Quantitative Methods

Linear Programming

Definitions • Linear Programming is one of the

important Techniques of OR

• It is useful in solving decision making problems which involves

• optimising a linear objective function

• subject to a set of linear constraints

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Examples• Selection of product mix which maximises

profits subject to production, material, marketing, personnel and financial constraints

• Determination of capital budget which maximises NPV of the firm subject to financial, managerial, environmental and other constraints

• Choice of mixing short term financing which minimises cost subject to certain funding constraints

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Formulation of LP problem DECISION VARIABLES

Are the variables whose optimum values are to be found out by applying LP technique

OBJECTIVE FUNCTION

The part of a linear programming model that expresses what needs to be either maximised or minimised depending on the objective for the problem

Formulation of LP problem

CONSTRAINTS• It is an inequality or equation that expresses

some restriction on the values that can be assigned to decision variables

• The constraints which represent non negativity conditions are called non negativity constraints

• The other constraints which represent restrictions on availability of resources etc are called structural constraints

Formulation of LP problem

FEASIBLE SOLUTION

• A solution represents specific combination of values of decision variables

• A feasible solution is one that satisfies all constraints whereas an infeasible solution violates at least one constraint

• The optimal solution is the best feasible solution according to the objective function

New Office Furniture Ltd• The new office furniture produces Desks, Chairs and

Moulded Steel with the profit and raw material usage per unit as given below. The total availability of raw material for production is 2000kg. To satisfy contract commitments at least 100 desks must be produced. Due to the availability of seat cushions no more than 500 chairs must be produced Find out the optimal product mix.

• Products Profit Raw Steel Used• Desks Rs500 7 kg per Desk• Chairs Rs300 3 kg Per Chair• Moulded Steel Rs60/ Kg 1.5 kg per Kg of moulded Steel

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OBJECTIVE FUNCTION

• D: amount of desks (number)

• C: amount of chairs (number)

• M: amount of moulded steel (Kgs)

Maximise Total Profit = 500 D + 300 C + 60 M

CCNSTRAINTS• New Office has only 2000 Kgs of raw steel

available for production.

7 D + 3 C + 1.5 M ≤ 2000• To satisfy contract commitments;

at least 100 desks must be produced.

D ≥ 100• Due to the availability of seat cushions,

no more than 500 chairs must be produced

C ≤ 500• No production can be negative;

D ≥ 0, C ≥ 0, M ≥09

Example Mathematical Model

• MAXIMIZE Z = 50 D + 30 C + 6 M (Total Profit)• SUBJECT TO: 7 D + 3 C + 1.5 M ≤ 2000 (Raw Steel)• D ≥100 (Contract)• C ≤500 (Cushions)• D, C, M ≥0 (Nonnegativity)• D, C are integers• Best or Optimal Solution of New Office Example• 100 Desks, 433 Chairs,• 0 Molded Steel• Total Profit:Rs180000

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Example

Graphical Solution Method

Graphical Solution Method

• It is applicable when there are two decision variables

• The decision variables are represented by horizontal & vertical axis

• Straight lines are used to demarcate the feasible region

• The feasible region shows the solutions that satisfy all constraints

• Optimal solution lies at one of the corner points

Example• A firm produces Two types of frames ,Type 1 and Type

2 .Each type 1 frame contributes a profit of Rs.225, whereas each type 2 frame contributes a profit of Rs.260.There are 4000 labor hours available. Each type 1 frame required 2 labor hours, and each type 2 frame requires 1 labor hour. There are 5000Kgs of metal available. Each type 1 frame requires 1Kg of metal and each type 2 frame requires 2 Kg of metal.

• Formulate the linear programming problem assuming that the demand exists for both the products.

• How many frames of each type should be produced to realize the optimal profit? (Use graphical method).What is the optimal profit?

Background Information• Each type 1 frame contributes a profit of

Rs.225, whereas each type 2 frame contributes a profit of Rs.260.

• The first constraint is a labor hour constraint. There are 4000 hours available. Each type 1 frame required 2 labor hours, and each type 2 frame requires 1 labor hour.

• Similarly, the second constraint is a metal constraint. There are 5000Kgs of metal available. Each type 1 frame requires 1Kg of metal and each type 2 frame requires 2 Kg of metal.

Model Formulation

• Let X1be number of frames of type I to be produced

• Let X2 be number of frames of type II to be produced

• The algebraic model is given below:

max 225x1 + 260x2 (profit objective)

subject to

2x1 + x2 4000 (labor constraint)

x1 + 2x2 5000 (metal constraint)

x1, x2 0 (non negativity constraint)

Solution

• The idea is to graph the constraints on a two-dimensional graph to see which points (x1, x2) satisfy all of the constraints. This set of points is labeled the feasible region. Then we see which point in the feasible region provides the largest profit.

