Transcript here
Lecture
5
Linear Programming (6S)
and
Transportation Problem (8S)
1
Linear Programming
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
George Dantzig – 1914 -2005
2
Product Mix Example (from session 1)
Type 1
Type 2
Profit per unit
$60
$50
Assembly time per
unit
4 hrs
10 hrs
Inspection time per
unit
2 hrs
Storage space per
unit
3 cubic ft
Resource
Amount available
Assembly time
100 hours
Inspection time
22 hours
Storage space
39 cubic feet
1 hr
3 cubic ft
3
Linear Programming Example
Variables
Maximize 60X1 + 50X2
Subject to
Objective function
4X1 + 10X2 <= 100
2X1 + 1X2 <= 22
Constraints
3X1 + 3X2 <= 39
Non-negativity Constraints
X1, X2 >= 0
What is a Linear Program?
• A LP is an optimization model that has
• continuous variables
• a single linear objective function, and
• (almost always) several constraints (linear equalities or inequalities)
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Linear Programming Model
Decision variables
Objective Function
unknowns, which is what model seeks to determine
for example, amounts of either inputs or outputs
goal, determines value of best (optimum) solution among all feasible (satisfy
constraints) values of the variables
either maximization or minimization
Constraints
restrictions, which limit variables of the model
limitations that restrict the available alternatives
Parameters: numerical values (for example, RHS of constraints)
Feasible solution: is one particular set of values of the decision
variables that satisfies the constraints
Feasible solution space: the set of all feasible solutions
Optimal solution: is a feasible solution that maximizes or minimizes
the objective function
There could be multiple optimal solutions
5
Another Example of LP: Diet Problem
Energy requirement : 2000 kcal
Protein requirement : 55 g
Calcium requirement : 800 mg
Food
Energy (kcal)
Protein(g)
Calcium(mg)
Oatmeal
110
4
2
Price per
serving($)
3
Chicken
Eggs
Milk
Pie
205
160
160
420
32
13
8
4
12
54
285
22
24
13
9
24
Pork
260
14
80
13
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Example 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 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: Constraints
Meet energy requirement
110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000
Meet protein requirement
4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55
Meet calcium requirement
2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800
Restriction on number of servings
0x14, 0x23, 0x32, 0x48, 0x52, 0x62
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So, how does a LP look like?
minimize 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6
subject to
110x1 + 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 Problem
Using LINDO 6.1
Food
Oatmeal
# of servings
4
Chicken
Eggs
Milk
Pie
0
0
6.5
0
Pork
2
Cost of diet = $96.50 per day
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Optimal Solution – Diet Problem
Using Management Scientist
Food
Oatmeal
# of servings
4
Chicken
Eggs
Milk
Pie
0
0
6.5
0
Pork
2
Cost of diet = $96.50 per day
12
Guidelines 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 Table
1
Factory
Warehouse
3
2
4
4
7
7
1
100
1
3
12
8
8
200
2
10
8
16
Factory 1
can
supply
100
units per
period
5
150
3
450
Demand
80
90
120
160
450
Warehouse B’s demand is 90
units per period
Total
supply
capacity
per
period
Total demand
per period
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LP Formulation of Transportation Problem
minimize
4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+
8x31+10x32+16x33+5x34
Minimize total cost of transportation
Subject to
x11+x12+x13+x14=100
Supply constraint for factories
x21+x22+x23+x24=200
x31+x32+x33+x34=150
x11+x21+x31=80
x12+x22+x32=90
Demand constraint of warehouses
x13+x23+x33=120
x14+x24+x34=160
xij>=0, i=1,2,3; j=1,2,3,4
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Solution 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 LINDO
Notice multiple
optimal solutions!
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Product 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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Product Mix Problem
Boat
Maximum
Builder
Cost
Speedhawk Sleekboat
Silverbird Sleekboat
Catman
Racer
Classy
Racer
$6000
$7000
$5000
$9000
Expected
Seating
3
5
2
6
Daily
Profit
$ 70
$ 80
$ 50
$110
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Product Mix Problem
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
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Product Mix Problem
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
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Product Mix Problem - Complete Formulation
Max 70x1 + 80x2 + 50x3 + 110x4
s.t.
