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Consider two microprocessors having 8- and 16-bit-wide external data buses, respectively. The two processors are identical otherwise and their bus cycles take just as long

a. Suppose all instructions and operands are two bytes long. By what factor do the maximum data transfer rates differ?

b. Repeat assuming that half of the operands and instructions are one byte long.

I was referencing a solution for this and for the second part it assumed that for 16 bit microprocessor for the 1 byte instructions/operands it will transfer only 8 bits in 1 bus cycle. I am not understanding why would that be the case since the 16 bit processor can send 16 bit i.e 2 bytes in one bus cycle and hence would be able to send 2 one byte instructions/data in one bus cycle. Can anyone solve this. Thanks

here for a istruction is of 2 bytes and also oprend is of two bytes and there are two microprocesor with the bus size 8 bytes and 16 bytes respectively .

so to fetch one instruction with oprend first microprocessor will take  the cycle of 4 (2 for instruction +2 for oprend at a time one bytes)

and for second microprocessor it takes 2 cycle(1 for instruction fetch and 1 for oprend here oprend and instruction can be fetch at a time of two bytes)

so for a rate  differ by two

and similarly for b=(0.5(1)+0.5(2)+0.5(1)+0.5(2))/2

this is for one instruction so rate will differ by 1.5