Example: 1. In this chapter we will discuss the transportation and assignment problems which are two special kinds of linear programming.

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1 Ch. 4 THE TRANSPORTATION AND ASSIGNMENT PROBLEMS In this chapter we will discuss the transportation and assignment problems which are two special kinds of linear programming. deals with transporting goods from their sources to their destinations. The assignment problem, on the other hand, deals with assigning people or machine to jobs. Example: 1 The amount of reinforcement steel available at three warehouses of a construction company is 2, 12, and 12 tons respectively. While, the amounts of reinforcement steel needed at four construction sites are 16, 7, 11, and 1 ton respectively. The following table shows the transportation cost in Egyptian Pounds (LE) between the different warehouses and the construction sites. It is required to draw the network flow diagram and determine how many tons of reinforcement steel should be shipped from each warehouse to each of the construction sites in order to minimize the total shipping cost. 1

2 Solution - Formulate the problem: Minimize: Z= 4X X 12 + X 13 +3X 14 +7X 21 + X 22 +2X 23 +3X 24 +2X 31 +6X 32 +5X 33 +4X 34 Subject to: X 11 +X 12 +X 13 +X 14 = 2 X 21 +X 22 +X 23 +X 24 = 12 X 31 +X 32 +X 33 +X 34 = 12 SUPPLY X 11 +X 21 +X 31 = 16 X 12 +X 22 +X 32 = 17 X 13 +X 23 +X 33 = 11 X 14 +X 24 +X 34 = 1 DEMAND 2

3 - This is an example of the transportation model. - This problem has a specific structure. All the coefficients of the variables in the constraint equations are 1 - Every variable appears in exactly two constraints - All constraint equations are formed using the equal sign not the or the signs - This is the special structure that distinguishes this problem as a transportation problem The Transportation Problem Model - the transportation model is concerned with distributing a commodity from a group of supply centers, called sources to a group of receiving centers, called destinations to minimize total cost - In general, source i (i = 1, 2, 3,.., m) has a supply of s i units, and destination j (j = 1, 2, 3,, n) has a demand for d j units - The cost of distributing units from source i to destination j, c ij, is directly proportional to the number of units distributed, x ij 3

4 The Transportation Problem Model - the transportation Table Cost per distributed unit Destination Supply 1 2 n 1 c 11 c 12 c 1n s 1 Source 2 c 21 c 22 c 2n s 2 : : : : : m c m1 c m2 c mn s m Demand d 1 d 2 d n The Transportation Problem Model - Let Z represents the total distribution cost and x ij (i = 1, 2,.., m; j = 1, 2,., n) be the number of units to be distributed from source i to destination j 4

5 The Transportation Problem Model - A necessary condition for a transportation problem to have any feasible solution is that: - This property may be verified by observing that the constraints require that both The Transportation Problem Model - Number of variables = m x n - Number of constraints = m + n - Number of basic variables = m + n 1 - This as any supply constraint equals the sum of demand constraints minus the sum of other supply constraints - Therefore, any basic feasible (BF) solution will have only m + n 1 basic variables (non-zero variables) while all other variables will have zero value. Also, the sum of allocations for each row or each column equals its supply or demand 5

6 The Transportation Problem Model - One important condition is that all supplies should equal all demands (balanced transportation problem) - If this is not true for a particular problem, dummy sources or destinations can be added to make it true - These dummy centers have zero distribution costs - If shipment is impossible between a given source and destination, a large cost of M is entered. This discourages the solution from using such cells The Transportation Problem Model - Constraint coefficients of the transportation problem x 11 x 12. x 1n x 21 x 22. x 2n. x m1 x m2. x mn

7 Vogel s Approximation Method - For each row and column remaining under consideration, calculate its difference, the arithmetic difference between the smallest and next-to-thesmallest unit cost c ij still remaining in that row or column. - In that row or column having the largest difference, select the variable having the smallest remaining unit cost - Ties for the largest difference, or for the smallest remaining unit cost, may be broken arbitrarily - Using Vogel s Approximation to solve for number of tons to be shipped from each warehouse: - First check that sources are equal to the demand: - Sources = = 44 = demand =

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9 - Substitute in the objective function to get Z Example: 2 Resolve problem no. 1 assuming the transportation table as follow with no reinforcement steel to be shipped from warehouse 1 to site 3 and from warehouse 3 to site 1. 9

10 Solution - Formulate the problem: Minimize: Z= 4X X 12 +3X 14 +7X 21 + X 22 +2X 23 +3X 24 +6X 32 +5X 33 +4X 34 Subject to: X 11 +X 12 +X 14 = 2 X 21 +X 22 +X 23 +X 24 = 12 X 32 +X 33 +X 34 = 12 X 11 +X 21 = 16 X 12 +X 22 +X 32 = 17 X 23 +X 33 = 11 X 14 +X 24 +X 34 = 1 SUPPLY DEMAND 1

11 - Using Vogel s Approximation to solve for number of tons to be shipped from each warehouse: - First check that sources are equal to the demand: - Sources = = 44 = demand =

