Section 6.2

Page 232 #1

Consider a stereo with CD player, FM–AM radio tuner, speakers (dual), and power amplifier (PA) components, as displayed with the reliabilities shown in Figure 6.11. Determine the system’s reliability. What assumptions are required in your model?

Define \(F(t)\) as the cumulative distrubution function of the failure rate,\(f(t)\), of a system or component, where \(t\) refers to time. The reliability \(R(t)\), of a system of component is the complement of \(F(t)\)

\(R(t) = 1-F(t)\)

rs <- c(0.95, 0.98, 0.97, 0.99, 0.99)
names(rs) <- c("PA","CD","Radio","SpeakerA","SpeakerB")

	rs
PA	0.95
CD	0.98
Radio	0.97
SpeakerA	0.99
SpeakerB	0.99

Given:

The diagram above, gives us the assumptions that the following subsystems operate in parallel.

Subsystem A: CD and FM/AM Radio
Subsystem B: Speaker A and Speaker B

In addition, it seems that the parallel subsystems run in serials with each other and the power amplifier.

Therefore, if any of the two subsystems or the Power Amplifier components fail, the system as a whole will fail.

Futhermore, we can assume that the individual components are independent of one another.

Lastly, we can assume that each component and subsystems are running in serials and have independent reliabilities.

Reliability Calculation

Reliability for Parallel Systems

\(R_s(t)\) can be calculated by the union of the reliabilty for the components by using the following formula:

\(\displaystyle R_s(t) = \bigcup_{i=1}^nR_i(t)\)

Calculating the respective reliability for each of the three parallel subsystems

# Subsystem A
r_s1 <- rs["CD"] + rs["Radio"] -  rs["CD"] * rs["Radio"]
names(r_s1) <- NULL

r_s1

## [1] 0.9994

# Subsystem B
r_s2 <- rs["SpeakerA"] + rs["SpeakerB"] -  rs["SpeakerA"] * rs["SpeakerB"]
names(r_s2) <- NULL

r_s2

## [1] 0.9999

Reliability for Systems in Series with Independent Components

\(\displaystyle R_s(t) = \prod_{i=1}^nR_i(t)\)

To calculate the reliability of the system as a whole, we have to assume the Power Amplifier runs in series with the two subsystems.

r_s_t <- rs["PA"] * r_s1 * r_s2

r_s_t

##        PA 
## 0.9493351

Based on the above model, the System has a 95% probability of properly functioning.

Section 6.2

Page 251 #2

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

items	protein	carbs	roughage	cost
hay	0.5	2.0	5.0	1.8
oats	1.0	4.0	2.0	3.5
feeding blocks	2.0	0.5	1.0	0.4
hpc	6.0	1.0	2.5	1.0
horse requirement	40.0	20.0	45.0	0.0

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

Solution

Variables

\(x_1=\mathrm{hay~per~bale}\\\)

\(x_2=\mathrm{oats~per~sack}\\\)

\(x_3=\mathrm{feeding~blocks~per~block}\\\)

\(x_4=\mathrm{high~protein~concentrate~per~sack}\)

Objective

\(\textrm{Minimize}~1.8 x_1+3.5x_2+0.4x_3+1.0x_4\)

Constraints

\(0.5x_1+x_2+2x_3+6x_4 \ge40\textrm{(protein)}\\\)

\(2x_1+4x_2+0.5x_3+x_4\ge20\textrm{(carbs)}\\\)

\(5x_1+2x_2+x_3+2.5x_4\ge45\textrm{(roughage)}\)

Page 264 #6

Maximize

\({10x+35y}\)

Subjected to

\(8x+6y\le48\textrm{(board-feet of lumber)}\\\)

\(4x+y\le20\textrm{(hours of capacity)}\\\)

\(y\ge5\textrm{(demand)}\\\)

\(x,y \ge0\textrm{(nonnegativity)}\)

obj.func <- function(x,y) 10*x + 35*y

s1 <- obj.func(0,5)
s2 <- obj.func(0,8)
s3 <- obj.func(9/4,5)

mydf <- data.frame(points=seq(1:3),x=c(0,0,9/4),y=c(5,8,5),obj_func = c(s1,s2,s3))

points	x	y	obj_func
1	0.00	5	175.0
2	0.00	8	280.0
3	2.25	5	197.5

Solution

The objective function is maximized at point (0,8) with a value of 280.

