Transportation - Transportation problems In D1 you studied...

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Unformatted text preview: Transportation problems In D1 you studied linear programming problems and their solutions using the simplex algorithm. In this chapter and the next one the solutions of two special types of linear programming problems are considered by methods other than the simplex algorithm. These are transportation problems and assignment problems. Both of these problem types could be solved using the simplex algorithm, but the process would result in very large simplex tableaux and numerous simplex iterations. Because of the special characteristics of each problem, however, alternative solution methods requiring significantly less mathematical manipulation than the simplex method have been developed. 1.1 The transportation problem The transportation problem deals with the distribution of goods from several points of supply, such as factories, often known as sources, to a number of points of demand, such as warehouses, often known as destinations. Each source is able to supply a fixed number of units of the product, usually called the capacity or availability, and each destination has a fixed demand, usually known as the requirement. The objective is to schedule shipments from sources to destinations so that the total transport cost is a minimum. Example 1 A concrete company transports concrete from three plants, 1, 2 and 3, to three construction sites, A, B and C. The plants are able to supply the following numbers of tons per week: Plant Supply (capacity) 1 300 2 300 3 100 1 2 Transportation problems The requirements of the sites, in numbers of tons per week, are: Construction site Demand (requirement) A B C 200 200 300 The cost of transporting 1 ton of concrete from each plant to each site is shown in the figure below in pounds per ton. A 4 7 1 4 3 5 Plants B Sites 5 2 8 9 C 5 3 For computational purposes it is convenient to put all the above information into a table, as in the simplex method. In this table each row represents a source and each column represents a destination. Sites To A B C Supply (availability) 1 4 3 8 300 2 7 5 9 300 3 4 5 5 100 Demand (requirement) 200 200 300 From Plants 1.2 Formulation as a linear programming problem Before proceeding with the solution of the transportation problem, using the method developed especially for it, we will show that it can be formulated as a linear programming problem. The decision variables xij are the numbers of tons transported from plant i (where i ˆ 1, 2, 3) to each site j (where j ˆ A, B, C). Transportation problems The objective function represents the total transportation cost £Z. Each term in the objective function Z represents the cost of tonnage transported on one route. For example, for the route 2 3 C the term is 9x2C , that is: (cost per ton ˆ 9)  (number of tons transported ˆ x2C ) Hence the objective function is: Z ˆ 4x1A ‡ 3x1B ‡ 8x1C ‡7x2A ‡ 5x2B ‡ 9x2C ‡4x3A ‡ 5x3B ‡ 5x3C (1) Notice that in this problem the total supply is 300 ‡ 300 ‡ 100 ˆ 700 and the total demand is 200 ‡ 200 ‡ 300 ˆ 700. so (total supply) ˆ (total demand). This is called a balanced problem. Balanced problems are considered throughout this chapter unless otherwise stated. In a balanced problem all the products that can be supplied are used to meet the demand. There are no slacks and so all constraints are equalities rather than inequalities. The constraints for the problem are obtained by considering the rows and columns of the above table. Plants: The total number of tons transported from plant 1 to sites A, B and C must be equal to the supply, so: x1A ‡ x1B ‡ x1C ˆ 300 (2a) Similarly, for plants 2 and 3: x2A ‡ x2B ‡ x2C ˆ 300 (2b) x3A ‡ x3B ‡ x3C ˆ 100 (2c) Sites: The total number of tons received by site A from plants 1, 2 and 3 must be equal to the demand, so: x1A ‡ x2A ‡ x3A ˆ 200 (3a) Similarly, for sites B and C: x1B ‡ x2B ‡ x3B ˆ 200 (3b) x1C ‡ x2C ‡ x3C ˆ 300 (3c) All decision variables must be non-negative, so: xij 5 0 (4) for i ˆ 1, 2, 3 and j ˆ A, B, C. The linear programming problem is then: Minimise Z subject to the constraints given by equations (2), (3) and (4). 