A digital computer has a memory unit with 24 bits per word. The instruction set

Question

A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different instructions. Allinstructions have an operation code part (opcode) and an addresspart (allowing for only one address). Each instruction isstored in one word of memory.
a. How many bits are needed for the opcode?
b. How many bits are left for the address part of theinstruction?
c. What is the maximum allowable size for memory?
d. What is the largest unsigned binary number that can beaccommodated in one
    word of memory?

Explanation / Answer

a) Instruction set has 150 instructions...     We know that opcode should differentiate theinstructions...therefore the number of bits allocated to opcodemust differentiate the 150 instructions..     If we allocate 8 bits to opcode..then it canaccommodate 28 = 256 instructions...if we allocate 7bits then it can only accommodate 27 = 128instructions.     As number of instructions is 150 >128 ....wehave to allocate 8 bits for theopcode b) Bits left for address part = 24 bits - bits for opcode                                          =24 -8                                           = 16 bits c) Address part should be accommodate using 16 bits...     Therefore there can be at most 216addresses.     Assuming each word has an address..there can be216 words of memory     Therefore Maximum allowable size of memory = 216words d) Largest unsigned binary integer that can be accommodated inmemory can be calculated as follows.     Each word has 24 bits..therefore the maximumpossible binary number is 1111......111 (24 times)     It is equivalent to 224 -1

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