ISDA609: Mathematical Modeling Techniques for Data Analytics \[ \ \] Assignment:06
Mohamed Elmoudni
March 12, 2016
Page 251, Question #2.
Nutritional Requirements – 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:
| Protein (lb) | Carbs (lb) | Roughage (lb) | Cost | |
|---|---|---|---|---|
| Hay (per bale) | 0.5 | 2.0 | 5.0 | $1.80 |
| Oats (per sack) | 1.0 | 4.0 | 2.0 | $3.50 |
| Feeding Blocks (per block) | 2.0 | 0.5 | 1.0 | $0.40 |
| High Protein Concentrate (per sack) | 6.0 | 1.0 | 2.5 | $1.00 |
| Requirements per horse (per week) | 40.0 | 20.0 | 45.0 |
Let’s \(x_1\) = the amount of Hay per bale
Let’s \(x_2\) = the amount of the amount of Oats per sack
Let’s \(x_3\) = the amount of Feeding Blocks (per block)
Let’s \(x_4\) = the amount of High Protein Concentrate (per sack)
Since each unit of cost of (Hay, Oats, Feeding Blocks, High Protein Concentrate) is (1.8, 3.5, 0.4, 1.0), then the total cost in dollars is:\[C= 1.8x_1 + 3.5x_2+0.4x_3+x_4\] which is what we want to minimize.
Next, we need to consider the minimum requirement amount of protein.
\[0.5x_1+x_2+2x_3+6x_4 >= 40\]
Next, we need to consider the minimum requirement amount of carbohydrates.
\[2x_1+4x_2+0.5x_3+x_4 >= 20\]
Next, we need to consider the minimum requirement amount of Roughage.
\[5x_1+2x_2+x_3+2.5x_4 >= 45\]
Hence the mathematical model that will help us determine how to meet the minimum nutrition requirements at minimum cost is as follow:
\[ minimize \qquad C= 1.8x_1 + 3.5x_2+0.4x_3+x_4\]
subject to: \[0.5x_1+x_2+2x_3+6x_4 >= 40\] \[2x_1+4x_2+0.5x_3+x_4 >= 20\] \[5x_1+2x_2+x_3+2.5x_4 >= 45\] where \[ x_1, x_2,x_3,x_4 >= 0\]
Page 264, Question 6
Minimize 10x + 35y
subject to:
\[ 8x + 6y \leq 48 \quad (Board-feet \ of \ lumber)\] \[ 4x + y \leq 20 \quad (hours \ of \ carpentry)\] \[ y \geq 5 \quad (demand)\] \[ x, y \geq 0 \quad ( nonnegativity) \]
Now let’s plot the main constraints
library(ggplot2)## Warning: package 'ggplot2' was built under R version 3.2.3library(reshape2)
x <- seq(0,10,by=0.5)
lumber= 8- (4/3) *x
labor = 20 - 4*x
demand = 5
data1 <- data.frame(x = x, lumber, labor, demand )
data2 <- melt(data1, id.vars="x", value.name="y")
ggplot(data2, aes(x=x, y=y, color=variable)) + geom_line()
Now lets find the convex set for the constraints
plot(c(0, 10), c(0, 20), type='n')
abline(8, -4/3)
abline(20,-4)
abline(5,0 )
conditions <- function(x,y) {
c1 <- (y <= 8 - x * (4/3))
c2 <- (y <= 20 - 4*x )
c3 <- (y >= 5)
return(c1 & c2 & c3 )
}
x <- seq(0,10,length=1000)
y <- seq(0,20,length=1000)
z <- expand.grid(x=x,y=y)
z <- z[conditions(z$x,z$y),]
points(z, col="red")
From the graph, the area colored in red shows three extreme points: (0,5), (0,8), and (2.4,5).
Hence for:
- (0,5) the objective function value is: 175
- (0,8) the objective function value is: 280
- (2.4, 5) the objective function value is: 199
Therefore, the maximum value happens when x = 0 and y = 8
Page 268, Question 6 Algebraic method
Minimize 10x + 35y
subject to:
\[ 8x + 6y \leq 48 \quad (Board-feet \ of \ lumber)\] \[ 4x + y \leq 20 \quad (hours \ of \ carpentry)\] \[ y \geq 5 \quad (demand)\] \[ x, y \geq 0 \quad ( nonnegativity) \]
Page 268, Answer 6 Algebraic method
First we need to add slack \(r_i\) and surplus \(s_i\) variable accordingly.
