Business Analytics Lab Worksheet 08
CME Group Foundation Business Analytics Lab
Richard Williams
April 1, 2019
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] 0Next 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 233333.3For 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+301A) 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.
ANSWER: THIS TEST INDICATES THAT THE LOWER SENSETIVITY LIMITS RADIO AND TV ARE -96.68 (FOR RADIO) AND -137.86 (FOR TV). THE TEST ALSO INDICATES THAT THE COEFFICIENTS HAVE NO UPPER LIMIT RELATED TO SENSETIVITY. THESE RESULTS INDICATE THAT THE COEFFICIENTS FOR RADIO AND TV CAN VARY IN A RANGE ABOVE -96.68 (FOR RADIO) AND -137.86 (FOR TV) WITHOUT CHANGING THE OPTIMUM SOLUTION OF THE MODEL. IF THE COEFFICIENTS DIP BELOW THESE LIMITS, THE OPTIMUM SOLUTION WILL CHANGE.
# 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+301B) 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.
ANSWER: THE TWO NON-ZERO ARE INDICATIONS OF BINDING CONSTRAINTS RELATED TO THE MODEL. AS SUCH ANY CHANGE TO THE BINDING CONSTRAINT WILL CHANGE THE OPTIMUM SOLUTION OF THE MODEL.
SO, IF THE BUDGET CONSTRAINT IS RAISED BY ONE DOLLAR (TO 350,001), THE OPTIMUM SOLUTON WILL INCREASE BY 124.12 DOLLARS.
THE OTHER RESULT ARE NON-BINDING CONSTRAINTS AND CAN THERFORE BE INCREASED WITHOUT AFFECTING THE OPTIMAL SOLUION.
To acquire a better understanding of the sensitivity results, and to confirm integrity of the calculations, independent tests can be conducted.
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).
AS OBSERVED IN THE SNESETIVITY TEST THE FIRST AND FOURTH CONSTRAINS ARE THE ONLY TWO THAT ARE BINDING AND THEREFORE THE ONLY TWO THAT WILL AFFECT THE OPTIMUM SOLUTION OF ALTERED. TO TEST THIS ASSUMPTION I WOULD RUN ANOTHER SOLVER, CHANGING THE BUDGET CONSTRAINT BACK TO 350000 THIS INCREASING THE 4TH CONSTRAINT BY $1. THIS SHOULD INCREASE THE OPTIMUM SOLUTION BY APROXIMATLEY $75 AS INDICATED BY THE RESULTS OF THE SENSETIVITY TEST CONDUCTED ABOVE.
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] 0.01965399# STD calculated from the random distribution of samples
std = sd(sims)
std## [1] 0.8887722# 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] 0.482A) 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.
# 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)
average1 = mean(sims1)
average1## [1] 0.08516007std1 = sd(sims1)
std1## [1] 0.9704709prob1 = sum(sims1 >=0.05)/runs1
prob1## [1] 0.511# number of simulations/samples
runs2 = 10000
# random number generator per defined normal distribution with given mean and standard deviation
sims2 = rnorm(runs2,mean=0.03,sd=0.97)
average2 = mean(sims2)
average2## [1] 0.01912972std2 = sd(sims2)
std2## [1] 0.973625prob2 = sum(sims2 >=0.05)/runs2
prob2## [1] 0.48662B) 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?
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 0.01965399 0.8887722 0.4800
## 2 1000 0.08516007 0.9704709 0.5110
## 3 10000 0.01912972 0.9736250 0.4866ANSWER: FROM OUR KNOWLEDGE OF STATISTICS, WE ASSUME THAT AS THE SAMPLE SIZE INCREASES TOWARDS THE SIZES OF THE ACTUAL POPULATION, THE CLOSER THE PREDICTED RESULTS GET TO THE ACTUAL RESULTS. AS SUCH, WE WOULD ASSUME run2, WITH 10000 SAMPLES WILL PREDICT THE ACTUAL DAILY RETURN OF THE S&P MORE CLOSELY THAT run AND run2 BECAUSE IT USES A LARGER SAMPLE SZE.
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]