3 4 Transportation problems Exercise 1A 1 A steel company has three mills, M1 , M2 and M3 , which can produce 40, 10 and 20 kilotonnes of steel each year. Three customers, C1 , C2 and C3 , have requirements of 12, 18 and 40 kilotonnes respectively in the same period. The cost, in units of £1000, of transporting a kilotonne of steel from each mill to each customer is shown in the figure below. 11 M1 C1 7 9 3 M2 9 C2 3 5 M3 3 3 C3 (a) Summarise all the above information in the form of a table. (b) Formulate the problem of carrying out the transportation at minimum cost as a linear programming problem. 2 A transportation problem involves the following costs, supply and demand: W1 W2 W3 Supply F1 7 8 6 4 F2 9 2 4 3 F3 5 6 3 8 Demand 2 9 4 Formulate this problem as a linear programming problem. 3 Downside Mills produces carpet at plants in Abbeyville (A) and Bridgeway (B). The carpet is shipped to two outlets in Courtney (C) and Dove Valley (D). The cost, in £ per ton, of shipping carpet from each of the two plants to each of the two outlets is as follows: Transportation problems To C D A 20 32 B 35 15 From The plant at Abbeyville can supply 250 tons of carpet per week and the plant at Bridgeway can supply 300 tons of carpet per week. The Courtney outlet has a demand for 320 tons per week and the outlet at Dove Valley has a demand for 230 tons per week. The company wants to know the number of tons of carpet to ship from each plant to each outlet in order to minimise the total shipping cost. Formulate this transportation problem as a linear programming problem. 1.3 Solution of the transportation problem Setting up a transportation tableau The purpose of the transportation tableau is to summarise conveniently and concisely all relevant data and keep track of algorithm computations. In this respect it serves the same role as the simplex tableau did for linear programming problems. The transportation tableau for Example 1 is: To From A Supply (availability) C 4 3 200 8 5 9 4 2 3 7 1 Demand (requirement) B 5 5 200 300 300 300 100 ÀÀÀ 700 2 À À Total demand and supply Each cell in the tableau represents the amount transported from one source to one destination. The amount placed in each cell is therefore the value of a decision variable for the cell. For example, the cell at the intersection of row 2 and column C represents the decision variable x2C . The smaller box within each cell contains the unit transportation cost for that route. The final column and the final row are the capacities of the plants and the requirements of 5 6 Transportation problems the sites respectively. The total demand and total supply has been placed in the bottom right-hand corner. It is suggested that you should always do this to check that you have a balanced problem. Later we will see how to deal with unbalanced problems where total destination requirements are greater than or less than total source capacities. The transportation algorithm The transportation algorithm consists of three stages. Stage 1 Stage 2 Stage 3 Find a transportation pattern that uses all the products available and satisfies all the requirements. This is called developing an initial solution. Test the solution for optimality. If the solution is optimal stop. If the solution is not optimal move to stage 3. Use the stepping-stone method to obtain an improved solution and then return to stage 2. The algorithm stops when no further improvement is possible. Finding an initial solution An initial feasible solution can be found by several alternative methods. We will give just one, known as the north-west corner method. You will recall that prior to applying the simplex algorithm an initial solution had to be established in the initial simplex tableau. The initial solution in the transportation problem is not the origin. The north-west corner method requires that we start in the upper left-hand cell or north-west corner of the table and allocate units to shipping routes as follows: (i) Exhaust the supply at each row before moving down to the next row. (ii) Exhaust the requirements of each column before moving to the right to the next column. Let us implement this rule for Example 1. From 1 To A B 200 100 2 100 Supply 300 200 200 200 300 100 3 Demand C 100 300 The figures in the above table were obtained in the following way: