Retailer Case Linear Optimization

Sensitivity analysis is the study of how the uncertainty in the output of a mathematical model or system (numerical or otherwise) can be apportioned to different sources of uncertainty in its inputs. We will use the lpSolveAPI R-package as we did in the previous lab.

In this lab assignment you are asked to first find the optimum solution for the problem defined below and then run a sensitivity analysis.

Problem

The Operations Manager of a retailer company, producing and selling jackets and parkas wants to identify the production rates for their famous jacket and parka that would maximize their profit. Based on past history he knows that per unit profits for their winter jacket and parka are $9 and $12.5, respectively.

. A medium size jacket requires 8.5 feet of fabric, while the parka requires 12.5. . The amount of machine time needed to produce a jacket is 1.5hours and 2hours for the parka. . A tailor usually spends about 2 hours on sewing and stitching for a winter jacket and about 3 hours for a parka. . The company has a monthly contract with a supplier to procure 4,000 feet of required fabric. . The amount of machine time allocated to producing two products is 650 hours/month . Each month several skilled tailors are assigned for the production of these products for around 900 hours.

They also know that historically they newer sold more than 150 parkas and 400 jackets per month.

Task 1: Formulate this problem into an LP model (1 point)

In order to conduct the sensitivity analysis, you will need to download the lpSolveAPI package unless you have it already installed in your R environment and make it available to this notebook.

# Require will load the package only if not installed 
# Dependencies = TRUE makes sure that dependencies are install
if(!require("lpSolveAPI",quietly = TRUE))
  install.packages("lpSolveAPI",dependencies = TRUE, repos = "https://cloud.r-project.org")
library(lpSolveAPI)

You can use the following empty chunk to define your lp model

# Define your lp model
lpmark = make.lp(0, 2)

Use the fallowing chunk to define the type of optimization, set the objective function, and add the constraints to your model object.

# Define type of optimization 
dump = lp.control(lpmark, sense="max") 

# Set the objective function
set.objfn(lpmark, c(9, 12.5))

# add constraints
add.constraint(lpmark, c(8.5, 12.5), "<=", 4000) #Fabric constraint
add.constraint(lpmark, c(1.5, 2), "<=", 650) #machine time constraint
add.constraint(lpmark, c(2, 3), "<=", 900) #tailor hours constraint
add.constraint(lpmark, c(1, 0), "<=", 400) #maximum jackets
add.constraint(lpmark, c(0, 1), "<=", 150) #maximum parkas

Finally we can explore and solve the model using the lpSolveAPI package.

# View the problem formulation in tabular/matrix form
lpmark
## Model name: 
##             C1    C2          
## Maximize     9  12.5          
## R1         8.5  12.5  <=  4000
## R2         1.5     2  <=   650
## R3           2     3  <=   900
## R4           1     0  <=   400
## R5           0     1  <=   150
## Kind       Std   Std          
## Type      Real  Real          
## Upper      Inf   Inf          
## Lower        0     0

Use the fallowing chunks to solve the optimization problem and display the optimum results.

# Solve 
solve(lpmark)
## [1] 0
# Display the objective function optimum value
get.objective(lpmark)
## [1] 3950
# Display the decision variables optimum values
get.variables(lpmark) 
## [1] 300 100

Task 2: Sensitivity Analysis (1 point)

Use the following chunk to get the sensitivity results for the coefficients of the objective function.

# display sensitivity to coefficients of objective function. 
get.sensitivity.obj(lpmark)
## $objfrom
## [1]  8.333333 12.000000
## 
## $objtill
## [1]  9.375 13.500
Q1) Explain in coincise manner what the coefficient sensitivity results represent in reference to the model (0.25 points)

Answer: The coefficient sensitivity results provide a range of values for the objective function coefficients (profits per unit for jackets and parkas) within which the current optimal solution remains unchanged.For the jacket’s profit per unit, the range is between approximately $8.33 and $9.38. For the parka’s profit per unit, the range is between approximately $12 and $13.50. Within these ranges, if the profits per unit for jackets and parkas remain within the specified bounds, the current optimal production quantities for jackets and parkas will not change. This means that as long as the profits per unit fall within these respective ranges, the recommended quantities of jackets and parkas for maximum profit will stay the same.

Use the following chunk to display the constraint sensitivity results

# display sensitivity to right hand side constraints. 
get.sensitivity.rhs(lpmark) 
## $duals
## [1] 0.0 4.0 1.5 0.0 0.0 0.0 0.0
## 
## $dualsfrom
## [1] -1.000e+30  6.375e+02  8.750e+02 -1.000e+30 -1.000e+30 -1.000e+30 -1.000e+30
## 
## $dualstill
## [1] 1.000000e+30 6.666667e+02 9.166667e+02 1.000000e+30 1.000000e+30
## [6] 1.000000e+30 1.000000e+30
Q2) Explain in coincise manner what the two non-zero sensitivity results represent. Distinguish the binding/non-binding constraints, the surplus/slack, and marginal values (0.25 points).

