Explain how we can find the address location of INT 0AH in Interrupt vector table
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1, Explain how we can find the address location of INT 0AH in Interrupt
2. What will be the content of AX register after execution of the instruction IMUL
CL, if CL = +1510 and AL = -13210..
3. Suppose that DS = 1300H, SS = 1280H, BP = 15A0H and SI = 01D0H.
Determine the address accessed by each of the following instructions and
mention their type of addressing mode.
i) MOV AX, [200H]
ii) MOV AL, [BP-SI+200H]
iii) ADD AL, [SI + 0100H]
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- 6. Suppose that the interrupt processing method of is to store the breakpoint in the address of 00000Q unit, and fetch the instruction from the 77777Q unit (that is the first instruction of the interrupt service routine) and execute it. Write the micro-operations sequence that completes this function.Assuming stall-on-branch and no delay slots, what is the new clock cycle time and execution time of this instruction sequence if beq address computation is moved to the MEM stage? What is the speedup from this change? Assume that the latency of the EX stage is reduced by 20 ps and the latency of the MEM stage is unchanged when branch outcome resolution is moved from EX to MEM.Th is exercise is intended to help you understand the relationship between forwarding, hazard detection, and ISA design. Problems in this exercise refer to the following sequence of instructions, and assume that it is executed on a 5-stage pipelined datapath:add r5,r2,r1lw r3,4(r5)lw r2,0(r2)or r3,r5,r3sw r3,0(r5)> If there is no forwarding or hazard detection, insert nops to ensure correct execution.> Repeat above step, but now use nops only when a hazard cannot be avoided by changing or rearranging these instructions. You can assume register R7 can be used to hold temporary values in your modifi ed code.> If the processor has forwarding, but we forgot to implement the hazard detection unit, what happens when this code executes?
- 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…Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. A - is for addressing mode type, e.g. direct B - physical address/es e.g. 19000-19001 C - content of the destination after execution e.g. if register: AX=1234 if memory (lower address first) : 12000=34;12001=12 GIVEN DS = 9CEFH SS = CB8DH DI = BACEH SI = 9EBCH BP = BAD2H AX = 6FCBH BX = 7BCFH CX = 5CADH DX = D1BCH SP = 4FABH MOV CX, [BP+DI+ACEH} Find the addressing mode: Physical address/es: Content of the destination:
- Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. A - is for addressing mode type, e.g. direct B - physical address/es e.g. 19000-19001 C - content of the destination after execution e.g. if register: AX=1234 if memory (lower address first): 12000=34,12001=12 GIVEN AX =AD6EH DX = 9ECFH SS = BEDAH SI = AECDH BX =AFECH CS = EBDCH ES = CFEDH BP = EFBAH CX =FADEH DS = AF0EH DI = BD8FH IP = AB8FHDetermine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. A - is for addressing mode type, e.g. direct B - physical address/es e.g. 19000-19001 C - content of the destination after execution e.g. if register: AX=1234 if memory (lower address first) : 12000=34;12001=12 GIVEN DS = 9CEFH SS = CB8DH DI = BACEH SI = 9EBCH BP = BAD2H AX = 6FCBH BX = 7BCFH CX = 5CADH DX = D1BCH SP = 4FABH MOV EDX, [BACEH] Find the Addressing mode: Physical Address/es: Content of the destination:Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. A - is for addressing mode type, e.g. direct B - physical address/es e.g. 19000-19001 C - content of the destination after execution e.g. if register: AX=1234 if memory (lower address first) : 12000=34;12001=12 GIVEN DS = 8EFDH SS = AFDEH DI = B8F9H SI = D3CDH BP = A5D8H AX = 5FFEH BX = B3DFH CX = CADEH DX = B2BCH SP = A4FBH MOV [BX+SI], BP Find the addressing mode: Physcial address/es: Content of the destination:
- Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. A - is for addressing mode type, e.g. direct B - physical address/es e.g. 19000-19001 C - content of the destination after execution e.g. if register: AX=1234 if memory (lower address first) : 12000=34;12001=12 GIVEN DS = ACF7H SS = BAC9H DI = ECABH SI = ABAEH BP = BAD2H AX = 6FCBH BX = 7BCFH CX = 5CADH DX =D1BCH SP = 4FABH MOV[BX], BX Find addressing mode: Physical Address/es: Content of the destination:Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. A - is for addressing mode type, e.g. direct B - physical address/es e.g. 19000-19001 C - content of the destination after execution e.g. if register: AX=1234 if memory (lower address first) : 12000=34;12001=12 GIVEN DS = 8EFDH SS = AFDEH DI = B8F9H SI = D3CDH BP = A5D8H AX = 5FFEH BX = B3DFH CX = CADEH DX = B2BCH SP = A4FBH MOV[BP+SI+9FEDH], EAX Find Addressing mode: Physical address/es: Content of the destination:Determine the specific type of addressing mode (SMALL LETTERS only) and compute for the address/es. If applicable, determine the content of the destination after the execution of the instruction. Otherwise, NA. For physical address and content of the destination, use CAPITAL LETTERS,NO SPACE/S in between and NO need to include "H" or the unit. A - is for addressing mode type, e.g. direct B - physical address/es e.g. 19000-19001 C - content of the destination after execution e.g. if register: AX=1234 if memory (lower address first): 12000=34,12001=12 GIVEN AX =AB8FH DX = D8ABH SS = C8EFH SI = A7BCH BX =B9DCH CS = DA9FH ES = BEFAH BP = B5EAH CX =CF05H DS = CAF8H DI = 8CDEH IP = BCFEH