Allocating Resources to Strategic Alternatives TP I
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Transcript of Allocating Resources to Strategic Alternatives TP I
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Allocating Resources toStrategic Alternatives
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Operations managers often prefer to
use mathematical model , as it is
quicker, inexpensive and morereliable
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Transportation Problem
The Transportation Problem is one of the
sub-classes of LPPs in which the objective
is to transport various quantities of a single
homogeneous commodity, that are initially
stored at various origins, to different
destinations in such a way that the total
transportation cost is minimum.
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The Transportation problem in LPP
The transportation problem is a specialcase of linear programming problem
This method is applicable in situations
involving the physical movement of goodsfrom plants to warehouses , warehouses towholesalers, wholesalers to retailers ,retailers to customers
A transportation problem can be eitherbalanced or unbalanced
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TransportationProblem
How much should be shipped from severalsources to several destinations
Sources: Factories, warehouses, etc.
Destinations: Warehouses, stores, etc.
Transportation models
Find lowest cost shipping arrangement
Used primarily for existing distribution
systems
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A Transportation Model Requires
The origin points, and the capacity or
supply per period at each
The destination points and the demand per
period at eachThe cost of shipping one unit from each
origin to each destination
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Consider a commodity which is produced at
various centers called SOURCES and is
demanded at various otherDESTINATIONS.
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The production capacity of each source
(availability) and the requirement of eachdestination are known and fixed.
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The cost of transporting one unit of the
commodity from each source to each
destination is also known.
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The commodity is to be transported from
various sources to different destinations in
such a way that the requirement of each
destination is satisfied and at the same timethe total cost of transportation in minimized.
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This optimum allocation of the commodity
from various sources to different destinations
is called TRANSPORTATION PROBLEM.
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A transportation problem can be statedmathematically as follows:
Let there be m SOURCES and n DESTINATIONS
Let ai : the availability at the ith source
bj : the requirement of the jth destination.
Cij : the cost of transporting one unit of
commodity from the ith source to
the jth destination
xij : the quantity of the commoditytransported from ith source to the jth
destination (i=1, 2, m; j=1,2, ..n)
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Transportation Table
Source Supply
Demand
1
2
:
m
a 1
a 2
:
a m
1 2 . . n
b 1 b 2 b n
Quantity demanded or
required
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Transportation Table
Destination
Source Supply
Demand
1
2
:
m
a1
a 2
:
a m
1 2 . . n
b 1 b 2 b n
x11
x12
. . x1n
x 21 x 22 . . 2n
: : : : : : :
x m1 x m2 . . x mn
:
xQuantity supplied fromsources to destinations
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Transportation Table
Destination
Source 1 2 . . n Supply
1x 11 c 11 x 12 c 12
. .x 1n c 1n a1
2 x 21 c 21 x 22 c 22 . . x 2n c 2n a2
: : : : : : : : : :
m x m1 c m1 x m2 c m2 . . x mn c mn am
Demand b 1 b 2 . . b n
Cost of supplying1 unit from sources to
destinations
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Transportation Table
ToFrom
Bangalore(A)
Mangalore(B)
Mysore
(C)Factory
Capacity
Chennai
(D)100
Delhi
(E)300
Pune
(F)
300
WarehouseRequirements
300 200 200 700
5
8
9 7
4
4 3
3
5
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Unbalanced transportation problem
When the total availability is equal to the
total requirement the problem (i.e. ai = bj)is said to be a balanced transportation
problem. If the total availability at different
sources is not equal to the total requirementat different destinations, (i.e. aibj), the
problem is said to be an unbalanced
transportation problem.
Steps to convert an unbalanced problem to a
balanced one are
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Ifai > bj i.e. the total availability is greaterthan the total requirement, a dummy
destination is introduced in the transportation
problem with requirement = ai - bj. Theunit cost of transportation from each source
to this destination is assumed to be zero.
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If ai < bj i.e. the total availability is less
than the total requirement, a dummy source
is introduced in the transportation problemwith requirement = bj - ai. The unit cost of
transportation from each destination to this
source is assumed to be zero.
After making the necessary modifications in thegiven problem to convert it to a balanced problem,
it can be solved using any of the methods.
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1. Destination
Source D1 D2 D3 D4 D5 Availability
S1 5 3 8 6 6 1100
S2 4 5 7 6 7 900
S3 8 4 4 6 6 700
Requirement 800 400 500 400 600
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2. Destination
SourceD1 D2 D3 D4 D5 Availability
S1 6 4 4 7 5 100
S2 5 6 7 4 8 125S3 3 4 6 3 4 175
Requirement 60 80 85 90 70
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Determination of the starting Solution
In any transportation model we determine a starting
BFS and then iteratively move towards the optimalsolution which has the least shipping cost.
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A solution where the row total of
allocations is equal to the
availabilities and the column total is
equal to the requirements is called a
Feasible Solution.
The solution with m+n-1 allocationsis called a Basic Solution.
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Different methods of obtaining initial basic
feasible solution to a balanced minimization
transportation problem are
LEAST COST METHOD or Matrix
Minima Method ( Best Method)
NORTH WEST CORNER RULE
VOGELS APPROXIMATION METHOD (
Penalty method orVAMS Method.
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LEAST COST Method of determining the
starting BFS.
In this method we start assigning as much as possible
to the cell with the least unit transportation cost (ties
are broken arbitrarily) and the associated amounts of
supply and demand are adjusted by subtracting the
allocated amount.
Cross out the row (column) with zero supply(zero demand) to indicate that no further
assignments can be made to that row(column).
