0 votes 0 votes $I_1 : MUL \ \ \ \ \ R_1, R_2, R_3$ // $R_1 \leftarrow R_2 \times R_3$ $I_2 : ADD \ \ \ \ \ R_4, R_4, R_1$ $I_3 : MUL \ \ \ \ \ R_1, R_5, R_6$ $I_4 : SUB \ \ \ \ \ R_4, R_4, R_1$ Sum of RAW, WAR and WAW dependencies is _____. CO and Architecture data-hazards data-dependency co-and-architecture numerical-answers + – Mk Utkarsh asked Oct 23, 2018 • retagged Aug 6, 2022 by Shubham Sharma 2 Mk Utkarsh 807 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments Shubhanshu commented Oct 23, 2018 reply Follow Share Is this question is self made? 0 votes 0 votes Mk Utkarsh commented Oct 23, 2018 reply Follow Share partially, i don't have answer key for this 0 votes 0 votes scholaraniket commented Jan 13, 2020 reply Follow Share Answer should be 7 I1 - I4 should not be considered for RAW dependency... 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Answer should be 7 RAW = 3 WAW = 2 WAW = 2 nephron answered Oct 23, 2018 nephron comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 3+2+2=7 RAW(1-2,2-4,3-4) WAR(2-3,2-4) WAW(1-3,2-4) Ram Swaroop answered Jul 14, 2019 Ram Swaroop comment Share Follow See all 0 reply Please log in or register to add a comment.