Chapter 19, Problem 2P

a)

Summary Introduction

To solve: The linear programming problem and answer the questions.

Introduction:

Linear programming:

Linear programming is a mathematical modelling method where a linear function is maximized or minimized taking into consideration the various constraints present in the problem. It is useful in making quantitative decisions in business planning.

Explanation of Solution

Given information:

$\begin{array}{c}MinimizeZ=1.80S+2.20T\\ Subjectto:\\ 5S+8T\ge 200gr\left(Potassium\right)\\ 15S+6T\ge 240gr\left(Carbohydrate\right)\\ 4S+12T\ge 180gr\left(\mathrm{Pr}otein\right)\\ T\ge 10gr\left(T\right)\\ S,T\ge 0\left(Non-negativity\right)\end{array}$

Calculation of coordinates for each constraint and objective function:

Constraint 1:

$5S+8T\ge 200gr\left(Potassium\right)$

$\text{Substituting }S=0\text{ to find }T\text{,}$

$\begin{array}{c}5\left(0\right)+8T=200\\ 8T=200\\ T=\frac{200}{8}\\ T=25\end{array}$

$\text{Substituting }T=0\text{ to find }S\text{,}$

$\begin{array}{c}5S+8\left(0\right)=200\\ 5S=200\\ S=\frac{200}{5}\\ S=40\end{array}$

Constraint 2:

$15S+6T\ge 240gr\left(Carbohydrate\right)$

$\text{Substituting }S=0\text{ to find }T\text{,}$

$\begin{array}{c}15\left(0\right)+6T=240\\ 6T=240\\ T=\frac{240}{6}\\ T=40\end{array}$

$\text{Substituting }T=0\text{ to find }S\text{,}$

$\begin{array}{c}15S+6\left(0\right)=240\\ 15S=240\\ S=\frac{240}{15}\\ S=16\end{array}$

Constraint 3:

$4S+12T\ge 180gr\left(\mathrm{Pr}otein\right)$

$\text{Substituting }S=0\text{ to find }T\text{,}$

$\begin{array}{c}4\left(0\right)+12T=180\\ 12T=180\\ T=\frac{180}{12}\\ T=15\end{array}$

$\text{Substituting }T=0\text{ to find }S\text{,}$

$\begin{array}{c}4S+12\left(0\right)=180\\ 4S=180\\ S=\frac{180}{4}\\ S=45\end{array}$

Constraint 4:

$T\ge 10gr\left(T\right)$

$\text{Therefore }T=10.$

Objective function:

The problem is solved with iso-cost line method.

$\text{Let }1.80S+2.20T=99$

$\text{Substituting }S=0\text{ to find }T\text{,}$

$\begin{array}{c}1.80\left(0\right)+2.20T=99\\ 2.20T=99\\ T=\frac{99}{2.20}\\ T=45\end{array}$

$\text{Substituting }T=0\text{ to find }S\text{,}$

$\begin{array}{c}1.80S+2.20\left(0\right)=99\\ 1.80S=99\\ S=\frac{99}{1.80}\\ S=55\end{array}$

Graph:

(1) Optimal value of the decision variables and Z:

The coordinates for the cost line is (45, 55). The cost line is moved towards the origin. The lowest at which the cost line intersects in the feasible region will be the optimum solution. The following equation are solved as simultaneous equation to find optimum solution.

$5S+8T=200$ (1)

$15S+6T=240$ (2)

Solving (1) and (2) we get,

$S=8,T=20$

The values are substituted in the objective function to find the objective function value.

$\begin{array}{c}MinimizeZ=1.80\left(8\right)+2.20\left(20\right)\\ =14.40+44\\ =58.40\end{array}$

Optimal solution:

$\begin{array}{l}S=8\\ T=20\\ Z=58.4\end{array}$

(2)

None of the constraints are having slack. All ≤ constraints are binding.

(3)

Protein and T constraint have surplus.

Protein:

$\begin{array}{l}4\left(8\right)+12\left(20\right)\ge 180gr\\ 32+240\ge 180\\ 272\ge 180\end{array}$

The surplus is 92 (272 – 180).

T:

$\begin{array}{l}T\ge 10\\ 20\ge 10\end{array}$

The surplus is 10 (20– 10).

