Let's go through the problem step by step as required. --- ## **1. Formulate an LP Model** Let: - \( x \) = number of STD bicycles produced - \( y \) = number of DEL bicycles produced ### **Objective Function:** Maximize **Total profit:** \[ \text{Maximize } Z = 100x + 125y \] ### **Constraints:** **Frames:** \[ x + y \leq 1400 \] **Add-ons (Hours):** \[ 1x + 2y \leq 2000 \] **STD Controls:** \[ x \leq 1000 \] **DEL Controls:** \[ y \leq 800 \] **Assembly (Hours):** \[ 0.33x + 1y \leq 900 \] **Non-negativity:** \[ x \geq 0,\, y \geq 0 \] --- ## **2. Assumptions** - **Linearity:** All relationships (profit and resources) are linear. - **Divisibility:** Can produce fractional bicycles (as per problem statement). - **Certainty:** All coefficients (profits, resource usage, availability) are known with certainty. - **Non-negativity:** Can't produce a negative number of bicycles. - **Resource usage:** Each type of bicycle uses the stated amount of each resource. --- ## **3. Solve Graphically** We'll find the feasible region and check the intersection points (since the number of variables is 2). ### **Step 1: List the constraints** 1. \( x + y \leq 1400 \) 2. \( x + 2y \leq 2000 \) 3. \( x \leq 1000 \) 4. \( y \leq 800 \) 5. \( 0.33x + y \leq 900 \) 6. \( x, y \geq 0 \) ### **Step 2: Find intersection points (corners)** Let’s solve for intersection points: #### **A. Intersection of \( x + y = 1400 \) and \( x + 2y = 2000 \):** Subtract the first from the second: - \( (x + 2y) - (x + y) = 2000 - 1400 \implies y = 600 \) - Put \( y = 600 \) in \( x + y = 1400 \implies x = 800 \) So, **Point 1:** \( (800, 600) \) #### **B. Intersection of \( x + y = 1400 \) and \( y = 800 \):** - \( x + 800 = 1400 \implies x = 600 \) - So, **Point 2:** \( (600, 800) \) #### **C. Intersection of \( x + 2y = 2000 \) and \( y = 800 \):** - \( x + 2(800) = 2000 \implies x = 400 \) - So, **Point 3:** \( (400, 800) \) #### **D. Intersection of \( x = 1000 \) and \( 0.33x + y = 900 \):** - \( 0.33(1000) + y = 900 \implies 330 + y = 900 \implies y = 570 \) - **Point 4:** \( (1000, 570) \) #### **E. Intersection of \( x + y = 1400 \) and \( 0.33x + y = 900 \):** Subtract the second from the first: - \( (x + y) - (0.33x + y) = 1400 - 900 \implies 0.67x = 500 \implies x = 746.27 \) - \( y = 1400 - 746.27 = 653.73 \) - **Point 5:** \( (746.27, 653.73) \) #### **F. Intersection of \( x = 1000 \) and \( x + y = 1400 \):** - \( 1000 + y = 1400 \implies y = 400 \) - **Point 6:** \( (1000, 400) \) #### **G. Intersection of \( x = 1000 \) and \( x + 2y = 2000 \):** - \( 1000 + 2y = 2000 \implies 2y = 1000 \implies y = 500 \) - **Point 7:** \( (1000, 500) \) #### **H. Intersection of \( x = 1000 \) and \( y = 800 \):** - **Point 8:** \( (1000, 800) \) (but check if it satisfies all constraints) #### **I. Intersection of \( y = 800 \) and \( 0.33x + y = 900 \):** - \( 0.33x + 800 = 900 \implies 0.33x = 100 \implies x = 303.03 \) - **Point 9:** \( (303.03, 800) \) #### **J. Intersection of \( x = 1000 \), \( x + y = 1400 \), \( x + 2y = 2000 \), \( y \leq 800 \), \( 0.33x + y = 900 \)** Check all intersection points above for feasibility. ### **Step 3: Check which points are feasible (i.e., satisfy all constraints)** Let's check the most promising ones: - **(800, 600):** - \( 800 + 600 = 1400 \leq 1400 \) ✔️ - \( 800 + 2*600 = 2000 \leq 2000 \) ✔️ - \( 800 \leq 1000 \) ✔️ - \( 600 \leq 800 \) ✔️ - \( 0.33*800 + 600 = 264 + 600 = 864 \leq 900 \) ✔️ - **(1000, 400):** - \( 1000 + 400 = 1400 \leq 1400 \) ✔️ - \( 1000 + 2*400 = 1800 \leq 2000 \) ✔️ - \( 1000 \leq 1000 \) ✔️ - \( 400 \leq 800 \) ✔️ - \( 0.33*1000 + 400 = 330 + 400 = 730 \leq 900 \) ✔️ - **(746.27, 653.73):** - \( 746.27 + 653.73 = 1400 \leq 1400 \) ✔️ - \( 746.27 + 2*653.73 = 746.27 + 1307.46 = 2053.73 \) **> 2000** ✖️ - **(303.03, 800):** - \( 303.03 + 800 = 1103.03 \leq 1400 \) ✔️ - \( 303.03 + 2*800 = 303.03 + 1600 = 1903.03 \leq 2000 \) ✔️ - \( 303.03 \leq 1000 \) ✔️ - \( 800 \leq 800 \) ✔️ - \( 0.33*303.03 + 800 = 99.9999 + 800 = 899.9999 \leq 900 \) ✔️ - **(400, 800):** - \( 400 + 800 = 1200 \leq 1400 \) ✔️ - \( 400 + 2*800 = 400 + 1600 = 2000 \leq 2000 \) ✔️ - \( 400 \leq 1000 \) ✔️ - \( 800 \leq 800 \) ✔️ - \( 0.33*400 + 800 = 132 + 800 = 932 > 900 \) ✖️ - **(1000, 570):** - \( 1000 + 570 = 1570 > 1400 \) ✖️ So, the **feasible points** are **(800, 600), (1000, 400), (303.03, 800)** ### **Step 4: Calculate objective function at these points** - (800, 600): \( 100*800 + 125*600 = 80,000 + 75,000 = 155,000 \) - (1000, 400): \( 100*1000 + 125*400 = 100,000 + 50,000 = 150,000 \) - (303.03, 800): \( 100*303.03 + 125*800 = 30,303 + 100,000 = 130,303 \) ### **Step 5: Conclude the optimal solution** **The optimal solution is:** - \( x = 800, y = 600 \) - **Maximum profit = €155,000** --- ## **4. Slack and Tight Constraints** At \( (800, 600) \): - **Frames:** \( 800 + 600 = 1400 \) → *Binding (tight)* - **Add-ons:** \( 800 + 2*600 = 2000 \) → *Binding (tight)* - **STD controls:** \( 800 \leq 1000 \), slack = 200 - **DEL controls:** \( 600 \leq 800 \), slack = 200 - **Assembly:** \( 0.33*800 + 600 = 864 \leq 900 \), slack = 36 **Tight constraints:** Frames, Add-ons **Slack constraints:** STD controls (200 left), DEL controls (200 left), Assembly (36 left) --- ## **5. Solve with Excel Solver** - **Set Objective:** Maximize \( 100x + 125y \) - **Variable cells:** \( x, y \) - **Constraints:** - \( x + y \leq 1400 \) - \( x + 2y \leq 2000 \) - \( x \leq 1000 \) - \( y \leq 800 \) - \( 0.33x + y \leq 900 \) - \( x, y \geq 0 \) Result should match previous solution: \( x = 800, y = 600 \), profit = €155,000 --- ## **6. Shadow Prices (Dual Values)** Shadow price = increase in objective value per unit increase in a tight constraint. - If you increase the availability of **Frames** or **Add-ons** by 1 unit, the optimal profit will increase by the shadow price for that constraint. (Found using Solver Sensitivity Report.) --- ## **7. Effect of Changing STD Profit** - If profit per STD bicycle is 75 EUR: - Objective function: \( 75x + 125y \) - If 50 EUR: - Objective: \( 50x + 125y \) - If 40 EUR: - Objective: \( 40x + 125y \) The lower profit means the solution will shift toward producing more DEL bicycles, possibly up to the DEL controls or assembly constraint. --- ## **8. Integer Solution with Supply Chain Constraint** Additional constraint: - Supply chain capacity: No more than 1400 STD bikes. But supply chain capacity of 4 STD = 3 DEL - So: \( x/4 + y/3 \leq 350 \) - Also, \( x \) and \( y \) must be integers. **Excel Solver:** - Add \( x/4 + y/3 \leq 350 \) - Set Integer constraint on \( x, y \) --- ## **9. Graphical Illustration** Plot the feasible region with the new supply chain constraint and integer values. --- ### **Summary Table** | Solution | STD (x) | DEL (y) | Profit (€) | |-----------------|---------|---------|------------| | Relaxed (LP) | 800 | 600 | 155,000 | | Integer & Supply| [found via Excel] | [found via Excel] | [found via Excel] | --- **If you need the Excel setup or the graphical plot, let me know!** If you want the integer solution for part 8 (with the supply chain constraint) worked out explicitly, just ask!