Let's solve the problem in logical steps, showing formulation and solution as if we were using the Simplex Method, and then interpret the shadow price and reduced cost. --- ### **Step 1: Formulate the Linear Program** #### **Variables:** Let \( x_1 \) = number of days Refinery 1 operates \( x_2 \) = number of days Refinery 2 operates #### **Objective:** Minimize Cost: \[ \text{Minimize } Z = 25,000x_1 + 30,000x_2 \] #### **Constraints:** - **High-grade oil:** \( 250x_1 + 250x_2 \geq 35,000 \) - **Medium-grade oil:** \( 300x_1 + 300x_2 \geq 30,000 \) - **Low-grade oil:** \( 150x_1 + 400x_2 \geq 40,000 \) - **Non-negativity:** \( x_1 \geq 0, \quad x_2 \geq 0 \) #### **Explanation:** We defined decision variables for how many days to run each refinery, created the objective function (total cost), and set constraints to ensure enough of each oil type is produced. --- ### **Step 2: Solve the Linear Program (Simplex Method or Excel Solver)** Let’s solve the constraints: #### **First, simplify the constraints:** - **High-grade:** \( x_1 + x_2 \geq 140 \) - **Medium-grade:** \( x_1 + x_2 \geq 100 \) (But the high-grade constraint is more restrictive: \( x_1 + x_2 \geq 140 \)) - **Low-grade:** \( 150x_1 + 400x_2 \geq 40,000 \) So, key constraints are: \[ \begin{align*} x_1 + x_2 &\geq 140 \\ 150x_1 + 400x_2 &\geq 40,000 \\ x_1, x_2 &\geq 0 \end{align*} \] #### **Let’s try to find the optimal solution (by corner points):** Let’s make the constraints equalities to find corner points. ##### **A) Set \( x_1 + x_2 = 140 \) and \( 150x_1 + 400x_2 = 40,000 \):** From \( x_1 + x_2 = 140 \), \( x_2 = 140 - x_1 \) Plug into second constraint: \[ 150x_1 + 400(140 - x_1) = 40,000 \\ 150x_1 + 56,000 - 400x_1 = 40,000 \\ -250x_1 = -16,000 \\ x_1 = 64 \] So \[ x_2 = 140 - 64 = 76 \] ##### **B) Test boundary values:** Let’s check if running only one refinery is feasible: If \( x_2 = 0 \), \( x_1 = 140 \): - Low-grade: \( 150 \times 140 = 21,000 < 40,000 \) (not sufficient) If \( x_1 = 0 \), \( x_2 = 140 \): - Low-grade: \( 400 \times 140 = 56,000 \geq 40,000 \) (OK) But cost: \( 30,000 \times 140 = \$4,200,000 \) Check if better than A) above. ##### **Calculate cost at A):** \[ Z = 25,000x_1 + 30,000x_2 = 25,000(64) + 30,000(76) = 1,600,000 + 2,280,000 = \$3,880,000 \] So, running both refineries for 64 and 76 days is cheaper. ##### **Check if running only refinery 2 for minimal days possible:** Set \( x_1 = 0 \), and \( x_2 \) such that all constraints are met. - High-grade: \( x_2 \geq 140 \) - Low-grade: \( 400x_2 \geq 40,000 \rightarrow x_2 \geq 100 \) So need at least 140 days. So, running only refinery 2 for 140 days costs \$4,200,000 (more expensive than previous solution). ##### **Check if running only refinery 1 for minimal days possible:** Not possible (low-grade constraint fails). --- ### **Final Answer** **Optimal Solution:** - Run Refinery 1 for **64 days** - Run Refinery 2 for **76 days** - **Minimum total cost:** \$**3,880,000** --- ## **Interpretation of Shadow Price and Reduced Cost** ### **Shadow Price (Dual Value):** - **Definition:** The shadow price for a constraint tells you **how much the total cost would decrease if you could increase the right-hand side of that constraint by one unit** (i.e., if you needed one more barrel of high-grade, medium-grade, or low-grade oil). - **In this scenario:** The **shadow price for the low-grade oil constraint** is **positive**, since the solution is determined by the intersection of the high-grade and low-grade constraints. This means **increasing the required low-grade oil by 1 barrel will increase the minimum total cost by the shadow price** (e.g., if the shadow price is \$75, then each extra barrel of low-grade oil ordered increases total cost by \$75). ### **Reduced Cost:** - **Definition:** The reduced cost for a variable tells you **how much the objective function would worsen (increase, in minimization) if you forced that variable to increase from zero**. - **In this scenario:** If either refinery was **not used at all** (i.e., its variable is zero), its reduced cost would indicate **how much more expensive it would be to start using it**. In our solution, both are used (variables are nonzero), so their reduced costs are zero. --- ## **Summary Table** | Refinery | Days Operated | Cost | |---------------|--------------|------------------| | Refinery 1 | 64 | \$1,600,000 | | Refinery 2 | 76 | \$2,280,000 | | **Total** | **140** | **\$3,880,000** | --- ### **Final Statement:** > **The company should operate Refinery 1 for 64 days and Refinery 2 for 76 days to meet all orders at minimum cost of \$3,880,000. The shadow price indicates how much more expensive it would be to increase the required oil by one barrel; the reduced cost shows the penalty of using a refinery not currently in use, which in this case is zero as both are used in the optimal solution.**