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Route Planning
Texas Transfer Corp (TTC) Case
1
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Linear programmingExample: Woodcarving, Inc.
Manufactures two types of wooden toysSoldiers sell for $27 and cost $24 to produceTrains sell for $21 and cost $19 to produceRequires two types of skilled laborFinishing (100 available hours each week)Carpentry (80 available hours each week)Soldiers require 2 hours of finishing labor and 1 hour of carpentry laborTrains require 1 hour of finishing labor and 1 hour of carpentry laborDemand for trains is unlimited, but no more than 40 soldiers are bought each weekObjective: maximize weekly profit
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Example: Woodcarving, Inc. (continued)
Decision Variables:X=number of soldiers to produce each week
Y=number of trains to produce each week
Objective Function:Maximize the total profit of soldier and train sales
Max 3X+2Y
Sell Price Production cost
Soldier $27 $24
Train $21 $19
→ Profit $3
→ Profit $2
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Example: Woodcarving, Inc. (continued)
Constraints:Labor hour constraints:
Total finishing hours must be 100 hours:
2X+Y 100 (1)
Total carpentry hours must be 80 hours:
X+Y 80 (2)
Demand constraint: do not produce more than 40 soldiers: X 40 (3)
Finishing Carpentry
Soldier 2 1
Train 1 1
Total available hours 100 80
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Linear programming problem
Decision variables• What decision must we make (e.g. production levels)?
Objective function•What is our measure of success? •What are we trying to maximize (e.g. profits) or minimize (e.g. cost)?
Constraints•What limits our success (e.g. available resources)?
Relationships (equations) are linear
40
80
1002 s.t.
23 Max,
X
YX
YX
YXYX
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FeasibleRegion
Graphical solution
X 40
2X + Y 100
X + Y 80
100
10000
Y = # trains
X = # soldiers
X=0, Y=80, profit=160
X=20, Y=60, profit=180 (optimal solution)
X=40, Y=20, profit=160
X=40, Y=0, profit=120
40
80
1002 s.t.
23 Max,
X
YX
YX
YXYX
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Use Excel Solver: Data-Solver
Objective Function
Decision Variables
Constraints
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The solution
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Easy rider toys
Easy Rider Toys (ERT) manufactures and markets toy vehicles, including cars and trucksCurrently ERT has10,000 cars and 12,000 trucks in inventory.Inventories can be sold in two different bundles:The Racer Set consists of 7 cars and 2 trucks, and is sold for $34.99. The Construction Set consists of 3 cars and 12 trucks, and is sold for $43.99.
How many sets of each bundle ERT should produce to maximize the profit?
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Easy rider formulation
X = # of Racer Sets to produceY = # of Construction Sets to produce
Problem formulation:
Cars Trucks Price
Racer Set 7 2 $34.99
Construction Set 3 12 $43.99
Available Inventory 10,000 12,000
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Transportation example
Philadelphia and Washington D.C. are supply nodesBuffalo, Pittsburgh, and Harrisburg are demand nodesTotal supply = total demand
Washington D.C. 150
Harrisburg 100$30
Philadelphia 100
$50
$20
$25
Buffalo 50
Pittsburgh 100
$30 $15$70
1
2
3
4
5
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Graphical representation of network flow problems
(5,10) (4,9)
(3,6)(1,3)
(4,8)4 6 2
5
4
Arc (i, j)
Cost of sending one unit from i to j
Minimum and Maximumallowable flows on arc (i,j)
i j
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Translating a graph to an LP
One decision variable Xi,j for each arc (i,j)
(the amount of flow to send on that arc)Two constraints for each arc (i,j)
Xi,j upper bound for flow on arc (i,j)
Xi,j lower bound for flow on arc (i,j)
One constraint for each node (flow-balance contr.)
flow into node i ═ flow out of node i
iji
ji
iih
ih XX
node leaving, arcs all
,
node entering, arcs all
,
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LP formulation
Objective function: Minimize the overall flow cost
,
all arcs ,in the network
cost to send one unit ofmin
flow along arc
s.t:
For each arc :
Upper bound for flow on arc ( )
Lower bound for flow on arc ( )
For each no
i ji j
i, j
i, j
Xi, j
(i, j)
X i, j
X i, j
, ,
Flow into node i out of node i
de :
h i i jall h all j
Flow
i
X X
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Decision variables
One decision variable for every arcInterpretation: How much flow (i.e. how many cars) should we send along that arc?
