1LINEAR PROGRAMMING IFORMULATING LINEAR

PROBLEMS

Lecture 3

Lecturer : Dr. Dwayne Devonish

MGMT 2012: Introduction to Quantitative MethodsLearning Objectives

Students should be able to:

To outline the key steps of formulatinglinear programming problemsmathematically,

To identify the various components of alinear problem,

To apply these linear programming modelformulations in real-life businesssituations.

What is Linear Programming? Many managerial decisions concern achieving

business objectives (being effective) but doing sowithin limited resources (being efficient).

These constrained resources can manifest astime, labour, energy, material, or money, etc.

Two major objectives of firms are:

Maximise (increase) profit

Minimise (reduce) costs

Managers who seek to either increase profits orminimise costs subject to constrained or limitedresources tend to utilise a scientific quantitativetechnique called linear programming to makesound decisions to achieve their objectives.

Linear Programming (LP) Linear programming is a mathematical technique that

managers use to solve linear programming problems these problems involve making decisions aboutmaximising profits or minimising costs withinconstrained conditions. See examples:

A manager seeking to maximise profits frommanufacturing wooden desks but is restricted by theamount of wood and number of labour hours available.

A financial analyst seeking to maximise return oninvestment but is constrained by the total amount ofinvestment funds available and the maximum amountsthat can be invested in each stock or bond.

A manager seeking to minimise production costs of twoproducts but must satisfy customer demand andoperate within limited processing time.

Components of a Linear Programming Problem

There are several components of a linear programmingproblem:

Decision variables: the quantity that a decision-makerhas control over; choices the decision-maker has interms of amounts. We represent them as x symbols(e.g. x1, x2, etc) which are unknown quantities.

Objective function: A linear mathematical relationshipthat describes the objective of the problem (i.e.maximising profit or minimising costs)

Constraints: Limitations that restrict thealternatives/choices of a decision-maker. Representlimited resources such as labour, time, money, etc, orspecific restrictive requirements.

Constraints Constraints may be based on limited availability or

capacity regarding necessary resources.

For example, one cannot manufacture unlimitedtables and chairs because the availability of wood (anecessary resource) within a plant is limited.

Hence, the quantity of wood used to make tablesand chairs cannot exceed the quantity of woodavailable, although the objective is to increase profits(produce as much output as possible).

Constraints may also specify a minimum requirementwhich says one must produce at least a certainquantity of some product; so this too affects anorganisations decisions (especially to reduce costs)

2Note on Constraints Constraints (contd): Constraints can be expressed

using symbols such as less than or equal to (), orgreater than or equal to (), or equal to (=).

A constraint suggests an upper limit on the amount ofsome scarce resource available for use. There are a lefthand side and right hand side of the constraint sign.For example, suppose one was making tables and chairsfrom certain type wood: the quantity of wood used tomake tables and chairs cannot exceed the quantity ofwood available. Lets say it takes 7 feet of wood to makeone table (where x1 = quantity of tables to make) and 4feet to make one chair (where x2 = quantity of chairs tomake), and there are only 100 feet of wood available.The constraint would be written mathematically as: 7X1 +4X2 100.

The left hand side is the total quantity of wood used (7x1+ 4x2) for tables and chairs, and right hand is the totalavailable (100) the left hand side cannot exceed righthand side.

Note on Constraint

A constraint suggests a minimum that must beachieved. For example, lets say a manager hasto be produce at least 100 tables for a customerorder. This is a minimum requirement he or shehas to produce at least 100 tables: X1 100.

A = constraint specifies exact number or amount.If manager has a specific quota to stick to: e.g.must produce exactly 50 chairs: X2 = 50.

Non-negativity constraints are special constraintsthat suggest that you cant produce negativeamounts/quantities (e.g. -200 cars doesnt makesense). You have to specify these constraints as:X1, X2 0.

Steps in LP Linear programming involves two general

phases:

Formulation where a decision problemrequiring linear programming is formulatedusing mathematical terms: identify andestablish decision variables, objective functionand linear constraints.

Solution apply graphical or mathematicalmethods to determine unknown quantities of thedecision variables (e.g. x1, x2, etc)

Steps in LP Step 1: Identify decision variables (i.e. x1, x2, etc)

Step 2: Formulate the objective function (i.e. profitmaximisation or cost minimisation)mathematically

Step 3: Identify and write out constraintsmathematically

Step 4: Solve LP problem (using graphical orcomputer software)

Step 5: Conduct sensitivity analysis

For this lecture, we will focus on formulationphase only which covers Steps 1 to 3.

Steps 4 and 5 deals with solution phase = nextweek.

Example of Linear Problem A potter is making cups and plates. It takes her 6

minutes to make a cup and 3 minutes to make aplate. Each cup uses 1lb. of clay and each plateuses 2 lbs. of clay. She has 20 hours available formaking the cups and plates, and has 250 lbs. ofclay on hand. She makes a profit of $2 on eachcup and $3 on each plate. How many cups andhow many plates should she make in order tomaximize her profit?