12 - Substitute in the objective function to get Z 12

13 Example: 3 Resolve problem no. 1 assuming that the reinforcing steel available at the three warehouses is 2, 12, and 18 tons respectively. While, the amounts of reinforcement steel needed at four construction sites are 16, 7, 11, and 1 ton respectively. - Using Vogel s Approximation to solve for number of tons to be shipped from each warehouse: - First check that sources are equal to the demand: - Sources = = 5, - While, demand = = 44 - So we add a dummy construction site C5 to accommodate the difference of 6 tons. 13

14 Solution - Formulate the problem: Minimize: Z= 4X X 12 + X 13 +3X 14 +7X 21 + X 22 +2X 23 +3X 24 +2X 31 +6X 32 +5X 33 +4X 34 Subject to: X 11 +X 12 +X 13 +X 14 + X 15 = 2 X 21 +X 22 +X 23 +X 24 + X 25 = 12 X 31 +X 32 +X 33 +X 34 + X 35 = 12 X 11 +X 21 +X 31 = 16 X 12 +X 22 +X 32 = 17 X 13 +X 23 +X 33 = 11 X 14 +X 24 +X 34 = 1 X 15 +X 25 +X 35 = 6 SUPPLY DEMAND 14

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16 - Substitute in the objective function to get Z A special case of the transportation problem when each supply is one and each demand is one As such, every supplier will be assigned one destination and every destination will have one supplier The Assignment Problem 16

17 Example The assignment problem A contractor owns three excavators of different capacity. He/she wants to dispatch these excavators into three different jobs. The performance of the excavators is measured as the time consumed to perform each job. The data of the time tests are shown as given in the table below. - Draw a network flow diagram. - Formulate a mathematical model to obtain the minimum time. - Solve this model using the Vogal s approximation model The assignment problem (S 1 =1) x 11 1 (d 1 =1) 7 x 12 X 31 (S 2 =1) 2 X x 13 X 22 X 32 2 (d 2 =1) (S 3 =1) X 23 X 33 3 (d 3 =1) 17

18 The assignment problem - Formulate the problem: Minimize: Z= 5X X X 13 +4X X 22 +8X 23 +1X 31 +6X 32 +3X 33 Subject to: X 11 +X 12 +X 13 = 1 X 21 +X 22 +X 23 = 1 X 31 +X 32 +X 33 = 1 X 11 +X 21 +X 31 = 1 X 12 +X 22 +X 32 = 1 X 13 +X 23 +X 33 = 1 SUPPLY DEMAND The Assignment Problem The Assignment Problem Model - In the assignment problem there are n resources or assignee (e.g., employee, machine) is to be assigned uniquely to a particular activity or assignment (e.g., task, job) - There is a cost c ij associated with assignee i (i = 1, 2,3,., n) assigned (performing) assignment j (j = 1, 2, 3,, n) - the objective is to determine how all the assignments should be made in order to minimize the total cost 18

19 The Assignment Problem The Assignment Problem Model - The number of assignees = the number of assignments (tasks) - Each assignee is to be assigned to exactly one task - Each task is to be performed by exactly one assignee - The decision variable x ij of assigning machine (assignee) i to job (task) j always take a value of one or zero The Assignment Problem Model Cost of assigning i to j Assignment Supply 1 2 n 1 c 11 c 12 c 1n 1 Assignee 2 c 21 c 22 c 2n 1 : : : : : n c n1 c n2 c nn 1 Demand

20 The Transportation Problem Model - Let Z represents the total assignment cost and x ij (i = 1, 2,.., n; j = 1, 2,., n) be a binary number representing assigning machine i to job j The assignment problem Using Vogel s Approximation 2

21 Optimality test TRANSPORTATION PROBLEM Example: A cement company has four factories A, B, C, and D. Its major distribution centers are located in three cities O, P, and R. The unit transportation cost from factories to the distribution centers is as follows: Factories O Distribution Centers P R A B C D Ch. 4 TRANSPORTATION PROBLEMS - The capacities of the four factories and the three distribution centers are as follow: Factory Production capacity (ton) Distribution center Distribution capacity (ton) A 85 O 15 B 11 P 12 C 6 R 75 D 45 The company needs to determine the best transportation program which will minimize the cost of transportation. 21

22 Ch. 4 TRANSPORTATION PROBLEMS Solution Factories A B C D Demand Col. Diff. Distribution Centers O P R Supply Row Diff X AP = 85 Eliminate Row A Ch. 4 TRANSPORTATION PROBLEMS Factories B C D Demand Col. Diff. Distribution Centers O P R Supply Row Diff X BO = 15 Eliminate Col. O Factories B C D Demand Col. Diff. Distribution Centers P R Supply Row Diff X BR = 5 Eliminate Row B 22

