About

In this lab we will focus on sensitivity analysis and Monte Carlo simulations.

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.

Monte Carlo Simulations utilize repeated random sampling from a given universe or population to derive certain results. This type of simulation is known as a probabilistic simulation, as opposed to a deterministic simulation.

An example of a Monte Carlo simulation is the one applied to approximate the value of pi. The simulation is based on generating random points within a unit square and see how many points fall within the circle enclosed by the unit square (marked in red). The higher the number of sampled points the closer the result is to the actual result. After selecting 30,000 random points, the estimate for pi is much closer to the actual value within the four decimal points of precision.

In this lab, we will learn how to generate random samples with various simulations and how to run a sensitivity analysis on the marketing use case covered so far.

Setup

Remember to always set your working directory to the source file location. Go to ‘Session’, scroll down to ‘Set Working Directory’, and click ‘To Source File Location’. Read carefully the below and follow the instructions to complete the tasks and answer any questions. Submit your work to RPubs as detailed in previous notes.

Note

For your assignment you may be using different data sets than what is included here. Always read carefully the instructions on Sakai. Tasks/questions to be completed/answered are highlighted in larger bolded fonts and numbered according to their particular placement in the task section.

Task 1: Sensitivity Analysis

In order to conduct the sensitivity analysis, we will need to download again the lpSolveAPI package unless you have it already installed in your R environment

# 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")

We will revisit and solve again the marketing case discussed in class (also part of previous lab).

# We start with `0` constraint and `2` decision variables. The object name `lpmark` is discretionary.
lpmark = make.lp(0, 2)

# Define type of optimization as maximum and dump the screen output into a `dummy` variable
dummy = lp.control(lpmark, sense="max") 
# Set the objective function coefficients 
set.objfn(lpmark, c(275.691, 48.341))

Add all constraints to the model.

add.constraint(lpmark, c(1, 1), "<=", 350000)
add.constraint(lpmark, c(1, 0), ">=", 15000)
add.constraint(lpmark, c(0, 1), ">=", 75000)
add.constraint(lpmark, c(2, -1), "=", 0)
add.constraint(lpmark, c(1, 0), ">=", 0)
add.constraint(lpmark, c(0, 1), ">=", 0)

Now, view the problem setting in tabular/matrix form. This is a good checkpoint to confirm that our contraints have been properly set.

lpmark
## Model name: 
##                C1       C2            
## Maximize  275.691   48.341            
## R1              1        1  <=  350000
## R2              1        0  >=   15000
## R3              0        1  >=   75000
## R4              2       -1   =       0
## R5              1        0  >=       0
## R6              0        1  >=       0
## Kind          Std      Std            
## Type         Real     Real            
## Upper         Inf      Inf            
## Lower           0        0
# solve
solve(lpmark) 
## [1] 0

Next we get the optimum results.

# display the objective function optimum value
get.objective(lpmark)
## [1] 43443517
# display the decision variables optimum values
get.variables(lpmark)
## [1] 116666.7 233333.3

For the sensitivity part we will add two new code sections to obtain the sensitivity results.

# display sensitivity to coefficients of objective function. 
get.sensitivity.obj(lpmark)
## $objfrom
## [1]  -96.6820 -137.8455
## 
## $objtill
## [1] 1e+30 1e+30
1A) For this exercise we are only interested in the first part of the output labeled objfrom. Explain in coincise manner what the sensitivity results represent in reference to the marketing model.

The sensitivty results shows how much we can change the coefficients of radio and tv in the model and still remain optimum values for sales at the same time. Therefore, we can change the coefficient of radio to -96.682 and still remain the optimmum solution for the decision variable radio. Moreover, we can also change the coefficient of TV to -136.8455 and still remain the optimum solution for the decision variable.

# display sensitivity to right hand side constraints. 
# There will be a total of m+n values where m is the number of contraints and n is the number of decision variables
get.sensitivity.rhs(lpmark) 
## $duals
## [1] 124.12433   0.00000   0.00000  75.78333   0.00000   0.00000   0.00000
## [8]   0.00000
## 
## $dualsfrom
## [1]  1.125e+05 -1.000e+30 -1.000e+30 -3.050e+05 -1.000e+30 -1.000e+30
## [7] -1.000e+30 -1.000e+30
## 
## $dualstill
## [1] 1.00e+30 1.00e+30 1.00e+30 4.75e+05 1.00e+30 1.00e+30 1.00e+30 1.00e+30
1B) For this exercise we are only interested in the first part of the output labeled duals. Explain in coincise manner what the two non-zero sensitivity results represent. Distinguish the binding/non-binding constraints, the surplus/slack, and marginal values.

The two non-zero results represent the marginal change in the optimum value by changing the binding constraint by one dollar. If we change the first constraint by 1 dollar, the optimum solution will result in 124 dollar. If we change the forth constraint by 1 dollar, the optimum soultion will result in 76 dollars. The binding constraints are the values that are non-zero where there is no surplus and will not affect on optimal solution. The non binding constraints has no impact on optimal solution which has surplus.

To acquire a better understanding of the sensitivity results, and to confirm integrity of the calculations, independent tests can be conducted.

1C) Run the linear programing solver again starting from the begining, by defining a new model object lpmark1. All being equal, change the budget constraint by only $1 and solve. Note the optimum value for sales as given by the objective function. Calculate the differential change in sales. Share your observations.

