Quantitative Methods

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Quantitative Methods

• Linear Programming

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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

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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

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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

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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

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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

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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 ProfitRs180000

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Example

• Graphical Solution Method

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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

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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?

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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.

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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)

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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.

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Graphical Solution
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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.

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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.

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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

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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

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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.

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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.

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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.

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Transportation, Assignment and Transshipment
Problems

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Applications of Network Optimization
Applications
Physical analogof nodes
Physical analogof arcs
Flow
Communicationsystems
phone exchanges, computers, transmission facilit
ies, satellites
Cables, fiber optic links, microwave relay
links
Voice messages, Data, Video transmissions
Hydraulic systems
Pumping stationsReservoirs, Lakes
Pipelines
Water, Gas, Oil,Hydraulic fluids
Integrated computer circuits
Gates, registers,processors
Wires
Electrical current
Mechanical systems
Joints
Rods, Beams, Springs
Heat, Energy
Transportationsystems
Intersections, Airports,Rail yards
Highways,Airline routes Railbeds
Passengers, freight, vehicles, operators
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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.

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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.

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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.

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

• 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

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2. Objective function

• Since we want to minimize the total cost of
  shipping from plants to cities
• Minimize Z 8X116X1210X139X14
• 9X2112X2213X237X24
• 14X319X3216X335X34

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3. Supply Constraints

• Since each supply point has a limited production
  capacity
• X11X12X13X14 lt 35
• X21X22X23X24 lt 50
• X31X32X33X34 lt 40

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4. Demand Constraints

• Since each demand point requires minimum supply
• X11X21X31 gt 45
• X12X22X32 gt 20
• X13X23X33 gt 30
• X14X24X34 gt 30

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5. Sign Constraints

• Since a negative amount of electricity can not be
  shipped all Xijs must be non negative
• Xij gt 0 (i 1,2,3 j 1,2,3,4)

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LP Formulation of Powercos Problem

• Min Z 8X116X1210X139X149X2112X2213X237X24
  
• 14X319X3216X335X34
• S.T. X11X12X13X14 lt 35 (Supply Constraints)
• X21X22X23X24 lt 50
• X31X32X33X34 lt 40
• X11X21X31 gt 45 (Demand Constraints)
• X12X22X32 gt 20
• X13X23X33 gt 30
• X14X24X34 gt 30
• Xij gt 0 (i 1,2,3 j 1,2,3,4)

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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.

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

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Balanced Transportation Problem

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

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Methods to find the bfs for a balanced TP

• There are three basic methods
• Northwest Corner Method
• Minimum Cost Method
• Vogels Method

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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).

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According to the explanations in the previous
slide we can set x113 (meaning demand of demand
point 1 is satisfied by supply point 1).
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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).
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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).
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Finally, we will have the following bfs, which
is x113, x122, x223, x232, x241, x342
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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

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• 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.

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An example for Minimum Cost MethodStep 1 Select
the cell with minimum cost.
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Step 2 Cross-out column 2
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Step 3 Find the new cell with minimum shipping
cost and cross-out row 2
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Step 4 Find the new cell with minimum shipping
cost and cross-out row 1
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Step 5 Find the new cell with minimum shipping
cost and cross-out column 1
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Step 6 Find the new cell with minimum shipping
cost and cross-out column 3
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Step 7 Finally assign 6 to last cell. The bfs is
found as X115, X212, X228, X315, X334 and
X346
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Step 7 Finally assign 6 to last cell. The bfs is
found as X115, X212, X228, X315, X334 and
X346
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3. Vogels 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.

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An example for Vogels MethodStep 1 Compute the
penalties.
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Step 2 Identify the largest penalty and assign
the highest possible value to the variable.
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Step 3 Identify the largest penalty and assign
the highest possible value to the variable.
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Step 4 Identify the largest penalty and assign
the highest possible value to the variable.
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Step 5 Finally the bfs is found as X110, X125,
X135, and X2115
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. 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.

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• Setup times
• (Also called the cost matrix)

Time (Hours) Time (Hours) Time (Hours) 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
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The Model

• According to the setup table Machincos problem
  can be formulated as follows (for i,j1,2,3,4)

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• For the model on the previous page note that
• Xij1 if machine i is assigned to meet the
  demands of job j
• Xij0 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.

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The Assignment Problem

• In general the LP formulation is given as
• Minimize

Each supply is 1
Each demand is 1
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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.

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