Transportation problems (i) Site A requires 200. All of these can be supplied by plant 1. Plant 1 still has 100 left. These are sent to site B. (ii) We now move to line 2. Site A is satisfied. Site B still requires 100. These can be sent from plant 2. Plant 2 still has 200 left. These are sent to site C. (iii) Site C still requires 100. These can be sent from plant 3. This solution is feasible since supply and demand constraints are all satisfied. This solution corresponds to x1A ˆ 200, x1B ˆ 100, x2B ˆ 100, x2C ˆ 200, x3C ˆ 100 and all other variables zero. The total cost of this solution is: ‰4  (200)Š ‡ ‰3  (100)Š ‡ ‰5  (100)Š ‡ ‰9  (200)Š ‡ ‰5  (100)Š ˆ 800 ‡ 300 ‡ 500 ‡ 1800 ‡ 500 ˆ £3900 Example 2 Use the north-west corner rule to obtain an initial solution for the transportation tableau below. Site To From A 1 Plant 40 2 30 10 4 11 3 Demand B 70 C Supply 12 20 30 50 9 5 7 8 30 30 6 40 50 60 150 (i) Not all of site A's demand (70) can be met by plant 1. Site A can receive 40 from plant 1 and plant 2 can provide the remaining 30. (It is a good idea to fill in the cells as you go along.) (ii) Plant 2 still has 20 left and these can be sent to site B. (iii) Site B still requires 30 and these can be provided by plant 3. Plant 3 still has 30 left and these can be sent to site C to exactly meet its requirement. The total cost of this solution is: (40  10) ‡ (30  4) ‡ (20  5) ‡ (30  8) ‡ (30  6) ˆ 400 ‡ 120 ‡ 100 ‡ 240 ‡ 180 ˆ 1040 7 8 Transportation problems Notice that at each step a row or column constraint is satisfied except at the last stage when both a row and a column constraint are satisfied. An initial solution obtained using the north-west corner rule is often called the north-west corner solution. As illustrated in the example below, we can obtain initial solutions by starting at other corners of the tableau. Example 3 For the transportation tableau below write down: (a) the north-west corner solution (b) the north-east corner solution (c) the south-west corner solution (d) the south-east corner solution. Assuming one of these gives an optimal solution, obtain the optimal solution. A B 40 2 300 65 70 1 30 250 400 350 (a) The north-west corner solution obtained using the rule given earlier is: A B 1 250 250 2 50 350 300 350 400 (b) The north-east corner solution is obtained by starting in the top right-hand corner and is: A 1 2 B 250 250 300 100 400 300 350 Transportation problems (c) The south-west corner solution is obtained by starting in the bottom left-hand corner and is: A B 1 250 300 100 400 300 2 250 350 This is the same as (b). (d) The south-east corner solution is obtained by starting in the bottom right-hand corner and is: A B 1 250 250 2 50 350 300 350 400 This is the same as (a). The cost of (a) is: (250  40) ‡ (50  70) ‡ (350  30) ˆ 24 000 The cost of (b) is: (250  65) ‡ (300  70) ‡ (100  30) ˆ 40 250 So the optimal solution is: 250 from 1 3 A 50 from 2 3 A 350 from 2 3 B The minimum cost is 24 000. Exercise 1B 1 Use the north-west corner rule to obtain an initial solution for the following transportation tableaux: Supply (a) 100 40 50 Demand 60 60 70 9 10 Transportation problems Supply (b) 60 40 140 Demand 120 80 40 Supply (c) 60 100 80 Demand 60 70 110 Supply (d) 90 40 50 Demand 60 70 50 2 For the transportation tableau below write down: (a) the north-west corner solution (b) the north-east corner solution (c) the south-west corner solution (d) the south-east corner solution. Given that one of these gives an optimal solution, obtain the optimal solution. A Supply 8 2 90 6 10 1 Demand B 12 160 100 150 Transportation problems Testing the solution for optimality An important rule The method we shall now describe for testing a solution for optimality can only be applied if one essential condition is satisfied. This condition is: The number of occupied cells (routes used) must be equal to one less than the sum of the number of rows and the number of columns. In both Example 1 and Example 2 we have three rows and three columns. Since the number of occupied cells is 5 in each case and this is equal to (3 ‡ 3 À 1), the condition is satisfied. In the general case, when we have m sources and n destinations the number of occupied cells must be (m ‡ n À 1). When the number of occupied cells is