Question 1161485: HiTec produces two models of electronic gadgets that use resistors, capacitors, and
chips. Let x1 and x2 be the amounts produced of Models 1 and 2, with profit pe unit Rs.3 and
Rs.4, respectively. The following are the LP model and its associated optimal simplex tableau.
Maximize z = 3x1 + 4x2
subject to
2x1 + 3x2 ≤ 1200 (Resistors)
2x1 + x2 ≤ 1000 (Capacitors)
4x2 ≤ 800 (Chips)
x1, x2 ≥ 0
( a )Find an optimum solution to the above problem.
( b )Determine the shadow prices for the resistors, capacitors, and chips constraints.
( c ) Determine the feasible ranges of the constraints RHS for the shadow prices obtained in (b).
( d )If the available number of resistors is increased to 1300 units, find the new optimum solution.
( e )If the available number of chips is reduced to 350 units, will you be able to determine the new
optimum solution directly from the given information? Explain.
( f ) A new contractor is offering to sell HiTec additional resistors at Rs.2 each, but only if HiTec
would purchase 200 units. Should HiTec accept this offer? Explain.
Answer by CPhill(2285)   (Show Source):
You can put this solution on YOUR website! To answer your questions accurately, we first need to establish the final optimal simplex tableau derived from your Linear Programming model.
### **LP Formulation**
$$\text{Maximize } z = 3x_1 + 4x_2$$
$$\text{Subject to:}$$
$$2x_1 + 3x_2 + s_1 = 1200 \quad \text{(Resistors constraint)}$$
$$2x_1 + x_2 + s_2 = 1000 \quad \text{(Capacitors constraint)}$$
$$4x_2 + s_3 = 800 \quad \text{(Chips constraint)}$$
$$x_1, x_2, s_1, s_2, s_3 \ge 0$$
At the optimal basic feasible solution, $x_1$ and $x_2$ are basic variables, and $s_3$ is basic:
* From Chips constraint: $4x_2 \le 800 \implies x_2 = 200$
* From Resistors constraint: $2x_1 + 3(200) = 1200 \implies 2x_1 = 600 \implies x_1 = 300$
* From Capacitors constraint: $s_2 = 1000 - 2(300) - 200 = 200$ (Capacitors have 200 unused units).
**Optimal Simplex Matrix Form ($B^{-1}$ matrix corresponding to slack variables $s_1, s_2, s_3$):**
$$B^{-1} = \begin{bmatrix} 1/2 & 0 & -3/8 \\ 0 & 1 & 0 \\ -1 & 0 & 1/4 \end{bmatrix}$$
---
### **( a ) Optimal Solution**
* **$x_1$ (Model 1 units):** $300$
* **$x_2$ (Model 2 units):** $200$
* **$s_1$ (Unused Resistors):** $0$
* **$s_2$ (Unused Capacitors):** $200$
* **$s_3$ (Unused Chips):** $0$
* **Maximum Profit ($z$):** $3(300) + 4(200) = \mathbf{Rs.\ 1700}$
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### **( b ) Shadow Prices (Dual Values)**
The shadow prices correspond to the marginal values (coefficients of the slack variables in the optimal $z$-row):
* **Resistors ($y_1$):** **Rs. 1.50** per resistor
* **Capacitors ($y_2$):** **Rs. 0.00** per capacitor (binding constraint has non-zero slack $s_2 = 200$)
* **Chips ($y_3$):** **Rs. 0.25** per chip
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### **( c ) Feasible RHS Ranges for Shadow Prices**
Using $B^{-1} \cdot b \ge 0$:
$$\begin{bmatrix} x_1 \\ s_2 \\ x_2 \end{bmatrix} = \begin{bmatrix} 1/2 & 0 & -3/8 \\ -1 & 1 & 1/4 \\ 0 & 0 & 1/4 \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$$
1. **Resistors RHS ($b_1 = 1200 + \Delta b_1$):**
* $x_1 = 300 + 0.5\Delta b_1 \ge 0 \implies \Delta b_1 \ge -600$
* $s_2 = 200 - 1.0\Delta b_1 \ge 0 \implies \Delta b_1 \le 200$
* **Feasible Range for Resistors ($b_1$):** ***
2. **Capacitors RHS ($b_2 = 1000 + \Delta b_2$):**
* $s_2 = 200 + \Delta b_2 \ge 0 \implies \Delta b_2 \ge -200$
* **Feasible Range for Capacitors ($b_2$):** *
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