# Topic â€“ 10 Transportation and Assignment Problems

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```					     Topic – 10
Transportation and
Assignment Problems
Transportation Problem
Major Characteristics:
 Several sources (plant) with given supply
capacity.
 Several destinations (warehouse) with given
demand.
 Cost Matrix: unit transportation cost from each
source to each destination.

Assumptions:
 No transshipment between any two
destinations/two sources.
 All demand/capacity constraints should be
attempted to meet.
Transportation Problem
Decisions: How much should be shipped from each
source to each destination so that total
transportation cost is minimized?

Why Develop a New Method? (LP can certainly be
applied)
 LP is still NP-complete to large size problems.
 Computational efficient methods are desirable to
large size problems.

Transportation Problem Model: Tabular
Presentation
Transportation Problem
Transportation Cost
Transportation Table
(Seek its initial solution on board)
LP Formulation:
Let: Xad – Amount to be shipped from D to A

Xcf – Amount to be shipped from F to C
(9 Variables)

Objective Function:

Min: Z = 5Xad + 4Xbd + 3Xcd + 8Xae
+ 4Xbe + 3Xce + 9Xaf + 7Xbf + 5Xcf
Sub. To:

Xad + Xbd + Xcd ≤ 100
Xae + Xbe + Xce ≤ 300   (Supply Constraints)
Xaf + Xbf + Xcf ≤ 300

Xad + Xae + Xaf ≥ 300
Xbd + Xbe + Xbf ≥ 200   (Demand Constraints)
Xcd + Xce + Xcf ≥ 200

(All Xij ≥ 0)
Steps in Solving
the
Transportation
Problem
Solution Procedure
1. Balancing Table: Make ∑di = ∑Si
– add a "Dummy" source when ∑ di > ∑ Si , or
– add a "Dummy" destination when ∑ Si > ∑di.
2. Finding an initial solution: (several methods)
– Upper-Left Corner (Northwest) Method
– Least-Cost (Large-Profit) Method
– VAM Method
3. Testing Optimality: (Can current solution be
improved?)
– Stepping-stone Procedure
– Modified-Distribution Method
4. Improving current solution toward a better
solution (Stepping-Stone method).
Solution Procedure
Special cases in Transportation Problem:
– Prohibited transportation route (take large "M"
as cost)
– Multiple optimal solution is possible.
– Transshipment allowed/Route shipment
capacity limits
(All special cases can be easily formulated and
solved by LP.)

Application Examples:
– Production Planning/Scheduling/Routing
Selection/....
Sup. P.10-18: Problem #6
Tabular form: (by northwest method-”left corner”)

to
W-1         W-2        W-3       W-4
from

Mine-1               8          13         9   500

Mine-2               11         14         5   1300

Dummy                0          0          0   600

Demand        1200        500        700       (2400)
Transportation problem
Tabular form: (by northwest method-”left corner”)

to
W-1         W-2        W-3       W-4
from

Mine-1       (1)    8          13         9   500
500

Mine-2       (2)    11   (3)   14   (4)   5   1300
700         500        100

Dummy               0          0    (5)   0   600
600

Demand       1200        500        700       (2400)

TC=8*500+11*700+14*500+5*100+0*600=19,200
Transportation problem
Tabular form: (by least-cost method)

to
W-1         W-2        W-3       W-4
from

Mine-1       500    8          13         9   500
700
Mine-2              11         14   600   5   1300
500        100
Dummy               0          0          0   600

Demand       1200        500        700       (2400)

