Design a simple, single instruction format and machine language for a computer w

Question

Design a simple, single instruction format and machine language for a computer with the following characteristics.

     The instructions for the ISA of this computer include only the Boolean operations And, Or, Not, and Exclusive Or. An instruction bit string specifies the location of each operand to be read from memory by holding a pointer to the desired memory location in a designated operand field. The same strategy is used to specify the location for the result to be written. Each operand and the result for each instruction is an 8-bit binary string. The computer memory has 210 storage locations. Each memory location holds a one-byte bit string. There are no registers, so all operands are sourced from memory and each result is stored to memory.

Clearly label the fields of your instruction format, and let the leftmost field be the Operation Code, or Opcode, field, followed by all operand fields, and ending with the result field. Next, using the minimum possible number of bits sufficient for each field to perform its function, specify how long each field in your instruction format is. For any field in the format for which the bit string is must be known before writing any programs for this computer, use your Designer’s Prerogative to choose the bit string that will fill that field and display your choice in your format for each instruction. Finally, for each field that is not needed by an instruction, add the word “Unused” after the bit string length entry for that field.

     Some entries on the table are already filled in to help you get started and to limit the number of correct answers.

Operation AND OR Opcode Operand 1 Operand 2 Result Unused

Explanation / Answer

For four instructions: AND , OR , NOT , XOR we require 2 binary bits (22 combinations) to formulate 4 opcodes.

As operands and results are memory addressable and we have 210 locations to address (27<210<28) hence we will be needing atleast 8 bits to address all memory locations. This results in the operand/result field size to be 8 bits.

Based on the above, the table can be filled as the following:

Hope this helps!

Operation Opcode Operand 1 Operand 2 Result AND 00 8 bit, Used 8 bit, Used 8 bit OR 01 8 bit, Used 8 bit, Used 8 bit NOT 10 8 bit, Used 8 bit, Unused 8 bit XOR 11 8 bit, Used 8 bit, Used 8 bit

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