23 Ch. 4 TRANSPORTATION PROBLEMS Solution Factories C D Demand Col. Diff. Distribution Centers P R Supply Row Diff X DR = 45 Eliminate Row D Factories C Demand Col. Diff. Distribution Centers P R Supply Row Diff. 6 X CP = 35, X CR = 25 Ch. 4 TRANSPORTATION PROBLEMS The preliminary solution Factories A B C D Demand O (3) (6) 15 (12) (13) 15 Distribution Centers P (4) 85 (9) (8) 35 (11) 12 R (2) (5) 5 (1) 25 (7) Supply Z min= 85 x x x x x x 7 = 184 L.E 23

24 LOOP METHOD for TESTING THE SOLUTION of TRANSPORTATION MODEL Test the solution Factories O Distribution Centers P R Supply A B C D (3) (6) 15 (12) (13) (4) 85 (9) (8) 35 (11) (2) (5) 5 (1) (7) Demand LOOP METHOD for TESTING THE SOLUTION of TRANSPORTATION MODEL - It is be noted that the quantity in cells X AO,X AR,X BP,X CO,X DO, and X DP have a zero value. - To test the obtained solution, the unoccupied cell will be tested using one unoccupied cell and the other cells are occupied. The cost of lost chance for the unoccupied cells will be determined. - To check the cell AR we can catch the cells AR, AP, CP, and CR. 24

25 LOOP METHOD for TESTING THE SOLUTION of TRANSPORTATION MODEL Cell AR Adding of one unit to cell AR will increase the cost by + 2 Subtracting of one unit from cell AP will decrease the cost by - 4 Adding of one unit to cell CP will increase the cost by + 8 Subtracting of one unit from cell CR will decrease the cost by - 1 : : Cost of lost chance Testing of other unoccupied cells will give positive values or zero values to the cost of lost chance for each unoccupied cell which mean increasing the value of cost for positive values and the solution doesn t change for zero values. LOOP METHOD for TESTING THE SOLUTION of TRANSPORTATION MODEL The final solution Factories A B C D Demand O (3) (6) 15 (12) (13) 15 Distribution Centers P (4) 6 (9) (8) 6 (11) 12 R (2) 25 (5) 5 (1) (7) Supply Z min = 6 x x x x x x 7 = 174 L.E 25

26 Optimality Test A basic feasible solution is optimal if and only if (c ij u i - v j ) for every (i, j) such that x ij is a non-basic Thus, the only work required by the optimality test is the derivation of the values of the u i and v j for the current basic solution and then the calculation of these (c ij u i - v j ) Since (c ij u i - v j ) is required to be zero if x ij is a basic variable, the u i and v j satisfy the set of equations Optimality Test c ij = u i + v j variable for each i and j such that x ij is a basic There are (m+n 1) basic variables, and accordingly (m+n-1) equations The number of unknowns (u i and v j ) is (m + n), one of these variables can be assigned a value arbitrarily Select the u i that has the largest number of allocations in its row and assign it the value of zero 26

27 TRANSPORTATION PROBLEMS Example: The following table shows the transportation cost of shipping one unit of a given product from the sources 1, 2 and 3 to destinations 1, 2, 3 and 4. The demand of each destination and the supply of each source are also given in the table Destinations Supply Sources Demand TRANSPORTATION PROBLEMS An initial basic feasible solution for this problem was determined as shown in the next table. Check if this solution is optimal or not and if not, determine the optimal solution. Destinations Sources x 11 =16 x 12 =4 2 x 22 =1 x 23 =11 3 x 32 =2 x 34 =1 27

28 TRANSPORTATION PROBLEMS Suppl y u i M M Demand Z=99 v j TRANSPORTATION PROBLEMS Check the optimality by checking the value (c ij u i - v j ) for all non-basic variables, these values should be positive or zero If any of these values are negative, then the solution is not optimal and another iterative is required Because one of these values (c 33 u 3 v 3 ) = -2 (negative), we conclude that, the current basic feasible solution is not optimal. Thus, the transportation simplex method must next go to the iterative step to find a better basic feasible solution 28

29 TRANSPORTATION PROBLEMS Supply u i M M M M Demand Z=99 v j Not optimal TRANSPORTATION PROBLEMS Find an entering basic variable, a leaving basic variable and then identifying the new basic feasible solution The entering basic variable must have negative (c ij u i v ) j If there more than one negative value exists, select the one having the largest absolute negative value In this example, x 33 is the entering basic variable Increasing the non-basic variable from zero sets off a chain reaction of changes in other basic variables in order to continue satisfying the supply and demand constraints The basic variable to be decreased to zero becomes the leaving basic variable. With x 33 being the entering basic variable, the chain reaction is shown in the next Table 29

30 TRANSPORTATION PROBLEMS Find an entering basic variable, a leaving Supply u i M M M M Demand v j TRANSPORTATION PROBLEMS Supply u i M M M M Demand Z=95 v j optimal 3

Chapter 7 TRANSPORTATION PROBLEM

Chapter 7 TRANSPORTATION PROBLEM Chapter 7 Transportation Transportation problem is a special case of linear programming which aims to minimize the transportation cost to supply goods from various sources