PLEASE HELP, and clearly identify the binary instruction in the response please. Data Path 20A Bookmark this pageWhich Insruction?Here is the data flow diagram for a particular instruction. The various values shown indicate the state of the machine after the action of this instruction.DRIGatePCGateMARMUXROXFBFCPC x:R1X00FEMARMUXTRAPVECTORR2XFFFCX:0:0PCMUXR3XFF01ZEXTR4X0FFFOFFSET6 ORR5X0F05+PC OR BASERR6хоBFCPCOFFSET[7:0]R7x0403ADDR2MUXADDRIMUXSR2SRI001[10:0]SEXTIMM5[8:0]SEXT1SEXTFINITESR2MUX[5:0]SEXTSTATE[4:0]МАCHINEAL selectArith/Logic OperationNP15IRA/L RESULTLOGICGateALUGateMDR1616MDRMARX16INPUTOUTPUTМЕМORYXFE02XFE06What is that instruction in binary? (You may use any number of spaces to organize the bits.)address instructionx30020000 100 0 0010 1101

Answered: Image /qna-images/answer/754959b2-9a07-46cf-bb44-481d819be2b3.jpg

Which Insruction? Here is the data flow diagram for a particular instruction. The various values shown indicate the state of the machine after GatePC DRI GateMARMUX RO XFBFC PC x R1 X00FE MARMUX TRAPVECTOR +1 R2 XFFFC X:0:0 PCMUX R3 XFF01 ZEXT R4 X0FFF OFFSET6 OR R5 X0F05

Which Insruction? Here is the data flow diagram for a particular instruction. The various values shown indicate the state of the machine after GatePC DRI GateMARMUX RO XFBFC PC x R1 X00FE MARMUX TRAPVECTOR +1 R2 XFFFC X:0:0 PCMUX R3 XFF01 ZEXT R4 X0FFF OFFSET6 OR R5 X0F05

