PersonalComputerProblemWithConstraint .sagews

# Optimization Problem

# Goal: To find the amount of each type of books to be sold in order to maximize profits

# Variables

# x = number of applied version of the book sold
# y = number of the traditional version of the book sold
# Cx = cost to produce the applied version of the book = 60
# Cy = cost to produce the traditional version of the book = 30
# C = total cost
# C = Cx + Cy
# sx = selling price of the applied version of the textbook
# sy = selling price of the traditional version of the textbook
# R = total revenue
# R = sx + sy
# P = profit
# P = R - C

# Assumptions

# Cx = 60*x
# Cy = 30*y
# C = 60*x + 30*y
# s = 450-0.01*x
# s = 450-0.002*y
# R = s(x) + s(y)
# P = R(x,y) - C(x,y)

# Constraint x + y <= 16000

# a- State the model for the profit as a function of the number of applied and traditional books sold

Cx(x) = 60*x
Cy(y) = 30*y
C(x,y) = Cx(x) + Cy(y)
s(x) = 450-0.01*x-0.002*y
s(y) = 250-0.002*x-0.003*y
R(x,y) = s(x)*x + s(y)*y
P(x,y) = R(x,y) - C(x,y)
show(P)

$\displaystyle \left( x, y \right) \ {\mapsto} \ x {\left(-0.00500000000000000 \, x + 250\right)} + {\left(-0.00200000000000000 \, x - 0.00300000000000000 \, y + 250\right)} y - 60 \, x - 30 \, y$

# b- Produce a contour plot of the profit function and sketch the feasible region

p1=contour_plot(P(x,y),(x,0,17000),(y,0,17000))
p2=implicit_plot(x+y==16000 ,(x,0,17000),(y,0,17000),color='blue')
plot(p1+p2)

# c- Determine the amount of each type of book to be sold in order to maximize profits

L1 = var('L1')
g(x,y) = x+y
eq = solve([diff(P,x)==L1*diff(g,x),diff(P,y)==L1*diff(g,y), g(x,y)==16000], x,y,L1)

show(eq)

[[$\displaystyle x = \left(\frac{8500}{3}\right)$, $\displaystyle y = \left(\frac{39500}{3}\right)$, $\displaystyle L_{1} = \left(\frac{406}{3}\right)$]]

P(eq[0][0].rhs(),eq[0][1].rhs()) # Maximum profit

2.80016666666667e6

# Test points
P(0,0)

0.000000000000000

P(16000,0)

1.76000000000000e6

P(8000,8000)

2.64000000000000e6

P(0,16000)

2.75200000000000e6

# Compute the sensitivity of the number of applied books sold to price elasticity of the applied book ($0.02)

Cx(x) = 60*x
Cy(y) = 30*y
C(x,y) = Cx(x) + Cy(y)
sa(x,a) = 450-0.01*x-a*y
s(y) = 250-0.002*x-0.003*y
R(x,y,a) = sa(x,a)*x + s(y)*y
Pa(x,y,a) = R(x,y,a) - C(x,y)
show(Pa)

$\displaystyle \left( x, y, a \right) \ {\mapsto} \ {\left(-a y - 0.0100000000000000 \, x + 450\right)} x + {\left(-0.00200000000000000 \, x - 0.00300000000000000 \, y + 250\right)} y - 60 \, x - 30 \, y$

L2=var('L2')
g(x,y) = x+y
eq1 = solve([diff(Pa,x)==L2*diff(g,x),diff(Pa,y)==L2*diff(g,y), g(x,y)==16000], x,y,L2)

show(eq1)

[[$\displaystyle x = \frac{1000 \, {\left(8000 \, a - 117\right)}}{1000 \, a - 11}$, $\displaystyle y = \frac{1000 \, {\left(8000 \, a - 59\right)}}{1000 \, a - 11}$, $\displaystyle L_{2} = -\frac{8 \, {\left(1000000 \, a^{2} - 34125 \, a + 229\right)}}{1000 \, a - 11}$]]

sxa(a)=diff(eq1 [0][0].rhs(),a)*a/eq1[0][0].rhs()

show(sxa)

$\displaystyle a \ {\mapsto}\ -\frac{1000 \, {\left(1000 \, a - 11\right)} a {\left(\frac{8000 \, a - 117}{{\left(1000 \, a - 11\right)}^{2}} - \frac{8}{1000 \, a - 11}\right)}}{8000 \, a - 117}$

sxa(0.02)

