allocating resources to strategic alternatives tp i

8/22/2019 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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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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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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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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satisfied then




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



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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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satisfied then




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


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.

allocating resources to strategic alternatives tp i

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