DMA madeeasy

Question

Here device transfer rate is 416.7 us and cpu executing an instruction in .5 us.

So how to decide the clock cycle time, to calculate the fraction of cpu slows down?

Comments

according to you . slow down = cpu time / Dma time = .5/ 416.6 = 0.0012= 0.12%

Answer 1

DMA can transffer in Cycle steeling Mode = 1 BYTE

 But can trasfer = $\frac{19200}{8}$ = 2400 B

Processor can transmit = 2 Million instruction. 

So slow down = $\frac{2400}{2M}$ = 1200 $\times$ 10^-6  = 0.0012 $\times$ 100= 0.12 %

Comments

Ok..so what is the meaning of 416.7us..

I thought...Device tranfer a byte in every 416.7us.

---

actually it is dma gets ready to transfer the byte after 416.7 micro sec

---

Actually, the question is not for %, it’s asking for the answer to be in micro sec.

Answer 2

here for every 1 byte it will steal one cycle of time

so for 1 byte how much time it will take 8/19200=416 micro sec

 

 

the cpu is fetching with 2 million per sec

1 sec ------------- 2 million instructions

1 instr-------------0.5 micro sec (cycle time)

in 416 micro sec it will steal one cycle (or one instruction of time) so total instructions will be 832 instructions

so the cpu slow down will be 1/832 * 100 = 0.12%