• The graphical solution appears on the next slide.

Graphical Solution

Solution -- continued• To produce the graph, we first locate the lines

where the constraints hold as equalities.• For example, the line for labor is 2x1 + x2 =

4000. The easiest way to graph this is to find the two points where it crosses the axes.

• Joining the points (0,4000) and (2000,0), we get the line where the labor constraint is satisfied exactly, that is, as an equality.

• All points below and to the left of this line are also feasible; there are these are the points where less than the maximum number of 4000 labor hours are used.

Solution -- continued

• We indicate the feasible side of the line by the short arrows pointing down to the left from the labor constraint line.

• Similarly the metal constraint line crosses the axes at the points (0,2500) and (5000,0), so we join these two points to find the line where all 5000 ounces of metal are used.

• Finally the points on or below both of these lines constitute the feasible region. These are the point below the heavy lines.

Solution -- continued• The crucial point, however, is that only three

points can be optimal: (2000,0), (0, 2500), or (1000, 2000), the three “corner” points (other than (0,0)) in the feasible region.

• To find out the best of these three optimal points calculate profit at each point and select that point which gives maximum profit

• It is found that profit is maximum at x1 = 1000 and x2 = 2000, with a corresponding profit of P = Rs.7450.

• Thus the optimal solution is to produce 1000 of type I frames and 2000 of type II frames

Solution -- continued

• You can think of the feasible region as all points on or inside the figure formed by four points: (0,0), (0,2500), (2000,0), and the point where labor hour and metal constraint lines intersect.

• The next step is to bring profit into the picture. We do this by constructing “isoprofit” lines – that is lines where total profit is a constant. Any such line can be written as 2.25x1 + 2.60x2 = P where P is a constant profit level. Solving for x2, we can put this equation in slope-intercept form: x2 = P/2.60 – (2.25/2.60)x1

Solution -- continued• This shows that any iso profit line has slope

–2.25/2.60, and it crosses the vertical axis at the value P/2.60. Three of these isoprofit lines appear in the chart as dotted lines.

• Therefore, to maximize profit, we want to move the dotted line up and to the right until it just barely touches the feasible region.

• Graphically, we can see that the last feasible point it will touch is the point indicated in the figure, where the labor hour and metal constraint lines cross.

Solution -- continued• We can then solve two equations in two

unknowns to find the coordinates of this point. They are x1 = 1000 and x2 = 2000, with a corresponding profit of P = Rs.7450.

• Note that if the slope of the isoprofit lines were much steeper, the the optimal point would be (2000,0). On the other hand,m if the slope were mush less steep, the optimal point would be (0,2500).These statements make intuitive sense.

• If the isoprofit lines are steep, this is because the unit profit from frame type 1 is large relative to the unit profit from frame type 2.

Solution -- continued

• The crucial point, however, is that only three points can be optimal: (2000,0), (0, 2500), or (1000, 2000), the three “corner” points (other than (0,0)) in the feasible region.

• The best of these depends on the relative slopes of the constraint lines and isoprofit lines in the graph.

Transportation, Assignment and Transshipment Problems

ApplicationsPhysical analog

of nodes Physical analog

of arcsFlow

Communicationsystems

phone exchanges, computers,

transmissionfacilities, satellites

Cables, fiber optic links, microwave

relay links

Voice messages, Data,

Video transmissions

Hydraulic systemsPumping stationsReservoirs, Lakes

PipelinesWater, Gas, Oil,Hydraulic fluids

Integrated computer circuits

Gates, registers,processors

Wires Electrical current

Mechanical systems JointsRods, Beams,

SpringsHeat, Energy

Transportationsystems

Intersections, Airports,

Rail yards

Highways,Airline routes

Railbeds

Passengers, freight,

vehicles, operators

Applications of Network Optimization

Description

A transportation problem basically deals with the problem, which aims to find the best way to fulfill the demand of n demand points using the capacities of m supply points. •While trying to find the best way, generally a variable cost of shipping the product from one supply point to a demand point or a similar constraint should be taken into consideration.

1 Formulating Transportation Problems

Example 1: Powerco has three electric power plants that supply the electric needs of four cities.•The associated supply of each plant and demand of each city is given in the table 1.•The cost of sending 1 million kwh of electricity from a plant to a city depends on the distance the electricity must travel.

