Transportation problem Operations research (OR) are concerned with scientifically deciding how to best design and operate people machine systems, usually under conditions requiring the allocation of scarce resources. (1Operations Research Society of America). Operations research tools and has been a decision-making aid in almost all manufacturing industries and in financial and service organizations. Key problem managers face is how to allocate scarce resources among various activities or projects. Linear programming, or LP, is a method of allocating resources in an optimal way. It is one of the most widely used In the term linear programming, programming refers to mathematical programming. One of the most important and successful applications of quantitative analysis to solving business problems has been in the physical distribution of products, commonly referred to as transportation problems. Basically, the purpose is to minimize the cost of shipping goods from one location to another so that the needs of each arrival area are met and every shipping location operates within its capacity. However, quantitative analysis has been used for many problems other than the physical distribution of goods. We could set up a transportation problem and solve it using the simplex method as with any LP problem (see using the Simplex Method to Solve Linear Programming Maximization However, the special structure of the transportation problem allows us to solve it with a faster, more economical algorithm than simplex. Problems of this type, containing thousands of variables and constraints, can be solved in only a few seconds on a computer. In fact, we can solve a relatively large transportation problem by hand. There are some requirements for placing an LP problem into the transportation problem category.
The transportation problem Linear programming is good at solving problems with zillions of options, and finding the optimal solution. Could it work for transportation problems? Costs are linear, and shipment quantities are linear, so maybe so. Since any transportation problem can be formulated as an LP, we can use the simplex method to find an optimal solution. Because of the special structure of a transportation LP, the iterations of the simple method have a very special form. The transportation simplex method is nothing but the original simplex method, but it streamlines the iterations given this special form. Introduction Consider a commodity which is produced at various centers called SOURCES and is demanded at various other DESTINATIONS. The production capacity of each source (availability) and the requirement of each destination are known and fixed. The cost of transporting one unit of the commodity from each source to each destination is also known. 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 time the total cost of transportation in minimized. This optimum allocation of the commodity from various sources to different destinations is called TRANSPORTATION PROBLEM. Transportation models deals with the transportation of a product manufactured at different plants or factories (supply origins) to a number of different warehouses (demand destinations). The objective to satisfy the destination requirements within the plants capacity constraints at the minimum transportation cost.
A typical transportation problem contains Inputs: Sources with availability Destinations with requirements Unit cost of transportation from various sources to destinations Objective: To determine schedule of transportation to minimize total transportation cost. Simple Network Representation A transportation problem can be stated mathematically as follows: Let there be m SOURCES and n DESTINATIONS Let a i : the availability at the i th source b j : the requirement of the j th destination. C ij : the cost of transporting one unit of commodity from the i th source to the j th destination
x ij : the quantity of the commodity transported from i th source to the j th destination (i=1, 2, m; j=1,2,..n) Source D 1 D 2 D 3 D 4 Availability S 1 C 11 C 12 C 13 C 14 a 1 S 2 C 21 C 22 C 23 C 24 a 2 S 3 C 31 C 32 C 33 C 34 a 3 Requirement b 1 b 2 b 3 b 4 Σa i = Σb j The problem is to determine the values of x ij such that total cost of transportation is minimized. We assume that the total quantity available is the same as the total requirement. i.e. Σa i = Σb j Balanced transportation problems Unbalanced transportation problems m m n a i b j i 1 j 1 n a i b j i 1 j 1 Include a dummy source or a dummy destination having a supply d or demand d to convert it to a balanced transportation problem. Where d= n m b a or i a bj respectively. j j 1 i 1 m n i i 1 j 1 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 allocations is called a Basic Solution.
Prototype Problem Holiday shipments of ipods to distribution centers Production at 3 facilities, A, supply 200k B, supply 350k C, supply 150k Distribute to 4 centers, N, demand 60k S, demand 140k E, demand 300k W, demand 200k Total demand vs. total supply Holiday shipments of ipods to distribution centers Production at 3 facilities, A, supply 200k B, supply 350k C, supply 150k Distribute to 4 centers, N, demand 60k S, demand 140k E, demand 300k W, demand 200k Total demand vs. total supply Prototype Problem When solving the transportation problem, the number of possible routes should be m+n-1.
