Linear Programming (Business)

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A company makes two products (bicycles and scooters) using two machines (A and B).

Each bicycle that is produced requires $\var{XtimeA}$ minutes processing time on machine A and $\var{XtimeB}$ minutes processing time on machine B.

Each scooter that is produced requires $\var{YtimeA}$ minutes processing time on machine A and $\var{YtimeB}$ minutes processing time on machine B.

Either Machine A or Machine B will be used given there is only one available operator.

At the start of the current week there are $\var{Xstock}$ bicycles and $\var{Ystock}$ scooters in stock.

Available processing time on machine A is forecast to be $\var{Atime}$ hours and on machine B is forecast to be $\var{Btime}$ hours.

The demand for bicycles in the current week is forecast to be $\var{Xforecast}$ units and for scooters is forecast to be $\var{Yforecast}$ units.

Company policy is to maximise the combined sum of the units of bicycles and the units of scooters in stock at the end of the week.

Formulate the problem of deciding how much of each product to make in the current week as a linear program.
Solve this linear program graphically.
Hence decide which machine is to be used and how many bicycles and scooters to produce on that machine.

(Author of original problem: J E Beasley: http://people.brunel.ac.uk/~mastjjb/jeb/or/rights.html)

Let $x,\;y$ be the number of bicycles and scooters produced respectively.

0,0

Stock left = 0.00

Stock left

P

What are the constraints on production given the forecasts for bicycles and scooters and the stock levels?

$x \ge\;$ $y \ge\;$ Expected answer:

Time inequality for Machine A: $x\;+\;$ $y\;\le\;$ Expected answer: (in minutes)

Time inequality for Machine B: $x\;+\;$ $y\;\le\;$ Expected answer: (in minutes)

Input the objective function which maximises Stock left.

Objective function: Expected answer:

Note that the objective function Stock left is displayed above for the values of $x$ and $y$ which just clear demand i.e. Stock left=0. Once you have input all of the constraints you can use the slider to move the objective function to the value of $x$ and $y$ in the Feasible region to maximise Stock left.

Using this diagram find values of the production of $x$ and $y$ which satisfy the constraints and maximise the amount of stock $x+y$ left at the end of the week.

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0,0

Machine B

Machine A

T

P

Q

R

Stock left = 25.70

Stock left

solutions:x= 58,y= 36.7

The constraints are:

$\var{xtimea}x + \var{ytimea}y \leq \var{atime}\times 60 =\var{atime*60}$ machine A time.

$\var{xtimeb}x + \var{ytimeb}y \leq \var{btime}\times 60 =\var{btime*60}$ machine B time.

$x \geq \var{xforecast} -\var{xstock}=\var{xforecast-Xstock}$

This ensures that production of bicycles $\geq$ demand ( $\var{xforecast}$ ) - initial stock ( $\var{xstock}$ ) and so meets demand.

$y \geq \var{yforecast} -\var{ystock}=\var{yforecast-ystock}$ .

This ensures that production of scooters $\geq$ demand ( $\var{yforecast}$ ) - initial stock ( $\var{ystock}$ ) and so meets demand.

The objective is: maximise $(x+\var{xstock}-\var{xforecast}) + (y+\var{ystock}-\var{yforecast}) = \simplify{x+y+{xstock+ystock-xforecast-yforecast}}$
i.e. to maximise the number of units left in stock at the end of the week

The Feasible region is the green area and we have moved the Stock left objective function line to the position where it intersects the Feasible region and maximises Stock left.

We find using the graphical method that using Machine : $x=\;$ , $y=\;$ 36.7, Stock left = 25.7

This occurs at the point R in the above diagram.

Your graphical solution may differ slightly from these as it is not always possible to get the objective line accurately through the desired point. Also, if these numbers are not whole numbers then you can easily find a whole number solution near these and any such reasonable solution near the above solutions is fine as long as it lies in the Feasible region.

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snapjacks and snaxz are two popular products.

There must be at least 9 batches of snapjacks and 6 batches of snaxz produced each day to meet ordered sales.

Each snapjack batch needs 23g porridge oats and 380g cocoa every day.

Each snaxz batch needs 52g porridge oats and 2900g cocoa every day.

However, the resources are limited:

1720g of porridge oats each day.

75900g of cocoa each day.

Each sale of a unit of snapjacks gives £2 profit. Each sale of a unit of snaxz gives £7 profit.

