g instruction sequence where registers R1,R2 and R3 are general purpose and MEMORY[X] denotes the content at the memory location X. InstructionMOV R1,(5000)MOV R2,(R3)ADDR2,R1MOV (R3),R2INC R3DEC R1BNZ 1004HALTSemanticsR1←MEMORY[5000]R2←MEMORY[R3]R2←R1+R2MEMORY[R3]←R2R3←R3+1R1←R1−1Branch if not zero to thegiven absolute addressStopInstruction Size (bytes)44242221 Assume that the content of the memory location 5000 is 10, and the content of the register R3 is 3000. The content of each of the memory locations from 3000 to 3020 is 50. The instruction sequence starts from the memory location 1000. All the numbers are in decimal format. Assume that

g instruction sequence where registers R1,R2 and R3 are general purpose and MEMORY[X] denotes the content at the memory location X. InstructionMOV R1,(5000)MOV R2,(R3)ADDR2,R1MOV (R3),R2INC R3DEC R1BNZ 1004HALTSemanticsR1←MEMORY[5000]R2←MEMORY[R3]R2←R1+R2MEMORY[R3]←R2R3←R3+1R1←R1−1Branch if not zero to thegiven absolute addressStopInstruction Size (bytes)44242221 Assume that the content of the memory location 5000 is 10, and the content of the register R3 is 3000. The content of each of the memory locations from 3000 to 3020 is 50. The instruction sequence starts from the memory location 1000. All the numbers are in decimal format. Assume that

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter4: Processor Technology And Architecture

Section: Chapter Questions

Problem 2PE: If a microprocessor has a cycle time of 0.5 nanoseconds, what’s the processor clock rate? If the...