If it is <m+n-1, it is called a degenerate solution. In such a case evaluation of the solution will not be possible. In order to evaluate the cells /routes (using the Modi-method or the stepping stone method) we need to imagine/introduce some used cells/routes carrying / transporting a very small quantity, say. That cell should be selected at the correct place. Solution of transportation problems Two phases: First phase: Find an initial feasible solution 2 nd phase: Check for optimality and improve the solution Find an initial feasible solution North west corner method Least cost method Vogel s approximation method Checking for optimality a) Stepping stone method b) Modified distribution (MODI) method. Steps in Solving the Transportation Problem How to solve? 1. Define the objective function to be minimized with the constraints imposed on the problem. 2. Set up a transportation table with m rows representing the sources and n columns representing the destination 3. Develop an initial feasible solution to the problem by any of these methods 4. a) The North west corner rule 5. b) Lowest cost entry method 6. c)vogel s approximation method
4. Examine whether the initial solution is feasible or not.( the solution is said to be feasible if the solution has allocations in ( m+n-1) cells with independent positions. 5. Test wither the solution obtained in the above step is optimum or not using a) Stepping stone method b) Modified distribution (MODI) method. 6. If the solution is not optimum, modify the shipping schedule. Repeat the above until an optimum solution is obtained. Applications To minimize shipping costs from factories to warehouses or from warehouses to retails outlets. To determine lowest cost location of a new factory, warehouse or sales office. To determine minimum cost production schedule that satisfies firm s demand and production limitations. North-West Corner Method Step1: Select the upper left (north-west) cell of the transportation matrix and allocate the maximum possible value to X11 which is equal to min(a1,b1). Sourc e Destination N S E W Supply A 16 13 22 17 200 B 14 13 19 15 350 C Dema nd v j 9 20 23 10 60 140 300 200 150 Z = 700
Step2: If allocation made is equal to the supply available at the first source (a1 in first row), then move vertically down to the cell (2,1). If allocation made is equal to demand of the first destination (b1 in first column), then mov e horizontally to the cell (1,2). If a1=b1, then allocate X11= a1 or b1 and move to cell (2,2). Step3: Continue the process until an allocation is made in the south-east corner cell of the transportation table. Advantages; it is simple and reliable. Easy to compute understand and interpret. Disadvantages: This method does not take into considerations the shipping cost, consequently the initial solution obtained by this method require improvement. Problem1: Obtain initial solution in the following transportation problem by using Northwest corner rule method Least Cost Method Step1: Select the cell having lowest unit cost in the entire table and allocate the minimum of supply or demand values in that cell. Problem1: Obtain initial solution in the following transportation problem by using LCM method. Source D 1 D 2 D 3 D 4 D 5 Availability S 1 5 3 8 6 6 1100 S 2 4 5 7 6 7 900 S 3 8 4 4 6 6 700 Requirement 800 400 500 400 600
Step2: Then eliminate the row or column in which supply or demand is exhausted. If both the supply and demand values are same, either of the row or column can be eliminated. In case, the smallest unit cost is not unique, then select the cell where maximum allocation can be made. Step3: Repeat the process with next lowest unit cost and continue until the entire available supply at various sources and demand at various destinations is satisfied. Vogel s Approximation Method Step1: Calculate penalty for each row and column by taking the difference between the two smallest unit costs. For each row and column, calculate its difference: = (Second smallest c ij in row/col) - (Smallest c ij in row/col) This penalty or extra cost has to be paid if one fails to allocate the minimum unit transportation cost. Step3: Adjust the supply and demand and eliminate the satisfied row or column. Eliminate any row/column with no supply / demand left from further steps. If a row and column are satisfied simultaneously, eliminate both the row and column. Step4:. Recompute the row and column difference for the reduced transportation table, omitting rows or columns crossed out in the preceding step. Step5: Repeat the process until all the supply sources and demand destinations are satisfied. Repeat until BFS found. Repeat the above procedure until the entire supply at factories are exhausted to satisfy demand at different warehouses. Problem1: Obtain initial solution in the following transportation problem by using VAM method Problem 2: Obtain initial solution in the following transportation problem by using VAM Algorithm of MODIFIED DISTRIBUTION (MODI) METHOD Step I: For an initial basic feasible solution with (m+n-1) occupied (basic) cells, calculate u i and v j values for rows and columns respectively using the relationship C ij = u i + v j for all allocated cells only. To start with assume any one of the u i or v j to be zero.
