1. Problem 3: Ford Corporation motor supply optimization. - Given production capacities for plants: Boston(50), Dallas(70), Los Angeles(60), St. Paul(80), Denver(100), Atlanta(100), Chicago(150) in thousands. - Demand at refrigerator locations: Boston(50), Dallas(70), Los Angeles(60), St. Paul(80) in thousands. - Profit per unit (1000 motors) for shipments from plants to locations is given. 2. Develop network model: - Nodes: Production Plants (7 nodes) and Refrigerator Locations (4 nodes). - Edges: Possible shipping routes with capacities and profit weights. 3. Formulate LP: - Decision variables: $x_{ij}$ = motors shipped from plant $i$ to location $j$ in thousands. - Maximize total profit: $$ \max \sum_{i,j} p_{ij} \cdot x_{ij} $$ where $p_{ij}$ is profit per 1000 motors from plant $i$ to location $j$. - Constraints: - Supply at plants: $\sum_j x_{ij} \leq$ capacity of plant $i$. - Demand at refrigerator locations: $\sum_i x_{ij} = $ demand at location $j$. - Non-negativity: $x_{ij} \geq 0$. 4. Use Excel Solver to input objective, variables, and constraints and find optimal $x_{ij}$ values. 5. Problem 4: Assign professors to courses to maximize ratings. - Let $y_{ij} = 1$ if professor $i$ assigned to course $j$, else 0. - Objective: $$ \max \sum_{i,j} r_{ij} y_{ij} $$ where $r_{ij}$ is rating. - Constraints: - Each professor assigned exactly one course: $\sum_j y_{ij} = 1$. - Each course assigned exactly one professor: $\sum_i y_{ij} = 1$. - $y_{ij} \in \{0,1\}$. 6. Solve using Excel's integer programming capabilities. 7. Problem 5: Shortest route from node 1 to 10 in road network. - Use Dijkstra's algorithm: - Initialize distances from node 1 with $d_1=0$, others $\infty$. - Visit neighbors updating distances by edge weights. - Continue until distance for node 10 finalized. 8. Calculation steps (abridged): - Start: Node 1 distance=0. - Update neighbors: 2(8),3(13),4(15),5(10). - From 2, update 3(8+5=13),7(8+15=23). - From 3, neighbors 4(13+6=19),6(13+5=18). - From 5, neighbors 4(10+4=14),6(10+9=19),9(10+12=22). - From 4, update 6(14+3=17). - From 6, update 7(17+4=21),8(17+2=19). - From 7, update 8(21+4=25),10(21+4=25). - From 8, update 9(19+5=24),10(19+7=26). - From 9, update 10(22+5=27). - Choose shortest distance to 10 is 25 via nodes: 1->2->3->6->7->10 or 1->5->4->6->7->10. Final shortest route: 1 to 5 (10 min), 5 to 4 (4 min), 4 to 6 (3 min), 6 to 7 (4 min), 7 to 10 (4 min); total 25 minutes. Thus: - Network formulation and LP setup provided for Problem 3 and 4. - Shortest path derived for Problem 5. Note: Detailed Excel solver numerical results are requested to be computed using MS Excel as per instructions.