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A(n) ________________ instruction always alters the instruction execution sequence. A(n) ______________ instruction alters the instruction execution sequence only if a specified Condition is true.
Consider the following store instruction: SW R1, 0x000F(R0). Assume that the registers R0 and R1 are initialized with 0x00000001 and 0x53A78BC Frespectively. A section of the MIPS byte addressable data memory is shown. Give the memory word of the following memory locations after the SW operation: (a). 0x00000015. (b). 0x00000014. (c). 0x00000013. (d) 0x00000012.(e). 0x00000011. (f). 0x00000010.
17. Consider the following hypothetical instruction: SubMem R1, mem1, mem2 This instruction works as follows: \[ \mathrm{R} 1 \leftarrow \text { [mem1] - [mem2] } \] In a multi-cycle datapath implementation, this instruction will: a. Use the MDR twice b. Use the ALU once c. Use the "shift to left" unit twice d. None of the above Answer: B 18. Consider the following hypothetical instruction: Mems mem1, R1, mem2 This instruction works as follows: \[ \text { [mem1] } \leftarrow \mathrm{R} 1 \text { - [mem2] } \] One of the following is correct about this instruction: a. It will not need theBregister b. It will require priting into MDR twice c. It will require writing into the ALUout three times d. None of the above Answer: A 19. By comparing the hypothetical instructions given in Questions (17) and (18), if we run these instructions on the same processor, then one of the following is correct: a. Both instructions have the same CPI b. Mems executes faster than SubMem c. SubMem executes…
Assume the usage of ARM assembly language instruction: STR R0,[R1],#0x8. Assume that Register R0 contains 0xA0, Register R1 contains 0x7008. Identify (i) contents of memory location 0x00A8 and (ii) R1 after executing the given instruction (i) 0xA0, (ii) 0x7008 (i) 0xA0, (ii) 0x7011 (i) 0xA2, (ii) 0x7008 (i) 0xA2, (ii) 0x7010
A CPU has an instruction pipeline with the following 4 segments. 1.F1(Fetch Instruction) 2.DA(Decode, Address) 3.FO(Fetch Operand) 4.EX(Execution) D1: Add R1,R2 //R1=R1+R2 Add R4,R1 //R4=R4+R1 Add R4,R3 //R4=R4+R3 Cmp R4,R5 // compare R4 and R5 Jg D1 //if Jump Greater (if R4>R5) Sub R2,R1 //R2=R2-R1 Sub R2,R3 //R2=R2-R3 Jmp D1
Consider a computer which has a memory which is capable of storing 4096 K words and each word in memory can be of size 32 bits. The computer supports a total of “8” addressing modes, and it has “50” computer registers. The computer supports instructions, where each instruction consists of following fields: Mode Operation code Register Register Memory Address Given that each instruction will be stored in one memory word, discuss with a suitable diagram the format of instruction by specifying number of bits for each field of instruction. Discuss each field of instruction
5. Consider the 2-address instruction SUB R1, X which subtract the contents of location X from the contents of register R1 and places the result in R1. The ALU can complement either of its inputs, but it cannot subtract. Integers are stored in two’s complement representation.List the sequence of micro-operations to implement the instruction. Assume processor registers PC, MAR, MBR, IR with their usual meanings
The following diagram shows some registers like processor registers R1 and R2, Program counter PC and Index Register XR along with their corresponding values. It also shows a memory with some instructions like instruction A and next instruction. The memory holds instruction B which consists of four fields as given above. First field of instruction represents the addressing Mode (I), second field specifies Opcode (operation code) ADD representing operation addition, the third field represents Address field 1 and the fourth field represents Address field 2. Consider the following addressing modes, evaluate the result of execution of above instruction by giving steps of evaluation for each addressing mode for the scenario given above. Ø Immediate Mode Ø Direct Mode Ø Register Ø Relative Mode Ø Index Mode Choose your own values for variables (v – w), T1, T2. Choose any one of the given value for T3 (200 or 300). V=700 W=800 T1=200 T2=200 T3=300
The following diagram shows some registers like processor registers R1 and R2, Program counter PC and Index Register XR along with their corresponding values. It also shows a memory with some instructions like instruction A and next instruction.The memory holds instruction B which consists of four fields as given above. First field of instruction represents the addressing Mode (I), second field specifies Opcode (operation code) ADD representing operation addition, the third field represents Address field 1 and the fourth field represents Address field 2. Consider the following addressing modes, evaluate the result of execution of above instruction by giving steps of evaluation for each addressing mode for the scenario given above. Immediate Mode  Direct Mode Register Relative Mode Index ModeNote: Choose your own values for variables k – w, T1, T2. Choose any one of the given value for T3 (200 or 300).
I want all staps for Consider a computer which has a memory which is capable of storing 4096 K words and each word in memory can be of size 32 bits. The computer supports a total of “6” addressing modes, and it has “8” computer registers. The computer supports instructions, where each instruction consists of following fields: • Mode • Operation code • Register • Register • Memory Address Given that each instruction will be stored in one memory word, discuss with a suitable diagram the format of instruction by specifying number of bits for each field of instruction. Discuss each field of instruction
Consider a Computer which has a memory which is capable of storing 4096 K words and each word in memory can be of size 32 bits. The computer supports a total of 6 addressingmodes, and it has 60 computer registers. The computer supports instructions, where each instruction consists of following fields: Mode Operation code Register Register Memory AddressGiven that each instruction will be stored in one memory word, discuss with a suitable diagram the format of instruction by specifying number of bits for each field of instruction. Discuss each field of instruction.
6. Assume that two numbers: dividend and divisor are saved in memory address M1 and M2 respectively. Quotient and remainder should be saved in R1 and R2 respectively. Write assembly language instructions and then list microoperations for each instruction and list the control signals required to be activated for each microoperation. MBR is used as buffer for any register to register transfer operation. Signal Description: Control signals operation Comments C0 MAR to RAM (through address bus) C1 PC to MBR C2 PC to MAR C3 MBR to PC C4 MBR to IR C5 RAM to MBR C6 MBR to ALU C7 Accumulator to ALU C8 IR to MAR C9 ALU to Accumulator C10 MBR to Accumulator C11 Accumulator to MBR C12 MBR to RAM (through data bus) C13 IR to Control Unit C14 MBR to R1 C15 MBR to R2 C16 MBR to R3 C17 MBR to R4…