Transcript Quantitative Methods

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
Varsha Varde
2
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
Varsha Varde
3
•
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
Varsha Varde
7
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 ≥0
9
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
Varsha Varde
10
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
Applications of Network
Optimization
Applications
Physical analog Physical analog
of nodes
of arcs
Flow
phone exchanges,
Cables, fiber optic Voice messages,
Communication
computers,
links, microwave
Data,
systems
transmission
relay links Video transmissions
facilities, satellites
Pumping stations
Hydraulic systems
Reservoirs, Lakes
Integrated
Gates, registers,
computer circuits
processors
Pipelines
Water, Gas, Oil,
Hydraulic fluids
Wires
Electrical current
Mechanical systems
Joints
Rods, Beams,
Springs
Heat, Energy
Transportation
systems
Intersections,
Airports,
Rail yards
Highways,
Airline routes
Railbeds
Passengers,
freight,
vehicles,
operators
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
Plant 3
Demand
(Million kwh)
$9
$14
45
$12
$9
20
$13
$16
30
$7
$5
30
50
40
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
i m j n
min  cijXij
i 1 j 1
j n
s.t. Xij  si (i  1,2,...,m)
j 1
i m
X
ij
 dj ( j  1,2,...,n)
i 1
Xij  0(i  1,2,...,m; j  1,2,...,n)
Balanced Transportation Problem
If Total supply equals to total demand,
the problem is said to be a balanced
transportation problem:
j n
i m
s  d
i
i 1
j
j 1
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
3
X
2
1
2
X
X
3
2
3
X
3
X
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
3
X
X
X
2
X
1
X
2
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 Method
Step 1: Select the cell with minimum cost.
2
3
5
6
5
2
1
3
5
10
3
12
8
8
4
4
6
6
15
Step 2: Cross-out column 2
6
5
3
2
5
5
3
1
2
2
8
12
X
6
4
8
3
4
6
15
Step 3: Find the new cell with minimum shipping
cost and cross-out row 2
2
3
5
6
5
2
1
3
5
X
2
8
3
10
8
X
4
4
6
6
15
Step 4: Find the new cell with minimum shipping
cost and cross-out row 1
2
3
5
6
X
5
2
1
3
5
X
2
8
3
5
8
X
4
4
6
6
15
Step 5: Find the new cell with minimum shipping
cost and cross-out column 1
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
4
6
5
X
X
4
6
10
Step 6: Find the new cell with minimum shipping
cost and cross-out column 3
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
5
4
6
4
X
X
X
6
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
X
5
2
1
3
5
X
2
8
3
8
5
4
4
X
X
6
6
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
X
5
2
1
3
5
X
2
8
3
8
5
4
4
X
X
6
6
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 crossout the row or column as in the previous
methods. Compute new penalties and use the
same procedure.
An example for Vogel’s Method
Step 1: Compute the penalties.
6
7
15
Demand
Column Penalty
Supply
Row Penalty
10
7-6=1
15
78-15=63
8
80
78
15
5
5
15-6=9
80-7=73
78-8=70
Step 2: Identify the largest penalty and assign the
highest possible value to the variable.
6
7
Supply
Row Penalty
5
8-6=2
15
78-15=63
8
5
15
Demand
Column Penalty
80
78
15
X
5
15-6=9
_
78-8=70
Step 3: Identify the largest penalty and assign the
highest possible value to the variable.
6
7
5
Column Penalty
Row Penalty
0
_
15
_
8
5
15
Demand
Supply
80
78
15
X
X
15-6=9
_
_
Step 4: Identify the largest penalty and assign the
highest possible value to the variable.
6
0
7
5
Supply
Row Penalty
X
_
15
_
8
5
15
80
78
Demand
15
X
X
Column Penalty
_
_
_
Step 5: Finally the bfs is found as X11=0, X12=5,
X13=5, and X21=15
6
0
7
5
Supply
Row Penalty
X
_
X
_
8
5
15
80
78
15
Demand
X
X
X
Column Penalty
_
_
_
. Assignment Problems
Example: 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 Model
According to the setup table Machinco’s problem
can be formulated as follows (for i,j=1,2,3,4):
min Z  14X 11  5 X 12  8 X 13  7 X 14  2 X 21  12X 22  6 X 23  5 X 24
 7 X 31  8 X 32  3 X 33  9 X 34  2 X 41  X 42  6 X 43  10X 44
s.t. X 11  X 12  X 13  X 14  1
X 21  X 22  X 23  X 24  1
X 31  X 32  X 33  X 34  1
X 41  X 42  X 43  X 44  1
X 11  X 21  X 31  X 41  1
X 12  X 22  X 32  X 42  1
X 13  X 23  X 33  X 43  1
X 14  X 24  X 34  X 44  1
Xij  0orXij  1
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
n
Minimize
n
 c
i 1 j 1
ij
xij
n
 xij  1, i  1,
,n
Each supply is 1
,n
Each demand is 1
j 1
n
x
i 1
ij
 1, j  1,
xij  0 or 1, ij
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.