1 lecture 5 linear programming (6s) and transportation problem (8s)
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Lecture5
Linear Programming (6S)and
Transportation Problem (8S)
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George Dantzig – 1914 -2005 Concerned with optimal allocation of limited
resources such as Materials Budgets Labor Machine time
among competitive activities under a set of constraints
Linear ProgrammingLinear Programming
George Dantzig – 1914 -2005
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Product Mix Example (from session 1)Product Mix Example (from session 1)
Type 1 Type 2
Profit per unit $60 $50
Assembly time per unit
4 hrs 10 hrs
Inspection time per unit
2 hrs 1 hr
Storage space per unit
3 cubic ft 3 cubic ft
Resource Amount available
Assembly time 100 hours
Inspection time 22 hours
Storage space 39 cubic feet
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Maximize 60X1 + 50X2
Subject to
4X1 + 10X2 <= 100
2X1 + 1X2 <= 22
3X1 + 3X2 <= 39
X1, X2 >= 0
Linear Programming ExampleLinear Programming ExampleVariables
Objective function
Constraints
What is a Linear Program?
• A LP is an optimization model that has
• continuous variables
• a single linear objective function, and
• (almost always) several constraints (linear equalities or inequalities)
Non-negativity Constraints
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Decision variables unknowns, which is what model seeks to determine for example, amounts of either inputs or outputs
Objective Function goal, determines value of best (optimum) solution among all feasible (satisfy
constraints) values of the variables either maximization or minimization
Constraints restrictions, which limit variables of the model limitations that restrict the available alternatives
Parameters: numerical values (for example, RHS of constraints)
Feasible solution: is one particular set of values of the decision variables that satisfies the constraints Feasible solution space: the set of all feasible solutions
Optimal solution: is a feasible solution that maximizes or minimizes the objective function
There could be multiple optimal solutions
Linear Programming ModelLinear Programming Model
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Another Example of LP: Diet Another Example of LP: Diet ProblemProblem
Energy requirement : 2000 kcal Protein requirement : 55 g Calcium requirement : 800 mgFood Energy (kcal) Protein(g) Calcium(mg) Price per
serving($)
Oatmeal 110 4 2 3
Chicken 205 32 12 24
Eggs 160 13 54 13
Milk 160 8 285 9
Pie 420 4 22 24
Pork 260 14 80 13
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Example of LP : Diet ProblemExample of LP : Diet Problem
oatmeal: at most 4 servings/day chicken: at most 3 servings/day eggs: at most 2 servings/day milk: at most 8 servings/day pie: at most 2 servings/day pork: at most 2 servings/day
Design an optimal diet plan which minimizes the cost per
day
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Step 1: define decision variablesStep 1: define decision variables
x1 = # of oatmeal servings x2 = # of chicken servings x3 = # of eggs servings x4 = # of milk servings x5 = # of pie servings x6 = # of pork servings
Step 2: formulate objective function• In this case, minimize total cost
minimize z = 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6
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Step 3: ConstraintsStep 3: Constraints
Meet energy requirement110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 Meet protein requirement4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55 Meet calcium requirement2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800 Restriction on number of servings0x14, 0x23, 0x32, 0x48, 0x52, 0x62
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So, how does a LP look like?So, how does a LP look like?
minimize 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6
subject to110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000
4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55
2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800
0x14
0x23
0x32
0x48
0x52
0x62
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Optimal Solution – Diet ProblemOptimal Solution – Diet ProblemUsing LINDO 6.1Using LINDO 6.1
Cost of diet = $96.50 per day
Food # of servings
Oatmeal 4
Chicken 0
Eggs 0
Milk 6.5
Pie 0
Pork 2
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Optimal Solution – Diet ProblemOptimal Solution – Diet ProblemUsing Management ScientistUsing Management Scientist
Cost of diet = $96.50 per day
Food # of servings
Oatmeal 4
Chicken 0
Eggs 0
Milk 6.5
Pie 0
Pork 2
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Guidelines for Model FormulationGuidelines for Model Formulation
Understand the problem thoroughly. Describe the objective. Describe each constraint. Define the decision variables. Write the objective in terms of the decision
variables. Write the constraints in terms of the decision
variables Do not forget non-negativity constraints
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A Transportation TableA Transportation Table
Warehouse
4 7 7 1100
12 3 8 8200
8 10 16 5150
450
45080 90 120 160
1 2 3 4
1
2
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Factory Factory 1can supply 100units per period
Demand
Warehouse B’s demand is 90 units per period Total demand
per period
Total supplycapacity perperiod
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LP Formulation of Transportation ProblemLP Formulation of Transportation Problem
minimize 4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+16x33+5x34
Subject to x11+x12+x13+x14=100 x21+x22+x23+x24=200 x31+x32+x33+x34=150 x11+x21+x31=80 x12+x22+x32=90 x13+x23+x33=120 x14+x24+x34=160 xij>=0, i=1,2,3; j=1,2,3,4
Supply constraint for factories
Demand constraint of warehouses
Minimize total cost of transportation
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Solution in Management ScientistSolution in Management Scientist
Total transportation cost = 4(80) + 7(0) + 7(10)+ 1(10) + 12(0) + 3(90) + 8(110) + 8(0) + 8(0) +10(0) + 16(0) +5 (150) = $2300
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Solution using LINDOSolution using LINDO
Notice multiple optimal solutions!