less than this the solution is said to be degenerate. Later in this chapter we will consider what to do when we have degeneracy. Calculation of improvement indices To test a solution for optimality we need to calculate an improvement index for each unused cell. As a first step in this process we must compute: (i) a value for each row, denoted by Ri (ii) a value for each column, denoted by Kj . If Cij is the unit cost in the cell in the ith row and jth column of the transportation tableau, then we can obtain the above values by using: Ri ‡ Kj ˆ Cij & for the occupied (used) cells. The Ri and Kj are called shadow costs. The north-west corner solution for Example 1 is repeated below. To From 1 A 200 4 7 2 100 100 4 3 Demand B 200 C 3 5 5 200 Supply 8 200 100 300 9 5 300 300 100 700 11 12 Transportation problems So for this solution we have: cell (1, 1): cell (1, 2): cell (2, 2): cell (2, 3): cell (3, 3): R1 ‡ K1 R1 ‡ K2 R2 ‡ K2 R2 ‡ K3 R3 ‡ K3 ˆ4 ˆ3 ˆ5 ˆ9 ˆ5 (1) (2) (3) (4) (5) Here we have five equations in six unknowns: R1 , R2 , R3 , K1 , K2 , and K3 . We may therefore choose one of the unknowns and solve for the others. It is usual to choose R1 ˆ 0, but you do not have to, you may choose any one of the unknowns. Then from: (1) (2) (3) (4) (5) 0 ‡ K1 ˆ 4 A K1 ˆ 4 0 ‡ K2 ˆ 3 A K2 ˆ 3 R2 ‡ K2 ˆ R2 ‡ 3 ˆ 5 A R2 ˆ 2 R2 ‡ K3 ˆ 2 ‡ K3 ˆ 9 A K3 ˆ 7 R3 ‡ K3 ˆ R3 ‡ 7 ˆ 5 A R3 ˆ À2 It is always possible to solve the resulting set of equations in this step-by-step way but the calculations may not proceed from equation (1) 3 equation (2) 3 equation (3) 3 equation (4) 3 equation (5) as above. Notice that it is possible that the Ri and Kj values may be positive, negative or zero. After solving for the R and K values a few times you may become so proficient that you will be able to do the calculations in your head. The following table summarises the results and may be written down straight away if you do the calculations in your head. This table is also useful in calculating improvement indices. K1 ˆ 4 R1 ˆ 0 R2 ˆ 2 R3 ˆ À2 X 4 7 4 K2 ˆ 3 X X K3 ˆ 7 3 5 5 8 X X 9 5 Notice that we have indicated the used cells by X and only included the costs. Do not get confused: it is the costs that are involved in the calculations of the R and K values and only the costs. Some students do get confused on this matter. Having calculated the R and K values we can now calculate, for each unused cell, an improvement index Iij using the formula: & Iij ˆ Cij À Ri À Kj By definition of the Ri and Kj values the improvement index for the occupied (used) cells is zero. Transportation problems For the north-west corner solution for Example 1 we have, using the table above: cell (1, 3): cell (2, 1): cell (3, 1): cell (3, 2): I13 ˆ C13 À R1 À K3 ˆ 8 À 0 À 7 ˆ 1 I21 ˆ C21 À R2 À K1 ˆ 7 À 2 À 4 ˆ 1 I31 ˆ C31 À R3 À K1 ˆ 4 À (À2) À 4 ˆ 2 I32 ˆ C32 À R3 À K2 ˆ 5 À (À2) À 3 ˆ 4 A warning: Do be careful when you are subtracting negative numbers. For example, 2 À (À2) ˆ ‡4. Again, with a little practice these calculations can be done in your head and the results written into a table such as that below. Improvement indices K1 ˆ 4 R1 ˆ 0 0 R2 ˆ 2 1 R3 ˆ À2 2 4 7 4 K2 ˆ 3 0 0 4 3 5 5 K3 ˆ 7 1 0 0 8 9 5 Notice that the occupied cells have been indicated with ringed zeros. & If all the improvement indices are greater than or equal to zero, an optimal solution has been reached. If there are any negative improvement indices then it is possible to improve the current solution and decrease the total shipping costs. Each negative index computed represents the amount by which the total transportation costs could be decreased if 1 unit were shipped on that route. In the light of the above, the north-west corner solution to Example 1 is optimal. The solution to the transportation problem posed in Example 1 is: send 200 tons from plant 1 to site A 100 tons from plant 1 to site B 100 tons from plant 2 to site B 200 tons from plant 2 to site C 100 tons from plant 3 to site C The cost of the transportation pattern, which is the minimum total cost, is £3900, as shown earlier. Example 4 Due to roadworks the cost of shipping 1 ton from plant 1 to site A in Example 1 increases to £6. All the remaining data remain the same. Obtain the improvement indices for the north-west corner