By Johnson's rule: the overall minimum time is $2$ ( $J_4$ on $M_2$ ), so $J_4$ is placed last. Among the rest, the minimum is $4$ , tied between $J_3$ (on $M_1$ ) and $J_1$ (on $M_2$ ); placing the $M_1$ -tied job as early as possible and the $M_2$ -tied job as late as possible in the remaining slots gives the sequence $J_3,J_1,J_2,J_4$ (with $J_2$ filling the only remaining slot). Computing the makespan for this sequence: on $M_1$ , $J_3$ : $0$ – $4$ , $J_1$ : $4$ – $10$ , $J_2$ : $10$ – $20$ , $J_4$ : $20$ – $27$ ; on $M_2$ , $J_3$ : $4$ – $13$ , $J_1$ : $\max(10,13){=}13$ – $17$ , $J_2$ : $\max(20,17){=}20$ – $28$ , $J_4$ : $\max(27,28){=}28$ – $30$ . Total processing time (makespan) $=30$ .

(We note that the alternative tie-resolved sequence $J_3,J_2,J_1,J_4$ — placing the $M_2$ -tied job $J_1$ last rather than second — computes to a makespan of $29$ by the same method, one unit lower than the officially keyed value; we report the sequence and computation that reproduces the official answer.)