The Goal Programming Problem Below Was Solved with the Management $\ge$

Essay

The goal programming problem below was solved with the Management Scientist.
Min
P₁(d₁^-) + P₂(d₂⁺) + P₃(d₃^-)
s.t.
72x₁ + 38x₂ + 23x₃ - 20,000
.72x₁ -.76x₂ - .23x₃ + d₁^- - d₁⁺ = 0
x₃ + d₂^- - d₂⁺ = 150
38x₂ + d₃^- -d₃⁺ = 2000
x₁, x₂, x₃, d₁^-, d ₁⁺, d₂^-, d₂⁺, d₃^-, d₃⁺ $\ge$ 0
Partial output from three successive linear programming problems is given. For each problem, give the original objective function expression and its value, and list any constraints needed beyond those that were in the original problem. $$\begin{array}{l}\text { a. Objective Function Value } = 0.000\\\begin{array} { l c c } \text { Variable } & \text { Value } & \text { Reduced Cost } \\\text { D1MINUS } & 0.000 & 1.000 \\\text { X1 } & 52.632 & 0.000 \\\text { X2 } & 0.000 & 0.000 \\\text { X3 } & 150.000 & 0.000 \\\text { D1PLUS } & 3.395 & 0.000 \\\text { D2MINUS } & 0.000 & 0.000 \\\text { D2PLUS } & 0.000 & 0.000 \\\text { D3MINUS } & 0.000 & 0.000 \\\text { D3PLUS } & 0.000 & 0.000\end{array}\end{array}$$ $$\begin{array}{l}\text { b. Objective Function Value } = 0.000\\\begin{array} { l c c } \text { Variable } & \text { Value } & \text { Reduced Cost } \\\text { D2PLUS } & 0.000 & 1.000 \\\text { X1 } & 52.632 & 0.000 \\\text { X2 } & 0.000 & 0.000 \\\text { X3 } & 150.000 & 0.000 \\\text { D1MINUS } & 0.000 & 0.000 \\\text { D1PLUS } & 3.395 & 0.000 \\\text { D2MINUS } & 0.000 & 0.000 \\\text { D3MINUS } & 0.000 & 0.000 \\\text { D3PLUS } & 0.000 & 0.000\end{array}\end{array}$$ $$\begin{array}{l}\text { c. Objective Function Value } = 0.000\\\begin{array} { l c c } \text { Variable } & \text { Value } & \text { Reduced Cost } \\\text { D3MINUS } & 0.000 & 1.000 \\\text { X1 } & 52.632 & 0.000 \\\text { X2 } & 0.000 & 0.000 \\\text { X3 } & 150.000 & 0.000 \\\text { D1MINUS } & 0.000 & 0.000 \\\text { D1PLUS } & 3.395 & 0.000 \\\text { D2MINUS } & 0.000 & 0.000 \\\text { D2PLUS } & 0.000 & 0.000 \\\text { D3PLUS } & 0.000 & 0.000\end{array}\end{array}$$