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Extension of LP

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    Extension of LP

    Nur Aini Masruroh

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    Outline

    Transportation problem

    Assignment problem

    Short-route problem

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    Transportation model

    Introduction

    Involves the shipment of some homogeneous commodity fromvarious sources (origins) of supply to a set of destinations (sinks)each demanding specified levels of the commodity.

    The model

    m sources, i = 1, 2,,m

    n destinations,j = 1, 2,,n

    sisupply units available at source i

    djdemand units available at destinationj cijunit transportation cost from source ito destinationj

    xijamount of goods shipped from source ito destinationj

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    Example 1

    A company has 3 plants from which it must supply 2 distributors. Thefirst distributor requires 2100 units of the product, while the secondrequires 2500. The data on the 3 plants are as follows:

    Plant Capacity Unit cost,$/unit Transportation cost, $/unitDistr 1 Distr 2

    1 2000 5,000 1,000 1,000

    2 1200 4,000 2,000 3,000

    3 1400 7,000 2,000 2,000

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    LP formulation for example 1

    Letxij = the amount of product shipped from Plant ito Distributor j

    3

    1

    2

    1

    mini j

    ijijxcz

    Subject to

    2,1;3,2,10

    3,2,1

    2,1

    2

    1

    3

    1

    jix

    isx

    jdx

    ij

    j

    iij

    i

    jij

    Demand of Distr j

    Supply of Plant i

    Unit transport cost

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    Example 2

    A caterer supplies napkins to a restaurant as follows:

    Day 1 2 3 4 5 Total

    Quantity 1000 700 800 1200 1500 5200

    He can buy them at $1.00 each (new) or send them for overnightwash at $0.20 each or 2-day wash at $0.10 each. How should hesupply the restaurant at minimum cost?

    Formulate this problem into LP!

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    LP Formulation of a m-Source, n-

    Destination Transportation ProblemAssumption: Total supplies = Total demands

    Letxij= the amount of product shipped from source ito destinationj

    jandiallforx

    misx

    njdx

    tosubject

    xcz

    ij

    i

    n

    j

    ij

    j

    m

    i

    ij

    m

    i

    n

    j

    ijij

    0

    ,...,1

    ,...,1

    min

    1

    1

    1 1

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    Dual of a m-Source, n-Destination LP

    Transportation ProblemLetui= the dual variable associated with supply constraint ivj= the dual variable associated with demand constraintj

    The dual problem is given by

    n

    j

    jj

    m

    i

    ii vdusw11

    max

    Subject to

    ui+ vj cij i=1,,m ;j=1,,n

    ui, vjunrestricted for all iandj

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    Transportation algorithm

    1. Set up the problem in the standard cost requirement tableform.

    2. Find an initial basic feasible solution.

    3. Determine whether a better feasible solution is possible. If

    yes, go to Step 4; otherwise, the optimal solution has beenfound, stop.

    4. Determine the improved basic feasible solution and then goto Step 3.

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    Initial solution procedures

    The North-West Corner Method

    The Least Cost Method

    The Vogels Approximation Method (VAM)

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    North west corner

    The method starts at the north-west corner cell of cost requirementmatrix

    The steps:1. Allocate as much as possible to the selected cell and adjust the

    associated amount of supply and demand by subtracting theallocated amount2. Cross out the row or column with zero supply or demand to

    indicate that no further assignments can be made in that row orcolumn. If both a row or a column net to zero simultaneously,cross out one only, and leave a zero supply (demand) in the

    uncross-out row (column)3. If exactly one row or column is left uncross-out, stop.

    Otherwise, move to the cell to the right if a column has justbeen crossed out or below if a row has been crossed out. Goto step 1

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    Least Cost Method

    Find the better starting solution byconcentrating on the cheapest routes

    It starts by assigning as much as possible to

    the cell with the smallest unit cost (ties arebroken arbitrarily)

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    The VAM method (contd)

    4. Check number of uncrossed out rows and columns

    a) If exactly one row and one column remain uncrossed out, stop.

    b) If only one row (column) with positive supply (demand) remainsuncrossed out, determine the basic variables in the row (column)

    by the Least Cost Method and then stop.c) If all uncrossed out rows and columns have zero supply and

    demand, determine the zero basic variables by the Least CostMethod and then stop; otherwise, re-compute the penalties forall the uncrossed out rows and columns and then go to Step 2.

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    How to determine optimum solution?

    Use the method of multipliers

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    The method of multipliers

    1. Associate with each Rowiof the transportation tableau amultiplierui.

    2. Associate with each Coljof the transportation tableau amultipliervj.

    3. For each basic variablexijin the current solution, write the

    equation: ui+ vj= cij. These equations give m+n-1 equations in m+n unknowns.

