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- Explain why, in MARIE, the MAR is only 12 bits wide while the AC is 16 bits wide. Hint: Consider the difference between data and addresses.Explain why, in MARIE,the MAR is only 12 bits wide while the AC is 16 biys wide.give the 9-bit microoperation field for the following micro- operations: a. AC←AC + 1, DR←DR + 1 b. PC←PC + 1, DR←M[AR] c. DR←AC, AC←DR
- Question 33334 Computer Architecture Use 4-bit version (result 8-bit), do the division with the given data 5ten (dividend) and 2ten (divisor), or 0101two and 0011two. Use Algorithm1 only that has 64-bit ALU.Assume that IEEE 754 floating point format is used and $f1 and $f2 store the floating numbers as below: $f1: 1000 1010 0110 1110 1001 1000 0000 0000 $f2: 1000 1010 0100 1111 0100 0010 0000 0000 What are the values of each register? What is the result computing for ($f1+$f2)What is the binary encoding of the following RISC-V assembly instruction? Please write your answer as an 8-digit hexadecimal number. slli x6, x7, 8
- Write an appropriate assembly language code for the following operation and interpret to Von Neumann IAS architecture Z=A/B Where A-greater than 40 bit data B-greater than 40 bit dataComputer Architecture Given the below IEEE single precision FP format: 1-bit Sign , 8-bit Biased Exponent , 23-bit Mantissa , Bias = 127 Convert (9.625)10 to IEEE 32-bit floating-point format.(RISC-V) Write an instruction that calculates the bitwise Exclusive OR of x3 and 466 (in decimal) and stores the result in x18.
