A digital computer has can memory unit with 26 bits per word

A digital computer has can memory unit with 26 bits per word. The instruction set consists of 50 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.

a)How many bits are needed for the opcode?

b)How many bits are left for the address part of the instruction?

c)What is the maximum allowable size for memory?

Solution

a) 2⁶ = 64 which is larger than 50, we take the next lower which is 2^5= 32 bits. Therefore the operation code needs 6 bits.

b) 26-6=20. There are 20 bits available for the address part of the instruction.

c) You can address at most 2²⁰= 1048576 words. Each word is 26 bits long so,

Maximum allowable size = 1048576*26= 27262976 bits= 3.407872 MB.