6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000
x1 + x2 + x3 + x4 > 50
Boat
# purchased
x1 + x2 - x3 - x4 = 0
Speedhawk
28
3x1 + 5x2 + 2x3 + 6x4 > 200
Silverbird
0
x1, x2, x3, x4 > 0
Catman
0
Classy
28
Daily profit = $5040
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Marketing Application: Media Selection
Advertising Media
# of potential
customers reached
Cost ($) 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
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
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Media Selection Formulation
Step 1: Define decision variables
Step 2: Write the objective in terms of the 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
Maximize 65DTV+90ETV+40DN+60SN+20R
Step 3: Write the constraints in terms of the decision variables
DTV
ETV
DN
SN
R
+
25
<=
4
<=
30
<=
30000
0
DN
25
SN
2
R
30
Exposure = 2370 units
Availability of
Media
Budget
+
ETV
>=
10
1500DTV
+
3000ETV
<=
18000
TV Constraints
1000DTV
+
2000ETV
>=
50000
Customers reached
+
100R
<=
ETV
DTV
2500SN
+
10
10
3000ETV
+
1000SN
<=
DTV
+
1500DN
+
15
Value
1500DTV
+
400DN
<=
Variable
300R
DTV, ETV, DN, SN, R >= 0
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Applications 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 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 Example
minimize
4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+16
x33+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 Example
maximize 2x1 + x2
subject to
-x1 +
x2 1
x1 - 2x2 2
x1 , x 2 0
x2 can be increased indefinitely without violating any
constraint
=> Objective function value can be increased indefinitely
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Multiple 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 Scheduling
Chapter 16
30
Scheduling
Establishing the timing of the use of equipment,
facilities and human activities in an
organization
Effective scheduling can yield
Cost savings
Increases in productivity
31
High-Volume Systems
Flow system: High-volume system with
Standardized equipment and activities
Flow-shop scheduling: Scheduling for highvolume flow system
Work Center #1
Work Center #2
Output
32
High-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
33
Scheduling 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
34
Gantt Load Chart
Figure 16.2
Gantt chart - used as a visual aid for loading
and scheduling
Work Mon. Tues. Wed. Thurs. Fri.
Center
1
Job 3
Job 4
2
Job 3 Job 7
3
Job 1
Job 6
Job 7
4
Job 10
35
More Gantt Charts
36
Assignment 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.
37
Cost Table for Assignment Problem
Aircraft (j)
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)
All assignment costs in thousands of $
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Management Scientist Solution
Pilot Assigned to
Cost
aircraft # (`000 $)
1
1
1
2
3
4
3
2
4
10
5
5
39
Formulation of Assignment Problem
minimize x11+4x12+6x13+3x14 + 9x21+7x22+10x23+9x24 +
4x31+5x32+11x33+7x34 + 8x41+7x42+8x43+5x44
subject to
Pilot Assigned to
Cost
x11+x12+x13+x14=1
aircraft # (`000 $)
x21+x22+x23+x24=1
1
1
1
x31+x32+x33+x34=1
x41+x42+x43+x44=1
2
3
10
3
2
5
x11+x21+x31+x41=1
4
4
5
x12+x22+x32+x42=1
Optimal Solution:
x13+x23+x33+x43=1
x11=1; x23=1; x32=1; x44=1; rest=0
x14+x24+x34+x44=1
Cost of assignment = 1+10+5+5=$21 (`000)
xij = 1, if pilot i is assigned to aircraft j, i=1,2,3,4; j=1,2,3,4
0 otherwise
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Sequencing
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
In-class example
Minimizes mean flow time
EDD - earliest due date
41
Performance 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
42
Scheduling 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
43
Minimizing Scheduling Difficulties
Set realistic due dates
Focus on bottleneck operations
Consider lot splitting of large jobs
44