Step II: For the unoccupied (non-basic) cells, calculate the cell evaluations or the net evaluations as Δ ij = C ij (u i + v j ). Step III: a) If all Δ ij > 0, the current solution is optimal and unique. b) If any Δ ij = 0, the current solution is optimal, but an alternate solution exists. c) If any Δ ij < 0, then an improved solution can be obtained; by converting one of the basic cells to a non basic cells and one of the non basic cells to a basic cell. Go to step IV. Step IV: Select the cell corresponding to most negative 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 at every corner. It may be noted that right angle turns in this path are permitted. Step V: Put a + sign in the entering cell and mark the remaining corners of the loop alternately with and + signs, with a plus sign at the cell being evaluated. 8.Determine the maximum number of units that should be shipped to this unoccupied cell. The smallest one with a negative position on the closed path indicates the number of units that can be shipped to the entering cell. This quantity is added to all the cells on the path marked with plus sign and subtract from those cells mark with minus sign. In this way the unoccupied cell under consideration becomes an occupied cell making one of the occupied cells as unoccupied cell. 9.Repeat the whole procedure until an optimum solution is attained i.e. Δ ij is positive or zero. Finally calculate new transportation cost. Problem 3 Origins D1 D2 D3 D4 Supply/capacity/a vailability O1 1 2 1 4 30 O2 3 3 2 1 50 O3 4 2 5 9 20 Demand/Requiremen ts 20 40 30 10 Special cases in Transportation
Unbalanced transportation Restricted routes Maximisation Unbalanced transportation problem When the total availability is equal to the total requirement the problem (i.e. Σa i = Σb j ) is said to be a balanced transportation problem. If the total availability at different sources is not equal to the total requirement at different destinations, (i.e. Σa i Σb j ), the problem is said to be an unbalanced transportation problem. Steps to convert an unbalanced problem to a balanced one are 1) If Σa i > Σb j i.e. the total availability is greater than the total requirement, a dummy destination is introduced in the transportation problem with requirement = Σa i - Σb j. 2) The unit cost of transportation from each source to this destination is assumed to be zero. 3) If Σa i < Σb j i.e. the total availability is less than the total requirement, a dummy source is introduced in the transportation problem with requirement = Σb j - Σa i. The unit cost of transportation from each destination to this source is assumed to be zero. After making the necessary modifications in the given problem to convert it to a balanced problem, it can be solved using any of the methods. Include a dummy source or a dummy destination having a supply d or demand d to convert it to a balanced transportation problem. Where d = n m b a or i a bj respectively. j j 1 i 1 m n i i 1 j 1
Problem.Holiday shipments of ipods to distribution centers Production at 3 facilities, A, supply 200k B, supply 350k C, supply 150k Distribute to 4 centers, N, demand 160k S, demand 140k E, demand 300k W, demand 200k Total demand total supply Obtain initial solution in the following transportation problem by using VAM method Source N S E W Supply A 16 13 22 17 200 B 14 13 19 15 350 C 9 20 23 10 150 Demand 160 140 300 200 v j u i
Restricted routes Sometimes in a transpiration problem some routes may not be available. This could be due to a variety of reasons like unfavorable weather condition or a strike on particular route etc. In such a situation there is a restrictions on route available for transportation. We assign a very large cost represented by M to each of such routes which are not available. The effect of adding a large cost element would be that such routes would automatically be eliminated in the final solutions. Problem1.The XYZ Tobacco company purchased and stores in warehouses located in the following four cities From\ To C1 C2 C3 A 7 10 5 B 12 9 4 C 7 3 11 D 9 5 7 Demand 120 100 110 Because of railroad construction, shipments are temporarily prohibited from warehouse at city A to company C1.i) Find the optimal distribution for XYZ tobacco Company.
