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LP - IV Lecture

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    (4)Transportation Method

    Special method.

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    Transportation Problems

    and Method to solvesuch problems

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    C11 C12 . . C1n

    C21 C22 . . C2n

    . . . . .

    . . . . .

    Cm1 . . . Cmn

    Destination (Warehouse)

    1 2 . . n

    1

    2

    .

    .

    m

    Origin(Plant)

    Cases

    Cij = Cost of shipping unit item from ith

    Origin to jth Destination

    Transportation Problem

    b1 b2 . . bn iabj

    a1

    a2

    .

    am

    bjia

    Cij

    Pij

    bjia

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    B. Unbalanced Minimization with Cij

    Categories of Transportation Problems

    bjia

    bjia

    bjia

    bjia

    A. Balanced Minimization with Cij

    D. Unbalanced Maximization with Pij

    C. Balanced Maximization with Pij

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    C11 C12 . . C1n

    C21 C22 . . C2n

    . . . . .

    . . . . .

    Cm1 . . . Cmn

    Destinations(Warehouses)

    1 2 . . n

    1

    2.

    .

    m

    Origins(Plants)

    Xij = No. of items shipped from ith Origin

    to jth Destination

    A.Balanced Minimization Problem

    b1 b2 . . bn bjia

    a1

    a2.

    am

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    Formulation of Transportation Problem

    as LPP

    1 1

    1

    1

    .

    / , ( )

    ( )

    0 & .

    m n

    i j

    n

    j

    m

    i

    Min Z Cij Xij

    s t Xij ai for all i I

    Xij bj for all j II

    All Xij for all i j

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    Methods to get IBFS :

    5. VAM: Vogels Approximate Method

    1. NWCM: North West Corner Method

    2. RMM: Row Minima Method

    3. CMM: Column Minima Method

    4. MMM: Matrix Minima Method

    (Least Cost Entry Method)

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 5

    1

    23

    3

    3 4 6 7 10

    5

    13

    12

    30

    1022 6 5

    2 2

    2 2

    5

    11

    10

    NWCM

    Z = 148

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 51

    23

    3

    3 4 6 7 10

    5

    13

    12

    30

    1022 6 5

    2

    Therefore NWCM solution is :

    Z = 148

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 51

    23 3

    3 4 6 7 10

    5

    13

    12

    30

    106

    21

    5

    3

    3 2

    1

    3

    6

    RMM

    3

    Z = 85

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 51

    23 3

    3 4 6 7 10

    5

    13

    12

    30

    106

    21

    5

    3

    Therefore RMM solution is :

    Z = 85

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 51

    23

    2

    3 4 6 7 10

    5

    13

    12

    30

    46

    7

    3

    2

    2 2

    11

    6

    CMM

    6

    4

    6

    Z = 108

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 51

    23

    2

    3 4 6 7 10

    5

    13

    12

    30

    46

    7

    3 2

    Therefore CMM solution is :

    6

    Z = 108

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 51

    23

    3 4 6 7 10

    5

    13

    12

    30

    2

    3

    10

    2

    3

    6

    MMM

    1

    5

    6

    13 33

    Z = 85

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 51

    23

    3 4 6 7 10

    5

    13

    12

    30

    2 10

    Therefore MMM solution is :

    5

    6

    133

    Z = 85

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 51

    23

    3 4 6 7 10

    5

    13

    12

    30

    2 0

    1

    1

    VAM

    0

    51 1

    10

    3

    PenaltyNumber

    PenaltyNumber

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    4 2 3 2

    5 4 5 2

    6 5 4 7

    1 2 3 4

    1

    2

    3

    3 4 6 7

    5

    3

    12

    202 0

    2

    1

    VAM

    04

    1

    1

    1

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    4 3 2

    6 4 7

    1 3 4

    1

    3

    3 6 4

    1

    12

    133

    2

    VAM

    1

    2 1 5

    1

    12

    3 6 3 63

    6 3

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    4 2 3 2 6

    5 4 5 2 1

    6 5 4 7 7

    1 2 3 4 51

    23

    3 4 6 7 10

    5

    13

    12

    30

    VAM solution is :

    10

    3

    3

    14

    63

    Z = 89

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    Optimality Test on IBFS

    Methods : 1. Stepping Stone Method2. MODI Method

    MODI Method Of Checking Optimality

    Condition To Be Satisfied :

    Filled Cells = m + n - 1

    Where m = No. of Rowsn = No. of columns

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    MODI Method for Optimality Test

    Get all ui & vj starting with any ui or vj as zero,

    such that ui + vj = cij in filled cells.

    Get vacant cell Evaluation of all vacant cells.

