The North-West Corner Method

3 Key excerpts on "The North-West Corner Method"

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  eBook - PDF

  Cost and Optimization in Government

  • Aman Khan(Author)
  • 2000(Publication Date)
  • Praeger

    (Publisher)

  We present here two such methods that have become quite common in heuristic solution methodologies: northwest corner rule, and greedy heuristic. Northwest Corner Rule. As the name implies, the northwest corner rule produces MATHEMATICAL PROGRAMMING 247 solutions to a heuristic programming problem that lie at the extreme northwest corner of a matrix, i.e., a table containing all the relevant information on the problem. The matrix is set up in such a way as to ensure that the northwest coiner always has the highest value. The solution starts by allocating all the resources to the cell in the northwest corner, then moving on to the corner with the next highest value and allocating the maximum available resources to the cell corresponding to that corner, and so on until all the resources have been fully allocated. It is important to note that since the allocation process begins by assigning all or most of the available resources to the cell with the highest corner values, it fails to take into account the cells with lower corner values, which may have a direct bearing on the solution. This is especially true in situations where the objective is to minimize costs or reduce services. Consequently, it tends to produce results that are not always optimal. Let us look at a simple problem to illustrate the procedure. Suppose the trans- portation department of a large metropolitan government wants to determine how best it could provide transit services to three different parts of the city that will produce the greatest amount of (cost) savings to the department. Table 7.5 presents information on the origin and destination points along with information on the cost of operation for each trip (transit service), actual demand for service, and the number of daily trips that could potentially be offered from each point of origin to their points of destination.

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  Linear Mathematics

  A Practical Approach

  • Patricia Clark Kenschaft(Author)
  • 2013(Publication Date)
  • Dover Publications

    (Publisher)

  Section 6.3 , but either would be time-consuming. This chapter presents a faster way.

  Northwest Corner Algorithm

  8.1.2 Example

  Find a feasible (but not necessarily optimal) solution to Example 8.1.1 using the northwest corner algorithm.

  Solution:

  The facts of the problem can be diagrammed in the following table, where the amounts the factories produce are written on the right, the amounts the warehouses can store are written on the bottom, and the black numbers in the boxes are the costs of shipping from the factory on the left to the warehouse above.

  The northwest corner algorithm first allocates as many widgets as possible to the upper left box (the northwest box). Next, proceed to the nearest box into which something can still be placed, and allocate as much as possible to that one. Then the process continues, each time moving either one box to the right, or one down, or one diagonally down, depending on how the shipments can be made.

  Since the 15 at the bottom of the first column is less than the 40 at the right of the first row, factory I can ship only 15 widgets to warehouse A, so we write a 15 in the upper left box. Then nothing else can go to warehouse A; that is, nothing else will be written in the boxes of the first column.

  Thus we move right from the 15, making x12 = 40 – 15 = 25. Now the capacity of factory I has been exhausted, so there can be no more numbers written in the first row. Moving down, we next set x22 = 45 – 25 = 20 to fill warehouse B. Now we must move right and set x23 = 50 – 20 = 30. The results are as follows:

  This table tells us we can ship 15 widgets from factory I to warehouse A, 25 widgets from factory I to warehouse B, 20 widgets from factory II to warehouse B, and 30 widgets from factory II to warehouse C. It is fairly obvious that this is not the optimum solution (that is, it is not the cheapest), but it is feasible. The total cost is C

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  Operations Research

  Operations Research: Theory and Practice

  • N.V.S Raju(Author)
  • 2019(Publication Date)
  • CRC Press

    (Publisher)

  This is the simplest method of all methods for finding IBFS of a TP. This method is independent of the cost or profit of transportation. The algorithm of NWCM goes as follows.

  Step 1:  

  Formulation: Given information is formulated in TP matrix.

  Step 2:  

  Standaradising (Minimisation form): Check whether the given TP is in standard form. A TP is said to be standard if the objective function is to minimise the total cost. However, if the TP is to maximise, it may be converted to minimisation form either of the two methods given below.

  1. (i)  From highest among all transportation profits, subtract all profits and put back in their respective cells (or)
  2. (ii)  Multiply profits in all cells by (−1).

  Note:   As this method is independent of casts or profits this step may be ignored also. But be careful while optimising the initial solution.

  Step 3:  

  Balancing: Check whether the TP is balanced. The balanced TP is that in which total availability (supply) is equal to total requirement (demand) of origins and destinations respectively. If the TP is balanced proceed to next step, otherwise balance by adding the defict amount of availability/requirement to a dummy origin/destination respectively with unit costs as zeros.

  Step 4:  

  Allocation at North West Corner:

  In the given TP matrix identify the cell at north west corner i.e., top left cell. At the starting of the problem it is the cell connecting origin-1 (O 1 ) to diestination-1 (D 1 ). Now check the availability of O 1 and requirement of D 1 and allocate as many units as possible (i.e., minimum of availability of O 1 and requirements of D 1 ) in this cell and indicate at one of the corners, usually at the top-right of the cell.

  Ex:

  Now, if availability of O 1 i.e.a 1 is less than requirement of D 1 i.e., b 1 then allocate a 1 units to the cell (C 11 ), otherwise (if b 1 <a 1 ) allocate b 1 units to the cell (C 1

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