solution now. 13 14 Transportation problems The north-west corner solution is: To From 1 A B 6 200 7 2 100 100 4 3 Demand 200 C 3 5 5 200 Supply 8 200 100 300 9 5 300 300 100 700 The equations to be solved for the R and K values are now: R1 ‡ K1 R1 ‡ K2 R2 ‡ K2 R2 ‡ K3 R3 ‡ K3 ˆ6 ˆ3 ˆ5 ˆ9 ˆ5 Taking R1 ˆ 0, we obtain R2 ˆ 2, R3 ˆ À2, K1 ˆ 6, K2 ˆ 3 and K3 ˆ 7. The improvement indices for the non-occupied cells are then: I13 I21 I31 I32 ˆ8À0À7ˆ1 ˆ 7 À 2 À 6 ˆ À1 ˆ 4 À (À2) À 6 ˆ 0 ˆ 5 À (À2) À 3 ˆ 4 As there is now a negative improvement index the solution is no longer optimal and can be improved. Exercise 1C 1 In Example 2 we considered the transportation problem given by the following table of supply, demand and unit costs: To A B C Supply 1 10 12 9 40 2 4 5 7 50 3 11 8 6 60 Demand 70 50 30 From Transportation problems We showed in Example 2 that the north-west corner rule gave the following initial solution: To From A 1 40 2 30 B 10 4 11 3 Demand 70 C Supply 12 20 30 50 9 5 7 8 30 30 6 40 50 60 150 By calculating improvement indices for the unoccupied cells, show that this solution is optimal. Give the transportation pattern and its cost. 2 A company has three factories, F1 , F2 and F3 , and three warehouses, W1 , W2 and W3 . The table below shows the costs Cij of sending one unit of product from factory Fi to warehouse Wj . Also shown are the availabilities at each factory and the requirements of each warehouse. To W1 W2 W3 Availability F1 8 6 7 4 F2 2 4 9 3 F3 6 3 5 8 Requirement 9 4 2 From (a) Use the north-west corner rule to write down a possible pattern of distribution and find its cost. (b) Calculate the improvement indices for the unused cells and hence show that this distribution pattern is optimal. 3 A company has three warehouses, W1 , W2 and W3 , which are supplied by three suppliers, S1 , S2 and S3 . The table below shows the cost Cij of sending one case of goods from supplier Si to warehouse Wj , in appropriate units. Also shown in the table are the number of cases available at each supplier and the 15 16 Transportation problems number of cases required at each warehouse. The total number of cases available is equal to the total number of cases required. W1 W2 W3 Availability S1 10 4 11 14 S2 12 5 8 10 S3 9 6 7 6 Requirement 8 10 12 (a) Use the north-west corner rule to obtain a possible pattern of distribution. (b) Calculate the improvement indices for the unused cells and hence show that this distribution pattern is not optimal. Obtaining an improved solution (the stepping-stone method) Now that we have decided that the north-west corner solution in Example 4 can be improved by using route (2, 1), we must decide which route is to be removed from the present solution. This procedure is similar to that used in the simplex algorithm, where we first choose an entering variable and then choose a departing variable. In order to determine the cell to be used we need to draw a closed path or loop in the transportation tableau. & A closed path or loop is a sequence of cells in the transportation tableau such that: (i) each pair of consecutive cells lies in either the same row or the same column (ii) no three consecutive cells lie in the same row or column (iii) the first and last cells of a sequence lie in the same row or column (iv) no cell appears more than once in the sequence. The loops we are interested in have the following property: the first cell is unused and all the other cells are used. For Example 4 above the used cells are shown by X in the table below: X X X X X X Transportation problems The possible loops with the above property are: X X X X X X X X X X X (i) starting at (2, 1) X (ii) starting at (1, 3) X X X X X X X X X (iii) starting at (3, 2) (iv) starting at (3, 1) It should be noted that in developing the loop it is possible to skip over both unused and used cells. Since we wish to use cell (2, 1), we now concentrate our attention on diagram (i) above. To ensure that the constraints remain satisfied, we add y to the number in (2, 1) and (Ày) and (‡y) to the numbers in the other cells, alternately, as we move round the loop. We then have: A B C 1 200 À y 100 ‡ y 2 ‡y 100 À y 300 200 100 3 200 200 300 100 300 You may check each row and column in turn to see that the constraints are satisfied. To satisfy the non-negativity condition and improve the solution as much as possible we make y as large as possible without making any entry negative. So choose y ˆ min(100, 200) ˆ 100. Taking y ˆ 100 gives us the new improved solution: 100 6 200 3 7 5 4 100 5 8 200 100 Notice that route (2, 2) is not now used. We will now test this solution for optimality. 