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  5. CS501 Assignment 2 Solution and Discussion
CS501 GDB 1 Solution and Discussion
zareenZ
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.
CS501 - Advance Computer Architecture
CS501 Assignment 3 Solution and Discussion
zareenZ
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
CS501 - Advance Computer Architecture
CS501 Assignment 2 Solution and Discussion
zareenZ
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
CS501 - Advance Computer Architecture
CS501 Assignment 1 Solution and Discussion
zareenZ
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CS501 - Advance Computer Architecture

CS501 Assignment 2 Solution and Discussion

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cs501assignment 2solutiondiscussionfall 2019
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  • zareenZ Offline
    zareenZ Offline
    zareen
    Cyberian's Gold
    wrote on last edited by
    #7

    Solution
    Addressing modes relate to the way architectures specify the address of the objects they access. These objects may be constants and registers, in addition to memory locations.
    bfd20701-e200-4cc6-b14d-577fcb582667-image.png

    Discussion is right way to get Solution of the every assignment, Quiz and GDB.
    We are always here to discuss and Guideline, Please Don't visit Cyberian only for Solution.
    Cyberian Team always happy to facilitate to provide the idea solution. Please don't hesitate to contact us!
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    • zareenZ Offline
      zareenZ Offline
      zareen
      Cyberian's Gold
      wrote on last edited by
      #8

      We have looked at the various jump instructions in our study of the FALCON-A. Using that knowledge, this problem can be solved easily.
      0ef8e85e-982c-4123-a722-38ba048256a6-image.png

      Discussion is right way to get Solution of the every assignment, Quiz and GDB.
      We are always here to discuss and Guideline, Please Don't visit Cyberian only for Solution.
      Cyberian Team always happy to facilitate to provide the idea solution. Please don't hesitate to contact us!
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      • zareenZ Offline
        zareenZ Offline
        zareen
        Cyberian's Gold
        wrote on last edited by
        #9

        Control Instruction
        5c133d07-ce7b-4c97-a9d4-70dc4adfbba5-image.png

        Discussion is right way to get Solution of the every assignment, Quiz and GDB.
        We are always here to discuss and Guideline, Please Don't visit Cyberian only for Solution.
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        • zareenZ Offline
          zareenZ Offline
          zareen
          Cyberian's Gold
          wrote on last edited by
          #10

          jump instruction
          b7008c25-f048-46f3-9b8c-c0bca40c0ba8-image.png

          Discussion is right way to get Solution of the every assignment, Quiz and GDB.
          We are always here to discuss and Guideline, Please Don't visit Cyberian only for Solution.
          Cyberian Team always happy to facilitate to provide the idea solution. Please don't hesitate to contact us!
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          • zareenZ zareen

            We can also find the machine code for these instructions. The machine code (in the hexadecimal representation) is given for these instructions in the given table.
            541367a2-d798-4e68-b10c-646873903d63-image.png

            Ahsan WaheedA Offline
            Ahsan WaheedA Offline
            Ahsan Waheed
            wrote on last edited by
            #11

            @zareen This is not a right assignment solution

            zareenZ 1 Reply Last reply
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            • Ahsan WaheedA Ahsan Waheed

              @zareen This is not a right assignment solution

              zareenZ Offline
              zareenZ Offline
              zareen
              Cyberian's Gold
              wrote on last edited by
              #12

              @Ahsan-Waheed said in CS501 Assignment 2 Solution and Discussion:

              @zareen This is not a right assignment solution

              this is not a solution just idea how to solve the question.

              Discussion is right way to get Solution of the every assignment, Quiz and GDB.
              We are always here to discuss and Guideline, Please Don't visit Cyberian only for Solution.
              Cyberian Team always happy to facilitate to provide the idea solution. Please don't hesitate to contact us!
              %(red)[NOTE: Don't copy or replicating idea solutions.]
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              • zareenZ Offline
                zareenZ Offline
                zareen
                Cyberian's Gold
                wrote on last edited by zareen
                #13

                @Ahsan-Waheed said in CS501 Assignment 2 Solution and Discussion:

                @zareen This is not a right assignment solution

                this is not a solution just idea how to solve the question.

