0 votes 0 votes Consider a device with 1MB per second transfer rate and operating in cycle stealing mode of DMA. It requires 0.5 microsecond to transfer the data (1 byte) when it is ready or prepared. Percentage of CPU blocked due to DMA? CO and Architecture co-and-architecture dma + – Ajit J asked Dec 9, 2018 • retagged Jul 30, 2022 by Shubham Sharma 2 Ajit J 1.3k views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply aambazinga commented Dec 9, 2018 reply Follow Share 0.5us/(0.5us+1us)*100=33.33% 1us is time required to transfer 1B of data from device to DMA 0 votes 0 votes Ajit J commented Dec 9, 2018 reply Follow Share My doubt here is why do we need to add 1us to the denominator, as it is cycle stealing method. As it uses pipelining, we shouldn't be adding 1us to the denominator, right? 0 votes 0 votes aambazinga commented Dec 9, 2018 reply Follow Share if you are not considering 1us in the denominator, then what reference are you using to find the %time blocked of CPU. 0 votes 0 votes Ajit J commented Dec 9, 2018 reply Follow Share Look at this question brother, then you'll get to know my query 0 votes 0 votes Satbir commented Dec 9, 2018 reply Follow Share how can u say that it is using pipelining ? 0 votes 0 votes Ajit J commented Dec 9, 2018 reply Follow Share I meant to say that it uses the concept of pipelining. Refer this pdf, which was shared by Bikram sir. 0 votes 0 votes aambazinga commented Dec 9, 2018 reply Follow Share ok yeah i missed that concept. it should be 0.5/1*100=50% 0 votes 0 votes Ajit J commented Dec 9, 2018 reply Follow Share So aambazinga brother 50 percent is the correct answer? 0 votes 0 votes Satbir commented Dec 9, 2018 reply Follow Share https://gateoverflow.in/250088/cycle-stealing-dma-doubt see this. in DMA case we check for interrupts at the end of every pipeline operation whereas for other interrupts only check at the end of each instructions. pipelining has nothing to do with this question i suppose 0 votes 0 votes Ajit J commented Dec 9, 2018 reply Follow Share but this question is based on cycle stealing 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Device transfer is given as 1 MBPS which means 1 MB per sec, So 1B is transferred in 1 microsec, Transfer Time is given as 0.5 microsec, So (0.5/1+0.5) x 100 % = 33.33% sakshipandey.exe answered Sep 21, 2022 sakshipandey.exe comment Share Follow See all 0 reply Please log in or register to add a comment.