Hence our systemof equation will be as follow:
\[ 8x + 6y + r_1 = 48 \quad (Board-feet \ of \ lumber)\] \[ 4x + y + r_2 = 20 \quad (hours \ of \ carpentry)\] \[ y - s_1 = 5 \quad (demand)\] \[ x, y, r_1, r_2, s_1 \geq 0 \quad ( nonnegativity) \]
1- let’s begin by assigning x and y to zero. our system becomes:
\[ 0 + 0 + r_1 = 48 \quad (Board-feet \ of \ lumber)\] \[ 0 + 0 + r_2 = 20 \quad (hours \ of \ carpentry)\] \[ 0 - s_1 = 5 \quad (demand)\] \[ x, y, r_1, r_2, s_1 \geq 0 \quad ( nonnegativity) \]
We get \(r_1 = 48,\) \(r_2= 20,\) \(s_1 = -5\)
Since \(s_1\) is less than zero, then x = 0 and y = 0 is not in feasible vertex.
2- Next we assign x= 0 and \(r_1=0\), our system becomes:
\[ 0 + 6y + 0 = 48 \quad (Board-feet \ of \ lumber)\] \[ 0 + y + r_2 = 20 \quad (hours \ of \ carpentry)\] \[ y - s_1 = 5 \quad (demand)\] \[ x, y, r_1, r_2, s_1 \geq 0 \quad ( nonnegativity) \]
We get, y = 8, \(r_2\)= 12, and \(s_1\)=3. since all variable are greater than zero, the point call it A=(x=0, y=8) is in feasible vertex. \(\qquad point: A\)
3- Next we assign x = 0 and \(r_2 =0\), our system becomes:
\[ 0 + 6y + r_1 = 48 \quad (Board-feet \ of \ lumber)\] \[ 0 + y + 0 = 20 \quad (hours \ of \ carpentry)\] \[ y - s_1 = 5 \quad (demand)\] \[ x, y, r_1, r_2, s_1 \geq 0 \quad ( nonnegativity) \]
We get, y = 20, \(s_1=-15\), and x =-72. since \(s_1\) and x ar eless than zero, then the point x,y is not in feasible vertex.
4- Next we assign \(r_1 = 0\) and \(r_2 =0\), our system becomes:
\[ 8x + 6y + 0 = 48 \quad (Board-feet \ of \ lumber)\] \[ 4x + y + 0 = 20 \quad (hours \ of \ carpentry)\] \[ y - s_1 = 5 \quad (demand)\] \[ x, y, r_1, r_2, s_1 \geq 0 \quad ( nonnegativity) \]
we get, y = 2, and \(s_1=-3\) since \(s_2\) is less than zero, then the point x,y is not in feasible vertex.
5- Next we assign \(r_1 = 0\) and \(s_1 =0\), our system becomes:
\[ 8x + 6y + 0 = 48 \quad (Board-feet \ of \ lumber)\] \[ 4x + y + r_2 = 20 \quad (hours \ of \ carpentry)\] \[ y - 0 = 5 \quad (demand)\] \[ x, y, r_1, r_2, s_1 \geq 0 \quad ( nonnegativity) \]
we get y = 5, x = 9/4, and \(r_2=6\). since all variable are greater than zero, then point call it B(x=9/4,y=5) is in feasible vertex. \(\qquad point: B\)
6- Next we assign \(r_2 = 0\) and \(s_1 =0\), our system becomes:
\[ 8x + 6y + r_1 = 48 \quad (Board-feet \ of \ lumber)\] \[ 4x + y + 0 = 20 \quad (hours \ of \ carpentry)\] \[ y - 0 = 5 \quad (demand)\] \[ x, y, r_1, r_2, s_1 \geq 0 \quad ( nonnegativity) \]
we get y = 5, x = 15/4, and \(r_1=-12\). since \(r_1\) is less than zero, then the point x,y is not in feasible vertex.