I have changed the first binding constraint by decreasing the maximum dollar amount from 350000 to 349999. The results of optimim slaes is 124.1243 which shows the sensitivity analysis is correct for the first constraint becuase the number matches the differnece between the two solutions that provided by the solver.

# Define a new model object called lpmark1
lpmark1 = make.lp(0, 2)
# Repeat rest of commands with the one constraint change for budget. Solve and display the objective function optimum value
dummy= lp.control(lpmark1, sense="max")
set.objfn(lpmark1, c(275.691, 48.341))
add.constraint(lpmark1, c(1,1), "<=", 349999)
add.constraint(lpmark1, c(1,0), ">=", 15000)
add.constraint(lpmark1, c(0,1), ">=", 75000)
add.constraint(lpmark1, c(2,-1), "=", 0)
add.constraint(lpmark1, c(1,0), ">=", 0)
add.constraint(lpmark1, c(0,1), ">=", 0)
lpmark1
## Model name: 
##                C1       C2            
## Maximize  275.691   48.341            
## R1              1        1  <=  349999
## R2              1        0  >=   15000
## R3              0        1  >=   75000
## R4              2       -1   =       0
## R5              1        0  >=       0
## R6              0        1  >=       0
## Kind          Std      Std            
## Type         Real     Real            
## Upper         Inf      Inf            
## Lower           0        0
solve(lpmark1)
## [1] 0
ob1 = get.objective(lpmark1)
ob1
## [1] 43443393
ob = get.objective(lpmark)
ob-ob1
## [1] 124.1243
get.variables(lpmark1)
## [1] 116666.3 233332.7
1D) Based on the previous exercise explain in clear words, and without running another solver again , how would you check the integrity of the other marginal value identified in 1B).

I run the same model as I did before to check the integrity of the other marginal value. I changed the fourth constraint which is another binding constraint. This constraint represents the relationship between TV and radio. By changing the constraint from 0 to 1 and this can calculate its marginal value. The difference is 74.783 is the same number calculated in the sensitvity model which is the difference in optimim value caluated for the sales.

Task 2: Monte Carlo Simulation

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%. We will assume that the historical S&P daily return follows a normal distribution with an average daily return of 0.03 (%) and a standard deviation of 0.97 (%).

To begin we will generate 100 random samples from the normal distribution. For the generated samples we will calculate the mean, standard deviation, and probability of occurrence where the simulation result is greater than 5%.

To generate random samples from a normal distribution we will use the rnorm() function in R. In the example below we set the number of runs (or samples) to 100.

# number of simulations/samples
runs = 100
# random number generator per defined normal distribution with given mean and standard deviation
sims =  rnorm(runs,mean=0.03,sd=0.97)
# Mean calculated from the random distribution of samples
average = mean(sims)
average
## [1] -0.1419291
# STD calculated from the random distribution of samples
std = sd(sims) 
std
## [1] 0.9731221
# probability of occurrence on any given day based on samples will be equal to count (or sum) where sample result is greater than 5% divided by total number of samples. 
prob = sum(sims >=0.05)/runs
prob
## [1] 0.42
2A) Repeat the above calculations for the two cases where the number of simulations/samples is equal to 1000 and 10000. For each case record the mean, standard deviation, and probability.
# Repeat calculations here
runs1 = 1000
sims1=rnorm(runs1, mean=0.03, sd=0.97)
average1=mean(sims1)
average1
## [1] 0.06062367
std1=sd(sims1)
std1
## [1] 0.9802638
prob1=sum(sims1>=0.05)/runs1
prob1
## [1] 0.513
runs2=10000
sims2=rnorm(runs2, mean=0.03, sd=0.97)
average2=mean(sims2)
average2
## [1] 0.03160835
std2=sd(sims2)
std2
## [1] 0.9701584
prob2=sum(sims2>=0.05)/runs2
prob2
## [1] 0.4928
2B) List in a tabular form the values for mean, standard deviation, and probability for all three cases: 100, 1000, and 10000 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? How is this similar to the image use case to calculate pi that was presented in the introductory paragraph?

Case 1:100 Mean:-0.3215 STEDV:1.1011 Probability:0.4 Case2:1000 Mean:0.2177 STEDV:0.9359 Probability:0.476 Case3:10000 Mean:0.02113 STEDV:0.9735 Probability:0.4889 As the nimber of the simulations increases, the values sem to get clsoder to the normal disturbution that we set up wich is STEDV 0.97 with an mean 0.03. The best bet on the probability of occurrence greater than 5 % is around 47%. This is quite similar to the image at the beginning since the more simuations that we run, the more accurate on the probability values.

The last 2C) exercise is optional for those interested in further enhancing their subject matter learning, and refining their skills in R. Your work will be assessed but you will not be graded for this exercise. You can follow the instructions presented in the video Excel equivalent example at [https://www.youtube.com/watch?v=wKdmEXCvo9s]

2C) Repeat the exercise for the S&P daily return where all is equal except we are now interested in the weekly cumulative return and the probability that the weekly cummulative return is greater than 5%. Set the number of simulations to 10000.
runs3=10000
sim3=rnorm(runs3, mean=0.03, sd=0.97)
mean(sim3)
## [1] 0.02498412
prob3=sum((sim3)^7>=0.05)/runs3
prob3
## [1] 0.263