1.49870801033592

# By analysing the elasticity, we can conclude that if the elasticity increases by 1%, the price of applied books will increase by 1.4987%

# e- Compute the sensitivity of the maximum profit to the assumption that the publisher can produce at most 16,000 books

# New constraint g(x+y) => 16000
# Let t = the total amount of textbooks produced
t = var('t')
L3 = var('L3')
eq3 = solve([diff(P,x)==L3*diff(g,x),diff(P,y)==L3*diff(g,y), g(x,y)==t],x,y,L3)
show(eq3)

[[$\displaystyle x = \frac{1}{3} \, t - 2500$, $\displaystyle y = \frac{2}{3} \, t + 2500$, $\displaystyle L_{3} = -\frac{7}{1500} \, t + 210$]]

Pmaxt = P(eq3[0][0].rhs(),eq3[0][1].rhs())

SPmaxt = Pmaxt.diff(t)*t/Pmaxt

show(SPmaxt)

$\displaystyle \frac{3 \, t {\left(-0.00466666666666667 \, t + 210.000000000000\right)}}{{\left(t - 7500\right)} {\left(-0.00166666666666667 \, t + 262.500000000000\right)} + 2 \, {\left(t + 3750\right)} {\left(-0.00266666666666667 \, t + 247.500000000000\right)} - 120 \, t + 225000}$

n(SPmaxt(16000))

0.773287304327123

# By analyzing the sensitivity of the maximum profit, we can conclude that if the production increase by 1%, the profit will increase by 0.773%

# d- What is the maximum amount the publisher should be willing to pay to increase production by 1 book?

# According to the answer of the question c- we have: L1 = 406/3

n(406/3)

135.333333333333

# The shadow price is equal to L1 = 135.333333333333. Thus the maximum the publisher should be willing to pay is $135.333333333

# 2 -  Suppose that a sudden loss in production materials was such that the publisher cannot produce more than 4,000 applied books. How does this affect the publisher’s maximal profit? Justify

# We are going to implement a new constraint which is x <= 4000

p3 = implicit_plot(x==4000,(x,0,17000),(y,0,17000),color = 'red')
plot(p1+p2+p3)

L4 = var('L4')
L5 = var('L5')
g(x,y) = x+y
eq4 = solve([diff(P,x)==(L4+L5)*diff(g,x),diff(P,y)==L4*diff(g,y), g(x,y)==16000,x==4000], x,y,L4,L5)

show(eq4)

[[$\displaystyle x = 4000$, $\displaystyle y = 12000$, $\displaystyle L_{4} = 140$, $\displaystyle L_{5} = \left(-14\right)$]]

# New maximal profit

P(4000,12000)

2.79200000000000e6

# The maximal profit without the new constraint was 2.80016666666667e6.

2.80016666666667e6 - 2.79200000000000e6

8166.66666666977

# The sudden loss in production materials will decrease the maximal profit by 8166.66666666977.



# Variables

# p = price of the manufacture
# b = budget for advertising
# To = total of unit sold
# R = Revenue
# C= cost
# P = Profit

# Assumptions
# R(p,b) = To(p,b) * p
# To(p,b) = 10000 + 200*(b-50000)/10000 + 5000*(950-p)/100
# C(p,b) = 700 * To(p,b) + b
# P(p,b) = R(p,b) - C(p,b)

# Constraints
# 0 <= p
# 50000 <= b
# b <= 100000
To(p,b) = 10000 + 200*(b-50000)/10000 + 5000*(950-p)/100
R(p,b) = To(p,b) * p
C(p,b) = 700 * To(p,b) + b
P(p,b) = R(p,b) - C(p,b)

p1 = contour_plot(P(p,b),(p,-10000,10000),(b,50000,100000))
p2 = implicit_plot(p==0, (p,-10000,10000),(b,50000,100000))
plot(p1+p2)

# When gradient = 0

solve([diff(P,p)==0, diff(P,b)==0],p,b)

[[p == 750, b == -825000]]

# constraint b = 50000
solve([diff(P,p)==0,diff(P,b)==L1,b==50000],p,b,L1)

[[p == 925, b == 50000, L1 == (7/2)]]