    (VCE)ij = cij(ui + vj)

    VAM S l ti

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    2 2

    2 1

    6 4 7

    -1

    2

    2

    7

    VAM Solution

    4

    3 6

    10

    1

    3

    3

    0-3 -10

    4 3 6

    5 4 55-2 7

    As (VCE)32 = -2, the solution under test

    is not optimal.

    Z = 89

    ui

    vj

    F difi i f i i l i

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    2 2

    2 1

    6 4 7

    -1

    2

    2

    7

    For modification of existing solution :

    4

    3 6

    10

    1

    3

    3

    0-3 -10

    4 3 6

    5 4 5

    5-2 7

    4 1

    3 3

    1 4

    ui

    vj

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    2 2

    2 1

    6 4

    Hence, improved solution will be :

    1

    3 6

    10

    4

    3

    3 5

    Z = 89(2) (3)

    Z = 83

    This solution is to be checked by MODI Method

    for optimality.

    i

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    2 2

    2 1

    6 4

    Improved solution :

    1

    3 6

    10

    4

    3

    3 5 0

    6 5 4

    - 3

    5

    - 3

    4

    4 3 6

    5 4 5

    7 7

    As all vacant cell evaluations are positive,the solution is optimal, giving Z = 83.

    ui

    vj

    MMM S l ti

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    6 5 4

    -1

    MMM Solution

    3

    1 10

    5

    2

    3 2

    2

    445

    1

    ui

    vj

    -1 -3

    0 2

    0

    6 63

    -1

    0

    4

    5

    2 3

    57

    6

    7

    Z = 85

    For modification of existing solution :

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    2 2

    2 1

    6 4 7

    For modification of existing solution :

    3 6

    10

    5

    2

    4 3 6

    5 4 5

    5 7

    1

    5

    23

    1 4

    ui

    vj

    -10

    -31

    3

    0-1

    -1

    0

    6 5 4 3 2

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    2 2

    2 1

    6 4

    Hence, improved solution will be :

    1

    3 6

    10

    4

    3

    3 5

    Z = 852

    Z = 83

    This solution is to be checked by MODI Method

    for optimality.

    I d l ti

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    2 2

    2 1

    6 4

    Improved solution :

    1

    3 6

    10

    4

    3

    3 5 0

    6 5 4

    - 3

    5

    - 3

    4

    4 3 6

    5 4 5

    7 7

    As all vacant cell evaluations are positive,the solution is optimal, giving Z = 83.

    ui

    vj

    F h l d fi i t E l

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    1

    For cases when closed figure is not Equal or

    Rectangle :

    7

    3

    5

    2

    4

    45

    2 3

    Problem :

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    The following table gives minimization transportation problem with

    given allocation, check whether given solution is optimal. If not, get

    the optimal solution.

    Problem :

    A

    B

    C

    D

    10 2225

    24 2810

    27 2515

    10 1520

    Demand 25 20 25

    20

    10

    18

    20

    25

    10

    15

    20

    X Y Z Supply

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    5

    10

    18

    15

    20

    18

    15

    2210

    10

    2025

    20

    10

    15

    -1

    -6

    -20

    Z = 1270

    515 0

    24 28

    2527

    10

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    20

    25 520

    20

    Hence, improved solution will be :

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    20

    5 5

    20

    Hence, improved solution will be :

    Z 870 (3) ( 870)

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    22

    10

    18

    12

    18

    10

    10

    205 20

    10

    15

    Z = 870(3) ( = 870

    10

    20

    -12-2 0

    22

    24

    27

    20

    28

    25

    15

    )

    Hence, this solution is Optimal.

    Problem :

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    12

    -20

    04

    2012

    8

    28120

    13530

    Z = 7100

    3216 4

    32 12

    3216

    160

    25

    -32

    Problem :

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    30

    120 95

    5525

    25

    Hence, improved solution will be :

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    12

    12

    04

    2012

    8

    2895

    -8

    13555

    Z = 7100(32) (25) = 6300

    016 4

    25

    32 12

    3216

    160

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    4

    4

    04

    2012

    8

    28

    40150

    Z = 6300(8) (95) = 5540816 4

    120

    32 12

    3216

    65 95

    Hence, this solution is Optimal.

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    Home Assignments :

    [ 1 ] Mr XYZ has suggested the following feasible solution

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    [ 1 ] Mr XYZ has suggested the following feasible solution

    for a minimizing transportation problem as given below.

    Check whether the solution is optimal. If not, get optimal

    solution.A B C D

    1 10 2 20 11

    2 12 7 9 20

    3 3 14 16 18

    15

    1015

    5

    [ 2 ] For following transportation (minimization) problem

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    [ 2 ] For following transportation (minimization) problem,

    present total cost of transportation is Rs. 3100. Is it possible to

    reduce this by proper scheduling ? What can be the saving ?