Graphs

x <- seq(0, 6, by=0.1)

data <- data.frame(x = x, board_feet = (48-8*x)/6, hours = 20-4*x, demand = 5)
tidy <- gather(data, variable,  "y", -x) 

ggplotly(ggplot(tidy, aes(x=x, y=y, color=variable)) + geom_line())

myint <- data.frame(x=c(0,0,2.25),y=c(8,5,5))

ggplot(data.frame(x=c(-5,10)),aes(x)) + 
  stat_function(fun=function(x) -4/3*x+8, geom="line", aes(col='y=-4/3x+8')) +
  stat_function(fun=function(x) -4*x + 20, geom="line", aes(col='y=-4x+20')) +
  stat_function(fun=function(x)5, geom="line", aes(col='y=5')) + 
  geom_vline(xintercept=0, aes(col= 'x=0')) + 
  geom_hline(yintercept= 0, aes(col='y=0')) + 
  geom_point(data=myint, aes(x,y)) + 
  annotate('text', x = 0, y = 9.2, label="(0, 8)", size=3 ) +
  annotate('text', x = 0, y = 3.8, label="(0, 5)", size=3 ) + 
  annotate('text', x = 2.25, y = 3.8, label="(9/4, 5)", size=3 )

Page 295 #3

Use the curve-fitting criterion to minimize the sum of the absolute deviations for the following models and data set:

\(y = ax\)
\(y = ax^2\)
\(y = ax^3\)

Model

\(y=ax\)

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
cv1 = a + (1 - r)*(b - a)
cv2 = a + r*(b -a)

cost <- function(a, df){
  sum(abs(df$y - a*(df$x)))
}

crit_val_1<- cost(cv1,df1)
crit_val_2<- cost(cv2,df1)

d1 <- data.frame(a=a,b=b,cv1=cv1,cv2=cv2,crit_val_1=crit_val_1,crit_val_2=crit_val_2)

t <- 0.02
d <- 1

while(d > t){
  if(crit_val_1<=crit_val_2){
    b <- cv2
    cv1 <- a+(1-r)*(b-a)
    cv2 <- a+r*(b-a)
    crit_val_1 <- cost(cv1,df1)
    crit_val_2 <- cost(cv2,df1)
    d1 <- rbind(d1,c(a,b,cv1,cv2,crit_val_1,crit_val_2))
    d <- b-a
  } else{
    a <- cv1
    cv1 <- a+(1-r)*(b-a)
    cv2 <- a+r*(b-a)
    crit_val_1 <- cost(cv1,df1)
    crit_val_2 <- cost(cv2,df1)
    d1 <- rbind(d1,c(a,b,cv1,cv2,crit_val_1,crit_val_2))
    d <- b-a
  }
}

l<- length(d1$a)

d1$a[l]

## [1] 7.066307

Calculate the Sum of the Absolute Deviations of the Model

d1$b[l]

## [1] 7.080309

cv1_final <- (d1$a[l]+ d1$b[l] ) /2
cv1_final

## [1] 7.073308

crit_val_1_optimal<- cost(cv1_final,df1)

crit_val_1_optimal

## [1] 199.5395

Model

\(y=ax^2\)

df2 <- data.frame(x=x, y=y, x2=x^2)

a<- 0
b<- 50
r<- ((1+sqrt(5))/2)-1
cv1 = a + (1 - r)*(b - a)
cv2 = a + r*(b -a)

cost2 <- function(a, df){
  sum(abs(df$y - a*(df$x2) ))
}

crit_val_1<- cost2(cv1,df2)
crit_val_2<- cost2(cv2,df2)

