The following LP formulation represents a transportation problem where raw material (in tons) is transported from Supplier =(1,2,3) to Planti. 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 of transporting 1 ton from Supplier 1 to Plant 1 becomes 2 and the cost of transporting 1 ton from Supplier 3 to Plant 3 becomes 2.5? Min Total cost = 1x11 + 3x12 + 5x13 + 35x21 + 4x2 + 4.8x23 + 3.5x31 + 3.6x32 + 3.2x33 Subject to x31 + x32 + x33 - 500 *11 + x21 + x31 = 500 x12 + x22 + x32 = 700 x13 + x23 + x33 600 xijc-200, fori (2,3) &j= (1.2.3) xij > 0, for i = (1,2,3) & j" (1,2,3) Variables Name X11 X12 Final Reduced Objective Allowable Allowable Value Cost Coefficient increase Decrease 500 0 1 25 1E130 700 0 06 16.301 200 0 5 11.30 0.2 0 2.5 1+30 2.5 0 4 15.30 1 200 0.2 16.30 0 3.5 15.30 25 0 3.6 16.30 0.6 200 O 3.2 16:30 X22 X23 31 X32 X33 OOOOO 8.0 Constraints Final Shadow Constraint Allowable Allowable Value Price RH Side Increase Decrease 200 O 15:30 300 O 0 200 15.30 200 Name TOTAL SUPPLY OF SUPPLIER 3 500 DANT Variables Name X11 X12 X13 X21 X22 X23 X31 X32 X33 Final Reduced Objective Allowable Allowable Value Cost Coefficient Increase Decrease 500 0 1 2.5 1E+30 700 3 0.6 1E+30 200 5 1E+30 0.2 0 3.5 1E+30 2.5 0 4 1E+30 1 200 0 4.8 0.2 1E+30 0 0 3.5 1E+30 245 e 3.6 1E+30 0.6 200 3.2 1.8 1E+30 olololo OO Constraints Name TOTAL SUPPLY OF SUPPLIER 3 SUPPLY FROM SUPPLER 2 TO PLANT 1 SUPPLY FROM SUPPUER 2 TO PLANT 2 SUPPLY FROM SUPPLIER 2 TO PLANT 3 SUPPLY FROM SUPPLIER 3 TO PLANT 1 SUPPLY FROM SUPPLER 3 TO PLANT 2 SUPPLY FROM SUPPLIER 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 200 500 1E30 300 0 200 1E-30 200 0 200 1E30 200 200 -0.2 200 200 200 0 0 200 1E+30 200 0 200 1E+30 200 200 -1.8 200 200 200 500 1 500 1E+30 500 700 3 700 1E-30 700 600 5 600 LE+30 200 OOOO00m