Business Analytics Lab Worksheet 08
CME Group Foundation Business Analytics Lab
Dehler Amoroso
04/05/2018
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.
PART A: 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+30TASK 1: 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 TASK 1: From the results found by R it may be concluded that the coeffecients of X1 & X2 have lower limits of -96.682 & -137.46. The Coeeffecient of X1 can be changed without affecting the optimal sultion until it reaches -96.682; the same can be said for X2 until it reaches -137.846. Both coeefecients have infinite upper limits, meaning that the optimum solution would never change as it continues to rise higher and higher.
# 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+30TASK 2: 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 TASK 2: The two non-zero sensitivity results represent binding contraints because any marginal change in the resource (right hand side constraints) will result in a change in the optimum value. These two are different than the 0-rendering results, as those are non-binding constraints. Differing from non-binding constraints, our binding constraints depicted here have no room for surplus/slack, as a $1 change in one or both of the resources (right hand constraints) will render a marginal $124.12 and/or a $75.78 change in the optimum value, respectively.
To acquire a better understanding of the sensitivity results, and to confirm integrity of the calculations, independent tests can be conducted.
TASK 3: 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.
# 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
dummy1 = lp.control(lpmark1, sense="max")
set.objfn(lpmark1, c(275.691, 48.341))
add.constraint(lpmark1, c(1, 1), "<=", 350001)
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)
solve(lpmark1) ## [1] 0get.objective(lpmark1)## [1] 43443641ANSWER TASK 4: lpmarksales=43443517 & lpmark1sales=43443641. By chaning the first constraint by $1 when dealing with the budget it would cause a marginal change of $124 in sales. This proves that the constraint of x1+x2<= 350,000 is a binding constraint. The change of $1 casused a $124 change in sales which is significant.
TASK 5: Running the linear programing solver again starting from the begining, by defining a new model object lpmark2.All being equal, change the constraint 2X1 - X2 = 0 by only $1 and solve. The new constraint will be 2X1 - X2 = 1. Note the optimum value for sales as given by the objective function.
# Define a new model object called lpmark2
lpmark2 = make.lp(0, 2)
# Repeat rest of commands with the above constraint changed. Solve and display the objective function optimum value
dummy2 = lp.control(lpmark2, sense="max")
set.objfn(lpmark2, c(275.691, 48.341))
add.constraint(lpmark2, c(1, 1), "<=", 350001)
add.constraint(lpmark2, c(1, 0), ">=", 15000)
add.constraint(lpmark2, c(0, 1), ">=", 75000)
add.constraint(lpmark2, c(2, -1), "=", 0)
add.constraint(lpmark2, c(1, 0), ">=", 0)
add.constraint(lpmark2, c(0, 1), ">=", 0)
solve(lpmark2) ## [1] 0get.objective(lpmark2)## [1] 43443641ANSWER TASK 6: lpmark2sales = 43443641. There is no marginal change from changing the $1 to $2 in the first constraint. The difference between the optimum values of lmpark and lpmark2 is still $124. This is Non-binding and has no impact when changing the constraint from $1 to $2; there is no marginal value between the change of lpmark2 & lpmark1.
PART B: 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] 0.09334629# STD calculated from the random distribution of samples
std = sd(sims)
std## [1] 0.9423356# 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] 0.53TASK 7: Repeat the above calculations for the case where the number of simulations/samples is equal to 1000. record the mean, standard deviation, and probability.
# Repeat calculations here
runs = 1000
sims = rnorm(runs, mean = 0.03, sd = 0.97)
average1 = mean(sims)
average1## [1] -0.01781955std1 = sd(sims)
std1## [1] 0.9395089prob1 = sum(sims >= 0.05)/runs
prob1## [1] 0.463TASK 8: Repeat the above calculations for the case where the number of simulations/samples is equal to 10000. record the mean, standard deviation, and probability.
# Repeat calculations here
runs = 10000
sims = rnorm(runs,mean=0.03,sd=0.97)
average2 = mean(sims)
average2## [1] 0.0451867std2 = sd(sims)
std2## [1] 0.976123prob2 = sum(sims >=0.05)/runs
prob2## [1] 0.4884TASK 9: List in a tabular form the values for mean, standard deviation, and probability for all three cases: 100, 1000, and 10000 simulations.
as.table(c(average, average1, average2))
A B C - 0.01666154 -0.02633712 0.03860424
as.table(c(std, std1, std2))
A B C1.092644 0.9733053 0.9620109
as.table(c(prob, prob1, prob2))
A B C 0.47 0.486 0.4962
TASK 10: 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?
ANSWER TASK 10: From the simulations/samples of all 3 cases the average stayed close to 0. The simulations taken from the probability showed that their is a 50% chance that an occurrence greater than 5% would occur.The similuarity between the pi in the beginning of the assignment shows that the larger amount of samples/simulations the more precise or accurate the data becomes.
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]