7- next we Next we assign x = 0 and \(s_1 =0\), our system becomes:
\[ 0 + 6y + r_1 = 48 \quad (Board-feet \ of \ lumber)\] \[ 0 + y + r_2 = 20 \quad (hours \ of \ carpentry)\] \[ y - 0 = 5 \quad (demand)\] \[ x, y, r_1, r_2, s_1 \geq 0 \quad ( nonnegativity) \] we get y = 5, \(r_2= 15\) , and \(r_1=18\). Since all variable are greater than zero, the point call it C(x=0,y=5) is in feasible vertex. \(\qquad point: C\)
Now, let’s find the cost function value at points:
- A(0,8) the objective function value is: 280
- B(2.4, 5) the objective function value is: 199
- C(0,5) the objective function value is: 175
Therefore, the maximum value happens at point A, when x = 0 and y = 8
Page_278_6_simplex<- readLines("C:/CUNY/Courses/IS609/Assignment609/Assignment06/simplex-289_99_3.txt", n=-1)
Page_278_6_simplex## [1] "The initial system is found by converting the inequality constraints into equality constraints by adding a slack variable:"
## [2] ""
## [3] "Maximize P = 10 x + 35y + 0s_1 + 0 s_2 + 0 s_3 subject to:"
## [4] "8 x + 6y + 1s_1 = 48"
## [5] "4 x + 1 y + 1s_2 = 20"
## [6] "0 x + 1 y -1s_3 = 5"
## [7] "x, y, s_1, s_2, s_3, ? 0"
## [8] " "
## [9] "Tableau 0"
## [10] "X y s1 s2 s3 P RHS Ratio"
## [11] "8 6 1 0 0 0 48 48/6 = 8"
## [12] "4 1 0 1 0 0 20 20/1= 20"
## [13] "0 1 0 0 -1 0 5 5/1=5"
## [14] "-10 -35 0 0 0 1 0"
## [15] ""
## [16] ""
## [17] "We have to move the smallest distance possible to stay within the feasible region. The pivot row is the row that has the smallest non-negative ratio. Row #3, with pivot 1"
## [18] "Hence, entries for pivot row (Row 3):"
## [19] "5 / 1 = 5"
## [20] "0 / 1 = 0"
## [21] "1 / 1 = 1"
## [22] "0 / 1 = 0"
## [23] "0 / 1 = 0"
## [24] "-1 / 1 = -1"
## [25] "1 / 1 = 1"
## [26] ""
## [27] ""
## [28] "Row 1:"
## [29] "48 - (6 * 5) = 18"
## [30] "8 - (6 * 0) = 8"
## [31] "6 - (6 * 1) = 0"
## [32] "1 - (6 * 0) = 1"
## [33] "0 - (6 * 0) = 0"
## [34] "0 - (6 * -1) = 6"
## [35] "0 - (6 * 1) = -6"
## [36] ""
## [37] "Row 2:"
## [38] "20 - (1 * 5) = 15"
## [39] "4 - (1 * 0) = 4"
## [40] "1 - (1 * 1) = 0"
## [41] "0 - (1 * 0) = 0"
## [42] "1 - (1 * 0) = 1"
## [43] "0 - (1 * -1) = 1"
## [44] "0 - (1 * 1) = -1"
## [45] ""
## [46] "Calculate the Z line:"
## [47] "(0) + (0 * 18) + (0 * 15) + (-35 * 5) = -175"
## [48] "(10) + (0 * 8) + (0 * 4) + (-35 * 0) = 10"
## [49] "(35) + (0 * 0) + (0 * 0) + (-35 * 1) = 0"
## [50] "(0) + (0 * 1) + (0 * 0) + (-35 * 0) = 0"
## [51] "(0) + (0 * 0) + (0 * 1) + (-35 * 0) = 0"
## [52] "(0) + (0 * 6) + (0 * 1) + (-35 * -1) = 35"
## [53] ""
## [54] ""
## [55] ""
## [56] ""
## [57] ""
## [58] ""
## [59] "Tableau 1"
## [60] "x y s1 s2 s3 P RHS Ratio"
## [61] "8 0 1 0 6 0 18 18/6 = 3"
## [62] "4 0 0 1 1 0 15 15/1= 15"
## [63] "0 1 0 0 -1 0 5 5/1=5"
## [64] "-10 0 0 0 -35 1 -175"
## [65] ""
## [66] ""
## [67] ""
## [68] ""
## [69] ""
## [70] "Pivot row (Row 1):"
## [71] "18 / 6 = 3"
## [72] "8 / 6 = 1.3333333333333"
## [73] "0 / 6 = 0"
## [74] "1 / 6 = 0.16666666666667"
## [75] "0 / 6 = 0"
## [76] "6 / 6 = 1"
## [77] ""
## [78] ""