0,0

Profit = 60.00

Profit

Resource 1

Resource 2

P

Let $x,\;y$ be the number of units of snapjacks and snaxz produced respectively.

What are the constraints on production ?

$x \ge\;$ Expected answer: $y \ge\;$ Expected answer:

Now input the constraints given by availability of porridge oats and cocoa:

Porridge oats: Expected answer: $x\;+\;$ Expected answer: $y\;\le\;$ Expected answer:

cocoa: Expected answer: $x\;+\;$ Expected answer: $y\;\le\;$ Expected answer:

Given the information on profits for snapjacks and snaxz input the objective function which is to be maximised: Expected answer:

Hence by moving the slider find an approximate value for the maximum profit and also solve the two equations which intersect at the maximum profit point and so find the amounts of snapjacks and snaxz to be produced to achieve this maximum profit.

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Advice

The solution

0,0

Resource 1

Resource 2

I

P

Q

R

Profit = 207.50

Profit

solutions:x= 22.2,y= 23.3

It is clear that the minimum profit is given by producing enough snapjacks and snaxz so that demand is met and no more. You can see this by moving the slider so that the objective function line goes through the point giving the demand. Note that this is given in the diagram in the statement of the question.

Solving the pair of equations given by the lines which intersect at point I we find that a solution for maximum profit is:

Number of snapjacks to be produced, $x=\;$ 22.2 (to one decimal place).

Number of snaxz to be produced, $y=\;$ 23.3 (to one decimal place).

If these are not whole numbers, then you will have to adjust them accordingly to get realistic solutions.

A chocolate manufacturer produces two types of chocolate bar: Asteroids and Blackholes.

Production of an Asteroid bar uses {xresource}g of cocoa and {xtime} minute(s) of machine time, whereas a Blackhole bar requires {yresource}g of cocoa and {ytime} minute(s) of machine time.

Altogether, {resource}kg of cocoa are available each day.

The company employs a single machine operator who works on the production of these bars for {timeavailable} hours per day.

The manufacturer must make at least {xconstraint} Asteroid and {yconstraint} Blackholes each day to keep up with demand.

The manufacturer makes {xprofit}p profit from each Asteroid bar and {yprofit}p profit from each Blackhole bar.

(a) Formulate the chocolate manufacturer’s situation as a linear programming problem and input this information in the fields below.

(b) After this input a suitable diagram will be drawn on screen to enable the problem to be solved graphically, but will give approximate solutions.

(c) Use your diagram to find an approximate value for the company’s minimum and maximum profit, £P.

(d) Using the information given by the diagram, once you have completed all inputs, you can identify and solve the pair of equations in order to find the production that achieves maximum profit. This give a more accurate solution, but you may meed to adjust to obtain whole numbers of products!

Let $x,\;y$ be the number of units of Asteroids, Blackholes produced respectively.

{userinput(xresource,yresource,xtime,ytime,xconstraint,yconstraint,resource,timeavailable,xprofit,yprofit)}

What are the constraints on production.?

$x \ge\;$ Expected answer: $y \ge\;$ Expected answer:

Now input the constraints given by availability of cocoa and time:

Cocoa: Expected answer: $x\;+\;$ Expected answer: $y\;\le\;$ Expected answer: (in grammes)

Time: Expected answer: $x\;+\;$ Expected answer: $y\;\le\;$ Expected answer: (in minutes)

Input the objective function which maximises profit.

Objective function : Expected answer:

The profit objective function is displayed above as a green dashed line. It is initially set at the profit if demand is just met.

Once you have input all of the constraints you can use the slider to move the objective function to the value of $x$ and $y$ in the Feasible region to maximise profit.

Using this diagram find values of the production of $x$ and $y$ which satisfy the constraints and maximise the profit in a day's production by identifying the equations you need to solve algebraically.

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Advice

{diagram(xresource,yresource,xtime,ytime,xconstraint,yconstraint,resource,timeavailable,xprofit,yprofit)}

It is clear that the minimum profit is given by producing enough toys so that demand is met and no more. You can see this by moving the slider so that the objective function line goes through the point giving the demand. Note that this is given in the diagram in the statement of the question.

Solving the pair of equations given by the lines which intersect at point we find that a solution for maximum profit is:

Number of Bears to be produced, $x=\;$ (to one decimal place).

Number of Cats to be produced, $y=\;$ (to one decimal place).

If these are not whole numbers, then you will have to adjust them accordingly to get realistic solutions.

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