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Product Mix ProblemProduct Mix Problem• Floataway Tours has $420,000 that can be used to
purchase new rental boats for hire during the summer. • The boats can be purchased from two different
manufacturers.• Floataway Tours would like to purchase at least 50 boats.• They would also like to purchase the same number from
Sleekboat as from Racer to maintain goodwill. • At the same time, Floataway Tours wishes to have a total
seating capacity of at least 200.
• Formulate this problem as a linear program
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Maximum Expected Daily
Boat Builder Cost Seating Profit
Speedhawk Sleekboat $6000 3 $ 70
Silverbird Sleekboat $7000 5 $ 80
Catman Racer $5000 2 $ 50
Classy Racer $9000 6 $110
Product Mix ProblemProduct Mix Problem
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Define the decision variables
x1 = number of Speedhawks ordered
x2 = number of Silverbirds ordered
x3 = number of Catmans ordered
x4 = number of Classys ordered Define the objective function Maximize total expected daily profit: Max: (Expected daily profit per unit) x (Number of units)
Max: 70x1 + 80x2 + 50x3 + 110x4
Product Mix ProblemProduct Mix Problem
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Define the constraints(1) Spend no more than $420,000:
6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000 (2) Purchase at least 50 boats: x1 + x2 + x3 + x4 > 50 (3) Number of boats from Sleekboat equals number
of boats from Racer: x1 + x2 = x3 + x4 or x1 + x2 - x3 - x4 = 0
(4) Capacity at least 200: 3x1 + 5x2 + 2x3 + 6x4 > 200
Nonnegativity of variables: xj > 0, for j = 1,2,3,4
Product Mix ProblemProduct Mix Problem
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Max 70x1 + 80x2 + 50x3 + 110x4
s.t.
6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000
x1 + x2 + x3 + x4 > 50
x1 + x2 - x3 - x4 = 0
3x1 + 5x2 + 2x3 + 6x4 > 200
x1, x2, x3, x4 > 0
Product Mix Problem - Complete FormulationProduct Mix Problem - Complete Formulation
Daily profit = $5040
Boat # purchased
Speedhawk 28
Silverbird 0
Catman 0
Classy 28
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Marketing Application: Media SelectionMarketing Application: Media Selection
Advertising budget for first month = $30000 At least 10 TV commercials must be used At least 50000 customers must be reached Spend no more than $18000 on TV adverts Determine optimal media selection plan
Advertising Media # of potential customers reached
Cost ($) per advertisement
Max times available per month
Exposure Quality Units
Day TV 1000 1500 15 65
Evening TV 2000 3000 10 90
Daily newspaper 1500 400 25 40
Sunday newspaper 2500 1000 4 60
Radio 300 100 30 20
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Media Selection FormulationMedia Selection Formulation Step 1: Define decision variables
DTV = # of day time TV adverts ETV = # of evening TV adverts DN = # of daily newspaper adverts SN = # of Sunday newspaper adverts R = # of radio adverts
Step 2: Write the objective in terms of the decision variables Maximize 65DTV+90ETV+40DN+60SN+20R
Step 3: Write the constraints in terms of the decision variables
DTV <= 15
ETV <= 10
DN <= 25
SN <= 4
R <= 30
1500DTV + 3000ETV + 400DN + 1000SN + 100R <= 30000
DTV + ETV >= 10
1500DTV + 3000ETV <= 18000
1000DTV + 2000ETV + 1500DN + 2500SN + 300R >= 50000
BudgetBudget
Customers Customers reachedreached
TV TV ConstraintConstraint
ss
Availability of Availability of MediaMedia
DTV, ETV, DN, SN, R >= 0DTV, ETV, DN, SN, R >= 0
Exposure = 2370 units
Variable Value
DTV 10
ETV 0
DN 25
SN 2
R 30
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Applications of LPApplications of LP
Product mix planning Distribution networks Truck routing Staff scheduling Financial portfolios Capacity planning Media selection: marketing
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Possible Outcomes of a LPPossible Outcomes of a LP
A LP is either Infeasible – there exists no solution which satisfies
all constraints and optimizes the objective function or, Unbounded – increase/decrease objective
function as much as you like without violating any constraint
or, Has an Optimal Solution Optimal values of decision variables Optimal objective function value
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Infeasible LP – An ExampleInfeasible LP – An Example minimize
4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+16x33+5x34
Subject to x11+x12+x13+x14=100 x21+x22+x23+x24=200 x31+x32+x33+x34=150
x11+x21+x31=80 x12+x22+x32=90 x13+x23+x33=120 x14+x24+x34=170
xij>=0, i=1,2,3; j=1,2,3,4
Total demand exceeds total supply
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Unbounded LP – An ExampleUnbounded LP – An Example
maximize 2x1 + x2
subject to