## [79] "Row 2:"
## [80] "15 - (1 * 3) = 12"
## [81] "4 - (1 * 1.3333333333333) = 2.6666666666667"
## [82] "0 - (1 * 0) = 0"
## [83] "0 - (1 * 0.16666666666667) = -0.16666666666667"
## [84] "1 - (1 * 0) = 1"
## [85] "1 - (1 * 1) = 0"
## [86] ""
## [87] ""
## [88] ""
## [89] ""
## [90] "Row 3:"
## [91] "5 - (-1 * 3) = 8"
## [92] "0 - (-1 * 1.3333333333333) = 1.3333333333333"
## [93] "1 - (-1 * 0) = 1"
## [94] "0 - (-1 * 0.16666666666667) = 0.16666666666667"
## [95] "0 - (-1 * 0) = 0"
## [96] "-1 - (-1 * 1) = 0"
## [97] ""
## [98] ""
## [99] "Row Z:"
## [100] "175 - (-35 * 3) = 280"
## [101] "-10 - (-35 * 1.3333333333333) = 36.666666666667"
## [102] "0 - (-35 * 0) = 0"
## [103] "0 - (-35 * 0.16666666666667) = 5.8333333333333"
## [104] "0 - (-35 * 0) = 0"
## [105] "-35 - (-35 * 1) = 0"
## [106] ""
## [107] "Tableau 2"
## [108] "x y s1 s2 s3 P RHS"
## [109] "4/3 0 1/6 0 1 0 3"
## [110] "4 0 -1/6 1 0 0 12"
## [111] "4/3 1 1/6 0 0 0 8"
## [112] "116/3 0 35/6 0 0 1 280"
## [113] ""
## [114] ""
## [115] ""
## [116] "As we can see there are no more negative coefficients in the last row. Hence, we have reached the "
## [117] "The optimal solution value is Z = 280 at:"
## [118] "Independent variables x = 0 and s_1 = 0. "
## [119] ""
## [120] "Hence replacing s_1=0 in 8x + 6y +s_1 = 48, we get 6y=48, hence y = 8. "
## [121] "Therefore, the optimal solution value is Z = 280"
## [122] "X1 = 0"
## [123] "X2 = 8"
## [124] ""
## [125] ""Page 295, question 3
Use the Curve Fitting Criterion to minimize the sum of the absolute deviations for the following models and data set:
\(y = ax\)
\(y = ax^2\)
\(y = ax^3\)
Page 295, Answer 3
Model \(y = ax\)
1- first define the interval a and b. i ll chos eteh interval 0, 50 2- set r = .618 and define th etest points x1 and x2 x1= a + ( 1- r)(b -a) x2 = a + r(b-a)
we wil find the function to be minimized
f(x) = abs(8 - 7c) + abs(41 - 14c) + abs(133 - 21c) + abs(250 - 28c) + abs(280 - 35c) + abs(297 - 42c)
x <- c(7,14,21,28,35,42)
y <- c(8,41,133,250,280,297)
df1 <- data.frame(x=x, y=y)
a<- 0
b<- 50
r<- ((1+sqrt(5))/2)-1
c1 = a + (1 - r)*(b - a)
c2 = a + r*(b -a)
cost <- function(a, df){
sum(abs(df$y - a*(df$x)))
}
f_c1<- cost(c1,df1)
f_c2<- cost(c2,df1)
table1 <- data.frame(a=a,b=b,c1=c1,c2=c2,f_c1=f_c1,f_c2=f_c2)
t <- 0.02
d <- 1
while(d > t){
if(f_c1<=f_c2){
b <- c2
c1 <- a+(1-r)*(b-a)
c2 <- a+r*(b-a)
f_c1 <- cost(c1,df1)
f_c2 <- cost(c2,df1)
table1 <- rbind(table1,c(a,b,c1,c2,f_c1,f_c2))
d <- b-a
} else{
a <- c1
c1 <- a+(1-r)*(b-a)
c2 <- a+r*(b-a)
f_c1 <- cost(c1,df1)
f_c2 <- cost(c2,df1)
table1 <- rbind(table1,c(a,b,c1,c2,f_c1,f_c2))
d <- b-a
}
}
table1## a b c1 c2 f_c1 f_c2