TC=8*500+11*700+5*600+0*600=14,700 <19,200
Problem formulation of LP
   X11-amount to be shipped from M-1to W-1
   X12-amount to be shipped from M-1to W-2
…    …………………………………. ... …
   X23-amount to be shipped from M-2to W-3
   Min z=  Ci X ij
       =8X11+13X12+9X13+11X21+14X22+5X23
   SUB TO: X11+X12+X13≤500
           X21+14X22+5X23≤1300
           X11+X21≥1200
           X12+X22≥500
           X13+X23≥700
   (Xij≥0, i,j=1…3)
Transportation Problem
Maximization Objectives
Problem Description
Klein Chemicals, Inc. manufactures a product at two plants (Clifton
Springs and Danville) and ships it to four different customers
(denoted D1, D2, D3 and D4)
Profit per unit for shipping from each plant to each customer.
Profit Per Unit:

Customers
D1     D2     D3    D4
Plant Clifton                  \$32 \$34 \$32 \$40
Springs
Danville                 \$34 430 \$28 \$38
Plant Capacities and Customer Orders:

Plant        Capacities   Customers     Orders
(Units)                    (Units)
Clifton      5000         D1            2000
Danville     3000         D2            5000
D3            3000
D4            2000
Total        8000         Total         12000
Capacity                  Orders
The Stepping-Stone Method
1.   Select any unused square to evaluate.
2.   Begin at this square. Trace a closed path back to the
original; square via square that are currently being
used (only horizontal or vertical moves allowed).
3.   Place + in unused square; alternate – and + on each
corner square of the closed path.
4.   Calculate improvement index: add together the unit
figures found in each square containing a +; subtract
the unit cost figure in each square containing a -.
5.   Repeat steps 1-4 for each unused square.
Stepping-Stone Method: Tracing a
Closed Path for the Des Moines to
Cleveland route
An Initial Solution by Northwest-Corner (VP- Corner Method)

Warehouse
Plant                                                              Capacity
Miami        Denver        Lincoln       Jackson
7             2             4             5
Chicago      70            30                                          100

3             1             5             2
Houston                    60            15                            75

6             9             7             4
Buffalo                                 30            50              80

Requirem                                                                   255
70            90            45            50
ent                                                                255
Total cost = 70(\$7) + 30(\$2) + 60(\$1) + 15(\$5) + 30(\$7) + 50(\$4) = \$1,095
Initial Solution by the “Least Cost Method”

Warehouse                                              VAM
Plant                                                              Capacity
Miami        Denver        Lincoln       Jackson                         Costs
7             2             4             5
Chicago    40            15            45                                100          \$2

3             1             5             2
Houston                  75                                              75           \$1

6             9             7             4
Buffalo    30                                        50                  80           \$2

Require-                                                                       255
70            90            45            50       255
ments

VAM
\$3            \$1            \$1            \$2
Costs

Total Cost = 40*7+15*2+45*4+75*1+30*6+50*4=945<(1095)
Optimal Solution

Warehouse
Plant                                                          Capacity
Miami        Denver        Lincoln    Jackson
7             2          4             5
Chicago    +\$3           55            45         \$3              100

3             1          5             2
Houston     40            35            +\$2        +\$1             75

6             9          7             4
Buffalo    30            +\$5           +\$1        50              80

Requirem                                                               255
70            90         45            50
ent                                                            255

Total Cost = 40(\$3) + 30(\$6) + 55(\$2) + 35(\$1) + 45(\$4) + 50(\$4) = \$825
Assignment Problem: A Special Case
100=1 track load (can’t be split)

warehouse
W-1       W-2       W-3       W-4       supply
plant

P-1              3         1         2             100
4

P-2              5
9         2         1   100

P-3              6         6         3         4   100

P-4                                  7             100
9         4                   2

demand     100       100       100       100       (400)
Assignment Problem
Major Characteristics:
– Finite number of Team-Job (or the like) to be assigned.
– Assignment must be made based on one-to-one basis.
– Payoff matrix: cost (profit) for each possible assignment.