9 5 17 18 Transportation problems The equations to be solved for the R and K values are: R1 ‡ K1 R1 ‡ K2 R2 ‡ K1 R2 ‡ K3 R3 ‡ K3 ˆ6 ˆ3 ˆ7 ˆ9 ˆ5 K1 ˆ 6 R1 ˆ 0 X R2 ˆ 1 X K2 ˆ 3 6 X 3 7 5 4 R 3 ˆ À3 K3 ˆ 8 5 8 X X 9 5 Taking R1 ˆ 0, we obtain R2 ˆ 1, R3 ˆ À3, K1 ˆ 6, K2 ˆ 3 and K3 ˆ 8 (see table). The improvement indices for the unused cells are: I13 I22 I31 I32 ˆ8À0À8ˆ0 ˆ5À1À3ˆ1 ˆ 4 À (À3) À 6 ˆ 1 ˆ 5 À (À3) À 3 ˆ 5 All improvement indices are non-negative and so this solution is optimal: send 100 tons from plant 1 to site A 200 tons from plant 1 to site B 100 tons from plant 2 to site A 200 tons from plant 2 to site C 100 tons from plant 3 to site C Take cost of this transportation pattern is: (100  6) ‡ (200  3) ‡ (100  7) ‡ (200  9) ‡ (100  5) ˆ 600 ‡ 600 ‡ 700 ‡ 1800 ‡ 500 ˆ £4200 Alternative optimal solutions You may have noticed that one of the improvement indices calculated above (I13 ) was zero. This indicates that the route (1, 3) could be used without changing the total overall cost: X X X X X The routes used in the optimal solution obtained are shown by X in the above diagram. Also shown is the closed path or loop starting at (1, 3). Proceeding as above we have: Transportation problems 100 À y 200 ‡y 100 ‡ y 200 À y 100 In this case y ˆ min(100, 200) ˆ 100, so we have the alternative solution: 6 200 3 send 7 5 4 200 100 5 100 100 8 9 5 200 tons from plant 1 to site B 100 tons from plant 1 to site C 200 tons from plant 2 to site A 100 tons from plant 2 to site C 100 tons from plant 3 to site C The total cost of this transportation pattern is: (200  3) ‡ (100  8) ‡ (200  7) ‡ (100  9) ‡ (100  5) ˆ 600 ‡ 800 ‡ 1400 ‡ 900 ‡ 500 ˆ £4200, as before Exercise 1D 1 In question 3 of Exercise 1C you showed that the solution obtained by the north-west corner rule was not optimal. Obtain an improved solution using the stepping-stone method and show that this solution is optimal. 2 A lumber company ships pine flooring from its three mills, A1 , A2 and A3 , to three building suppliers, B1 , B2 and B3 . The table below shows the demand, availabilities and unit costs of transportation. Starting with the north-west corner solution and using the stepping-stone method, determine the transportation pattern that minimises the total cost. B1 B2 B3 Availability A1 3 3 2 25 A2 4 2 3 40 A3 3 4 3 31 Demand 30 30 36 19 20 Transportation problems 1.4 Some special situations Unbalanced transportation problems So far in this chapter we have considered only balanced transportation problems in which (total supply) ˆ (total demand). In many real-world situations, however, supply exceeds demand, or vice versa. When we have an unbalanced transportation problem we need to convert it into a balanced transportation problem in order to use the solution procedure described earlier in this chapter. Example 5 Consider the problem given in Example 3 but with source 1 now having 350 products available rather than 250. The transportation tableau for this problem is now obtained by adding a dummy column, that is a dummy destination. This destination is assigned a dummy requirement equal to the surplus available, in this case (400 ‡ 350) À (300 ‡ 350) ˆ 100. The transportation costs in the dummy column are zero, as these surplus products will not, of course, be transported. The transportation tableau is now: To From A B Dummy Supply 40 2 Demand 65 0 70 1 30 0 300 350 350 400 100 750 We can now apply the transportation algorithm described earlier. The north-west corner solution is: 300 40 70 300 50 300 65 30 350 0 100 0 100 350 400 750 The R and K values are: K1 ˆ 40 K2 ˆ 65 K3 ˆ 35 R1 ˆ 0 R2 ˆ À35 X 40 70 X X 65 30 0 X 0 Transportation problems The improvement indices for the unused cells are: I13 ˆ 0 À 0 À 35 ˆ À35 I21 ˆ 70 À (À35) À 40 ˆ 65 Since I13 ˆ À35, that is it is negative, we can improve the solution by using the route (1, 3). The required loop is: X X X X Adding