transportation + assignment models (1)
TRANSCRIPT
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Transportation and
Assignment Models
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Specialized Problems
Transportation Problem Distribution of items from several sources
to several destinations. Supply capacities
and destination requirements known. Assignment Problem
One-to-one assignment of people to jobs,etc.
Special ized algo ri thm s
save t ime!
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Transportation Problem
Goal ~To determine the delivery routes tobe used & units delivered so as tominimise total transportation cost
Initial sol. technique ~ North west cornerrule, Vogels approx., Lowest cost,
Final sol. ~ Stepping Stone, MODI,
Involves step by step process
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Example of transportation problem
Masjid Tanah(D)
Banting(E)
Taiping(F)
Kuantan(A)
Johor Bahru(B)
Gombak(C)
Factories
(Sources)
Retail-shops
(destinations)
Capacities requirementsroutes
100units
300units
300units
300units
200units
200units
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Transportation Problem
Des Moines
(100 units)
capacity
Cleveland
(200 units)
required
Boston(200 units)
requiredEvansville
(300 units)
capacity
Ft. Lauderdale
(300 units)
capacity
Albuquerque(300 units)
required
D
A
E
C B
F
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Transportation Costs
From
(Sources)
To(Destinations)
Albuquerque(A)Boston(B)
Cleveland(C)
Des Moines(D)
Evansville(E)
FortLauderdale(F)
$5
$8
$9
$4
$4
$7
$3
$3
$5
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Unit Shipping Cost:1Unit, Factory toWarehouse
Des Moines
(D)
Evansville
(E)
Fort
Lauderdale (F)
Warehouse
Req.
Albuquerque
(A) Boston (B)Cleveland
(C) Factory
Capacity5 4 3
57
48
9
3
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Total Demand and Total Supply
Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
100
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ranspor a on a e or xecu veFurniture Corp.
Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
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Initial Solution Using the NorthwestCorner Rule
Start in the upper left-hand cell andallocate units to shipping routes asfollows:
Exhaust the supply (factory capacity) ofeach row before moving down to the nextrow.
Exhaust the demand (warehouse)requirements of each column beforemoving to the next column to the right.
Check that all supply and demand
requirements are met.
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Initial SolutionNorth West Corner Rule
Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
100
200 100
100 200
Total Transportation cost = 100x5 + 200x8 + 100x4 + 100x7 + 200x5= $4200
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1. Find the difference between the two lowest unittransportation costs for each row & column
2. Determine the row or column with the greatest
difference3. Assign as many units as possible to the cost
square in the row or column selected
4. Eliminate any row or column that has just beencompletely satisfied by the assignment made
5. Repeat steps 1-4
Initial SolutionVogels Approximation steps (refer p 424)
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Vogels Approximation1. For each row/column of table, find
difference between two lowest costs.(Opportunity cost)
2. Find greatest opportunity cost.
3. Assign as many units as possible tolowest cost square in row/column with
greatest opportunity cost.
4. Eliminate row or column which has beencompletely satisfied.
4. Begin again, omitting eliminated
rows/columns.
I iti l S l ti
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Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
Total Transportation cost = $3900
Initial SolutionVogels Approximation (refer p 427)
100
200
200 100
100
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Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
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Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
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Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
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The Stepping-Stone Method
1. Select any unused square to evaluate.
2. Begin at this square. Trace a closed path back tothe original square via squares that are currentlybeing used (only horizontal or vertical movesallowed).
3. Place + in unused square; alternate - and + oneach corner square of the closed path.
4. Calculate improvement index: add together theunit cost figures found in each square containing a
+; subtract the unit cost figure in each squarecontaining a -.
5. Repeat steps 1 - 4 for each unused square.
If all indices computed > = 0, optimal sol.
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Stepping-Stone Method - The DesMoines-to-Cleveland Route
Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
200
100
100
100 200
- +
-
+
+
-
Start
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Stepping-Stone MethodAn Improved Solution
Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
100
100
200
200100
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Third and Final Solution
Des Moines
(D)
Evansville
(E)
Ft Lauderdale
(F)
Warehouse
Req.
Albuquerque(A)
Boston(B)
Cleveland(C)
FactoryCapacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
100
200
100200
100
I iti l S l ti
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Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
Using stepping stone:Each improvement index in the unoccupied cell >=0, optimal solutionTotal Transportation cost = $3900
Initial SolutionVogels Approximation (refer p 427)
100
200
200 100
100
+2 +2
+1
+1
MODI Method: 5 Steps
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MODI Method: 5 Steps1. Compute the values for each row and column: set Ri+ Kj= Ci jfor
those squares current ly used o r occu pied.2. After writing all equations, set R1 = 0.
3. Solve the system of equations forRiand Kjvalues.4. Compute the improvement index for each unused square by the
formula improvement index:
Ii j = Ci j- Ri- Kj5. Select the largest negative index and proceed to solve the problem
as you did using the stepping-stone method.
