10 votes 10 votes Consider the following instructions. $I_1:R_1=100$ $I_2:R_1=R_2+R_4$ $I_3:R_2=R_4+25$ $I_4:R_4=R_1+R_3$ $I_5:R_1=R_1+30$ Calculate sum of ($\text{WAR, RAW and WAW}$) dependencies the above instructions. $10$ $12$ $6$ $8$ CO and Architecture co-and-architecture data-dependency + – Deepak Sharma 1 asked Dec 7, 2015 edited Aug 8, 2021 by soujanyareddy13 Deepak Sharma 1 6.5k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Digvijaysingh Gautam commented Oct 19, 2016 reply Follow Share Ans is 11 ? 0 votes 0 votes KISHALAY DAS commented Oct 19, 2016 reply Follow Share They have given 7 but i m getting 10.could you explain how r u getting 11? 0 votes 0 votes Digvijaysingh Gautam commented Oct 19, 2016 reply Follow Share 1) Consider for I1 between I1 and I2 :- WAW = 1 between I1 and I3 :- no between I1 and I4 :- RAW = 1 between I1 and I5 :- no since ID phase of I5 is after WB of I1 2) Consider for I2 between I2 and I3 = WAR =1 between I2 and I4 = WAR =1, RAW = 1 between I2 and I5 = WAW = 1, RAW = 1 3) Consider for I3 between I3 and I4 = WAR =1 between I3 and I5 = no 4) Consider for I4 between I4 and I5 = WAR =1 So total =9 0 votes 0 votes Please log in or register to add a comment.
Best answer 20 votes 20 votes WAW dependence: 1. I1-I2 2. I2-I5 3. I1-I5 RAW dependence: 1. I2-I4 2. I2-I5 3. I1-I4 4. I1-I5 The dependences 3 and 4 are special in the sense that they never cause a hazard. So, if hazards are asked never include them. For dependences as per definition we can include them. WAR dependence: 1. I2-I3 2. I2-I4 3. I3-I4 4. I4-I5 Totally 11. Arjun answered Dec 4, 2016 selected May 13, 2021 by Arjun Arjun comment Share Follow See all 13 Comments See all 13 13 Comments reply Show 10 previous comments KUSHAGRA गुप्ता commented Dec 21, 2019 reply Follow Share @`JEET Depends on what ? Which answer we should go with @srestha mam or @Arjun sir on RAW dependencies ? 0 votes 0 votes nvs16 commented Sep 20, 2020 reply Follow Share Is the answer 11 or 9(in the accepted answer) ? 0 votes 0 votes Raj_81 commented Oct 4, 2020 reply Follow Share Link not found :( 0 votes 0 votes Please log in or register to add a comment.
24 votes 24 votes RAW or Flow (TRUE) dependence happens when a READ to a register follows a WRITE to the same. Here, RAW is there between, $I_2- I_4$ $I_2 -I_5$ I1-I4 I1-I5 The dependences 3 and 4 are special in the sense that they never cause a hazard. So, if hazards are asked never include them. For dependences as per definition we can include them. For finding RAW dependence we have to see all register usages, and the just preceding write to the same (need not be adjacent instructions). WAR or anti dependence happens when a WRITE follows a READ from a register. Here, it happens between, $I_2-I_3$ $I_2-I_4$ $I_3-I_4$ $I_4-I_5$ WAW or Output dependence happens when a WRITE to a register follows another WRITE. Here it happens between $I_1-I_2$ $I_2-I_5$ $I_1-I_5$ Total 11 dependencies srestha answered Dec 7, 2015 edited Jan 19, 2019 by Arjun srestha comment Share Follow See all 16 Comments See all 16 16 Comments reply Show 13 previous comments Xylene commented Oct 25, 2017 reply Follow Share If the above answer is correct then the answer to this previous year question https://gateoverflow.in/3622/gate2006-it-78 should be wrong. which definition should we follow? @Arjun Sir , Please see this ! 1 votes 1 votes Amit Tiwari 5 commented Jul 30, 2020 reply Follow Share I1-I5 wont be a dependecy if there's a prob b/w I1 I2 doesn't matter there is a prob b/w I1 I5 and if there is no prob b/w I1 I2 there cant be a prob b/w I1 I5 so here is only 1 dependency 0 votes 0 votes Akash 1234Upadhyay commented Feb 6, 2021 reply Follow Share ITS CORRECT 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes yeah,bro..and wrong answer is given by madeeasy. WAW - 3 I1I2 , I1I5, I2I5 RAW - 4 I1I4 , I1I5 , I2I4, I2I5 WAW - 4 I2I3 , I2I4 , I3I4, I4I5 total = 11.. It will be the right answer. resuscitate answered Jan 3, 2016 resuscitate comment Share Follow See all 3 Comments See all 3 3 Comments reply tiger commented Jan 3, 2016 reply Follow Share can u explain method what need to consider in RAW, WAR, WAW ? 0 votes 0 votes resuscitate commented Jan 3, 2016 reply Follow Share raw-read after write war- write after read waw- write after write just follow the meaning and check them between 2 instructions 0 votes 0 votes asu commented May 16, 2016 reply Follow Share LAST ONE IS THE WAR DEP. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Assume I and J are two instrucitons and J follows I in the given question. For RAW : (check only adjacent instructions I and J ) check for commom registers between input registers of instruction J and and output reguster of instruction I So, no of RAW dependencies : 0 For WAR : (check all previous instructions (I's) before instruction J) check for commom registers between output register of instruction J and input registers of instriction I So, WAR dependencies : I3 - I2 (R2) I4 - I3 (R4) I4 - I2 (R4) I5 - I4 (R1) Total WAR = 4 For WAW : (check all previous instructions (I's) before instruction J) check for commom registers between output register of instruction J and output registers of instriction I I2 - I1 (R1) I5 - I2 (R1) I5 - I1 (R1) So, Total WAW = 3 dd answered Aug 29, 2016 dd comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments Arjun commented Aug 29, 2016 reply Follow Share This is not correct. 0 votes 0 votes nimesh kumar commented Aug 30, 2016 reply Follow Share sir please give correct solution. 0 votes 0 votes Arjun commented Aug 30, 2016 reply Follow Share You can see the accepted answer. 0 votes 0 votes Please log in or register to add a comment.