Business Analytics Lab Worksheet 08
CME Group Foundation Business Analytics Lab
Maura Walde
March 29, 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: The sensitivity result represents the lower boundary for the radio and tv coefficient values. The first value shows the lower boundary for radio and the second displays the lower boundary for tv. Therefore, the optimum solution will not change unless the coefficients for radio and tv go below the ‘objfrom’ values.
# 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 values display the binding constraints, the first and fourth constraints. The first value displays what will happen if we increase the first constraint. If we increase the 350,000 to 350,001, the sales value will increase by 124.12433. The second value, 75.7833, demonstrates that if the fourth constraint is increased from 0 to 1 the sales value will increase by 75.78333. Thus, the binding constraints are one and four. The zeros represent non-binding constraints, two, three, five and six. The zero, non-binding values also represent the constraints that have surplus/slack. The surplus for constraint 2 is 101,666.7. The surplus for constraint 3 is 158,333.3. The surplus for constrain 5 is 116666.7 and for constraint 6 it is 233333.3. The binding values do not have any surplus or slack. Non-binding values, the zeros, have no marginal value. The marginal values are 124.12433 and 75.78333, the values that correspond to the binding constraints.
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
dummy = 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)
lpmark1 ## Model name:
## C1 C2
## Maximize 275.691 48.341
## R1 1 1 <= 350001
## 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 0solve(lpmark1) ## [1] 0get.objective(lpmark1) ## [1] 43443641get.variables(lpmark1)## [1] 116667 233334Optimum value for sales is 43443641. ##### TASK 4: Calculate the differential change in sales. Share your observations. ## ANSWER TASK 4: 43443641- 43443517= 124. This proves that increasing the first constraint by 1 increases sales by 124 as predicted in task 2.
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
dummy = lp.control(lpmark2, sense="max")
set.objfn(lpmark2, c(275.691, 48.341))
add.constraint(lpmark2, c(1, 1), "<=", 350000)
add.constraint(lpmark2, c(1, 0), ">=", 15000)
add.constraint(lpmark2, c(0, 1), ">=", 75000)
add.constraint(lpmark2, c(2, -1), "=", 1)
add.constraint(lpmark2, c(1, 0), ">=", 0)
add.constraint(lpmark2, c(0, 1), ">=", 0)
lpmark2 ## Model name:
## C1 C2
## Maximize 275.691 48.341
## R1 1 1 <= 350000
## R2 1 0 >= 15000
## R3 0 1 >= 75000
## R4 2 -1 = 1
## R5 1 0 >= 0
## R6 0 1 >= 0
## Kind Std Std
## Type Real Real
## Upper Inf Inf
## Lower 0 0solve(lpmark2) ## [1] 0get.objective(lpmark2) ## [1] 43443592get.variables(lpmark2)## [1] 116667 233333Optimum value for sales= 43443592. #### TASK 6: Calculate the differential change in sales. Share your observations. ## ANSWER TASK 6: 43443592- 43443517= 75. This proves that increasing the fourth constraint by 1 increases sales by 75 as predicted in task 2.
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] 0.003371263# STD calculated from the random distribution of samples
std = sd(sims)
std## [1] 0.8024776# 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] 0.45TASK 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 # Repeat calculations here
runs1000 = 1000
sims1000 = rnorm(runs1000, mean=0.03, sd=0.97)
average = mean(sims1000)
average ## [1] 0.01086403std = sd(sims1000)
std ## [1] 0.9695666prob = sum(sims1000 >=0.05)/runs1000
prob## [1] 0.478Mean= 0.04905686 Standard deviation= 1.007553 Probability= 0.486 ##### TASK 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)
average = mean(sims)
average ## [1] 0.02986333std = sd(sims)
std ## [1] 0.9720028prob = sum(sims >=0.05)/runs
prob## [1] 0.4883Mean= 0.04809597 Standard deviation= 0.9700968 Probability= 0.4991 ##### TASK 9: List in a tabular form the values for mean, standard deviation, and probability for all three cases: 100, 1000, and 10000 simulations. 100 = 0.1316492, 0.9888453, 0.48 1000 = 0.04905686, 1.007553, 0.486 10000 = 0.04809597, 0.9700968, 0.4991 ##### 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: The mean gets smaller as the number of simulations is increased. The standard deviation increases and then decreases as the number of simulations increases. The probability increases as the number of simulations increases. The 10,000 simulation has the highest probability at 0.4991. Therefore it is most likely that there will be an occurance greater than 5% in this simulation than in the others. This is similar to the image in the introductory paragraph because in the image, the higher the number of sampled points, the closer the estimate of pi to the actual pi number.
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]