Available Solution Techniques to Assignment Problem:
– Linear Programming
– Dynamic Programming
– Integer IP (Branch-and-Bound)
– Complete Enumeration
– Transportation Model (special case with all Si = di = 1)
– Hungrain Method (A "smart" method for large size
problems)
Assignment Problem
Special Cases in Assignment Problem:
– Prohibited Assignment (assign a large
"M" as cost)
– It is possible to have multiple optimal
solutions.
– Special arrangement for a fixed
assignment or One-to-More assignment.
Special Solution Procedure
Hungrain Method: (Tabular Form)
– Principle: Add (or subtract) a constant to all
cells of a row (or column) in the table will not
change final optimal solution.
– Four-Step Procedure
– For Max Problem: (Transforming into Min
Problem)
 Change the sign ("+" or "-") of each cell in
the table.
 Add the largest cell value to all cells.
(Examples)
Special Solution Procedure
Summary:
– Both Transportation Problem and Assignment
Problem are special cases of LP problem, and
can be solved by Simplex method.
– Tabular solution technique mainly provides a
more efficient way for some large size practical
problems.
– Many practical managerial problems can be
formulated and solved by Transportation or
Assignment problem techniques.
The Assignment Method Algorithm
1. Subtract the smallest number in each row from every number in that row.
Subtract the smallest number in each column from every umber in that
column.

2. Draw the minimum number of vertical and horizontal straight lines
necessary to cover zeros in the table. If the number of lines equals the
number of rows or columns, then one can make an optimal assignment
(Step-4).

3. If the number of lines does not equal the number of rows or columns:
Subtract the smallest number not covered by a line from every other
uncovered number.
Add the same number to any number lying at the intersection of any two
lines.

4. Make optimal assignments at locations of zeros within the table.


Service costs of different team
assignments (\$ in thousand)
Service
zone   Z1   Z2      Z3
team

S1               20   15      31

S2               17   16      33

S3               18   19      27
Service costs of different team
assignments (\$ in thousand)
Service
zone   Z1       Z2       Z3
team

S1               20   5   15   0   31   16

S2               17   1   16   0   33   17

S3               18   0   19   1   27
9
Service costs of different team
assignments (\$ in thousand)
5     0     16      5     0     7

1     0     17      1     0     8

0     1     9       0     1     0
Problem 7- Supplement, p.10-18

mach   (M)    (N)    (O)     (P)
job           1      2      3       4
A       6      7      5       9

B      8      5       6      7

C      10     8       6      6

dummy        0      0       0      0
Problem 7- Supplement
   LP formulation
   Let x11,x22,….x34(12 variables)=job A assigned to
machine 1
   Min: Z=6X11+7X12+5X13+9X14+…+6X34
   Sub to: X11+X12+X13+X14=1
             X21+X22+X23+X24=1
             X31+X32+X33+X34=1
             X11+X21+X31=1
             X12+X22+X32=1
             X13+X23+X33=1
             X13+X23+X33=1       all Xij=1 or 0
Written Assignment-7 is due
on next –
Sup. P.10-22: Northeastern blood bank Case

 A) identify demand/supply     Total cost
 Type: A: (6+2)=8(Pints)       8*30=240
6*36=216
       B: (2+4)=6(Pints)
5*35=175
       A: (2+3)=5(Pints)       5*40=200
\$831
       A: (3+2)=5(Pints)
Case I: Northeastern blood bank

 Supply:
A   :      4 (pints)
B   :      10 (pints)
O    :     11 (pints)
 AB :      7 (pints)
Case I: Northeastern blood bank

 Identify cost matrix:
 1) allowable blood transfusion pattern
 (supply)                             B           \$1000
A
While A
A                            O
\$30     AB                         \$1000
(Assign a large case to prevent
A    \$1000
Unallowable transfusion) B
B              While     B
AB                   O    \$1000
\$36
A    \$1000
AB     B   \$1000
AB       AB
O
\$1000
A        B        O        AB    dummy       supply

A            30   1000 1000          30           0     4

B        1000         36   1000      36           0     10

O            35       35       35    35           0     11

AB       1000     1000 1000          40           0     7

demand       8        6        5     5      (8)        (32)

24

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