b$.$ Computer A executes the MIPS ISA and has a $3$Ghz clock frequency. Computer $8$ executes the $\backslash$times $86$ and has a $4$Ghz clock frequency. On average, programs execute $2$ times as many $\backslash$times $86$ instructions than MIPS instructions. For Program P$1,$ Computer A has a CPT of $3$ and Computer $8$ has a CPT of $4.$

Which computer is faster for P$1?$ What is the speedup? $[2$ Marks$)$