Washington D.C. 150
Harrisburg 100
Philadelphia 100Buffalo 50
Pittsburgh 100
1
2
3
4
5X23
X13X34
X35
X45 X54 X12
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Objective function
Minimize total transportation costs
70×X12 + 30×X13 + 50×X23 + 20×X34 + 25×X35 + 30×X45 + 15×X54
Interpretation: We will ship at the lowest cost possible while still satisfying demand.
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Constraints
One constraint for every nodeOutflow = InflowPhiladelphia: X12 + X13 = 100Washington: X23 = 150 + X12Harrisburg: X13 + X23 = 100 + X34 + X35Buffalo: X34 + X54 = 50 + X45Pittsburgh: X45 + X35 = 100 + X54
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The shortest path problem
Texas Transfer Corp. is an express package delivery company
Services 10 cities in Texas
Must plan fastest routes
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The shortest path problem
The flow limits on origin and destination arcs are (1,1)
The flow limits on all other arcs are (0,1)
Assign a flow of 1 along an arc if it is used in the route; otherwise the flow is 0.
Minimize the overall flow cost (distance).
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Decision variables
One decision variable for each arc in the network
Example:
X12: A flow from City 1 to City 2
X12 is either 0 or 1
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User input variables
Oi = 1 if the origin is city i, 0 otherwiseExample: O1 = 1, O2 = … = O10 = 0
Dj = 1 if the destination is city j, 0 otherwiseExample: D10 = 1, D1 = … = D9 = 0
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Travel time
City 1 → City 2
122 / 51 + 0 = 2.39
22
Distance / Avg. speed + Delay
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23
Min 2.39X12 + 6.08X13 + … + 3.41X10_9
Excel function:
=SUMPRODUCT( , )
Objective function
Column Column
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Constraints
Flow-balance constraints
City Constraint
(1) Amarillo O1 + X21 + X31 = D1 + X12 + X13
(2) Lubbock O2 + X12 + X42 + X52 = D2 + X21 + X24 + X25
(3) Fort Worth O3 + X13 + X53 + X83 + X93 = D3 + X31 + X35 + X38 + X39
(4) El Paso O4 + X24 + X54 + X64 = D4 + X42 + X45 + X46
(5) Abilene O5 + X25 + X35 + X45 + X65 = D5 + X52 + X53 + X54 + X56
(6) San Angelo O6 + X46 + X56 + X76 = D6 + X64 + X65 + X67
(7) San Antonio O7 + X67 + X87 + X97 + X10_7 = D7 + X76 + X78 + X79 + X7_10
(8) Austin O8 + X38 + X78 = D8 + X83 + X87
(9) Houston O9 + X39 + X79 + X10_9 = D9 + X93 + X97 + X9_10
(10) Corpus Christi
O10 + X7_10 + X9_10 = D10 + X10_7 + X10_9
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Flow-balance constraints
Example: City 1
O1 + X21 + X31 = D1 + X12 + X13
Excel Function: SUMIF
O1 + SUMIF( , , )
D1 + SUMIF( , , )
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Column 1 Column
Column 1 Column
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Other constraints
We also include non-negativity constraints.
We can ignore the upper bounds on the decision variables (e.g. X21≤1) since they are redundant.
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Results
Shortest route:
1→3→9→10
Travel time:
13.81 hours
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