    Centres

    A B C Supply

    X 25 35 10 150

    Y 20 5 80 100

    Demand 50 50 150

    Factories

    [ 3 ] A m f t t t hi 8 l d f hi d t

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    [ 3 ] A manufacturer wants to ship 8 loads of his product as

    shown in table below. The matrix gives km from origin

    O to destination D. The shipping costs are Rs. 10 per load per

    km. Find the optimal solution.

    D1 D2 D3

    O1 50 30 220 1

    O2 90 45 170 3

    O3 250 200 50 4

    4 2 2

    [ 4 ] Solve the following minimization Transportation

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    [ 4 ] Solve the following minimization Transportation

    problem.

    D1 D2 D3 D4 Supply

    O1 2 3 11 7 6

    O2 1 1 6 1 1

    O3 5 8 15 10 10

    Demand 7 5 3 2

    [ 5 ] In following transportation problem X, Y, Z are

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    [ ] g p p , ,

    warehouses and P, Q, R, S are customers. The capacity at

    the warehouses and the demand from customers are

    shown around the perimeter. Per unit transportation costsare shown in the cells.

    P Q R S

    X 7 8 11 10 30

    Y 10 12 5 4 45

    Z 6 10 11 9 35

    20 28 17 33

    ( i ) Find VAM solution. Is this solution optimal ? If not,

    get optimal solution.

    ( ii ) By how much should YP cost need to be reduced in

    order to make shipments along this route worth while ?

    [ 6 ] Using NWC rule method get initial feasible solution for

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    [ 6 ] Using NWC rule method, get initial feasible solution for

    following minimization transportation problem. Is this

    solution optimal ? If not, get optimal solution and

    corresponding optimal value of objective function.

    A B C Plant

    capacity

    W 4 8 8 55

    X 16 24 16 25

    Y 8 16 24 35

    Project Requirement 35 45 35

    [ 7 ] A company has three plants at P1 P2 P3 which supply to

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    [ 7 ] A company has three plants at P1, P2, P3 which supply to

    warehouses W1, W2, W3, W4 and W5. Monthly plant capacities

    are 800, 500 and 900 units respectively. Monthly warehouse

    requirements are 100, 400, 500, 400 and 800 units respectively.The unit transportation costs in rupees are given in table below.

    W1 W2 W3 W4 W5

    P1 5 8 6 5 3P2 4 7 7 3 6

    P3 8 4 6 4 2

    ( i ) Determine the optimal allocation for the company.

    ( ii ) If the warehouse W1 is closed because of reduced

    demand and the corresponding units are shipped to warehouse

    W3 by raising its capacity find the total cost of shipment. Is it

    optimal schedule ? If not, get optimal schedule.

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    Transshipment Problems :

    Transportation Problem :

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    7 3

    9 3

    3 3

    5

    3 8

    6

    5

    O1

    O2

    D1 D2

    Z = 45

    Transportation Problem :

    Transshipment Problem :

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    Transshipment Problem :

    0 2 7 3

    2 0 9 3

    7 8 0 4

    2 2 2 0

    O1

    O1

    O2

    O2

    D1

    D1

    D2

    D2

    11 11 + 6

    11 + 5

    11

    11 11

    11

    11 + 3 11 + 8

    11

    11

    3

    6

    5

    8

    Z = 39

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    O1

    D2

    6

    D13

    O2

    5

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    Home Assignments :

    [ 1 ]

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    0 1 3 4

    1 0 2 4

    3 2 0 1

    4 4 1 0

    O1

    O1

    O2

    O2

    D1

    D1

    D2

    D2

    5

    25

    20 10

    [ 2 ]

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    0 2 2 1

    1 0 2 3

    2 2 0 2

    1 3 2 0

    O1

    O1

    O2

    O2

    D1

    D1

    D2

    D2

    8

    3

    7 4

    [ 3 ]

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    0 4 20 5 25 12

    10 0 6 10 5 20

    15 10 0 8 45 7

    20 25 10 0 30 6

    20 18 60 15 0 10

    10 25 30 23 4 0

    O1

    O1

    O2

    O2

    D1

    D1

    D2

    D2

    O3

    O4

    O3 O4

    Th k

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    Thank you

    For any Query or suggestion :Contact :Dr. D. B. Naik

    Professor, Training & Placement SectionSardar Vallabhbhai National Institute ofTechnology, Surat

    Ichchhanath, Surat395007, Gujarat.

    Email: [email protected]


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