Transportation Problem, Lecture notes of Operational Research

Download Transportation Problem and more Lecture notes Operational Research in PDF only on Docsity! Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 1 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. Transportation and Assignment Problems The transportation model is a special class of linear programs. It received this name because many of its applications involve determining how to optimally transport goods. However, some of its important applications (eg production scheduling) actually have nothing to do with transportation. The second type of problem is assignment problem. It involves such applications as assigning people to tasks. Although its applications appear to be quite different from those for the transportation, we shall see the assignment problem can be viewed as a special type of transportation problem. Application of the transportation and assignment problem tend to require a very large number of constraints and variables, so straightforward computer applications of simplex method may require an exorbitant computational effort. Fortunately, a key characteristic of these problems is that most of the aij coefficient in the constraints are zeros, and the relatively few non zero coefficient appear in a distinctive pattern. As a result, it has been possible to develop special streamlined algorithms that achieve dramatic computational savings by exploiting this special structure of the problem. Therefore, it is important to become sufficiently familiar with these special types of problems that you can recognise them when they arises and apply the proper computational procedure. Transportation problem As pointed out above that many applications of TP deals with transportation of a commodity from source to destination. Therefore I will use prototype example to explain the TP. Example- Suppose XYZ Pvt Ld produce good X. The production take place at 3 place- S1, S2 and S3. From these production places, company supplies the X to its warehouses, which are located near its demand centre. The shipping cost or transportation cost from different production place to warehouses is given table 1 Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 2 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. Table 1 Shpping cost per unit W1 W2 W3 W4 Output S1 464 513 654 867 75 S2 352 416 690 791 125 S3 995 682 388 685 100 Allocation 80 65 70 85 Si represents the production place(sources) and Wi represents the warehouses (destinations). We can represent the above problem with network diagram as following In above fig, the arrows shows the possible routes for transportation, where the number next to each arrow is the shipping cost per unit for that route. A square bracket net to each location gives the number of units to be shipped out of that location (so that the allocation into each warehouse is given as a negative number) Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 5 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. In transportation model we make following two assumption.  The requirement assumption- Each sources has a fixed supply of units, where this entire supply must be distributed to the destinations. Similarly, each destination has a fixed demand for units, where this entire demand must be received from the sources.  The cost assumptions- the cost of distributing units from any particular source to any particular destination is directly proportional to the number of units distributed. Therefore, the cost is just the unit cost of distribution times the number of units distributed. Note that the resulting table of constraint coefficients has the special structure shown in following table. Table 3 So any problem (whether involving transportation or not) fits the model for a transportation problem if it can be described completely in terms of a parameter table like above table and it satisfies both the requirement assumptions and the cost assumption. The objective is to minimise the total cost of distributing the units. All the parameters of the model are included in this table. Any linear programming problem that fits this special formulation is of the transportation problem type, regardless of its physical context. Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 6 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. Solution Properties The feasible solution property- a transportation problem will have feasible solutions if and only if ∑si=∑dj ⇒ model is balanced. i.e. total supply must be equal to total demand. Integer solution property- for transportation problems where every si and dj have an integer value, all the basic variables (allocations) in every basic feasible solution (including an optimal one) also have integer values. Basic Terminology 1. Rim Requirement- the supply and demand requirements at various source and destinations is called Rim requirement. 2. Feasible Solution- A feasible solution to a transportation problem is a set of non-negative allocations, xij, that satisfies the rim (row and column) restrictions. 