The optimum solution (shown by the underscored zeros) calls for assigning child 1 to chore 1, child 2 to chore 3, child 3 to chore 2, and child 4 to chore 4. The associated optimal cost is 1 + 10 + 5 + 5 = $21. The same cost is also determined by summing the p/s, the q/s, and the entry that was subtracted after the shaded cells were determined-that is, (1 + 7 + 4 + 5) +(0 + 0 + 3 + 0) + (1) = $21.