Maximisation Problem A Transpiration Tableau contains unit profits instead of unit costs and the objective function be the maximization of profits. To convert maximization problem to minimization all the values of profit matrix are subtracted from the highest profit value in the matrix The objective function is determined with reference to the original profit matrix If a maximization type of transportation problem is unbalanced then it should be balanced by introducing necessary dummy row or column before converting it into maximization problem. Similarly if such a problem has prohibited route, then the pay of element for such a route should be submitted by M before proceeding to convert to maximization type. Prob.Solve the following transportation problem for maximum profit. X A B C D Availability X 12 18 6 25 200 Y 8 7 10 18 500 Z 14 3 11 20 300 Demand 180 320 100 400 Thank you krishmandya@gmail.com
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Module 7.1: Structure of the Transportation Problem Introduction Ah finally after 14 weeks my favorite of all quantitative method applications the transportation problem. For seventeen years in the Air Force, most of my career was spent figuring out how to efficiently move troops and their equipment, weapons and weapon systems, communication equipment, and medical supplies from point A to point B. Then, when I moved to the Pentagon to finish my career as Chief of Air Force Transportation Programs, I worked for three years on how to effectively deploy troops and their equipment, weapons and weapon systems, communication equipment, and medical suppliers from many ports of embarkation to multiple ports of debarkation. After I retired from the Air Force and joined the faculty at the University of South Florida, my initial applied research was working with supply chain managers at Johnson & Johnson and 3M to figure out how to effectively move J&J medical products and 3M consumer products through their supply chains. For 3M, this included using quantitative methods to help decide where intermediate distribution centers should be located, especially to meet European expansion in the early 1990's. Whether in military or commercial applications, the transportation problem is that problem which addresses what origin point should ship to what final destination point over which route so as to minimize transportation costs while meeting the problem constraints. The problem constraints are staying within the available supply at the origins, and meeting customer demand at the destinations. The problem can become quickly complex by adding intermediate transshipment points, such as distribution centers, which add constraints such as whatever is shipped
into transshipment points must be shipped out. The transshipment problem is the subject of Module 7.2 Notes. Complexity is also added by placing capacity constraints on the routes or the transshipment points. You will see that the transportation problem is simply another application of linear programming, but it is such a widespread application that this and other quantitative texts devote a chapter just to this application. Companies like 3M that have multiple manufacturing sites, many distribution centers and warehouses, and multiple consumer demand locations find the transportation problem to be so complex that linear programming applications are in common use. Illustration of the General Transportation Problem Let's start this module with a simplified example of the problem facing logistics staffs at the Pentagon. How many troops should be deployed from aerial and water ports of embarkation to aerial and water ports of debarkation so as to minimize total deployment time. For example, we could use Fort Bragg, North Carolina and Fort Hood, Texas as aerial ports of embarkation and Adana, Turkey; Dhahran, Saudi Arabia; and Wheelus, Libya as aerial ports of debarkation. Please understand that this is not an actual deployment scenario; but rather an example for illustration. The constraints include "supply" and "demand." Supply constraints are used to ensure that the number of troops deployed do not exceed the number available at the two ports of embarkation. Demand constraints are used to ensure that the number of troops deployed meet the need for troops at the ports of debarkation. Before continuing, please understand that the two origins could be Detroit and Memphis automobile production plants. The three destinations could be customers (automobile dealerships) in Philadelphia, Washington DC and Miami. Instead of moving 20,000 troops, we may be moving 20,000 new cars out of Detroit and Memphis to meet dealer demand at the three destinations. The following table presents a picture of this transportation problem. The table includes the nodes (origins and destinations) and the arcs (deployment or shipping routes). Table 7.1.1 Origin (Troops Available) Fort Bragg (14,000) To Ada --> To Dha --> To Whl --> -> 7 Days From FtB -> 10 Days From FtH Destination (Troops Needed) Adana (5,000) -> 8 Days From FtB Dhahran