Transportation tableau

A transportation problem is specified by the supply, the demand, and the shipping costs. So the relevant data can be summarized in a transportation tableau. The transportation tableau implicitly expresses the supply and demand constraints and the shipping cost between each demand and supply point.

Table 1. Shipping costs, Supply, and Demand for Powerco Example

From To

City 1 City 2 City 3 City 4 Supply (Million kwh)

Plant 1 $8 $6 $10 $9 35

Plant 2 $9 $12 $13 $7 50

Plant 3 $14 $9 $16 $5 40

Demand (Million kwh)

45 20 30 30

Transportation Tableau

Solution

1. Decision Variable:

Since we have to determine how much electricity is sent from each plant to each city;

Xij = Amount of electricity produced at plant i and sent to city j

X14 = Amount of electricity produced at plant 1 and sent to city 4

2. Objective function

Since we want to minimize the total cost of shipping from plants to cities;

Minimize Z = 8X11+6X12+10X13+9X14

+9X21+12X22+13X23+7X24

+14X31+9X32+16X33+5X34

3. Supply Constraints

Since each supply point has a limited production capacity;

X11+X12+X13+X14 <= 35

X21+X22+X23+X24 <= 50

X31+X32+X33+X34 <= 40

4. Demand Constraints

Since each demand point requires minimum supply;

X11+X21+X31 >= 45

X12+X22+X32 >= 20

X13+X23+X33 >= 30

X14+X24+X34 >= 30

5. Sign Constraints

Since a negative amount of electricity can not be shipped all Xij’s must be non negative;

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

LP Formulation of Powerco’s Problem

Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24

+14X31+9X32+16X33+5X34

S.T.: X11+X12+X13+X14 <= 35 (Supply Constraints)

X21+X22+X23+X24 <= 50

X31+X32+X33+X34 <= 40

X11+X21+X31 >= 45 (Demand Constraints)

X12+X22+X32 >= 20

X13+X23+X33 >= 30

X14+X24+X34 >= 30

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

General Description of a Transportation Problem

1. A set of m supply points from which a good is shipped. Supply point i can supply at most si units.

2. A set of n demand points to which the good is shipped. Demand point j must receive at least di units of the shipped good.

3. Each unit produced at supply point i and shipped to demand point j incurs a variable cost of cij.

Xij = number of units shipped from supply point i to demand point j

),...,2,1;,...,2,1(0

),...,2,1(

),...,2,1(..

min

1

1

1 1

njmiX

njdX

misXts

Xc

ij

mi

i

jij

nj

j

iij

mi

i

nj

j

ijij

Balanced Transportation Problem

If Total supply equals to total demand, the problem is said to be a balanced transportation problem:

nj

j

j

mi

i

i ds11

Methods to find the bfs for a balanced TP

There are three basic methods:

1. Northwest Corner Method

2. Minimum Cost Method

3. Vogel’s Method

1. Northwest Corner Method

To find the bfs by the NWC method:

Begin in the upper left (northwest) corner of the transportation tableau and set x11 as large as possible (here the limitations for setting x11 to a larger number, will be the demand of demand point 1 and the supply of supply point 1. Your x11 value can not be greater than minimum of this 2 values).

According to the explanations in the previous slide we can set x11=3 (meaning demand of demand point 1 is satisfied by supply point 1).

5

6

2

3 5 2 3

3 2

6

2

X 5 2 3

After we check the east and south cells, we saw that we can go east (meaning supply point 1 still has capacity to fulfill some demand).

3 2 X

6

2

X 3 2 3

3 2 X

3 3

2

X X 2 3

After applying the same procedure, we saw that we can go south this time (meaning demand point 2 needs more supply by supply point 2).

3 2 X

3 2 1

2

X X X 3

3 2 X

3 2 1 X

2

X X X 2

Finally, we will have the following bfs, which is: x11=3, x12=2, x22=3, x23=2, x24=1, x34=2

3 2 X

3 2 1 X

2 X

X X X X

2. Minimum Cost Method

The Northwest Corner Method dos not utilize shipping costs. It can yield an initial bfs easily but the total shipping cost may be very high. The minimum cost method uses shipping costs in order come up with a bfs that has a lower cost. To begin the minimum cost method, first we find the decision variable with the smallest shipping cost (Xij). Then assign Xij its largest possible value, which is the minimum of si and dj

After that, as in the Northwest Corner Method we should cross out row i and column j and reduce the supply or demand of the noncrossed-out row or column by the value of Xij. Then we will choose the cell with the minimum cost of shipping from the cells that do not lie in a crossed-out row or column and we will repeat the procedure.