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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The following byte sequence is the machine code of a program function compiled with the Y86- 64 instruction set. The memory address of the first byte is 0x300. Note that the byte sequence is written in hex-decimal form, i.e., each number/letter is one hex-decimal number representing 4 binary bits, and two numbers/letters represent one byte. 630030F3020000000000000030F11E000000000000007023030000000 00000601061316211761F0300000000000090- Please write out the assembly instructions (in Y86-64 instruction set) corresponding to the machine codes given by the above bytes sequence, and explain what this program function is computing. The machine has a little-endian byte ordering.
MUL & DIV instruction in Assembly Language Table 1 EAX EBX ECX EDX 10H 20H 2H 0H Refer to Table 1 above for the following tasks.Write your own code in Visual Studio 2019 to execute the instruction, then fill in the blanks with the correct value of the registers. Also, note the changes. **Note: Always start with the initial value of the registers given in Table 1 above for each question (a-f) This means you have to execute only one instruction per debugging session e.g. execute MUL BX and then debug and get the values of the registers and then stop. Then start again with MUL CX. Answer all questions in 32bit format XXXX XXXX e.g. 0000 1000 with only one white space in between a. MUL BX EAX EBX ECX EDX Answer Answer Answer Answer b. MUL CX EAX EBX ECX EDX c. MUL AX EAX EBX ECX EDX d. DIV BX EAX EBX ECX EDX e. DIV CX EAX EBX ECX EDX f. DIV AX…
E In the following code block(Reference:Q11), you will a set of assembly instructions with corresponding line numbers (line numbers are for informational purpose only and they are not part of the source code). 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 mov edx, 5 dec ecx jmp LABEL1 mov eax, 1 LABEL1: mul edx jmp ecx mov edx, 0678h sub edx, eax jmp DWORD PTR [edx] neg ebx add ecx, ebx mov eax, 0 For each of the conditions/scenario listed below, indicate the corresponding line number (that cause or is associated with the condition/scenario). Enter 0 (Zero) if the condition is not caused by the block of code. 1) Memory indirect jump: type your answer... type your answer... type your answer... type your answer... 2) Register indirect jump: 3) Relative short jump: type your answer... 5) Two's complement type your answer... 4) Relative near jump: 6) Unreachable code
int i = 5; is a statement in a C program. A. during execution, value of i may change but not its address B. during execution both the address and value may change C. repeated execution may result in different addresses for i D. i may not have an associated address
MIPS Simulator QtSpim: You are to have a complete program in MIPS assembly language that behaves exactly as the included C program. This program contains four functions in addition to the main() one. Your solution must contain all five C routines as they have been coded in the example. Make sure to run the program in MIPS and show the same output on MIPS as well to make sure there are no errors. Below is the five C routines and attached is the image of what the output must print out on QtSpim. #include <stdio.h> int getMax(int arr[], int n){int mx = arr[0];for (int i = 1; i < n; i++)if (arr[i] > mx)mx = arr[i];return mx;}void countSort(int arr[], int n, int exp){int output[n];int i, count[10] = { 0 };for (i = 0; i < n; i++)count[(arr[i] / exp) % 10]++;for (i = 1; i < 10; i++)count[i] += count[i - 1];for (i = n - 1; i >= 0; i--) {output[count[(arr[i] / exp) % 10] - 1] = arr[i];count[(arr[i] / exp) % 10]--;}for (i = 0; i < n; i++)arr[i] = output[i];}void…
Upload answer sheets COVISHIELD dose carton holds 50 vaccine vials (100 doses). The dimension of the carton is 18.5 x 9.5 x 4.0 cm. If 25 such cartons are transported, compute the size of the container and the number of doses that would be transported. Write a program to carry out the computations using 8087 instruction set.
Part A For each byte sequence listed, determine the Y86 instruction sequence it encodes. If there is some invalid byte in the sequence, show the instruction sequence up to that point and indicate where the invalid value occurs. For each sequence, the starting address, then a colon, and then the byte sequence are shown. 0x100: 30f3fcfffff40630008000000000000 0x100: 30f3fcfffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rrmovq %rsi,0×80A(%rcx) O0x115: 00 halt Ox100: 30f3fcffffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox114: 00 halt 0x100: 30f3fcfffffffff rrmovq $-8,%rbx Ox109: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) 0x200: a06f800c020000000000000030f30a00000000000000 0x113: 00 halt 0x100: 30f3fcffffffffff irmovq $-4,%rbx 0x200: a06f pushq %rsi 0x10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox202: 800c02000000000000 call proc Ox116: 00 halt 0x20b: 00 halt 0x20c: proc: Submit Request Answer 0x20c: 30f30a00000000000000 | irmovq $10,%rbx…
* The LOOP # will affect only CX O All CS, CX and IP Both CX and IP non of those only IPO if BX=1000, DS=0400, and AL=EDH, for the following instruction: MOV [BX] + 1234H, AL. the physical address is 6243H O 6324H 4244H O 6234H 4234H
Assignment for Computer Architecture Instructions: The assignment is to create a program that adds the number 1/2 to itself a large number of times and adds the number 1/3 to itself a large number of times separately first using type float and then type double. It is to then compare the values of adding the numbers to multiplying 1/2 time the number of times added to compute the “same sum” in a different way. The program will also multiply 1/3 times the number of times 1/3 was added to itself to compute the “same sum” in a different way. The program will then compare these two methods at arrive for the same value and output the difference. Hint, the value for the ½’s will be the same for the smaller numbers of times, the 1/3’s will never be the same. The output from your program is to be to a *.txt file which you are to turn in along with your code. The program must first add the ½’s and 1/3’s using type float and compare to the value obtain using multiplication instead of addition.…
A computer with a 32-bit word uses an instruction format that includes direct and indirect addressing of 8 megabytes and one of 16 registers. The highest order bits are used for the opcode, followed by the bits indicating the register, followed by the indirect/direct bit, followed by the bits indicating the memory address. Draw the instruction word's format, showing how many bits are used for all four fields (Make sure to show here how many bits for each and the order in which they appear. You can submit a drawing of the word's format separately
4. The Single Instruction Computer (SIC) has only one instruction that can do all operations our RISC-V does (you did a homework problem). The instruction has the following format sbn a, b, c # Mem[a]=Mem[a]- Mem[b]; if (Mem[a]- Mem[b]<0) go to PC+c For example, here is the program to copy a number from location a to location b: sbn temp, temp, 1 sbn temp, a, 1 sbn b,b, 1 sbn b, temp 1 Start: Design a single cycle datapath and control for this instruction set architecture.
assembly language programs for the 8086 microprocessors to perform multiplication of two matrices Am*n and Bn*p. The value of m = 4, n = 2, and p =3. Assembly lan+ Assume all the elements in matrices A, B, and answer of the multiplication are 8-bit numbers. Flow chart Pseudocode Assembly Code with brief describe comments (EMU8086 emulator).