-> 7 Days From FtH (10,000) Fort Hood (6,000) To Ada --> To Dha --> To Whl --> -> 8 Days from FtB -> 5 Days From FtH Wheelus (5,000) The table shows that 14,000 troops are available for deployment out of Fort Bragg. It takes 7 days to deploy troops from Fort Bragg to Adana, which has a demand for 5,000 troops. So, one alternative is to deploy 5,000 troops out of Fort Bragg to Adana, leaving 9,000 troops available at Fort Bragg for Dhahran or Wheelus. The "cost" of this move would be 5,000 troops times 7 days giving 35,000 troop deployment days. Another way of satisfying the demand at Adana is to deploy 5,000 troops from Fort Hood. Note it takes 10 days to deploy troops from Fort Hood to Adana, which would give 5,000 times 10 days or 50,000 troop deployment days. We could continue to try different combinations of origins and destinations to minimize total troop deployment days while meeting demand and staying within troop availability constraints. For a small problem, it would not be too difficult to try all of the combinations of solutions to find the optimal solution. But if there were many origins, such as 10 or more, and many destinations, such as 15 or more, the problem would be too cumbersome to work "by hand." I should note here that in the parallel automobile example, the 7, 10 and other deployment day coefficients might be $700, $1000 and other freight charges. The objective for the commercial application would be to minimize transportation costs while meeting demand and staying within the available supply constraints. Linear programming is an excellent quantitative method for application to the transportation problem. Recall from Module 6, that to formulate a linear program we need to decide on the decision variables, create the objective function as a linear equation, and then formulate the constraints as linear equations. For the transportation problem, the decision variables are: FtB_Ada = Nbr of troops to deploy from Fort Bragg to Adana FtB_Dha = Nbr of troops to deploy from Fort Bragg to Dhahran FtB_Whl = Nbr of troops to deploy from Fort Bragg to Wheelus FtH_Ada = Nbr of troops to deploy from Fort Hood to Adana FtH_Dha = Nbr of troops to deploy from Fort Hood to Dhahran FtH_Whl = Nbr of troops to deploy from Fort Hood to Wheelus Note that the number of variables for the standard transportation problem is the number of origins times the number of destinations. For this problem, there are two origins and three destinations which gives 2 times 3 or 6 decision variables. The decision variables then represent the units shipped over the deployment or shipping routes. Also note that I used a code for naming the variables. The text uses X12 to represent the number of
units shipped from origin one to destination 2. I like to use a more descriptive code to represent the route, and abbreviate names to keep within the 8 character variable name restrictions of The Management Scientist. The objective function is to minimize troop deployment days, where days ("costs") are the coefficients multiplied times the number of troops routing decision variables. Minimize Z = 7 FtB_Ada + 8 FtB_Dha + 8 FtB_Whl +10 FtH_Ada + 7 FtH_Dha + 5 FtH_Whl To show how this equation works, let's compute the deployment days for the solution FtB_Ada = 5,000; FtB_Dha = 5,000; Fth_Whl = 4,000; Fth_Dha = 5,000 and FtH_Whl = 1,000. Minimize Z = 7 (5,000) + 8 (5,000) + 8 (4,000) + 7 (5,000) + 5 (1,000) = 147,000 troop deployment days This is a feasible solution but we do not know if it is optimal until we try all combinations (or finish the constraint set and let the computer software run the combinations). Now for the constraints which include staying within the available supply at each origin node. Fort Bragg Supply: FtB_Ada + FtB_Dha + FtB_Whl < 14,000 Fort Hood Supply: FtH_Ada + FtH_Dha + FtH_Whl < 6,000 The constraints also include meeting the customer demand: Adana Demand: FtB_Ada + FtH_Ada = 5,000 Dhahran Demand: FtB_Dha + FtH_Dha = 10,000 Wheelus Demand: FtB_Whl + FbH_Whl = 5,000 Note that I made the demand constraints strict equalities, but allowed for slack in the supply constraints. This general formulation works as long as supply = demand or supply is greater than demand. If supply is greater than demand, the slack constraint will indicate which origin should have the excess supply. I will talk about how to handle the special case of demand being greater than supply after we look at the solution.