Answer to Question #217980 in Algorithms for Chuzzy

Question

 A computer uses a memory unit with 256k words of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts, an indirect bit, an operation code, a register code part to specify one of 64 registers, and an address part i. How many bits are there in the operation code, the register code part, and the address ii. Draw the instruction word format and indicate the number of bits in each part. iii. How many bits are there in the data and address inputs of the memory? 

Answer 1

1. Register code needs log2(64) = 6 bits, address needs log2(256k) = 8+10 = 18 bits, so for operation code remains 32(word lenth) - 1(indirect bit) - 6(register code) - 18(address) = 7 bits

1. Data input has 32 bits, address input has log2(256k) = 8+10 = 18 bits