The diagram illustrates a transportation network with nodes and arrows. The nodes represent cities: Jefferson City, Omaha, Des Moines, Kansas City, and St. Louis. The arrows between nodes indicate possible routes for transportation with associated costs per unit.**Details:**- **Supplies and Demands:** - Jefferson City has a supply of 30 units. - Omaha has a supply of 20 units. - Des Moines has a demand of 20 units. - Kansas City has a demand of 20 units. - St. Louis has a demand of 10 units.- **Transportation Costs:** - Jefferson City to Des Moines: $14/unit - Jefferson City to Kansas City: $16/unit - Jefferson City to St. Louis: $2/unit - Omaha to Des Moines: $10/unit - Omaha to Kansas City: $8/unit - Omaha to St. Louis: $5/unit**Optimal Solution Table:**| | Des Moines | Kansas City | St. Louis | Supply ||----------------|------------|-------------|-----------|--------|| Jefferson City | 20 | 0 | 10 | 30 || Omaha | 0 | 20 | 0 | 20 |**Demand:**| | Des Moines | Kansas City | St. Louis ||-------|------------|-------------|-----------|| Total | 20 | 20 | 10 |- **Total Cost:** $550.Students are encouraged to find an alternative optimal solution to minimize the cost under the given constraints. If necessary, answers should be rounded to the nearest whole number. If there's no cost, enter "0".Another table is provided for entering alternative solutions:| | Des Moines | Kansas City | St. Louis ||----------------|------------|-------------|-----------|| Jefferson City | | | || Omaha | | | |- To calculate the total cost for any alternative solution, fill in the table appropriately and compute the resulting total based on the transportation costs.

Linear programming, often known as linear optimization, is a method for obtaining the best result (such as the highest profit or the lowest cost) in a mathematical model using linear criteria. Linear programming is a method for determining the most optimal solution to a problem given constraints. Using linear programming, we convert our real-world issue into a mathematical model. It is made up of a goal function, linear inequalities, and constraints. Given data,Denote Jefferson city and Omaha as 1 and 2 respectively. Also, denote Des Moines, Kansas City and St. Louis as 1,2, and 3 respectively. x11 is the quantity transported from Jefferson City to Des Moines. x12 is the quantity transported from Jefferson City to Kansas City. The decision variables are x11, x12,...... the costs of these are to be minimized. The objective function is minimize Z=14x11+16x12+7x13+8x21+10x22+5x23 Constraints: Supply 1:x11+x12+x13≤30Supply 2:x21+x22+x23≤20Demand 1:x11+x21≥20Demand 2:x12+x22≥20Demand 3: x13+x23≥10 Optimal solution is: x12=20, x22=0, and x23=0 and the total minimum cost is $550. Alternative optimal solution can be derived using the "Excel Solver" and inputting the given objective and constraints. Enter the following formulas: Set the objective and constraints: Result:Final answer: