title: “INFS 343 Business Analytics” author: John Hynes date: 4/24/2024 output: word_document: default html_document: default html_notebook: default pdf_document: default subtitle: Lab Assignment 05 —
About
In this lab we will focus on sensitivity analysis.
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, we will learn 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")## Warning: package 'lpSolveAPI' was built under R version 4.2.3We 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
lp.control(lpmark, sense="max")## $anti.degen
## [1] "fixedvars" "stalling"
##
## $basis.crash
## [1] "none"
##
## $bb.depthlimit
## [1] -50
##
## $bb.floorfirst
## [1] "automatic"
##
## $bb.rule
## [1] "pseudononint" "greedy" "dynamic" "rcostfixing"
##
## $break.at.first
## [1] FALSE
##
## $break.at.value
## [1] 1e+30
##
## $epsilon
## epsb epsd epsel epsint epsperturb epspivot
## 1e-10 1e-09 1e-12 1e-07 1e-05 2e-07
##
## $improve
## [1] "dualfeas" "thetagap"
##
## $infinite
## [1] 1e+30
##
## $maxpivot
## [1] 250
##
## $mip.gap
## absolute relative
## 1e-11 1e-11
##
## $negrange
## [1] -1e+06
##
## $obj.in.basis
## [1] TRUE
##
## $pivoting
## [1] "devex" "adaptive"
##
## $presolve
## [1] "none"
##
## $scalelimit
## [1] 5
##
## $scaling
## [1] "geometric" "equilibrate" "integers"
##
## $sense
## [1] "maximize"
##
## $simplextype
## [1] "dual" "primal"
##
## $timeout
## [1] 0
##
## $verbose
## [1] "neutral"# Set the objective function coefficients
set.objfn(lpmark, c(273.691, 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 273.691 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] 43210183# 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] 48.3410 -136.8455
##
## $objtill
## [1] 1.00000e+30 2.73691e+021A) 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.
1A) Your explanation for the coefficient sensitivity analysis:
The coefficients for TV and Radio can decrease until they reach their lower limit withing affecting the optimum solution. It will only affect it if they go below the lower limits.
# 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] 123.45767 0.00000 0.00000 -75.11667 0.00000 0.00000 0.00000
## [8] 0.00000
##
## $dualsfrom
## [1] 1.125e+05 -1.000e+30 -1.000e+30 -4.750e+05 -1.000e+30 -1.000e+30 -1.000e+30
## [8] -1.000e+30
##
## $dualstill
## [1] 1.00e+30 1.00e+30 1.00e+30 3.05e+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.
1B) Your Answer:
If the budget increases by 1 the optimul solution will go up by 123.45.
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