-x1 + x2 1
x1 - 2x2 2
x1 , x2 0
x2 can be increased indefinitely without violating any constraint
=> Objective function value can be increased indefinitely
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Multiple Optima – An ExampleMultiple Optima – An Example
maximize x1 + 0.5 x2
subject to
2x1 + x2 4
x1 + 2x2 3
x1 , x2 0• x1= 2, x2= 0, objective function = 2
• x1= 5/3, x2= 2/3, objective function = 2
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Operations SchedulingChapter 16
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Establishing the timing of the use of equipment, facilities and human activities in an organization
Effective scheduling can yield
Cost savings
Increases in productivity
Scheduling Scheduling
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High-Volume SystemsHigh-Volume Systems
Flow system: High-volume system with Standardized equipment and activities
Flow-shop scheduling: Scheduling for high-volume flow system
Work Center #1 Work Center #2 Output
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High-Volume Success FactorsHigh-Volume Success Factors
Process and product design
Preventive maintenance
Rapid repair when breakdown occurs
Optimal product mixes
Minimization of quality problems
Reliability and timing of supplies
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Scheduling Low-Volume SystemsScheduling Low-Volume Systems
Loading - assignment of jobs to process centers
Sequencing - determining the order in which jobs will be processed
Job-shop scheduling Scheduling for low-volume
systems with many variations in requirements
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Gantt Load ChartGantt Load Chart
Gantt chart - used as a visual aid for loading and scheduling
WorkCenter
Mon. Tues. Wed. Thurs. Fri.
1 Job 3 Job 42 Job 3 Job 73 Job 1 Job 6 Job 74 Job 10
Figure 16.2
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More Gantt ChartsMore Gantt Charts
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Assignment ProblemAssignment Problem Objective: Assign n jobs/workers to n machines
such that the total cost of assignment is minimized Special case of transportation problem
When # of rows = # of columns in the transportation tableau
All supply and demands =1 Plenty of practical applications
Job shops Hospitals Airlines, etc.
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Cost Table for Assignment ProblemCost Table for Assignment Problem
1 2 3 4
1 $1 $4 $6 $3
2 $9 $7 $10 $9
3 $4 $5 $11 $7
4 $8 $7 $8 $5
Pilot (i)
Aircraft (j)
All assignment costs in thousands of $
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Management Scientist SolutionManagement Scientist Solution
Pilot Assigned to aircraft #
Cost (`000 $)
1 1 1
2 3 10
3 2 5
4 4 5
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Formulation of Assignment ProblemFormulation of Assignment Problem minimize x11+4x12+6x13+3x14 + 9x21+7x22+10x23+9x24 +
4x31+5x32+11x33+7x34 + 8x41+7x42+8x43+5x44
subject to x11+x12+x13+x14=1 x21+x22+x23+x24=1 x31+x32+x33+x34=1 x41+x42+x43+x44=1
x11+x21+x31+x41=1 x12+x22+x32+x42=1 x13+x23+x33+x43=1 x14+x24+x34+x44=1
xij = 1, if pilot i is assigned to aircraft j, i=1,2,3,4; j=1,2,3,4 0 otherwise
Pilot Assigned to aircraft #
Cost (`000 $)
1 1 1
2 3 10
3 2 5
4 4 5Optimal Solution:
x11=1; x23=1; x32=1; x44=1; rest=0
Cost of assignment = 1+10+5+5=$21 (`000)
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SequencingSequencing Sequencing: Determine the order in which jobs at a
work center will be processed. Workstation: An area where one person works,
usually with special equipment, on a specialized job. Priority rules: Simple heuristics used to select the
order in which jobs will be processed.
FCFS - first come, first served
SPT - shortest processing time
Minimizes mean flow time
EDD - earliest due date
In-class example
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Performance MeasuresPerformance Measures Job flow time
Length of time a job is at a particular workstation
Includes actual processing time, waiting time, transportation time etc.
Lateness = flow time – due date
Tardiness = max {lateness, 0}
Makespan
Total time needed to complete a group of jobs
Length of time between start of first job and completion of last job
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Scheduling DifficultiesScheduling Difficulties
Variability in Setup times Processing times Interruptions Changes in the set of jobs
No method for identifying optimal schedule Scheduling is not an exact science Ongoing task for a manager
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Minimizing Scheduling DifficultiesMinimizing Scheduling Difficulties
Set realistic due dates
Focus on bottleneck operations
Consider lot splitting of large jobs