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If both a row and a column are simultaneously
satisfied then
if exactly one row or column is left uncrossed
make the obvious allocations and stop. Else cross
out one only (either the row or the column) and
leave a zero supply(demand) in the uncrossed outrow(column).
Next look for the uncrossed out cell with the
smallest unit cost and repeat the process until no
further allocations are to be made.
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Consider the transportation table:
3 7 6 45
2 4 3 2
2
4 3 8 53
3 3 2 2
Destination
1 2 3 4 Supply
Source
1
2
3
Demand
Total shipping cost = 36
2
1
1 4
3
0
220 2
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1
A B C D Supply
E 1 2 3 4 6
F 4 3 2 0 8
G 0 2 2 1 10Demand 4 6 8 6
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2
D E F G Available
A 11 13 17 14 250
B 16 18 14 10 300
C 21 24 13 10 400
Demand 200 225 275 250
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NORTH-WEST Corner Method for
determining a starting BFS
The method starts at the north-west corner cell
(i.e. cell (1,1)).
Step 1. We allocate as much as possible to theselected cell and adjust the associated amounts of
supply and demand by subtracting the allocated
amount.Step 2. Cross out the row (column) with zero
supply (zero demand) to indicate that no further
assignments can be made to that row(column).
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If both a row and a column are simultaneously
satisfied then
if exactly one row or column is left uncrossed
make the obvious allocations and stop. Else cross
out one only (either the row or the column) and
leave a zero supply(demand) in the uncrossed out
row(column).
Step 3. If no further allocation is to be made,stop. Else move to the cell to the right (if a
column has just been crossed out) or to the cell
below if a row has just been crossed out. Go to
Step 1.
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Consider the transportation table:
3 7 6 45
2 4 3 2
2
4 3 8 53
3 3 2 2
Destination
1 2 3 4 Supply
Source
1
2
3
Demand
32
2
1
1 11
1
1 22
Total shipping cost = 48
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1. Destination
Source D1 D2 D3 D4 D5 Availability
S1 5 3 8 6 6 1100
S2 4 5 7 6 7 900
S3 8 4 4 6 6 700
Requirement 800 400 500 400 600
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2. Destination
SourceD1 D2 D3 D4 D5 Availability
S1 6 4 4 7 5 100
S2 5 6 7 4 8 125S3 3 4 6 3 4 175
Requirement 60 80 85 90 70
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1
A B C D Supply
E 1 2 3 4 6
F 4 3 2 0 8
G 0 2 2 1 10Demand 4 6 8 6
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2
D E F G Available
A 11 13 17 14 250
B 16 18 14 10 300
C 21 24 13 10 400
Demand 200 225 275 250
3 Vogels approximation method (VAM)
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3. Vogels approximation method (VAM)
Step 1. For each row (column) remaining under
consideration, determine a penalty by subtracting thesmallest unit cost in the row (column) from the next
smallest unit cost in the same row(column). ( If two
unit costs tie for being the smallest unit cost, then
the penalty is 0).
Step2. Identify the row or column with the largest
penalty. Break ties arbitrarily. Allocate as much as
possible to the cell with the least unit cost in the
selected row or column.(Again break the ties
arbitrarily.) Adjust the supply and demand and cross
out the satisfied row or column.
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If both a row and a column are simultaneously
satisfied then
if exactly one row or column is left uncrossed
make the obvious allocations and stop. Else cross
out one only (either the row or the column) and
leave a zero supply(demand) in the uncrossed outrow(column). (But omit that row or column for
calculating future penalties).
Step 3. If all allocations are made, stop. Else go tostep 1.
D ti ti
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3 7 6 45
2 4 3 22
4 3 8 53
3 3 2 2
Destination1 2 3 4 Supply Row Penalties
So
u
rc
e
1
2
3
Demand
Total shipping cost = 32
Column
Penalties
1
0
1
1 1 3 2
2
0
1
-
1
1 4 - 1
3
0
3 0 0 2
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2
D E F G Available
A 11 13 17 14 250
B 16 18 14 10 300
C 21 24 13 10 400
Demand 200 225 275 250
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All three Methods
Origin Destination Supply
D1 D2 D3
O1 2 7 4 5
O2 3 3 1 8
O3 5 4 7 7
O4 1 6 2 14
Demand 7 9 18 34
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Moving Towards Optimality
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Algorithm of MODIFIED DISTRIBUTION(MODI) METHOD
Step I: For an initial basic feasible solutionwith (m+n-1) occupied (basic) cells,calculate ui and vj values for rows andcolumns respectively using the relationship ui+ vj = Cij for all allocated cells only. To startwith assume any one of the ui or vj to bezero.
Step II: For the unoccupied (non-basic)cells, calculate the cell evaluations or thenet evaluations as ij = (ui + vj)Cij
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Step III:
a) If all ij < 0, the current solution is optimaland unique.
b) If any ij = 0, the current solution is
optimal, but an alternate solution exists.c) If any ij > 0, then an improved solutioncan be obtained; by converting one of the
basic cells to a non basic cells and one of thenon basic cells to a basic cell. Go to step IV.
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Step IV: Select the cell corresponding to
most Positive cell evaluation. This cell is
called the entering cell. Identify a closed
path or a loop which starts and ends at the
entering cell and connects some basic cells atevery corner.
Step V: Put a + sign in the entering cell andmark the remaining corners of the loop
alternately withand + signs.
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Step VI: From the cells marked withsign,
select the smallest quantity (say ). Add
to each quantity of the cell marked with +
sign and subtract from each quantity of
the cell marked withsign. In case of a tie,make zero allocation to any one of the cells.
This will make one non-basic cells as basic
and vice-versa.
Step VII: Return to step I.