    4. Set u1=0 and solve the m+n-1 equations in the remainingm+n-1 unknown multipliers.

    5. Evaluate cij bar= cij ui-vjfor each non-basic variablexij.

    6. Select the most negative cij bar as the entering variable toenter the basic.

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    The method of multipliers

    7. Determine the leaving variable by constructing a loopassociated with the selected entering variable as follows:

    a) Identify a closed loop which starts and ends at the selected non-basic variable.

    b) The loop consists of successive horizontal and vertical segments

    whose end points must be basic variables, except for the twosegments starting and ending at the non-basic variable.c) The leaving variable is selected from among the corner variables

    of the loop which will decrease when the entering variableincreases above zero level. It is selected as the one having thesmallest value, since it will be the first to reach zero value andany further decrease will cause it to be negative

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    Irregular transportation problem

    Imbalanced transportation problem (Supply

    demand)

    Add dummy row/column

    Degeneration Balanced transportation table can be solved if the number

    of allocated cells = number of raw + number of column 1

    If not, add artificial allocation with the amount of 0 so thatthe stepping stone path can be performed

    Forbidden route

    Assign M (big value) to the respective route

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    Application of transportation problems

    1. Transportation of goods from supply points todemand centers to minimize total transportationcost.

    2. Refuse collection The refuse collection includes the collection of refuse at

    the various sources and transporting them to the disposalpoints. The operation may be divided into the followingactivities:

    a) Storage at or near the sources b) Collection and haulage of the refuse to the disposal sites c) Disposal

    The problem can be modeled as a transportation problem.

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    Application of transportation problems

    (contd)

    3. Recruitment of staff

    Suppose an organization has vacancies on n types ofjobs. Each type of jobjrequires djpersons to fulfill thevacancy. After advertising, the organization receives

    applications from m applicants to apply for these posts.Let cijdenote the cost of assigning applicant ito jobj. Theproblem can be modeled as a transportation problem.

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    Application of transportation problems

    (contd)

    4. Production planning

    Given the demand djforn periods and the regular timeand overtime production rates as well as production costsand inventory holding costs, an optimal production

    schedule to minimize the total of production and inventorycosts can be obtained by modeling this as a transportationproblem.

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    Assignment problem

    Given

    njobs

    n machines

    cij: the cost of doing jobjon machine I

    The objective of the problems is to assign the jobs to themachines (one job per machine) at the least total cost.

    It is a special case of transportation problem with

    m = n

    si = 1

    dj = 1

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    LP formulation of assignment problem

    Consider a njobs and n machines assignment problem

    n1,...,jn ;1,...,i1or

    sconstraintMachine

    sconstraintJob

    tosubject

    otherwise

    jobwithassignedismachineIfL et

    0

    ,...,11

    ,...,11

    min

    0

    1

    1

    1

    1 1

    ij

    n

    i

    ij

    n

    j

    ij

    n

    i

    n

    j

    ijij

    ij

    x

    njx

    nix

    xcz

    jix

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    Assignment probleman example

    We have 3 jobs and 3 workers. The 3 workers have different levelsof expertise necessary to do the three jobs. Labor costs will varydepending on which worker is assigned to each job. The cost matrixis given below:

    Job

    Worker

    1 2 3

    1 7 15 15

    2 9 11 13

    3 11 13 13

    Problem: Which job to assign to each worker so as to minimizeour cost?

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    Hungarian method

    The steps:

    1. For the original cost matrix, identify each rowsminimum, and subtract it from all the entries of therow

    2. For the matrix resulting from step 1, identify eachcolumns minimum, and subtract it from all theentries of the column

    3. Identify the optimal solution as the feasible

    assignment associated with the zero elements ofthe matrix obtained in step 2

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    Shortest route problem

    Given a network G with n nodes and m edges and acost cijassociated with each edge (i,j) in G.

    Problem: Find the shortest route from node 1 to noten in G. The cost of the route is the sum of costs ofall the edges in the route.

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    Example 1

    What is the shortest route from node 1 to node 7?

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    Example 2

    There are six time periods during which a piece of equipment mustbe used.

    Let cij= The cost of renting an equipment at the beginning of periodi, maintaining it until the beginning of periodjand then trading itfor another new equipment.

    The objective is to find the shortest route from node 1 to node 7.Each node entered in this path represents a replacement.

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    LP formulationexample 1

    min z = 6x12 +9x13 + 3x14 + 4x24 + 3x26 + 10x27 +2x34 + 4x35 + 6x45 +x56+ 4x57 + 6x67subject toNode 1:x12 +x13 +x14 = 1Node 2:x24 +x26 +x27 -x12 = 0Node 3:x34 +x35 -x13 = 0

    Node 4:x45 (x14 +x24+x34) = 0Node 5:x56+x57 (x35+x45) = 0Node 6:x67 (x26+x56) = 0Node 7: (x27 +x57+x67) = -1xij 0, for all (i,j) G

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    Shortest route algorithm

    1. Label node 1 with L1 = 0

    2. Continue to label the remaining nodes according to thefollowing formula: Lj= min {Li+cij(i,j)G}

    3. When the last node, n, has been labeled, Lnis the shortest

    distance from node 1 to node n.4. The route is determined by tracing backward from node n to

    all nodes so that

    Lj= Li+cij

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    Shortest route algorithmexample 1

    L1 = 0

    L2 = min(L1+6) = 6L3 = min(L1+9) = 9

    L4 = min(L2+4, L1+3, L3+2) = min(10,3,11) = 3

    L5 = min(L3+4, L4+6) = min(13,9) = 9

    L6

    = min(L2

    +3, L5

    +1) = min(9,10) = 9

    L7 = min(L2+10, L5+4 , L6+6) = min(16,13,15) = 13

    Shortest path = 1 4 5 7


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