3=-1, $9=0, $1035, th 3=0, $9=-1, 3=5, $9=5, $10=0, the 3=5, $9=-1, $10=-1, th $10-5, th ions are wrong
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- Let us assume that the MIPS instruction below is executed, in this case what is the value of $t0?sub $t0, $s0, $s1$s0: 0 x 9 0 0 0 0 0 0 0$s1: 0 x A 0 0 0 0 0 0 0d) Is the result in $t0 the desired result, or has there been overflow?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``?* <!-- response -->Joe wants to calculate the perimeter of an octagon using the most efficient MIPS native instruction. Assuming that register $s0 contains the length of each side of the octagon and Joe wants the perimeter to be stored in register $s1, fill the blanks below to help Joe. __, $s1,$s0,__ I know the answer but please explain why it is sll and 3!!
- What happens if an instruction is not accepted and it writes to VA page 30? In the following scenarios, a software-managed TLB would be faster than a hardware-managed TLB:Single instruction computer (SIC) has only one instruction that for all operations our MIPS does. The instruction has the following format. sbn a, b, c # Mems[a]=Mem[a]- Mem[b]; if (Mem[a]<0) go to PC+c For example, here is the program to copy a number from location a to location b: Start: sbn temp, temp, 1 sbn temp, a, 1 sbn b, b, 1 sbn b, temp, 1 So build SIC program to add a and b, leaving the result in a and leaving b unmodified.Suppose we add the following instruction to MARIE’s ISA:This instruction increments the value with effective address “Operand,” and if this newly incremented value is equal to 0, the program counter is incremented by 1. Basically, we are incrementing the operand, and if this new value is equal to 0, we skip the next instruction. Show how this instruction would be executed using RTN.
- Write an assembly program such that: a. When the least significant bit of registers R4 and R5 have the value 1, the register R9 gets the value 0FF0h. b. When only one of the least significant bits of either register R4 or R5 has the value 1, the one’s complement of the value in register R9 will be saved in register R10. please use assembly code (ex: mov.w, and.w )Consider the following portions of three different programs running at the same time on three processors in a symmetric multicore processor (SMP). Assume that before this code is run, Total is 15, val_1 is 45, val_2 is 25, and val_3 is 5. Core 1: Total = Total + val_1;Core 2: Total = Total - val_2;Core 3: Total = Total + val_3; 1. Show the assembly code for both instructions. Assume that Total, var_1, and var_2 are stored at 0x0100, 0x0120, and 0x0130 respectively.2. What are all the possible resulting values of Total? For each possibleoutcome, explain how we might arrive at those values. You will need to show all possible interleavings of instructions.3. How could you make the execution more deterministic so that only one set of values is possible?In MIPS assembler code below, how many times is instruction memoryaccessed? How many times is data memory accessed? (Count only accesses to memory, not registers.) lw $v1, 0($a0)lw $v0, 4($a0)addi $v1, $v1, 1sw $v1, 0($a0)addi $a0, $a0, 4sub $v0, $v0, $v1sw $v0, 0($a0)sw $v0, 4($a0
- You are given snapshots of the MIPS pipeline from two successive cycles. For each figure, fill in all the relevant information in each stage, corresponding to the instruction in that stage at that cycle. Remember to specify clearly how control signals are set. Assume $x contains x * 100 (e.g. $1 contains 100). MIPS instructions used: add rd, rs, rt rd = rs + rt or rd, rs, rt rd = rs or rt lw rt, I(rs) rt = Mem[rs + I] (load word) sw rt, I(rs) Mem[rs + I] = rt (store word)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.Suppose we add the following instruction to MARIE’s ISA: JumpIOffset X This instruction will jump to the address calculated by going to address X, then adding the value found there to the value in the AC.Show how this instruction would be executed using RTN.