3. Basic Feasible solution- A feasible solution is called a basic feasible solution if it contains no more than m+n-1 non negative allocations where m is the number of rows and n is the number of columns of the transportation problem. 4. Degenerated basic feasible- A basic solution that contains less than m+n-1 non-negative allocations in independent positions. The allocation are said to be independent positions, if it is not possible to form a closed path (loop) means by allowing horizontal and vertical lines and if all the corner cells are occupied. (explanation in class) 5. Non-degenerated basic feasible- A basic solution that contains exactly m+n-1 non-negative allocations is called degenerated basic feasible solution. 6. Optimal Solution- A feasible solution (non necessarily basic) is said to be optimal if it minimises (maximises) the transportation cost (profit). 7. Balance and Unbalance Transportation Problem-If the total demand is equal to total supply then TP is called balance TP, otherwise it is unbalance TP 8. Occupied and Unoccupied cells- the allocated cells in the transportation cells occupied cells and empty cells are called non-occupied cells. Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 7 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. The Transportation Algorithm The transportation algorithm follows the exact steps of the simplex method. However, instead of using the regular simplex tableau, we take advantage of the special structure of the transportation model to organise the computations in a more convenient form. Steps- the steps of the transportation algorithm are exact parallels of the simplex algorithm, Step 1- Determine a starting basic feasible solution, and Step 2- use the optimality condition of the simplex method to determine the entering variable from among all the non basic variables. If the optimality condition is satisfied, stop. Otherwise go to step 3 Step 3- use the feasibility condition of the simplex method to determine the leaving variable from among all the current basic variables, and find the new basic solution. Return to step 2. Step 1: Determination of the starting solution A general transportation model with m sources and n destinations has m+n constraint equations, one for each source and each destination. However, because the transportation model is always balanced (sum of the supply=sum of the demad), one of these equations is redundant. Thus, the model has m+n-1 independent constraint equations, which means that the starting solution basic solution consists of m+n- 1 independent equations, which means that the starting basic solution consists of m+n-1 basic variables. The special structure of the transportation problem allows securing a nonartificial starting basic solution using one of three methods (1) Least cost method (2) Vogel approximation method. Above methods differ in the quality of the starting basic solution they produce, in the sense that a better starting solution yields a smaller objective value. Algorithm- A process or set of rules to be followed in calculations or other problem solving operations Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 10 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. Stepping Stone Method This is a procedure for determining the potential of improving upon each of the non-basic variables in terms of the objective function. To determine this potential, each of the non-basic variables is considered one by one. For each such cell, we find what effect on the total cost would be if one unit is assigned to this cell. With this information, then, we come to know whether the solution is optimal or not. If not, we improve that solution. We can summarise the Stepping Stone method in following steps 1. Construct a transportation table with a given unit cost of transportation along with the rim conditions 2. Determine a initial basic feasible solution (allocation) using a suitable method as discussed earlier 3. Evaluate all unoccupied cells for the effect of transferring one unit from an occupied cell to the unoccupied cell. This transfer is made by forming a closed path that retains the SS and DD condition of the problem 4. Check the sign of each of the net change in the unit transportation costs. If the net changes are plus or zero, then the an optimal solution has been arrived at, otherwise go to step 5 5. Select the unoccupied cell with most negative net change among all unoccupied cells. 6. Assign an many units as possible to unoccupied cell satisfying rim conditions. The maximum number of units to be assigned are equal to the smaller circled number among the occupied cells with the minus value in a closed path. 7. Go to step 3, and repeat the problem until all unoccupied cells are evaluated and the net change result in positive or zero. Numerical Example – in Class Modified Distribution (MODI) method MODI method is based on the concept of duality. To explain this, we will use prototype example. Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 11 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. Suppose the unit cost of transportation of XYZ company is given in following table. Destination Supply W1 W2 Source A1 c11 c12 S1 A2 c21 c22 S2 Demand d1 d2 The simplex formulation of this problem will be Min Z = c11x11 + c12x12 +c21x21+c22x22 s.t x11 + x12 =s1 x21 + x22 = s2 x11 + x21 =d1 x12 +x22 =d2 For duality, introduce ui for supply constraint and vj for demand constraint. Dual Max Z=u1s1 +u2s2 +v1d1 +v2d2 s.t. u1 +v1 ≤ c11 u1 + v2 ≤ c12 u2 + v1 ≤ c21 u2 + v2 ≤ c22 so in case of general transportation model: Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 12 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. Min Z = ∑i∑j cijxij Subject to ∑i xij =si ∑j xij = dj .xij ≥0 for all i and j The dual will be Min Z= ∑i uisi + ∑j vjdj ui + vj ≤ cij for all i and j ui , vj ≥ 0 for all i and j from economic interpretation, ui + vj = implied cost (ui denotes the location rent and vj denotes the market price) and cij = actual cost The variable xij form an optimal solution to the given transportation provided i. Solution xij is feasible for all (i,j) with respect to original transportation model, ii. (cij - ui - vj) xij =0 for all i and j. This condition is known as complementary slackness for transportation problem. The last condition indicates that a. If xij > 0 and is feasible then cij - ui - vj =0 so cij = ui + vj for each occupied cell b. If xij = 0 and is feasible then cij - ui - vj ≠ 0 so cij ≠ ui + vj for each occupied cell. Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 15 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. Xij ≥0 for all i and j. The first set of functional constraints specifies that each assignee is to perform exactly one task, whereas the second set requires each task to be performed by exactly one assignee. TP & AP In comparison to TP, in AP we have following restrictions 1. Number of sources (m) = number of destinations (n) 2. Each supply (Si) = 1 3. Each demand (di) = 1 Because assignment model is special case of transportation model so we can solve assignment problem directly as regular transportation model. Nevertheless, the fact that all the supply and demand amounts equal 1 has led to the development of a simple solution algorithm called the Hungarian Method. The use of transportation Model technique in assignment model has following drawbacks- 1. Wasted iterations- because in assignment model, the number of basic variable is n and all basic variables are binary variable so there always are n-1 degenerate basic variables. As we know that degenerate basic variable do not cause any major complication in the execution of the algorithm. However they do frequently cause wasted iterations. 2. Transportation simplex method is purely a general purpose algorithm for solving all transportation problems. Therefore, it does nothing to exploit the additional special structure in this special type of transportation problem (n=m, si=1 and di=1). Hungarian Method Hungarian Method is for assigning jobs by a one-for-one matching to identify the lowest-cost solution. Each job must be assigned to only one machine. It is assumed that every machine is capable of handling every job, and that the costs or values associated with each assignment combination are known and fixed. The number of rows and columns must be the same. 1. Subtract the smallest number in each row from every number in the row. This is called row reduction Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) 16 http://rajkumar2850.weebly.com/, Rajkumar, Assistant Professor, College of Vocational Studies, Delhi. Email- rajkumar_dse@outlook.com Note-This notes are not substitutes of Text books. Please refer to your text book. 2. Subtract the smallest number in each column of the new table from every number in the column. This is called column reduction. 3. Test whether an optimal assignment can be made. You do this by determining the minimum number of lines to cover all zeros. If the number of lines equals the number of rows, an optimal set of assignment is possible. Otherwise go on to step 4 4. If the number of lines is less than the number of rows, modify the table in the following way (a) Subtract the smallest uncovered number from every uncovered number in the table (b) Add the smallest uncovered number to the numbers at intersections of covering lines (c) Numbers crossed out but at the interactions of cross out lines carry over unchanged to the next table 5. Repeat step 3 and 4 until an optimal set of assignments is possible. 6. Make the assignments one at a time in positions that have zero elements. Begin with rows or columns that have only one zero. Since each row and each column needs to receive exactly one assignment, cross out both the row and the column involved after each assignment is made. Then move on to the rows and such row or column that are not yet crossed out to select the next assignment, with preference again given to any such row or column that has only one zero that is not crossed out. Continue until every row or column has exactly one assignment and so has been crossed out. Numerical Example----in class Sources  Taha H A, Natarajan A M etc, Operation Research: An introduction, 8 th Edition  Hillier F S, Lieberman G J etc, Introduction to operation research, 9 th edition  Sharma J K, Operation Research  Vohra, N D, Quantitative Techniques for Management.