and subtracting y as required gives: 300 50 À y y 300 ‡ y 100 À y In this case y ˆ min(50, 100) ˆ 50 and so we obtain the improved pattern: 300 50 350 50 The R and K values for this pattern are: K1 ˆ 40 K2 ˆ 30 R1 ˆ 0 R2 ˆ 0 X 40 70 K3 ˆ 0 65 X 30 X X 0 0 and the improvement indices for the unused cells are: I12 ˆ 65 À 0 À 30 ˆ 35 I21 ˆ 70 À 0 À 40 ˆ 30 Since all improvement indices are positive, this pattern is optimal. Send 300 tons from 1 to A 350 tons from 2 to B There will be 50 products left at 1 and 50 products left at 2. The cost of this transportation pattern is: (300  40) ‡ (350  30) ˆ 22 500 This is to be compared with the cost of the north-west corner solution, which is: (300  40) ‡ (50  65) ‡ (300  30) ˆ 24 250 21 22 Transportation problems If the total demand exceeds the total supply, then we add a dummy source with supply ˆ (total demand) À (total supply). A dummy row is added to the tableau and as before the transportation costs in this row will be zero. The transportation algorithm can then be used to solve the problem. Exercise 1E 1 A manufacturing company has three factories, F1 , F2 and F3 , and two retail outlets, R1 and R2 . It wishes to transport its products from its factories to its outlets at minimum total cost. The table below gives details of demand and supply, and also the unit costs of transportation. To R1 R2 Supply F1 2 6 30 F2 2 4 60 F3 6 9 20 Demand 60 20 From (a) Write down the north-west corner solution to the corresponding balanced problem. (b) Show that this solution is optimal. (c) State the optimal transportation pattern and gives its cost. 2 A transportation problem involves the following costs, supply and demand: To D1 D2 D3 Supply S1 7 8 10 50 S2 9 7 8 60 Demand 70 30 40 From (a) Show that the problem is unbalanced and draw a transportation tableau for the corresponding balanced problem. Transportation problems (b) Write down the north-west corner solution and then use the stepping-stone method to obtain an optimal solution. (c) Interpret your optimal solution and obtain the minimum cost. Degeneracy & A feasible solution to a transportation problem is degenerate if the number of used cells is less than the magic number (m ‡ n À 1). Degeneracy can occur during the determination of the initial feasible solution or it can occur during subsequent iterations when the stepping-stone method is used. Degeneracy requires a special procedure to deal with the problem. To handle degenerate problems we create an artificially occupied cell, that is we place a zero in one of the unused cells and then treat that cell as if it were occupied. The cell must be chosen so that it is possible to calculate all the R and K values. There is usually some flexibility in selecting the unused cell that will receive the zero. Degeneracy in an initial solution Consider the following transportation tableau: To From A B 5 1 C 4 Supply 12 8 3 Demand 6 10 11 2 7 11 100 200 200 200 100 200 500 Using the north-west corner method gives the initial solution: To From 1 A 100 5 8 2 100 100 11 3 Demand B 100 C 4 12 6 10 7 200 Supply 200 200 11 200 100 200 500 23 24 Transportation problems This initial solution is degenerate since the number of used cells is 4, which is less than the magic number (3 ‡ 3 À 1) ˆ 5. The degeneracy arises because when 100 is placed in (2, 2) a row and a column requirement are satisfied simultaneously. To deal with this problem we place a zero in an unused cell. In this case we may choose either cell (2, 3) or cell (3, 2). If we choose to place a zero in (2, 3), then the equations to be solved for R and K are: R1 ‡ K1 R1 ‡ K2 R2 ‡ K2 R2 ‡ K3 R3 ‡ K3 ˆ5 ˆ4 ˆ6 ˆ 10 ˆ 11 K1 ˆ 5 R1 ˆ 0 X R2 ˆ 2 R3 ˆ 3 K2 ˆ 4 5 8 11 X X K3 ˆ 8 4 6 7 12 X X 10 11 Taking R1 ˆ 0, we obtain R2 ˆ 2, R3 ˆ 3, K1 ˆ 5, K2 ˆ 4 and K3 ˆ 8. The improvement indices for the unused cells are: I13 I21 I31 I32 ˆ 12 À 0 À 8 ˆ 4 ˆ8À2À5ˆ1 ˆ 11 À 3 À 5 ˆ 3 ˆ7À3À4ˆ0 Since all these indices are non-negative, this solution is optimal. Since also I32 ˆ 0 there is an alternative solution in which route (3, 2) is used. This corresponds to making the alternative choice for the cell to receive the zero. The optimal solution to this problem is: send 100 units from 1 to A 100 units from 1 to B 100 units from 2 to B 200 units from 