List all assumptions you got directly from the question description in order to complete (a) - (c). List and justify all additional assumptions you needed to make. I have attached the answers of (a) - (c) along with the questions. Please answer it from mathematics viewpoints correctly.

\f1. Paul Bergey is in charge of loading cargo ships for International Cargo Company (ICC) at the port in Brisbane. Paul is preparing a loading plan for an ICC freighter destined for US. An agriculture cargos dealer wants to transport the following cargo products aboard this ship. Paul learned that this ship has space to take 480,000 cubic feet of cargo, and also has a cargo weight limit of 13,000 tons. Paul can elect to load any of the available cargos shown in the table below. Volume per Ton of the Amount cargo Prot per Ton Car 0 Available tons cubic feet $ 1 4800 40 70 2 5000 35 50 3 1200 60 60 4 1700 55 80 Paul plans to solve the following optimization problem: Maximize 70 X1 + 50 X2 + 60 X3 + 80 X4 Subject to: X1 +X2 +X3 +X4 S 13,000 40 X1 + 35 X2 + 60 X3 + 55 X4 S 480,000 X1 S 4800 X2 S 5000 X3 S 1200 X, S 1700 X1, X2, X3, X4 2 0 Where X1, X2, X3, and X4 represent the amount (tons) to load to the ship for Cargos 1, 2, 3 and 4, respectively. (a) (10 marks) Solve this Linear Programming problem in Excel using Solver and report the results. (b) (5 marks) Present a table showing how sensitive Paul's loading strategy is to the prot per ton for each type of cargo. Which cargo's prot per ton is the loading strategy most sensitive to? Why? (c) (5 marks) How much of each type of cargo should Paul accept, if the prot per ton for Cargo 1 decreases to $52? Why? (b) To calculate the sensitivity coefficients, we can use Solver in Excel to solve the linear programming model with different profit per ton values for each cargo category. By observing how the optimal quantities change in response to variations in profit per ton, we can determine the sensitivity coefficients. A considerable shift in the loading strategy's optimum amount of a given cargo type may result from even the slightest change in the yield per ton for that cargo type. In this case, the loading strategy of Cargo 3 is the most sensitive as it has the smallest range between 85.714 and 0, where an allowable increase is 25.714 and an acceptable decrease is 60, accounting for the given objective co-efficient 60. (C) Cargo 1 now has a per-ton prot of $70. It is suggested that $52 be made per ton. Thus, $18 is indicated as a reduction in price. The optimal solution remains unchanged when the change in objective coefficient is within the range of optimality signified by the allowable increase and decrease. The ideal approach for Paul would be accepting 4800 tons of Cargo 1, 5000 tons of Cargo 2, 325 tons of Cargo 3, and 1700 tons of Cargo 4, with a revenue stream of about $655,100, as $18 falls under the range between $12 and positive innite values. I) 6 Variable Cells 7 Final Reduced Objective Allowable Allowable 8 Cell Name Value Cost Coefcient Increase Decrease 9 $B$6 Number to Make X1 4800 0 52 1E+30 12 10 $C$6 Number to Make X2 5000 0 50 1E+30 15 11 SD$6 Number to Make X3 325 0 60 18 60 12 SE56 Number to Make X4 1700 O 80 1E+30 25 13 In terms of binding constraints, every available resource is supposed to be utilized, resulting in more profits for its additional units. In contrast, this theory is habituated to show reverse results when considering non-binding constraints, as there are leftovers or slacks that can hardly cause profitability growth. 4..) 14 Constraints 15 Final Shadow Constraint Allowable Allowable 16 Cell Name Value Price R.H. Side Increase Decrease 17 $F$10 ConstraintlLHS 11825 0 13000 1E+3O 1175 18 $F$11 ConstraintZLHS 480000 1 480000 52500 19500 19 $F$12 ConstraintsLI-Is 4800 12 4800 487.5 1312.5 20 $F$13 Constraint4LHS 5000 15 5000 557.1428571 1500 21 $F$14 ConstraintSLHS 325 0 1200 1E+3O 875 22 $F$15 Constraint6LHS 1700 25 1700 3545454545 954.5454545 23 Constraint No. Increasing Resources Optimal function Cargo No. by1 Unit development 2 + $1 Any mentioned Cargo 3 + $12 1 4 + $15 2 6 + $25 4