Answer: The two non-zero sensitivity results in the context of the problem represent that the machine time constraint is binding within its allowable range (between 637.5 and 650 hours). If additional machine time were available beyond this range, the profit would increase by $4 per additional hour. This means that the company is utilizing machine time efficiently within the range. Surplus/Slack: These sensitivity results indicate the surplus or slack in the constraints. The sensitivity result of 0.0 indicates that the fabric constraint is non-binding. Even if additional fabric was made available in the rance, the optimal profit would not change. This means that the company is not fully utilizing the available fabric and there is a surplus in the constraint. Marginal Values: The sensitivity values (dual values) indicate the marginal impact of relaxing or tightening the constraints. Larger sensitivity values suggest a higher impact on the objective function with changes in the constraint values.

Q3) Interpret the sensitivity result for the amount of fabric (0.25 points).

Answer: The sensitivity result for the fabric constraint being 0.0 suggests that this constraint is not affecting the optimal solution within the range. It implies that the company is not utilizing the entire available fabric (4000 feet) and is likely operating with surplus or slack in fabric procurement. Even if more fabric were supplied within the specified limit, it wouldn’t impact the optimal profit, indicating an excess in available fabric.

Q4) Interpret the sensitivity result for the machine time (0.25 points)

Answer: The sensitivity result of 4.0 for machine time indicates that this constraint significantly impacts the optimal solution. Specifically, within the range of 637.5 to 650 hours (the active range for this sensitivity), an increase in available machine time would directly increase the profit by $4 per additional hour. This means the company is operating at maximum efficiency within the given machine time range, and any additional machine time within this range would lead to increased profitability.

Task 3: Monte Carlo Simulation (1 point)

For this task we will be running a Monte Carlo simulation to calculate the probability that the daily return from S&P will be > 5%. From the historic data we are given that the S&P daily return follows a normal distribution with an average daily return of 0.03 (%) and a standard deviation of 0.97 (%).

First generate 100 random samples from the normal distribution. For the generated samples calculate the mean, standard deviation, and probability of occurrence where the simulation result is greater than 5%. (0.25 points)

# define the number of simulations/samples
runs = 100
# Use the defined normal distribution with given mean and standard deviation to create simulations
sims =  rnorm(runs,mean=0.03,sd=0.97)
# Calculate the mean from the random distribution of samples
average = mean(sims)
average
## [1] -0.02822005
# Calculate the standard deviation from the random distribution of samples
std = sd(sims) 
std
## [1] 0.9185057
# Calculate the probability of occurrence 
prob = sum(sims >=0.05)/runs
prob
## [1] 0.51
Repeat the above calculations for the two cases where the number of simulations is equal to 1000 and 100000. For each case record the mean, standard deviation, and probability (0.50 points)
# Repeat calculations here
# number of simulations/samples
runs1 = 1000

# random number generator per defined normal distribution with given mean and standard deviation
sims1 =  rnorm(runs1,mean=0.03,sd=0.97)

# Calculate the mean and the standard deviation from the random distribution of samples
average1 = mean(sims1)
average1
## [1] 0.04982056
#the standard deviation 
std1 = sd(sims1)
std1
## [1] 0.9803717
# Calculate the probability of occurrence
prob1 = sum(sims1 >=0.05)/runs1
prob1
## [1] 0.508
# Repeat calculations here
# number of simulations/samples
runs2 = 100000
# random number generator per defined normal distribution with given mean and standard deviation
sims2 =  rnorm(runs2,mean=0.03,sd=0.97)

# Calculate the mean and the standard deviation from the random distribution of samples
average2 = mean(sims2)
average2
## [1] 0.02931639
#the standard deviation
std2 = sd(sims2)
std2
## [1] 0.9701367
# Calculate the probability of occurrence 
prob2 = sum(sims2 >=0.05)/runs2
prob2
## [1] 0.49095
List in a tabular form the values for mean, standard deviation, and probability for all three cases: 100, 1000, and 100000 simulations. Describe how the values change/behave as the number of simulations is increased. What is your best bet on the probability of occurrence greater than 5% and why? (0.25 points)
# Display your results in a tabular format
myresults <- data.frame (Runs = c("100", "1000", "10000"), Average = c(average, average1, average2), STD = c(std, std1, std2), Probability = c(prob, prob1, prob2))
myresults
##    Runs     Average       STD Probability
## 1   100 -0.02822005 0.9185057     0.51000
## 2  1000  0.04982056 0.9803717     0.50800
## 3 10000  0.02931639 0.9701367     0.49095
Your answer:

As we increase the number of sampled points, our results tend to converge closer to the actual result. Therefore, with a larger sample size, we anticipate our findings to align more closely with the daily returns from the S&P. Hence, the probability derived from the larger sample size would be the preferable choice for making predictions or estimations.