1 Question 1

Find the maximum solution to \[Z=4x+3y\] Suppose that the objective function is subject to the following constraints: \[x≥0\] \[y≥2\] \[2y≤25-x\] \[4y≤2x-8\] \[y≤2x-5\]

##Answer

## Success: the objective function is 55
## [1] 7 9
## [1]  1.5 -2.0
## [1] 1e+30 8e+00
## [1]  2  0 -1  0  0
## [1]  2.500e+00 -1.000e+30 -2.375e+01 -1.000e+30 -1.000e+30
## [1] 1.00e+30 1.00e+30 1.25e+01 1.00e+30 1.00e+30

Solution: The maximum z value (and thus, the optimum) that can be obtained while satisfying the given constraints is 55, where x = 7 and y = 9. The sensitivity coefficients go from 1.5 and -2.0 to 1e+30 and 8e+00. The shadow/dual prices of the constraints are 2, 0 and -1, while for the decision variables are 0 and 0, respectively. The shadow/dual prices lower limits of the constraints are 2.500e+00, -1.000e+30 and -2.375e+01, while for the decision variables are -1.000e+30 and -1.000e+30, respectively. Finally, the shadow/dual prices upper limits of the constraints are 1.00e+30, 1.00e+30 and 1.25e+01, while for the decision variables are 1.00e+30 and 1.00e+30, respectively.

2 Question 2

Let say you are working as consultant for a boutique car manufacturer, producing luxury cars. They run on one-month (30 days) cycles, we have one cycle to show we can provide value. There is one robot, 2 engineers and one detailer in the factory. The detailer has some holiday off, so only has 21 days available.The 2 cars need different time with each resource: \[Car A = x\] \[Car B = y\] \[Robot = 3x + 4y ≥ 30\] \[Engineer = 5x + 6y ≥ 60\] \[Detailer = 1.5x +3y ≥ 21\] Car A provides $ 30,000 profit, whilst Car B offers $45,000 profit. At the moment, they produce 4 of each car per month, for $300,000 profit. Not bad at all, but we think we can do better for them.

2.1 Answer

## List of 28
##  $ direction       : int 1
##  $ x.count         : int 2
##  $ objective       : num [1:2] 30000 45000
##  $ const.count     : int 3
##  $ constraints     : num [1:4, 1:3] 3 4 1 30 5 6 1 60 1.5 3 ...
##   ..- attr(*, "dimnames")=List of 2
##   .. ..$ : chr [1:4] "" "" "const.dir.num" "const.rhs"
##   .. ..$ : NULL
##  $ int.count       : int 2
##  $ int.vec         : int [1:2] 1 2
##  $ bin.count       : int 0
##  $ binary.vec      : int 0
##  $ num.bin.solns   : int 1
##  $ objval          : num 330000
##  $ solution        : num [1:2] 2 6
##  $ presolve        : int 0
##  $ compute.sens    : int 0
##  $ sens.coef.from  : num 0
##  $ sens.coef.to    : num 0
##  $ duals           : num 0
##  $ duals.from      : num 0
##  $ duals.to        : num 0
##  $ scale           : int 196
##  $ use.dense       : int 0
##  $ dense.col       : int 0
##  $ dense.val       : num 0
##  $ dense.const.nrow: int 0
##  $ dense.ctr       : num 0
##  $ use.rw          : int 0
##  $ tmp             : chr "Nobody will ever look at this"
##  $ status          : int 0
##  - attr(*, "class")= chr "lp"
## [1] 0
## x y 
## 2 6
## [1] "Total Profit: 330000"

Solution : The total production of Car A = 2 and Car B = 6 to get maximum profit 330.000

3 Question 4

Please visualize a contour plot of the following function: \[f(x,y)=x^2+y^2+3xy\]

## $minimum
## [1] 1.499924
## 
## $objective
## [1] -1.25

the minimum value is obtain 1.499924 and value of objective -1.25

3.0.1 How coordinate descent works

Coordinate descent successively minimizes along coordinate directions to find the minimum of a function. At each iteration, the algorithm determines a coordinate, then minimizes over the corresponding hyperplane while fixing all other coordinates.

3.0.2 How Steepest descent works

The method of steepest descent or stationary-phase method or saddle-point method is an extension of Laplace’s method for approximating an integral, where one deforms a contour integral in the complex plane to pass near a stationary point (saddle point), in roughly the direction of steepest descent or stationary phase.Depending on the starting value, the steepest descent algorithm could take many steps to wind its way towards the minimum.

3.0.3 How Newton Method works

The Newton-Raphson method (also known as Newton’s method) is a way to quickly find a good approximation for the root of a real-valued function \(f(x)=0f(x)=0\). It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.

3.0.4 How Conjugate Gradient works

Conjugate gradient methods represent a kind of steepest descent approach “with a twist”. With steepest descent, we begin our minimization of a function \(f\) starting at \(x_0\) by traveling in the direction of the negative gradient \(−f′(x_0)\). In subsequent steps, we continue to travel in the direction of the negative gradient evaluated at each successive point until convergence.