#### Question

In this problem, we discuss how shadow prices can be interpreted for blending problems. If we define x6J = pounds of grade 6 oranges in juice x9J = pounds of grade 9 oranges in juice x6B = pounds of grade 6 oranges in bags x9B = pounds of grade 9 oranges in bags then the appropriate formulation is $\max z=0.45\left(x_{6J}+x_{9J}\right)+0.30\left(x_{6 B}+x_{9 B}\right)\\ \text{s.t.} x_{6J}+x_{6 B} \leq 120,000 \quad \text{(Grade 6 constraint)}\\ x_{9J}+x_{9B} \leq 100,000 \quad \text{(Grade 9 constraint)}\\ \frac{6 x_{6J}+9 x_{9J}}{x_{6J}+x_{9J}} \geq 8 \text{(Orange juice constraint)} \quad \text{(1)}\\ \frac{6 x_{6B} +9 x_{9B}} {x_{6 B}+x_{9 B}} \geq 7 \text{(Bags constraint)} \quad \text{(2)}\\ x_{6J}, x_{9J}, x_{6 B}, x_{9 B} \geq 0$ Constraints (1) and (2) are examples of blending constraints, because they specify the proportion of grade 6 and grade 9 oranges that must be blended to manufacture orange juice and bags of oranges. It would be useful to determine how a slight change in the standards for orange juice and bags of oranges would affect profit. At the end of this problem, we explain how to use the shadow prices of Constraints (1) and (2) to answer the following questions: a. Suppose that the average grade for orange juice is increased to 8.1. Assuming the current basis remains optimal, by how much would profits change? b. Suppose the average grade requirement for bags of oranges is decreased to 6.9. Assuming the current basis remains optimal, by how much would profits change? The shadow price for both (1) and (2) is −0.15. The optimal solution is x6J = 26,666.67, x9J = 53,333.33, x6B = 93,333.33, x9B = 46,666.67. To interpret the shadow prices of blending Constraints (1) and (2), we assume that a slight change in the quality standard for a product will not significantly change the quantity of the product that is produced. Now note that (1) may be written as $6 x_{6 J}+9 x_{9J} \geq 8(x_{6 J}+ x_{9J})$, or $-2 x_{6J}+x_{9 J} \geq 0$ If the quality standard for orange juice is changed to 8 + Δ, then (1) can be written as $6 x_{6 J}+9 x_{9J} \geq(8+\Delta)\left(x_{6J}+x_{9J}\right)$ or $-2 x_{6J}+x_{9 J} \geq \Delta\left(x_{6J},+x_{9J}\right)$ Because we are assuming that changing orange juice quality from 8 to 8 + Δ does not change the amount produced, x6J + x9J will remain equal to 80,000, and (1) will become $-2 x_{6 J}+x_{9 J} \geq 80,000 \Delta$ Using the definition of shadow price, now answer parts (a) and (b).

#### Solution

Verified#### Step 1

1 of 6$\text{\textcolor{#c34632}{\textbf{a:}}}$

From the LP:

$\begin{align*} B = \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ -2 & 1 & 0 & 0 \\ 0 & 0 & -1 & 2 \end{bmatrix} \end{align*}$

Since $rank(B) = 4$ , $\text{BV} = {x_{6J},x_{9J},x_{6B},x_{9B}}$.

$\begin{align*} c_{BV} = \begin{bmatrix} 0.45 & 0.45 & 0.3 & 0.3 \end{bmatrix} \quad \text{and } \quad b= \begin{bmatrix} 120,000\\ 100,000\\ 0\\ 0 \end{bmatrix} \end{align*}$

$\begin{align*} B^{-1} = \begin{bmatrix} \frac{-1}{3} & \frac{2}{3} & \frac{-2}{3} & \frac{-1}{3}\\ \frac{-2}{3} & \frac{4}{3} & \frac{-1}{3} & \frac{-2}{3}\\ \frac{4}{3} & \frac{-2}{3} & \frac{2}{3} & \frac{1}{3}\\ \frac{2}{3} & \frac{-1}{3} & \frac{1}{3} & \frac{2}{3} \end{bmatrix} \end{align*}$

Therefore

$\begin{align*} c_{BV} \cdot B^{-1} = \begin{bmatrix} \frac{3}{20} & \frac{3}{5} & \frac{-3}{20} & \frac{-3}{20} \end{bmatrix} \end{align*}$

Fort constraint (1):

$\begin{align*} 6x_{6J}+9x_{9J} & \geq 8.1\cdot (x_{6J}+x_{9J})\\ -2.1x_{6J}+0.9x_{9J} &\geq 0 \end{align*}$

So

$\begin{align*} a_{x_{6J}}= \begin{bmatrix} 1\\ 0\\ -2.1\\ 0 \end{bmatrix} \quad \text{, } a_{x_{9J}}= \begin{bmatrix} 0\\ 1\\ 0.9\\ 0 \end{bmatrix} \quad \text{and } c_{x_{6J}}=c_{x_{9J}}=0.45 \end{align*}$

Princing out $x_{6J}$:

$\begin{align*} \overline{c_{x_{6J}}} &= c_{BV} \cdot B^{-1} \cdot a_{x_{6J}} - c_{x_{6J}} &= \begin{bmatrix} \frac{3}{20} & \frac{3}{5} & \frac{-3}{20} & \frac{-3}{20} \end{bmatrix} \cdot \begin{bmatrix} 1\\ 0\\ -2.1\\ 0 \end{bmatrix} -0.45\\ &= \frac{3}{20}-2.1\cdot \frac{3}{5}-0.45 = -1.56 \$ \end{align*}$