YesIs nextNoinstructionMAR E PCNo memoryIBR2accessrequiredMBR + MOMARYes BR E MERN9)IR E MBR(0:7)MAR E MBR(8:19)|NoLeftIR E IBR (0:7)MAR E IBR (8:19)IR E MBR(20.27)MAR + MBR(28:39)instructioneguiregPC + PC+1MemorLeft instructionRight instructiony locationLOAD M(300)SUB M(301)LOAD MQ , M(302)STOR M(303)STOR M(304)100101STOR M(305)MULT M(305)LOAD MQ102103104HALT300301302303304305Suppose you start your program from Memory location 100, and we execute theinstruction with program counter is 103. What is the content of IBR?FETCH CYCLE

Name: YesIs nextNoinstructionMAR E PCNo memoryIBR2accessrequiredMBR + MOMAR
Uploaded: 2024-03-20T06:51:23.000Z
Channel: Bartleby
Description: YesIs nextNoinstructionMAR E PCNo memoryIBR2accessrequiredMBR + MOMARYes BR E MERN9)IR E MBR(0:7)MAR E MBR(8:19)|NoLeftIR E IBR (0:7)MAR E IBR (8:19)IR E MBR(20.27)MAR + MBR(28:39)instructioneguiregPC + PC+1MemorLeft instructionRight instructiony lo

Burden M(300) loads esteem 7 SUB M(301) : 7 - 2 = 5 STORE M(305) : Value 5 is put away at memory…

Engineering

Computer Science

snest No struction Yeu MAR E PC Nommory IBR access required BR E MOMAR Yes ER E MERN) IR E MBR0.7) MAR + MBR8 19) |No Left IRE MBR(2027) MAR E MBRCE39) IRE IBR (0:7) instruction MAR E IBR (819) quireg PC+ PC+1 Memor Left instruction Right instruction y location LOAD M(300) STOR M(305) MULT M(305) LOAD MQ SUB M(301) LOAD MQ , M(302) STOR M(303) STOR M(304) 100 101 102 103 104 HALT 300 301 302 2 303 304 305 Suppose you start your program from Memory location 100, and we execute the instruction with program counter is 103. What is the content of IBR? FETCH CYCLE

snest No struction Yeu MAR E PC Nommory IBR access required BR E MOMAR Yes ER E MERN) IR E MBR0.7) MAR + MBR8 19) |No Left IRE MBR(2027) MAR E MBRCE39) IRE IBR (0:7) instruction MAR E IBR (819) quireg PC+ PC+1 Memor Left instruction Right instruction y location LOAD M(300) STOR M(305) MULT M(305) LOAD MQ SUB M(301) LOAD MQ , M(302) STOR M(303) STOR M(304) 100 101 102 103 104 HALT 300 301 302 2 303 304 305 Suppose you start your program from Memory location 100, and we execute the instruction with program counter is 103. What is the content of IBR? FETCH CYCLE

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

See similar textbooks

Related questions

Q: 3) – Find the contents of R20 after each instruction of the following codes is executed: (a) LDI…

Q: CPU Datapath The following figure shows the overall datapath of the simple 5-stage CPU we have…

A: ANSWER: a) MUX 1 Input 1: Register A input 2:Register B b) MUX 2 Input 1: Register C input…

Q: Micro8086 In imul what makes the answer 0002 what changes it I know it’s negative number can you…

A: The x86 assembly language's IMUL instruction executes signed multiplication. When IMUL CL is used,…

Q: 3- What is the difference between: a- MOV BX,[1234H] and LEA BX,[1234H] b- LDS AX,[200H] and LES…

A: - According to the guideline we are allowed to answer only the first question. - The given…

Q: Will you fill in the missing code in Marie Assembly language.

A: Solution- An complete marie code with their instuction is given as-

Q: At the start "STKPTR = 0x40" and then the following assembly code is executed. Address Instruction…

A: Given that STKPTR is 0x40 in decimal is 64 then executed the code line by line push means add pc…

Q: What is the final value of BX after the bitwise XOR instruction executes? MOV BX, 29D6h XOR BX,…

A: Here, we are given BX value and then perform XOR instruction between the content of BX and value…

Q: What is the final value of BX after the bitwise XOR instruction executes? MOV BX, 29D6h XOR BX,…

A: 1) MOV BX, 29D6h moves 29D6 into BX register 2) XOR BX, 8181h perform bit wise XOR operation between…

Q: What does the "jal" instruction do? O First saves the address of the next instruction in a…

A: JAL is the jump and link instruction (jal) to call a function.

Q: 21. The program that follows implements what is known as a delay loop. MOV CX, 1000H DLY: DEC CX NOP…

A: Here is the explanation regarding the delay loop:

Topic Video

Question

Yes
Is next
No
instruction
MAR E PC
No memory
IBR2
access
required
MBR + MOMAR
Yes BR E MERN9)
IR E MBR(0:7)
MAR E MBR(8:19)
|No
Left
IR E IBR (0:7)
MAR E IBR (8:19)
IR E MBR(20.27)
MAR + MBR(28:39)
instruction
eguireg
PC + PC+1
Memor
Left instruction
Right instruction
y location
LOAD M(300)
SUB M(301)
LOAD MQ , M(302)
STOR M(303)
STOR M(304)
100
101
STOR M(305)
MULT M(305)
LOAD MQ
102
103
104
HALT
300
301
302
303
304
305
Suppose you start your program from Memory location 100, and we execute the
instruction with program counter is 103. What is the content of IBR?
FETCH CYCLE

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.

Similar questions

10.- Consider the following code and pictureLoop1MOVLW 0x32MOVWF REG2DECFSZ REG2,FGOTO LOOP1 If the system clock frequency is aMHz and each machine cycle consumes 4 cycles of it. Calculate the delay time of the LOOP1 loop. Include the execution difference time of the DECFSZ instruction in the last cycle. Select an answerA) 0.6.sB) 6msC)6usD) 60us
In lab this week you wrote code that simulates a simple computer; that code could be turned into a circuit. Assume that we have set up the following: • an 8-bit pc register, with 8-bit input pc_in and 8-bit output pc_out. • memory returns the instruction on the 8-bit output code_byte. Your code used a get_bits function we provided to retrieve a 2-bit part of the instruction (code_byte) as the index into the register file for register A. In a physical chip, the simplest correct form of the 2-bit extraction would be to: A. Connect 2 of the 8 wires that make up the code_byte to the register file as needed (no gates) B. A right-shift circuit to move the 2 bits for A to the least significant bits and use those two wires C. A mux with 8 bits of the code_byte as inputs and the 2 bits we want as the selector D. An even larger circuit than the ones above
Indirect Addressing Mode Instruction 002A J@ RETADR 3E20030030 RETADR RESW 1 1.Instruction Starts with 002A2.Opcode of J is 3C3.Object code is 3E2003Please show the details
In the MIPS micro-architecture, the PC is incremented for each instruction, during the "Instruction Fetch" stage. The value of the PC may also be updated later within the "Memory Access" phase. * If the value of the PC is: 0x0000 2F1C, what is the value of the PC after you complete the execution of the instruction: ``addi $t0, $s0, $ra``?* 
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