data609hw6

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2. A rancher has determined that the minimum weekly nutritional requirements for an average-sized horse include 40lb of protein, 20lb of carbohydrates, and 45lb of roughage. These are obtained from the following sources in varying amounts at the prices indicated:

table

Formulate a mathematical model to determine how to meet the minimum nutritional requirements at minimum cost.

Amounts of all the different feed types are paramaters in the model.

Minimize \(Cost(H, T, F, P) = 1.8H + 3.5T + 0.4F + 1.0P\)

This is subjec to following:

Protein: \(0.5H + 1.0T + 2.0F + 6.0P >= 40.0\)

Carbs: \(2.0H + 4.0T + 0.5F + 1.0P >= 20.0\)

Roughage: \(5.0H + 2.0T + 1.0F + 2.5P >= 25.0\)

\(H,T,F,P >=0\)

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6. Use graphical analysis to Maximize 10x + 35y subject to:

8x + 6y <= 48 (board-feet of lumber)

4x = y <= 20 (hours of carpentry)

y >= 5 (demand)

x,y, >= 0 (nonnegativity)

library(ggplot2)

dfC1 <- data.frame(x=c(0,6), y=c(8,0))
dfC2 <- data.frame(x=c(0,5), y=c(20,0))
dfC3 <- data.frame(x=c(0,6), y=c(5,5))
dfC4x <- data.frame(x=c(0,0), y=c(0,8))
dfC4y <- data.frame(x=c(0,6), y=c(0,0))
region <- data.frame(x=c(0, 0, 2.25), y=c(5, 8, 5))

g1 <- ggplot() + geom_line(data=dfC1, aes(x=x, y=y), colour="red") + 
  geom_line(data=dfC2, aes(x=x, y=y), colour="blue") + 
  geom_line(data=dfC3, aes(x=x, y=y), colour="green") + 
  geom_line(data=dfC4x, aes(x=x, y=y), colour="purple") + 
  geom_line(data=dfC4y, aes(x=x, y=y), colour="purple") + 
  geom_polygon(data=region, aes(x=x, y=y), fill="orange") 

g1

The extreme points are:

The intersection of constraint 1 and 5 at point (0,8)

The intersection of constraint 1 and 3 at point (2.25, 5)

The intersection of constraint 3 and 5 at point (0,5)

Plug them into the objective equation : \((10 * x) + (35 * y)\) to find the maximum which is (0,8) with value of 280.0.

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6. Using the method in section 7.3, Maximize 10x + 35y subject to:

8x + 6y <= 48 (board-feet of lumber)

4x + y <= 20 (hours of carpentry)

y >= 5 (demand)

x,y, >= 0 (nonnegativity)

Decision Variables to format xi

8x1 + 6x2 <= 48 (board-feet of lumber)

4x1 + x2 <= 20 (hours of carpentry)

x2 >= 5 (demand)

x1,x2 >= 0 (nonnegativity)

Add yi variables

8x1 + 6x2 + y1 = 48 (board-feet of lumber)

4x1 + x2 + y2 = 20 (hours of carpentry)

x2 = 5 + y3(demand)

x1,x2, y1, y2, y3 >= 0 (nonnegativity)

Iterate through the variable combinations, set 4x1 + x2 + y2 = 20 to 0.

x1,x2 = 0, x2 violates constraint 3

when x1,y1 = 0

6x2 = 48 (board-feet of lumber)

x2 + y2 = 20 (hours of carpentry)

x2 = 5 + y3 (demand)

Results:

x2 = 8 y2 = 12 y3 = 3

The feasible point at (0,8)

When x1, y2 = 0

6x2 + y1 = 48

x2 = 20

x2 = 5

Results: x2 = 20 y3 =15 y1 = -52.

Solution is not possible, negative

When x1,y3=0

6x2 + y1 = 48

x2 + y2 = 20

x2 = 5

Results:

x2 = 5 y2 = 15 y1 = 18

Feasible points at (0,5)

x2,y1 = 0, constraint 3 is violated

y1,y2 = 0. consraint 3 is violated

When y1,y3 = 0

8x1 + 6x2 = 48

4x1 + x2 + y2 =20

x2 = 5

Results:

x2= 5 x1 = 2.25 y2=6

Feasible points at (2.25,5)

When y2,y3 = 0

8x1 + 6x2 + y1 = 48

4x1 + x2 = 20

x2 = 5

Results:

x2 = 5 x1 = 3.75 y1 = -12

Solution is not possible, negative

Enter three possible solutions into \((10 * x) + (35 * y)\)

Get the same answer, find the maximum which is (0,8) with value of 280.0.