a) How large must the mode field be? b) How large must the register field be? c) How large must the address field be? d) How large is the opcode field?
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The memory unit of a computer has 1M words of 32 bits each. The computer has an
instruction format with 4 fields: an opcode field; a mode field to specify 1 of 6 addressing
modes; a register address field to specify one of 28 registers; and a memory address field.
Assume an instruction is 32 bits long. Answer the following:
a) How large must the mode field be?
b) How large must the register field be?
c) How large must the address field be?
d) How large is the opcode field?
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- 8- The memory unit of a computer has 2.00E+20 words. The computer has instruction format with four fields; 3- An operation code field, a mode field to specify one of 4 addressing modes, a register address field to specify one of 65 processor registers, and a memory address. i- ii- Specify the number of bits in each field if the instruction occupies one memory word of 32 bits. Opcode Mode Register AR Specify the size of the memory word and the number of bits in each field if the available number of opcodes is increased to 32. word Opcode Mode Register AR iii- Find the size of the new memory in K Bytes (1K=1024 Bytes) Memory size in K Bytes Memory size in K Bytes8- The memory unit of a computer has 2.00E+20 words. The computer has instruction format with four fields: 3-An operation code field, a mode field to specify one of 4 addressing modes, a register address field to specify one of 65 processor registers, and a memory address. - Specify the number of bits in each field if the instruction occupies one memory word of 32 bits. Opcode Mode Register AR ii-Specify the size of the memory word and the number of bits in each field if the available number of opcodes is increased to 32. word Opcode Mode Register AR Find the size of the new memory in K Bytes (1K-1024 Bytes) Memory size in KBytes Memory size in KBytesThe memory unit of a computer has 2M words of 32 bits each. The computer has an instruction format with 4 fields: an opcode field; a mode field to specify 1 of 4 addressing modes; a register address field to specify one of 9 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following: d) How large is the opcode field?
- The memory unit of a computer has 256K words of 32 bits each. The computer has an instruction format with four fields: an opcode field; a mode field to specify one of seven addressing modes; a register address field to specify one of 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:Q.) How large must the address field be?A computer with a 32-bit word uses an instruction format that includes direct and indirect addressing of 8 megabytes and one of 16 registers. The highest order bits are used for the opcode, followed by the bits indicating the register, followed by the indirect/direct bit, followed by the bits indicating the memory address. Draw the instruction word's format, showing how many bits are used for all four fields (Make sure to show here how many bits for each and the order in which they appear. You can submit a drawing of the word's format separatelyAssume a CPU with a fixed 32-bit instruction length has the following instruction format:opcode mode [operand1] [operand2] [operand3]The mode encodes the number of operands and each operand’s mode. For instance, one mode indicates three registers, another indicates two registers and an immediate datum, another indicates a main memory reference, etc. Assume there are 94 instructions and 22 modes. Answer the following.a. One mode indicates three registers. How many registers can be referenced in this mode?b. One mode indicates two registers and an immediate datum in two’s complement. Assuming there are 32 registers, what is the largest immediate datum that can be referenced?c. One mode has a destination register and a source memory address (an unsignednumber). Assuming 16 registers, what is the largest memory reference available?d. One mode has two memory addresses, both using base displacement. In both, the basesare stored in index registers and the displacements are specified in the…
- A program consists of 100,000 instructions as follows: Integer arithmetic Data transfer Floating-point arithmetic Control transfer Instruction Type M1 - 400 MHz M2 - 800 MHz Processor Type of Instruction, IC, and CPI Instruction Count decimal places) 45,000 decimal places) 36,000 10,000 9,000 Execution Time Determine the program execution time, the effective CPI for the machine, and the MIPS rate for the following processors. (each box is 2 points except CPI) seconds (round to 5 4 seconds (round to 5 5 10 3 Cycles per Instruction Type of instruction, IC, and CPI CPI (round to 2 decimal places) MIPS rate (Whole number only with round up) (Whole number only with round up)In simple words, describe the final data pointer register.Part A For each byte sequence listed, determine the Y86 instruction sequence it encodes. If there is some invalid byte in the sequence, show the instruction sequence up to that point and indicate where the invalid value occurs. For each sequence, the starting address, then a colon, and then the byte sequence are shown. 0x100: 30f3fcfffff40630008000000000000 0x100: 30f3fcfffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rrmovq %rsi,0×80A(%rcx) O0x115: 00 halt Ox100: 30f3fcffffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox114: 00 halt 0x100: 30f3fcfffffffff rrmovq $-8,%rbx Ox109: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) 0x200: a06f800c020000000000000030f30a00000000000000 0x113: 00 halt 0x100: 30f3fcffffffffff irmovq $-4,%rbx 0x200: a06f pushq %rsi 0x10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox202: 800c02000000000000 call proc Ox116: 00 halt 0x20b: 00 halt 0x20c: proc: Submit Request Answer 0x20c: 30f30a00000000000000 | irmovq $10,%rbx…
- In the memory of Computer, if we say 256 * 8, it means 256 is size and 8 is Data bus sizeA digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory.1. a) How many bits are needed for the opcode?2. b) How many bits are left for the address part of the instruction?3. ◆c) What is the maximum allowable size for memory?4. d) What is the largest unsigned binary number that can be accommodated in one word of memory?The memory unit of a computer has 256K words of 32 bits each. The computer has an instruction format with four fields: an operation code field, a mode field to specify one of seven addressing modes, a register address field to specify one of 60 processor registers, and a memory address. Specify the instruction format and the number of bits in each field of the instruction if the instruction is in one memory word.