The following LP formulation represents a transportation problem where raw material (in tons) is transported from Supplier i, i={1,2,3} to Plant j.j={1,2,3). The problem is solved and the optimal objective function value is 5200. Using the sensitivity report shown below, what would be the optimal total cost if the cost transporting 1 ton from Supplier 2 to Plant 3 becomes 5? Min Total cost = 1x11 + 3x12 + 5x13 +3.5x21 +4x22 +4.8x23 +3.5x31 +3.6x32 +3.2x33 Subject to 231 +232 + x33 = 0, for i = (1,2,3} &j= {1,2,3) Variables Final Reduced Objective Allowable Allowable Name Value Cost Coefficient increase Decrease X11 500 a 1 2.5 1E+30 X12 700 0 3 0.6 1E+30 X13 200 5 1E+30 0.2 X21 0 3.5 1E+30 2.5 X22 o 4 1E30 1 X23 200 4.8 0.2 1E+30 X31 0 3.5 1E+30 2.5 X32 3.6 1E430 0.6 OOOOOOO Variables Name Final Reduced Objective Allowable Allowable Value Cost Coefficient Increase Decrease 500 0 1 2.5 1E+30 700 0 3 0.6 1E+30 200 0 5 1E+30 0.2 0 0 3.5 1E+30 0 0 4 1E+30 1 200 0 4.8 0.2 1E+30 0 3.5 1E+30 2.5 0 0 3.6 1E430 0.6 200 0 3.2 1.8 1E+30 X11 X12 X13 X21 X22 X23 X31 X32 X33 2.5 Constraints Name TOTAL SUPPLY OF SUPPLIER 3 SUPPLY FROM SUPPLIER 2 TO PLANT 1 SUPPLY FROM SUPPLIER 2 TO PLANT 2 SUPPLY FROM SUPPLER 2 TO PLANT 3 SUPPLY FROM SUPPLIER 3 TO PLANT 1 SUPPLY FROM SUPPLIER 3 TO PLANT 2 SUPPLY FROM SUPPUER 3 TO PLANT 3 Demand of Plant 1 Demand of Plant 2 Demand of Plant 3 Final Shadow Constraint Allowable Allowable Value Price R.H. Side Increase Decrease 200 0 500 1E+30 300 0 0 200 1E+30 200 0 0 200 1E+30 200 200 -0.2 200 200 200 0 200 1E+30 200 0 200 1E430 200 200 -1.8 200 200 200 500 1 500 1E+30 500 700 3 700 1E+30 700 600 600 1E+30 200