                f85aff3a-865b-4bc6-b7a7-6fe281c2e3ac-image.png
                ceb1695a-5822-4980-8e8c-a3b4e94fa4cd-image.png

                cs501 assignment 2 idea solution…docx

                Discussion is right way to get Solution of the every assignment, Quiz and GDB.
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                • zareenZ Offline
                  zareenZ Offline
                  zareen
                  Cyberian's Gold
                  wrote on last edited by zareen
                  #14

                  Solution # 1

                  Instruction Processor Hexadecimal Code Behavioral RTL
                  JPL R5, [26] Falcon - A 10000 101 00011010, 1000 0101 0001 1010, 851A (R[5] ≥ 0): PC ← PC+ (26-PC);
                  STS R7, R2 (100) Falcon - E 00101 111 010, 000000000000001100100, 0010 1111 0100 0000, 0000 0000 0110 0100, 2F40 0064 M[R[2]+100] ← R [7]
                  STACC R4, 36 Modified EAGLE 10111 100 00100100, 1011 1100 0010 0100, BC24 M[R[4] + (8@C<7>)©C] ← ACC C represents the constant 36
                  DIV R2 EAGLE 10000 010, 1000 0010, 82 R[0] ← R[0]/R[2] R[2] ← R[0]%R[2]
                  SHIFTL R5, R2, 7 FALCON - A 01100 101 010 00111, 0110 0101 0100 0111, 6547 R[5]<15…0> ← R [2]<(15- N)…0>©(N@0) N represents constant 7

                  Solution # 2

                  Instruction Value of Destination Operand Data Bus <15…0> Address Bus (15…0>
                  LDACC B 55CEh 55CEh 3320h
                  SUB R1 55B9h N/A N/A
                  LDACC C 2015h 2015h AB0Eh
                  ADD R2 45E0h N/A N/A
                  STACC A 45E0h 45E0h AB10h

                  Table 3 Data Bus and Address Bus Contents for Modified Eagle

                  Calculation Steps (Instruction-By-Instruction)

                  • LDACC B
                  LDACC stands for load accumulator. In LDACC, the destination operand is accumulator and source operand is the memory location labelled as B. The memory label B points to the memory address 3320h. When this instruction is executed, the value stored at memory address 3320h will be read and loaded in Accumulator register. This address 3320h will be copied into Address Bus which will then read its contents from memory and load the contents at data bus. The operand size in Modified EAGLE is 2-byte. Therefore, the values stored at addresses 3320h and 3321h will be loaded at data bus. These values are CEh and 55h respectively. Since, Modified EAGLE employs Little endian notation hence, the 2-bye value will be read as 55CEh. The value of data bus will also be 55CEh and this will be loaded into destination operand Accumulator.
                  • SUB R1
                  SUB R1 means to subtract the value of source operand register R1 which is 0015h, from the destination operand Accumulator (ACC) register which contains 55CEh. The result of subtraction will be stored back into ACC. After subtraction, the value stored in ACC will be 55B9h. Because SUB is not a memory instruction, we are not concerned with the contents of Data Bus or Address Bus because the values we need to execute the instruction are already available in registers. Hence, Data Bus and Address Bus values will be labelled as N/A.
                  • LDACC C
                  When this instruction is executed, the value stored at memory address labelled with C is read and loaded in Accumulator register ACC. In this case, the address of the C is AB0Eh which is also the value of address bus. The operands in Modified EAGLE are 2-byte values. The contents at addresses AB0Eh and AB0Fh will be copied into data bus which will then be loaded into ACC register. These contents are 15h and 20h. Due to Little endian notation, the 2-bye value will be 2015h. The value of data bus will be 2015h and same will be loaded in destination register ACC.
                  • ADDR2
                  When ADD is executed, the value of register R2 is added to ACC register. Hence, after the execution, the ACC register will hold 45E0h. AS usual, ADD is not a memory instruction, so we are not concerned with the values of Data Bus and Address Bus and both are labelled as N/A.
                  • STACC A
                  STACC stands for Store ACC. There is one destination operand which is a memory label A. When the instruction is executed, the value of Accumulator register ACC is stored at the memory address labelled by A. The destination memory address will be AB10h. The value of ACC is 45E0h will be stored as address AB10h. However, due to Little-Endian notation, the address will be stored as E0h at memory location AB10h and then 45h at memory location AB11h.

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