Let's break this problem into clear, step-by-step parts, as a professional tutor would do, and then write the code for it. --- ## **Step 1: Understanding the Problem** - **Objective:** Place ambulances in districts so that every district can reach an ambulance within **3 minutes** (or an ambulance can reach every district within 3 minutes). - **Goal:** Find the minimum number of ambulances needed and their locations. - **Formulation:** This is a **Set Covering Problem** (specifically, a facility location problem). --- ## **Step 2: Integer Linear Programming (ILP) Formulation** ### **Variables** - \( y_j \): Binary. 1 if an ambulance is placed in district \( j \), 0 otherwise. - \( a_{ij} \): Binary parameter. 1 if district \( i \) can be covered by ambulance in district \( j \) (i.e., \( t_{ij} \leq 3 \)), 0 otherwise. ### **Objective** Minimize the number of ambulances: \[ \text{Minimize} \quad Z = \sum_{j} y_j \] ### **Constraints** Every district \( i \) must be covered by at least one ambulance: \[ \sum_{j} a_{ij} y_j \geq 1,\quad \forall i \] \( y_j \) are binary variables. --- ## **Step 3: Data Preparation** The binary coverage matrix \( a_{ij} \) is given in the second table in your image. --- ## **Step 4: Implementation (Python + PuLP)** Let's write Python code using the `PuLP` library to solve this ILP. ```python import pulp # Coverage matrix a_ij (from second table; 1 if district i can be covered by ambulance at j) a = [ [1,1,1,0,0,0,0,0], [1,1,1,1,0,0,0,0], [1,1,1,1,1,0,0,0], [0,1,1,1,1,1,0,0], [0,0,1,1,1,1,1,0], [0,0,0,1,1,1,1,1], [0,0,0,0,1,1,1,1], [0,0,0,0,0,1,1,1] ] n = 8 # number of districts # Define the problem prob = pulp.LpProblem("Ambulance_Placement", pulp.LpMinimize) # Decision variables: y_j = 1 if ambulance placed at district j y = [pulp.LpVariable(f'y_{j+1}', cat='Binary') for j in range(n)] # Objective: Minimize total ambulances prob += pulp.lpSum(y) # Constraints: Each district must be covered by at least one ambulance for i in range(n): prob += pulp.lpSum(a[i][j]*y[j] for j in range(n)) >= 1 # Solve prob.solve() # Output print("Minimum number of ambulances needed:", int(pulp.value(prob.objective))) for j in range(n): if pulp.value(y[j]) == 1: print(f"Place ambulance in district {j+1}") ``` --- ## **Step 5: Solution Explanation** - **What this does:** For each district, it checks which other districts can cover it in ≤3 mins (from `a`), and ensures at least one ambulance is placed in those. - **It minimizes** the number of ambulances needed. - **Prints** which districts to put ambulances in. --- ## **Step 6: Final Answer** **The ILP and the code above will tell you:** - The minimum number of ambulances needed. - The specific districts to place them in. **Run the code to get the actual answer (should be 2 ambulances, as mentioned).** --- ### **Summary Table** | Step | Description | |------|-------------| | 1 | Define binary variables for ambulance placement | | 2 | Create coverage matrix from travel times | | 3 | Formulate ILP: Minimize sum of placements, cover all districts | | 4 | Solve using code (Python+PuLP) | | 5 | Interpret solution: districts to place ambulances in | --- If you need the explanation in a different programming language, let me know!