We must formulate linear programming problem torepresent this scenario. Lets follow these steps.

Linear Programming Table for Maximization Problem

Decision Variables

Constraints

(No. of cups made) X1

(No. of plates made) X2

Total Availability

Time

(minutes) 6 3

20 hrs

(i.e. 1200 mins)

Clay (lbs) 1 2 250 lbs

Profit $2 $3

Objective: To maximize profit

3Steps in Formulation Step 1: Identify decision variables: The decision

variables represent (1) the number of cups and (2)the number plates to produce. The quantities tobe produced can be represented as:

X1 = Number of cups to be produced X2 = Number of plates to be produced Step 2: Write out the objective function: The

objective of the potter is to maximise profit. Profitderived from cups is determined by multiplying theunit profit ($2) by the number of cups produced(x1). Likewise, profit derived from plates isdetermined by multiplying the unit profit ($3) by thenumber of plates produced (x2). So we have towrite out the objective function as:

Steps in Formulation Maximise profit = 2X1 + 3X2 : recall unit profit

($2, $3) multiplied by number of units produced(x1, x2).

Step 3: Identify and write out constraints ofdecision variables There are two (2) resourcesthat have limited availability: time (hours)constraint and amount of clay constraint. Giventhat these resources are limited, we have to use constraints (i.e. they have an upper limit).

In terms of time, a cup takes 6 minutes so the timetaken to produce 1 cup is 6X1 minutes, and a platetakes 3 minutes so the time taken to produce 1plate is 3X2

Hence, the total number of hours for producingboth products = 6X1 + 3X2

Steps in Formulation However, the total number of hours (6X1 + 3X2) is

limited to 20 hours (or converted to minutes= 1200minutes)

Hence, the time constraint (Constraint 1): 6X1 + 3X2 1200 minutes : This suggests that

the number of cups and plates that can be usedmust generate hours that are less than or equal to20 hrs (1200 minutes).

The clay constraint is defined similarly. Each cupuses 1lb. of clay and each plate uses 2 lbs. of clay.250 lbs of clay is the maximum amount of clayavailable.

Hence, the clay constraint (Constraint 2): 1X1 + 2X2 250 lbs

Problem formulation Hence, the full linear programming problem is

written as: Maximise profit: 2x1 + 3x2

Subject to: Time in hours : 6X1 + 3X2 1200 Amount of clay : 1X1 + 2X2 250 Non-negativity : X1, X2 0

This is the problem formulation stage we willconsider how to solve this later (i.e. determineoptimal number of cups and plates to produce thatdo not violate any of these constraints). Recall thatthe optimal numbers of products here (x1 and x2)must satisfy all constraints.

Linear Programming Table for Maximization Problem

Decision Variables

Constraints

(No. of cups made) X1

(No. of plates made) X2

Total Availability

Time

(minutes) 6 3

20 hrs

(i.e. 1200 mins)

Clay (lbs) 1 2 250 lbs

Profit $2 $3

Objective: To maximize profit

Lets do another example A farmer owns 45 acres of land. He plants wheat and

corn. An acre of wheat planted consumes 1 acre ofland, and each acre of corn planted consumes 1 acreof land. Each acre of wheat requires 3 hours oflabour, and each acre of corn requires 2 hours oflabour. The total number of labour hours per weekavailable is 100 hours. Each acre of wheat requires 2tons of fertilizer, and each acre of corn requires 4 tonsof fertilizer. The farmer has a total of 120 tons offertilizer available. Each acre planted with wheatgenerates a profit of $200, whereas each acre plantedwith corn generates a profit of $300. How many acresof wheat and corn must be planted for the farmer tomaximise profit?

4 A farmer owns 45 acres of land. He plants wheat andcorn. An acre of wheat planted consume 1 acre ofland, and each acre of corn planted consume 1 acre ofland. Each acre of wheat requires 3 hours of labour,and each acre of corn requires 2 hours of labour. Thetotal number of labour hours per week available is 100hours. Each acre of wheat requires 2 tons of fertilizer,and each acre of corn requires 4 tons of fertilizer. Thefarmer has a total of 120 tons of fertilizer available.Each acre planted with wheat generates a profit of$200, whereas each acre planted with corn generatesa profit of $300. How many acres of wheat and cornmust be planted for the farmer to maximise profit?

Linear Programming Table

Constraints

Decision Variables

Total

Availability

X1 (No. of acres of wheat

X2 (No. of acres of

corn

Land (acres)

1 1 45

Labour (hours)

3 2 100

Fertilizer (tons)

2 4 120

Profit $200 $300

Linear Programming Formulation

Hence, the problem:

Let x1 = number of acres of wheat

Let x2 = number of acres of corn

Maximise profit: 200X1 + 300X2Subject to:

Land constraint: 1X1 + 1X2 45 acres

Labour hours constraint: 3X1 + 2X2 100 hrs

Fertilizer constraint: 2X1 + 4X2 120 tons

X1, X2 0

So far we have looked at examples of maximization, now lets look at a minimization problem.

