Implement and Compute the total memory traffic in bytes for both instruction fetch and instruction execution for the code that implements the expression a=(b*c) d on 3-, 2-, 1-, and 0- address machines. Assume opcode occupy one byte, address occupy two bytes and data values also occupy two bytes. The word length is one byte.
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.Implement and Compute the total memory traffic in bytes for both instruction fetch and instruction execution for the code that implements the expression a=(b*c) d on 3-, 2-, 1-, and 0- address machines. Assume opcode occupy one byte, address occupy two bytes and data values also occupy two bytes. The word length is one byte.
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- Consider the following fragment of C code for (i=0; i<=100; i=i+1) { a[i] = b[i] + c; } Assume that a and b are arrays of words and the base address of a is in register x17 and the base address of b is in x18. Register x19 is associated with variable i and register x20 with c. Write the code for RISC-V. How many instructions are executed during the running of the code? How many memory data references will be made during execution?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.Implement and conpute the total memory traffic in bytes for both instruction fetch and instruction execution for the code that implements the expression a-(b*c)+d on 3-,2-,1- and 0- adress machines. Assume opcode occupy 1byte , adress occupy two bytes and data values also occupy 2 bytes. The word length is one byte
- 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=300The 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).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…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 instructionPlease look at the entire text below. Please solve and show all work. Thank you. What is the corresponding MIPS assembly code for the following C statement? Assume that the variables f, g, h, i, and j are assigned to register $s0, $s1, $s2, $s3, and $s4, respectively. Assume that the base address of the arrays A and B are in registers $s6 and $s7, respectively. B[8] = A[i−j] Translate the following C code to MIPS. Assume that the variables f, g, h, i, and j are assigned to register $s0, $s1, $s2, $s3, and $s4, respectively. Assume that the base address of the arrays A and B are in registers $s6 and $s7, respectively. Assume that the elements of the arrays A and B are 8-byte words: B[8] = A[i] + A[j] Assume that registers $s0 and $s1 hold the values 0x80000000 and 0xD0000000, respectively. What is the value of $t0 for the following assembly code? add $t0, $s0, $s1 Is the result in $t0 the desired result, or has there been an overflow? For the contents of registers $s0 and $s1 as…
- Q1: Suppose the hypothetical processor has two I/O instructions: (3+3+3)0011=Load AC from I/O0111=Store AC to I/OIn this case, the 12-bit address identifies a particular external device. Show the program execution using figure for the following program:a) Load AC from device 6b) Add contents of memory location 880c) Store AC to device 7 (Note: Question is to be solved similar to the pictures attached with minimum explaination of a line or two with the steps and SHOULD include the memory location 880 as stated in the question)For the MIPS assembly instructions below, what is thecorresponding C statement? Assume that the variables f, g, h, i, and j areassigned to registers $s0, $s1, $s2, $s3, and $s4, respectively. Assume thatthe base address of the arrays A and B are in registers $s6 and $s7,respectively. Note: for each line of MIPS code below, write the respective Ccode. After that, write the corresponding C code for the MIPS.sll $t0, $s0, 2add $t0, $s6sll $t1, $s1, 2 add $t1, $s7, $t1lw $s0, 0($t0)addi $t2, $t0, 4lw $t0, 0($t2)add $t0, $t0, $s0sw $t0, 0($t1)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…