# constraint b = 100000
solve([diff(P,p)==0,diff(P,b)==L1,b==100000],p,b,L1)

[[p == 935, b == 100000, L1 == (37/10)]]

# Test of Points
P(0,50000)
P(0,100000)
P(925,50000)
P(935,100000)

-40300000 -41050000 2481250 2661250

# The manufacturer of the personal computer reachs the maximum profit when he balances advertising to $100000 and price to $935

#b Determine the sensitivity of the decision variables to the price

To(p,b,s) = 10000 + 200*(b-50000)/10000 + s*(950-p)
R(p,b,s) = To(p,b,s) * p
C(p,b,s) = 700 * To(p,b,s) + b
P(p,b,s) = R(p,b,s) - C(p,b,s)
# constraint b = 100000
L2 = var('L2')
eq = solve([diff(P,p)==0,diff(P,s)==L2, b==100000],p,b,L2)

show(eq)

[[$\displaystyle p = \frac{275 \, {\left(3 \, s + 20\right)}}{s}$, $\displaystyle b = 100000$, $\displaystyle L_{2} = \frac{15625 \, {\left(s^{2} - 1936\right)}}{s^{2}}$]]

Sp(s) = diff(eq[0][0].rhs())*s/(eq[0][0].rhs())

n(Sp(50))

-0.117647058823529

# Compute the sensitivity of the decision variable to the advertising agency's estimateof 200 new sales each time

To(p,b,g) = 10000 + g*(b-50000)/10000 + 5000*(950-p)/100
R(p,b,g) = To(p,b,g) * p
C(p,b,g) = 700 * To(p,b,g) + b
P(p,b,g) = R(p,b,g) - C(p,b,g)

L3 = var('L3')
eq = solve([diff(P,p)==0,diff(P,b)==L3,b==100000],p,b,L3)

show(eq)

[[$\displaystyle p = \frac{1}{20} \, g + 925$, $\displaystyle b = 100000$, $\displaystyle L_{3} = \frac{1}{200000} \, g^{2} + \frac{9}{400} \, g - 1$]]

Sg(g) = diff(eq[0][0].rhs())*g/(eq[0][0].rhs())

n(Sg(200))

0.0106951871657754

# The value of the multiplier is L1 in the question (a) is 3.7. it is representative of the monetary value of advertising. It means that if the manufacturer increases $1 in his budget for aadvertising, he will increase the profit of $3.70


## RISK MANAGEMENT PROJECT ##

# Using a large dataset to understand, apply and plot the logarithm of return #

# Application of the relative frequency to understand how the the variance of the price might affect the decision making process

```{r,warning=FALSE,message=FALSE}
getwd()
library (qrmdata)
library(xts)
data <- read.csv(file="MCD_PriceDaily.csv",head=TRUE,sep=",")
class(data)
data.first.5 <- data[1:5,]
data.first.5
names(data)
adjPrice <- data[,7]
class(adjPrice)
adjPrice

n <- 1177
netReturn <- adjPrice[2:n]/adjPrice[1:(n-1)]-1
netReturn
logReturn = diff(log(adjPrice))
logReturn
plot(netReturn,logReturn,pch=20,cex=0.2,col="blue",main="Log returns vs. Net Returns")
abline(h=0,v=0,col="lightblue",lwd=0.5)

# the correlation between the net returns and the log returns is positive
```
```{r}
meanNetReturn <- mean(netReturn)
meanNetReturn
meanLogReturn <- mean(logReturn)
meanLogReturn
sdNetReturn <- sd(netReturn)
sdNetReturn
sdLogRreturn <- sd(logReturn)
sdLogRreturn
```

```{r}
hist(netReturn, col = "blue")
hist(logReturn, col = "red")

```

```{r}
P<-86 # initial stock value
mu<-meanLogReturn/253 # daily mean of log return
sigma<-sdLogRreturn/sqrt(253) # daily SD of log return
event<-function(){
  r<-rnorm(20,mean=mu,sd=sigma) # random Normal numbers
  stock_20days<-P*exp(cumsum(r)) # vector of stock values over 20 days
  flag<-any(stock_20days<85) # TRUE if any stock value is below 85
  return(flag) # return TRUE or FALSE
}
sample<-replicate(1e4,event()) # replicate random exp
mean(sample) # relative frequency of falling below 85