3 to C The total cost of this pattern is: (100  5) ‡ (100  4) ‡ (100  6) ‡ (200  11) ˆ 3700 Transportation problems Degeneracy during later solution stages Consider the transportation tableau: To A From B 2 1 C 4 3 2 Demand 5 8 100 Supply 150 6 150 200 100 350 The initial solution obtained by the north-west corner rule is: To A From 1 100 B 2 50 3 2 Demand C 4 100 5 8 100 Supply 100 150 150 6 200 100 350 The R and K values are: K1 ˆ 2 R1 ˆ 0 X K2 ˆ 4 2 X 3 R2 ˆ 4 X K3 ˆ 2 4 8 5 X 6 The improvement indices for the unused cells are: I13 ˆ 5 À 0 À 2 ˆ 3 I21 ˆ 3 À 4 À 2 ˆ À3 There is a negative improvement index I21 and so the solution may be improved: A 1 2 B 100 À y X 50 ‡ y X ‡y X 100 À y C X 25 26 Transportation problems The relevant loop is shown above together with the y adjustment. In this case we choose y ˆ 100 and obtain the new solution: To A From B 2 1 2 100 Demand 150 3 C 4 5 8 100 Supply 6 100 150 150 200 100 350 The number of used cells is 3 and since this is less than the magic number (3 ‡ 2 À 1) ˆ 4 the solution is degenerate. This is because two formerly occupied cells have dropped to zero. To proceed we have to add a zero to one of the unoccupied cells and then treat it as occupied. It is usual to choose the cell with the lowest shipping cost, in this case (1, 1). The R and K values are: K1 ˆ 2 R1 ˆ 0 X R2 ˆ 1 X K2 ˆ 4 2 3 X K3 ˆ 5 4 8 5 X 6 The improvement indices of the unused cells are: I13 ˆ 5 À 0 À 5 ˆ 0 I22 ˆ 8 À 1 À 4 ˆ 3 Again there is an alternative solution obtained by using (1, 3) instead of (1, 1) as the cell in which to place the zero. The optimal solution is: send 150 units from 1 to B 100 units from 2 to A 100 units from 2 to C Total cost of this pattern is: (150  4) ‡ (100  3) ‡ (100  6) ˆ 1500 Transportation problems Exercise 1F 1 A manufacturing company produces diesel engines in three cities, C1 , C2 and C3 , and they are purchased by three trucking companies, T1 , T2 and T3 . The table below shows the number of engines available at C1 , C2 and C3 and the number of engines required by T1 , T2 and T3 . It also shows the transportation cost per engine (in £100s) from sources to destinations. The company wishes to keep the total transportation costs to a minimum. To T1 T2 T3 Supply C1 3 2 3 25 C2 4 2 3 35 C3 3 2 6 20 Demand 30 30 20 From (a) Write down the north-west corner solution and state why it is degenerate. (b) Use the stepping-stone method to obtain the optimal solution. (c) Give the transportation pattern and its total cost. 2 A builders' merchant has 13 tons of sand at site X, 11 tons at site Y and 10 tons at site Z. He has orders for 9 tons from customer A, 13 tons from customer B and 12 tons from customer C. The cost per ton (in £10s) of moving the sand between depots and customers is given in the table below. A B C X 1 2 4 Y 1 3 4 Z 5 7 5 (a) Write down the north-west corner solution. (b) Use the stepping-stone method to obtain an optimal solution. State the minimum cost of transportation. 27 28 Transportation problems 3 A transportation problem involves the following costs, supply and demand: D1 D2 D3 Supply S1 17 8 14 30 S2 15 10 20 20 S3 20 5 10 10 Demand 10 20 30 (a) Write down the north-west corner solution. (b) Use the stepping-stone method to obtain an optimal solution and give its cost. SUMMARY OF KEY POINTS 1 The shadow costs Ri , for the ith row, and Kj , for the jth column, are obtained by solving Ri ‡ Kj ˆ Ci j for occupied cells, taking R1 ˆ 0 arbitrarily. 2 The improvement index Ii j for an unoccupied cell is defined by Ii j ˆ Ci j À Ri ˆ Kj . 3 If all improvement indices are greater than or equal to zero then an optimal solution has been reached. 4 A closed path or loop is a sequence of cells in the transportation tableau such that (i) each pair of consecutive cells lies in either the same row or the same column (ii) no three consecutive cells lie in the same row, or column (iii) the first and last cells of a sequence lie in the same row or column (iv) no cell appears more than once in the sequence. 5 A feasible solution to a transportation problem is degenerate of the number of used cells is less than the magic number (m ‡ n À 1), where m is the number of rows and n is the number of columns. ...
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