(4)

The protein constraint is redundant because, it does not intersect at any point in the feasible region.

b)

Summary Introduction

To solve: The linear programming problem and answer the questions.

Introduction:

Linear programming:

Linear programming is a mathematical modelling method where a linear function is maximized or minimized taking into consideration the various constraints present in the problem. It is useful in making quantitative decisions in business planning.

Explanation of Solution

Given information:

$\begin{array}{c}MinimizeZ=2x_1+3x_2\\ Subjectto:\\ 4x_1+2x_2\ge 20\left(D\right)\\ 2x_1+6x_2\ge 18\left(E\right)\\ x_1+2x_2\le 12\left(F\right)\\ x_1,x_2\ge 0\left(Non-negativity\right)\end{array}$

Calculation of coordinates for each constraint and objective function:

Constraint 1:

$4x_1+2x_2\ge 20\left(D\right)$

$\text{Substituting }{x}_1=0\text{ to find }{x}_2\text{,}$

$\begin{array}{c}4\left(0\right)+2x_2=20\\ 2x_2=20\\ x_2=\frac{20}{2}\\ x_2=10\end{array}$

$\text{Substituting }{x}_2=0\text{ to find }{x}_1\text{,}$

$\begin{array}{c}4x_1+2\left(0\right)=20\\ 4x_1=20\\ x_1=\frac{20}{4}\\ x_1=5\end{array}$

Constraint 2:

$2x_1+6x_2\ge 18\left(E\right)$

$\text{Substituting }{x}_1=0\text{ to find }{x}_2\text{,}$

$\begin{array}{c}2\left(0\right)+6x_2=18\\ 6x_2=18\\ x_2=\frac{18}{6}\\ x_2=3\end{array}$

$\text{Substituting }{x}_2=0\text{ to find }{x}_1\text{,}$

$\begin{array}{c}2x_1+6\left(0\right)=18\\ 2x_1=18\\ x_1=\frac{18}{2}\\ x_1=9\end{array}$

Constraint 3:

$x_1+2x_2\le 12\left(F\right)$

$\text{Substituting }{x}_1=0\text{ to find }{x}_2\text{,}$

$\begin{array}{c}\left(0\right)+2x_2=12\\ 2x_2=12\\ x_2=\frac{12}{2}\\ x_2=6\end{array}$

$\text{Substituting }{x}_2=0\text{ to find }{x}_1\text{,}$

$\begin{array}{c}{x}_1+2\left(0\right)=12\\ x_1=12\\ x_1=\frac{12}{1}\\ x_1=12\end{array}$

Objective function:

The problem is solved with iso-cost line method.

$\text{Let }2x_1+3x_2=24$

$\text{Substituting }{x}_1=0\text{ to find }{x}_2\text{,}$

$\begin{array}{c}2\left(0\right)+3x_2=24\\ 3x_2=24\\ x_2=\frac{24}{3}\\ x_2=8\end{array}$

$\text{Substituting }{x}_2=0\text{ to find }{x}_1\text{,}$

$\begin{array}{c}2x_1+3\left(0\right)=24\\ 2x_1=24\\ x_1=\frac{24}{2}\\ x_1=12\end{array}$

Graph:

(1) Optimal value of the decision variables and Z:

The coordinates for the cost line is (12, 8). The cost line is moved towards the origin. The lowest at which the cost line intersects in the feasible region will be the optimum solution. The following equation are solved as simultaneous equation to find optimum solution.

$4x_1+2x_2=20$ (1)

$2x_1+6x_2=18$ (2)

Solving (1) and (2) we get,

$x_1=4.2,x_2=1.6$

The values are substituted in the objective function to find the objective function value.

$\begin{array}{c}MinimizeZ=2\left(4.2\right)+3\left(1.6\right)\\ =8.4+4.8\\ =13.2\end{array}$

Optimal solution:

$\begin{array}{l}{x}_1=4.2\\ x_2=1.6\\ Z=13.2\end{array}$

(2)

Constraint F is having slack as shown below.

$\begin{array}{c}\left(4.2\right)+2\left(1.6\right)\le 12\\ 4.2+3.2\le 12\\ 7.4\le 12\end{array}$

The slack is 4.6 (12 – 7.4).

(3)

There are no surplus. D and E constraints with ≥ are binding.

(4)

There are no redundant constraints