To find the total execution time for this program on $1$ processor, we can use the following formula $-$total execution time $=($number of instructions$)*($CPI$)*($clock cycle time$)$By substituting the given values, we get $-$total execution time $=(2\mathrm{.}56$E$9$ arithmetic instructions $+1\mathrm{.}28$E$9$ load$/$store instructions $+256$ million branch instructions$)*(1$ arithmetic CPI $+12$ load$/$store CPI $+5$ branch CPI$)*(1/2$ GHz$)$This simplifies to:total execution time $=(2\mathrm{.}56$E$9+1\mathrm{.}28$E$9+256$ million$)*(1+12+5)*(1/2$ GHz$)$total execution time $=4\mathrm{.}096$E$9*18*(1/2$ GHz$)$total execution time $=7\mathrm{.}366$E$9/2$ GHztotal execution time $=3\mathrm{.}683$ secondsTo find the total execution time for this program on $2$ processors,We can use the same formula as above, but we need to divide the number of arithmetic and load$/$store instructions by $0\mathrm{.}7*2=1\mathrm{.}4,$ since the program is being parallelized over $2$ processors.We can use the following formula $-$total execution time $=((2\mathrm{.}56$E$9/1\mathrm{.}4)$ arithmetic instructions $+(1\mathrm{.}28$E$9/1\mathrm{.}4)$ load$/$store instructions $+256$ million branch instructions$)*(1$ arithmetic CPI $+12$ load$/$store CPI $+5$ branch CPI$)*(1/2$ GHz$)$This simplifies to:total execution time $=(2\mathrm{.}56$E$9/1\mathrm{.}4)*18*(1/2$ GHz$)$total execution time $=3\mathrm{.}257142857142857$ secondsThe relative speedup of the $2$ processor result relative to the single processor result is$=3\mathrm{.}683/3\mathrm{.}257142857142857=1\mathrm{.}127946127946128$To find the total execution time for this program on $4$ processors,we can use the same formula as above, but we need to divide the number of arithmetic and load$/$store instructions by $0\mathrm{.}7*4=2\mathrm{.}8,$since the program is being parallelized over $4$ processors.We can use the following formula $-$total execution time $=((2\mathrm{.}56$E$9/2\mathrm{.}8)$ arithmetic instructions $+(1\mathrm{.}28$E$9/2\mathrm{.}8)$ load$/$store instructions $+256$ million branch instructions$)*(1$ arithmetic CPI $+12$ load$/$store CPI $+5$ branch CPI$)*(1/2$ GHz$)$This simplifies to:total execution time $=(2\mathrm{.}56$E$9/2\mathrm{.}8)*18*(1/2$ GHz$)$total execution time $=2\mathrm{.}8$ secondsThe relative speedup of the $4$ processor result relative to the single processor result is$=3\mathrm{.}683/2\mathrm{.}8=1\mathrm{.}3121428571428571$To find the total execution time for this program on $8$ processors,we can use the same formula as above, but we need to divide the number of arithmetic and load$/$store instructions by $0\mathrm{.}7*8=5\mathrm{.}6,$ since the program is being parallelized over $8$ processors.We can use the following formula:total execution time $=((2\mathrm{.}56$E$9/5\mathrm{.}6)$ arithmetic instructions $+(1\mathrm{.}28$E$9/5\mathrm{.}6)$ load$/$store instructions $+256$ million branch instructions$)*(1$ arithmetic CPI $+12$ load$/$store CPI $+5$ branch CPI$)*(1/2$ GHz$)$This simplifies to:total execution time $=(2\mathrm{.}56$E$9/5\mathrm{.}6)*18*(1/2$ GHz$)$total execution time $=1\mathrm{.}6$ secondsThe relative speedup of the $8$ processor result relative to the single processor result is$=3\mathrm{.}683/1\mathrm{.}6=2\mathrm{.}306875$