d2 <- data.frame(a=a,b=b,cv1=cv1,cv2=cv2,crit_val_1=crit_val_1,crit_val_2=crit_val_2)

t <- 0.02
d <- 1

while(d > t){
  if(crit_val_1<=crit_val_2){
    b <- cv2
    cv1 <- a+(1-r)*(b-a)
    cv2 <- a+r*(b-a)
    crit_val_1 <- cost2(cv1,df2)
    crit_val_2 <- cost2(cv2,df2)
    d2 <- rbind(d2,c(a,b,cv1,cv2,crit_val_1,crit_val_2))
    d <- b-a
  } else{
    a <- cv1
    cv1 <- a+(1-r)*(b-a)
    cv2 <- a+r*(b-a)
    crit_val_1 <- cost2(cv1,df2)
    crit_val_2 <- cost2(cv2,df2)
    d2 <- rbind(d2,c(a,b,cv1,cv2,crit_val_1,crit_val_2))
    d <- b-a
  }
}

l<- length(d2$a)

d2$a[l]

## [1] 0.2232466

Calculate the Sum of the Absolute Deviations of the Model

d2$b[l]

## [1] 0.2372483

cv2_final <- (d2$a[l]+ d2$b[l] ) /2

cv2_final

## [1] 0.2302474

crit_val_2_optimal<- cost2(cv2_final,df2)

crit_val_2_optimal

## [1] 219.5671

Model

\(y=ax^3\)

df3 <- data.frame(x=x, y=y, x3=x^3)

a<- 0
b<- 50
r<- ((1+sqrt(5))/2)-1

cv1 = a + (1 - r)*(b - a)
cv2 = a + r*(b -a)

cost3 <- function(a, df){
  sum(abs(df$y - a*(df$x3)))
}

crit_val_1<- cost3(cv1,df3)
crit_val_2<- cost3(cv2,df3)

d3 <- data.frame(a=a,b=b,cv1=cv1,cv2=cv2,crit_val_1=crit_val_1,crit_val_2=crit_val_2)

t <- 0.02
d <- 1

while(d > t){
  if(crit_val_1<=crit_val_2){
    b <- cv2
    cv1 <- a+(1-r)*(b-a)
    cv2 <- a+r*(b-a)
    crit_val_1 <- cost3(cv1,df3)
    crit_val_2 <- cost3(cv2,df3)
    d3 <- rbind(d3,c(a,b,cv1,cv2,crit_val_1,crit_val_2))
    d <- b-a
  } else{
    a <- cv1
    cv1 <- a+(1-r)*(b-a)
    cv2 <- a+r*(b-a)
    crit_val_1 <- cost3(cv1,df3)
    crit_val_2 <- cost3(cv2,df3)
    d3 <- rbind(d3,c(a,b,cv1,cv2,crit_val_1,crit_val_2))
    d <- b-a
  }
}

l<- length(d3$a)

d3$a[l]

## [1] 0

Calculate the Sum of the Absolute Deviations of the Model

d3$b[l]

## [1] 0.01400168

cv3_final <- (d3$a[l]+ d3$b[l] ) /2

cv3_final

## [1] 0.00700084

crit_val_3_optimal <- cost3(cv3_final,df3)

crit_val_3_optimal

## [1] 433.7104

Solution

## [1] "The Optimal Value for the Model y=ax = 199.53946343722"

## [1] "The optimal value for the model y=ax^2 = 219.567063518652"

## [1] "The optimal value for the model y=ax^3 = 433.710399871649"

Data 609 Assignment #3

Joseph Simone

2/19/2020

Section 6.2

Page 232 #1

Given:

Reliability Calculation

Reliability for Parallel Systems

Reliability for Systems in Series with Independent Components

Section 6.2

Page 251 #2

Solution

Variables

Objective

Constraints

Page 264 #6

Maximize

Subjected to

Solution

Graphs

Page 295 #3

Model

Calculate the Sum of the Absolute Deviations of the Model

Model

Calculate the Sum of the Absolute Deviations of the Model

Model

Calculate the Sum of the Absolute Deviations of the Model

Solution