1. Set up equation for each occupied celllet R1=0 then solve for K1=5, R2=3, K2=1, R3=6, and K3= -1
2. Compute improvement index for each unoccupied cellI_EA = 8 - 3 - 5 = 0I_DB = 4 - 0 - 1 = 3I_DC = 3 - 0 - (-1) = 4I_FB = 7 - 6 - 1 = 0
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Stepping-Stone Method - The DesMoines-to-Cleveland Route
Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)
Warehouse
Req.
Albuquerque
(A)Boston
(B)
Cleveland
(C)Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
200
100
100
100 200
Kj K1 K2 K3
Ri
R1
R2
R3
3 4
1
-2
Each improvement index Iij in the unoccupied cell >=0, optimal solutionI_FA isve (not optimal yet). Total Transportation cost = $4200
To
From
+
-
-
+
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Special Problems in TransportationMethod
Unbalanced Problem
Demand Less than Supply
Demand Greater than Supply
Degeneracy (one or more Iij =0)
More Than One Optimal Solution
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Unbalanced ProblemDemand Less than Supply
Factory 1
Factory 2
Factory 3
Customer
Requirements
Customer1 Customer 2
DummyFactory
Capacity
150 80 150 380
80
130
1708 5 0
0
09
1015
3
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Unbalanced ProblemSupply Less than Demand
Factory 1
Factory 2
Dummy
Customer
Requirements
Customer1
Customer2
Customer3
FactoryCapacity
150 80 150 380
80
130
1708 5 16
7
00
1015
0
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Degeneracy
Factory 1
Factory 2
Factory 3
Customer
Requirements
Customer1
Customer2
Customer3
FactoryCapacity
100 100 100 300
80
120
1005 4 3
3
57
48
9
100
100
80
20
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Degeneracy - Coming Up!
Factory 1
Factory 2
Factory 3
Customer
Requirements
Customer1
Customer2
Customer3
FactoryCapacity
150 80 50 280
80
130
708 5
7
9
16
10
1015
3
70
80
50
50
30
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Stepping-Stone Method - The DesMoines-to-Cleveland Route
Des Moines
(D)
Evansville
(E)
Fort
Lauderdale(F)
Warehouse
Req.
Albuquerque(A)
Boston
(B)
Cleveland(C)
Factory
Capacity
300 200 200 700
300
300
1005 4 3
3
57
48
9
Start
200
100
100
100 200
- +
-
+
+
-
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The Assignment Method
1. subtract the smallest number in each row fromevery number in that row
subtract the smallest number in eachcolumn from every number in that column
2. draw the minimum number of vertical andhorizontal straight lines necessary to coverzeros in the table
if the number of lines equals the number ofrows or columns, then one can make anoptimal assignment (step 4)
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The Assignment Method - continued
3.if the number of lines does not equal thenumber of rows or columns
subtract the smallest number not covered bya line from every other uncovered number
add the same number to any number lying atthe intersection of any two lines
return to step 2
4. make optimal assignments at locations ofzeros within the table
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The Assignment Problem
Project
Person 1 2 3
Adams $11 $14 $6
Brown $8 $10 $11
Cooper $9 $12 $7
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Hungarian Method
Initial TablePerson Project
1 2 3
Adams 11 14 6
Brown 8 10 11
Cooper 9 12 7
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Hungarian Method
Row ReductionPerson Project
1 2 3
Adams
Brown
Cooper
5 8 0
0 2 3
2 5 0
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Hungarian Method
Column Reduction
Person Project
1 2 3
Adams 5 6 0
Brown 0 0 3
Cooper 2 3 0
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Hungarian Method
Person Project
1 2 3
Adams
Brown
Cooper
5 6 0
0 0 3
2 3 0
Testing CoveringLine 2
CoveringLine
1
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Hungarian Method
Person Project
1 2 3
Adams 3 4 0
Brown 0 0 5
Cooper 0 1 0
Revised Opportunity Cost Table
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Hungarian Method
Person Project
1 2 3
Adams 3 4 0
Brown0 0 5
Cooper0 1 0
TestingCovering Line1
Covering
Line 2
Covering Line 3
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Hungarian Method
AssignmentsPerson Project
1 2 3
Adams 6
Brown 10
Cooper 9
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Maximization Assignment Problem
Project
1 2 3 Dummy
Adams $11 $14 $6 $0
Brown $8 $10 $11 $0
Cooper $9 $12 $7 $0
Davis $10 $13 $8 $0
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Maximization Assignment Problem
Project
1 2 3 Dummy
Adams $32 $0 $8 $14
Brown $6 $4 $3 $14
Cooper $5 $2 $7 $14
Davis $4 $1 $6 $14
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Part A ~ 14 October 2004
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