Continuing in the same manner, row 2 will produce the highest penalty (= 11), and we as sign X23 = 15, which crosses out column 3 and leaves 10 units in row 2. Only column 4 is left, and
it has a positive supply of 15 units. Applying the least-cost method to that column, we successively
assign X14 = 0, X34 = 5, and X24 = 10 (verify!). The associated objective value for this solution i