An example for Minimum Cost MethodStep 1: Select the cell with minimum cost.

2 3 5 6

2 1 3 5

3 8 4 6

5

10

15

12 8 4 6

Step 2: Cross-out column 2

2 3 5 6

2 1 3 5

8

3 8 4 6

12 X 4 6

5

2

15

Step 3: Find the new cell with minimum shipping cost and cross-out row 2

2 3 5 6

2 1 3 5

2 8

3 8 4 6

5

X

15

10 X 4 6

Step 4: Find the new cell with minimum shipping cost and cross-out row 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

X

X

15

5 X 4 6

Step 5: Find the new cell with minimum shipping cost and cross-out column 1

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5

X

X

10

X X 4 6

Step 6: Find the new cell with minimum shipping cost and cross-out column 3

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4

X

X

6

X X X 6

Step 7: Finally assign 6 to last cell. The bfs is found as: X11=5, X21=2, X22=8, X31=5, X33=4 and

X34=6

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4 6

X

X

X

X X X X

Step 7: Finally assign 6 to last cell. The bfs is found as: X11=5, X21=2, X22=8, X31=5, X33=4 and

X34=6

2 3 5 6

5

2 1 3 5

2 8

3 8 4 6

5 4 6

X

X

X

X X X X

3. Vogel’s Method

Begin with computing each row and column a penalty. The penalty will be equal to the difference between the two smallest shipping costs in the row or column. Identify the row or column with the largest penalty. Find the first basic variable which has the smallest shipping cost in that row or column. Then assign the highest possible value to that variable, and cross-out the row or column as in the previous methods. Compute new penalties and use the same procedure.

An example for Vogel’s MethodStep 1: Compute the penalties.

Supply Row Penalty

6 7 8

15 80 78

Demand

Column Penalty 15-6=9 80-7=73 78-8=70

7-6=1

78-15=63

15 5 5

10

15

Step 2: Identify the largest penalty and assign the highest possible value to the variable.

Supply Row Penalty

6 7 8

5

15 80 78

Demand

Column Penalty 15-6=9 _ 78-8=70

8-6=2

78-15=63

15 X 5

5

15

Step 3: Identify the largest penalty and assign the highest possible value to the variable.

Supply Row Penalty

6 7 8

5 5

15 80 78

Demand

Column Penalty 15-6=9 _ _

_

_

15 X X

0

15

Step 4: Identify the largest penalty and assign the highest possible value to the variable.

Supply Row Penalty

6 7 8

0 5 5

15 80 78

Demand

Column Penalty _ _ _

_

_

15 X X

X

15

Step 5: Finally the bfs is found as X11=0, X12=5, X13=5, and X21=15

Supply Row Penalty

6 7 8

0 5 5

15 80 78

15

Demand

Column Penalty _ _ _

_

_

X X X

X

X

. Assignment ProblemsExample: Machineco has four jobs to be completed. Each machine must be assigned to complete one job. The time required to setup each machine for completing each job is shown in the table below. Machinco wants to minimize the total setup time needed to complete the four jobs.

Setup times

(Also called the cost matrix)

Time (Hours)

Job1 Job2 Job3 Job4

Machine 1 14 5 8 7

Machine 2 2 12 6 5

Machine 3 7 8 3 9

Machine 4 2 4 6 10

The ModelAccording to the setup table Machinco’s problem can be formulated as follows (for i,j=1,2,3,4):

10

1

1

1

1

1

1

1

1..

10629387

5612278514min

44342414

43332313

42322212

41312111

44434241

34333231

24232221

14131211

4443424134333231

2423222114131211

ijij orXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXX

XXXXts

XXXXXXXX

XXXXXXXXZ

For the model on the previous page note that:

Xij=1 if machine i is assigned to meet the demands of job j

Xij=0 if machine i is not assigned to meet the demands of job j

In general an assignment problem is balanced transportation problem in which all supplies and demands are equal to 1.

The Assignment Problem

In general the LP formulation is given as

Minimize 1 1

1

1

1 1

1 1

0

, , ,

, , ,

or 1,

n n

ij iji j

n

ijj

n

iji

ij

c x

x i n

x j n

x ij

Each supply is 1

Each demand is 1

Comments on the Assignment Problem

• The Air Force has used this for assigning thousands of people to jobs.

• This is a classical problem. Research on the assignment problem predates research on LPs.

• Very efficient special purpose solution techniques exist. – 10 years ago, Yusin Lee and J. Orlin solved a

problem with 2 million nodes and 40 million arcs in ½ hour.