## 1 0.000000 50.000000 19.098301 30.901699 1798.4502 3533.5498
## 2 0.000000 30.901699 11.803399 19.098301 726.0996 1798.4502
## 3 0.000000 19.098301 7.294902 11.803399 204.1929 726.0996
## 4 0.000000 11.803399 4.508497 7.294902 437.6078 204.1929
## 5 4.508497 11.803399 7.294902 9.016994 204.1929 316.4982
## 6 4.508497 9.016994 6.230590 7.294902 256.7881 204.1929
## 7 6.230590 9.016994 7.294902 7.952683 204.1929 218.0063
## 8 6.230590 7.952683 6.888371 7.294902 211.0326 204.1929
## 9 6.888371 7.952683 7.294902 7.546152 204.1929 209.4692
## 10 6.888371 7.546152 7.139621 7.294902 200.9320 204.1929
## 11 6.888371 7.294902 7.043652 7.139621 201.2499 200.9320
## 12 7.043652 7.294902 7.139621 7.198933 200.9320 202.1776
## 13 7.043652 7.198933 7.102964 7.139621 200.1622 200.9320
## 14 7.043652 7.139621 7.080309 7.102964 199.6865 200.1622
## 15 7.043652 7.102964 7.066307 7.080309 199.8227 199.6865
## 16 7.066307 7.102964 7.080309 7.088962 199.6865 199.8682
## 17 7.066307 7.088962 7.074960 7.080309 199.5742 199.6865
## 18 7.066307 7.080309 7.071655 7.074960 199.5048 199.5742l<- length(table1$a)
table1$a[l] ## [1] 7.066307table1$b[l]## [1] 7.080309c1_optimal <- (table1$a[l]+ table1$b[l] ) /2
c1_optimal## [1] 7.073308f_c1_optimal<- cost(c1_optimal,df1)
f_c1_optimal## [1] 199.5395##############Model \(y = ax^2\)
We wil find the function to be minimized:
f(x) = abs(8 - 7^2 c) + abs(41 - 14^2 c) + abs(133 - 21^2 * c) + 250 - 28^2 c) + abs(280 - 35^2 c) + abs(297 - 42^2 *c)
x <- c(7,14,21,28,35,42)
y <- c(8,41,133,250,280,297)
df2 <- data.frame(x=x, y=y, x2=x^2)
a<- 0
b<- 50
r<- ((1+sqrt(5))/2)-1
c1 = a + (1 - r)*(b - a)
c2 = a + r*(b -a)
cost2 <- function(a, df){
sum(abs(df$y - a*(df$x2) ))
}
f_c1<- cost2(c1,df2)
f_c2<- cost2(c2,df2)
table2 <- data.frame(a=a,b=b,c1=c1,c2=c2,f_c1=f_c1,f_c2=f_c2)
t <- 0.02
d <- 1
while(d > t){
if(f_c1<=f_c2){
b <- c2
c1 <- a+(1-r)*(b-a)
c2 <- a+r*(b-a)
f_c1 <- cost2(c1,df2)
f_c2 <- cost2(c2,df2)
table2 <- rbind(table2,c(a,b,c1,c2,f_c1,f_c2))
d <- b-a
} else{
a <- c1
c1 <- a+(1-r)*(b-a)
c2 <- a+r*(b-a)
f_c1 <- cost2(c1,df2)
f_c2 <- cost2(c2,df2)
table2 <- rbind(table2,c(a,b,c1,c2,f_c1,f_c2))
d <- b-a
}
}
l<- length(table2$a)
table2$a[l] ## [1] 0.2232466table2$b[l]## [1] 0.2372483c2_optimal <- (table2$a[l]+ table2$b[l] ) /2
c2_optimal## [1] 0.2302474f_c2_optimal<- cost2(c2_optimal,df2)
f_c2_optimal## [1] 219.5671##############Model \(y = ax^3\)
We wil find the function to be minimized:
f(x) = abs(8 - 7^3 c) + abs(41 - 14^3 c) + abs(133 - 21^3 * c) + 250 - 28^3 c) + abs(280 - 35^3 c) + abs(297 - 42^3 *c)
x <- c(7,14,21,28,35,42)
y <- c(8,41,133,250,280,297)
df3 <- data.frame(x=x, y=y, x3=x^3)
a<- 0
b<- 50
r<- ((1+sqrt(5))/2)-1
c1 = a + (1 - r)*(b - a)
c2 = a + r*(b -a)
cost3 <- function(a, df){
sum(abs(df$y - a*(df$x3)))
}
f_c1<- cost3(c1,df3)
f_c2<- cost3(c2,df3)
table3 <- data.frame(a=a,b=b,c1=c1,c2=c2,f_c1=f_c1,f_c2=f_c2)
t <- 0.02
d <- 1
while(d > t){
if(f_c1<=f_c2){
b <- c2
c1 <- a+(1-r)*(b-a)
c2 <- a+r*(b-a)
f_c1 <- cost3(c1,df3)
f_c2 <- cost3(c2,df3)
table3 <- rbind(table3,c(a,b,c1,c2,f_c1,f_c2))
d <- b-a
} else{
a <- c1
c1 <- a+(1-r)*(b-a)
c2 <- a+r*(b-a)
f_c1 <- cost3(c1,df3)
f_c2 <- cost3(c2,df3)
table3 <- rbind(table3,c(a,b,c1,c2,f_c1,f_c2))
d <- b-a
}
}
l<- length(table3$a)
table3$a[l] ## [1] 0table3$b[l]## [1] 0.01400168c3_optimal <- (table3$a[l]+ table3$b[l] ) /2
c3_optimal## [1] 0.00700084cost3(c3_optimal,df3)## [1] 433.7104##############