Phils Chemicals produces 2 products: Glow-one andGlow-two. Based on analysis of inventory and customerdemand levels for the upcoming month, the companymust ensure that they manufacture at least 350 gallonsof Glow-one and Glow-two combined. Separately, amajor customers order of 125 gallons of Glow-onemust be satisfied. Glow-one requires 2 hours ofprocessing time per gallon, and Glow-two needs 1 hourof processing time, and there are 600 hours ofprocessing time available for the next month. Thecompany wants to satisfy these requirements at aminimum total production cost. Production costs forGlow-one are $5 per gallon, and $4 per gallon forGlow-two. What is the optimal number of gallons ofeach product to produce under a minimum costschedule?

Lets define steps again

Step 1: Define decision variables:

X1 = number of gallons of Glow-one

X2 = number of gallons of Glow-two

Step 2: Write objective function: Given the cost ofGlow-one is $5 per gallon, and Glow-two is at $4per gallon. The objective function is:

Minimise Cost: 5X1 + 4X2

Lets define steps again

Step 3: Identify and write out each constraint:

Constraint 1: The company must produce at least350 gallons of the two products combined, hence:

1X1 + 1X2 350

Constraint 2: The company must satisfy thecustomer demand for 125 gallons of Glow-one,hence at least 125 gallons of Glow-one must beproduced.

1X1 125

5Lets define steps again

Step 3: Identify and write out each constraint:

Constraint 3: Finally, the company has limitedprocessing time to make the products: 600 hours;recall that Glow-one requires 2 hrs, and Glow-tworequires 1 hr; so:

2X1 + 1X2 600

Recall that the final constraint for these problemsis that you cant make negative gallons of eitherproduct so:

X1, X2 0

Linear program for the problem

Minimise cost: 5X1 + 4X2Subject to:

Total production constraint: 1X1 + 1X2 350

Demand for Glow-one constraint: 1X1 125

Processing time constraint: 2X1 + 1X2 600

Non-negativity constraint: X1, X2 0

Another example

Grans Meat Packaging Company produces a hotdog mixture in 1,000 pound batches. The mixturecontains two ingredients chicken and beef. Thecost per pound of chicken is $3, and $5 per poundof beef. Each 1000-pound batch must have atleast 500 pounds of chicken, and at least 200pounds of beef. The company wants to know theoptimal mixture of ingredients that will minimizecost.

Formulation:

Step 1: Define decision variables:

X1 = lb of chicken

X2 = lb of beef

Step 2: Write objective function: Given the cost ofa pound of chicken is $3, and beef is $5 perpound. The objective function is:

Minimise Cost: 3X1 + 5X2

Lets define steps again

Step 3: Identify and write out each constraint:

Constraint 1: Recall that each batch must consistof 1000 pounds of mixture, hence:

1X1 + 1X2 = 1000

Constraint 2: Each batch must consist of at least500lb of chicken.

1X1 500

Constraint 3: Each batch must also consist of at least 200 lb of beef.

1X2 200

Linear program for the problem

Minimize cost: 3X1 + 5X2Subject to:

1X1 + 1X2 = 1000

1X1 500

1X2 200

X1, X2 0

6Different types of constraints

Ratio Constraints Constraints can be expressed in different ways

given the nature of restrictions. For example, lets go back to last example with

the chicken and beef mixture. Lets say for each 1,000 batch produced, the

ratio of chicken (X1 ) to beef (X2) must be at least 2 to 1. Hence, twice as much as chicken as beef. This constraint is expressed as a ratio.

X1 / X2 2 / 1 You have now to cross-multiply this constraint,

and you will get 1X1 2X2

Ratio Constraints

1X1 2X2 You have now to ensure that the right-side of the

equation is a constant (a number; usually a zero), and all variables in a constraint must be on the left side of the inequality sign.

To obtain this, you have to subtract the variable amount on the right side from both sides. This creates a zero on the right side andL

1X1 2X2 0

Extension of programL

Minimize cost: 3X1 + 5X2Subject to:

1X1 + 1X2 = 1000

1X1 500

1X2 200

1X1 2X2 0

X1, X2 0

Percentage Constraints A constraint can also specify a percentage for one

or more variables relative to one or more othervariables.

Lets say we need beef to be no more than 70% ofthe overall mixture (i.e. chicken and beef mixture).

So mathematically: X2 .70(X1 + X2) or

X2 .70X1 + .70X2 We have to move all variables to the left side, so

we can leave a constant (i.e. 0) on the right only.So subtract the right side from both sides.

So X2 .70X1 + .70X2 Answer : -.70X1 + .30X2 0

Extension of programL

Minimize cost: 3X1 + 5X2Subject to:

1X1 + 1X2 1000 1X1 500 1X2 200

1X1 2X2 0 -.70X1 + .30X2 0

X1, X2 0

7END OF LECTURE

Download tutorial assignment forformulating linear programming problems.

Read Chapter 7 (Linear programming) focus on formulation (not on solving) aswell as Chapter 8 of Render et. al.

Next week: Solving Linear ProgrammingProblems