (A)
The formula to calculate the execution time : Execution Time = IC CPI T
IC for Load / Store Instructions
IC for ALU instructions
IC for Control instructions
= Total Instructions Ratio of Load / Store Instructions = 200 0.35
= 70 instructions
= Total Instructions Ratio of ALU Instructions = 200 0.55
= 110 instructions
= Total Instructions Ratio of Control Instructions = 200 0.10
= 20 instructions
Now, we will calculate the total clock cycles required to execute each type of instructions
Total Clock Cycles for Load / Store
Total Clock Cycles for ALU
Total Clock Cycles for Control
= IC for Load / Store CPI for Load / Store = 70 2.5
= 175 clock cycles
= IC for ALU CPI for ALU = 110 1.25
= 137.5 clock cycles
= IC for control CPI for control = 20 3
= 60 clock cycles
Time required (in seconds) for each clock cycle (T)=1/CPU frequency 1 / 500 106 = 0.002 10−6 seconds
= 210−9 seconds
Now finally, we will calculate the execution time
Execution Time (ET ) = Total Clock Cycles 1/ CPU Frequency
= (175 + 137.5 + 60) (1/ 500 106 ) seconds
= 372.5210−9seconds 1/500106 =210−9seconds = 745 10−9 seconds
= 745 nanosecondsSolution (B)
If decrease the average CPI for ALU by 20%, the new average CPI
New CPI for ALU = 1.25 (100−20)/100 = 1.25 0.8
= 1 CPI
If average CPI for Load / Store instruction is increased by 10%, new average CPI New CPI for Load / Store = 2.5 (100 +10)/100
= 2.5 1.1 = 2.75 CPI
Hence, new execution time will be
ExecutionTime(E.T) = (702.75+1101+203)x(1/500106)seconds
= (192.5+110+60)/(5108)seconds
= 362.5 2 10−9 seconds = 725 10−9 seconds
= 725 nanosecondsQ. 2 Solution:
Solution A (0-Address Code)
PUSH B
PUSH C
ADD ; gives B+C PUSH A
MUL ; gives A(B+C) PUSH 2
PUSH A
MUL ; gives 2A PUSH C
MUL ; gives 2AC
PUSH B
DIV ; gives 2AC/B
SUB ; gives A(B+C) - 2AC/B
PUSH C
PUSH C
MUL ; gives C2
ADD ; gives POP D
A(B+C) - 2AC/B + C2Solution A (1-Address Code)
LDA C MULA C STA X
LDA A MULA C MULA 2
; loads the value stored at memory location C in Accumulator ; gives C2
; stores C2 at memory location X
; loads the value stored at memory location A in Accumulator ; gives AC
; gives 2AC
DIVA B ADDA X STA Y
LDA B ADDA C MULA A SUB Y STA D
; gives 2AC/B
; adding 2AC/B with C2 stored in X gives 2AC/B + C2 ; stores 2AC/B + C2 at memory location Y
; loads the value stored at memory location B in Accumulator ; gives (B+C)
; gives A(B+C)
; subtracts 2AC/B + C2 from A(B+C)
; stores the result at memory location D -
Solution (A)
The formula to calculate the execution time : Execution Time = IC CPI T
IC for Load / Store Instructions
IC for ALU instructions
IC for Control instructions
= Total Instructions Ratio of Load / Store Instructions = 200 0.35
= 70 instructions
= Total Instructions Ratio of ALU Instructions = 200 0.55
= 110 instructions
= Total Instructions Ratio of Control Instructions = 200 0.10
= 20 instructions
Now, we will calculate the total clock cycles required to execute each type of instructions
Total Clock Cycles for Load / Store
Total Clock Cycles for ALU
Total Clock Cycles for Control
= IC for Load / Store CPI for Load / Store = 70 2.5
= 175 clock cycles
= IC for ALU CPI for ALU = 110 1.25
= 137.5 clock cycles
= IC for control CPI for control = 20 3
= 60 clock cycles
Time required (in seconds) for each clock cycle (T)=1/CPU frequency 1 / 500 106 = 0.002 10−6 seconds
= 210−9 seconds
Now finally, we will calculate the execution time
Execution Time (ET ) = Total Clock Cycles 1/ CPU Frequency
= (175 + 137.5 + 60) (1/ 500 106 ) seconds
= 372.5210−9seconds 1/500106 =210−9seconds = 745 10−9 seconds
= 745 nanosecondsSolution (B)
If decrease the average CPI for ALU by 20%, the new average CPI
New CPI for ALU = 1.25 (100−20)/100 = 1.25 0.8
= 1 CPI
If average CPI for Load / Store instruction is increased by 10%, new average CPI New CPI for Load / Store = 2.5 (100 +10)/100
= 2.5 1.1 = 2.75 CPI
Hence, new execution time will be
ExecutionTime(E.T) = (702.75+1101+203)x(1/500106)seconds
= (192.5+110+60)/(5108)seconds
= 362.5 2 10−9 seconds = 725 10−9 seconds
= 725 nanosecondsQ. 2 Solution:
Solution A (0-Address Code)
PUSH B
PUSH C
ADD ; gives B+C PUSH A
MUL ; gives A(B+C) PUSH 2
PUSH A
MUL ; gives 2A PUSH C
MUL ; gives 2AC
PUSH B
DIV ; gives 2AC/B
SUB ; gives A(B+C) - 2AC/B
PUSH C
PUSH C
MUL ; gives C2
ADD ; gives POP D
A(B+C) - 2AC/B + C2Solution A (1-Address Code)
LDA C MULA C STA X
LDA A MULA C MULA 2
; loads the value stored at memory location C in Accumulator ; gives C2
; stores C2 at memory location X
; loads the value stored at memory location A in Accumulator ; gives AC
; gives 2AC
DIVA B ADDA X STA Y
LDA B ADDA C MULA A SUB Y STA D
; gives 2AC/B
; adding 2AC/B with C2 stored in X gives 2AC/B + C2 ; stores 2AC/B + C2 at memory location Y
; loads the value stored at memory location B in Accumulator ; gives (B+C)
; gives A(B+C)
; subtracts 2AC/B + C2 from A(B+C)
; stores the result at memory location D -