قالىا سبحانك ال علم لنا إال ما علمتنا صدق هللا العظيم. Lecture 5 Professor Sayed Fadel Bahgat Operation Research

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1 قالىا سبحانك ال علم لنا إال ما علمتنا إنك أنت العليم الحكيم صدق هللا العظيم 1

2 والصالة والسالم علي اشرف خلق هللا نبينا سيدنا هحود صلي هللا عليه وسلن سبحانك اللهم وبحمدك اشهد أن ال هللا إال أنت استغفرك وأتىب اليك 2

3 sayed_fadel2006@yahoo.com 3

4 Text Book : Problems in Operations Research ( Principles and Solutions) P.K. GUPTA And Dr. D. S. HIRA 4

5 Chapter 10 Page ( 406 ) Transportation Model 5

6 Definition of Transportation Model: Transportation problems are expressed in matrix form. The matrix consists of squares called cells. The cell located at the intersection of a row and a column is designated by its row and column headings. Ex. Thus the cell at intersection of row A and column 3 is called (A,3). Unit costs are placed in each cell. 6

7 Transportation models deal with problems concerning as to what happens to the effectiveness function when we associate each of a number of origins ( sources ) with each of a possibly different number of destinations ( jobs ). The total movement from each origin and the total movement to each destination is given and it is desired to find how the associations be made subject to the limitations on total. 7

8 suppose that there are: m sources and n destinations. Let a i be the number of supply units available at source i ( i=1,2,3,.,m). Let b j be the number of demand units required at destination j (j=1,2,3,,n). Let C ij represent the per unit transportation cost for transporting the units from source i to destination j 8

9 The objective is to determine the number of units to be transported from source I to destination j so that the total transportation cost is minimum. In addition, the supply limits at the sources and the demand requirements at the destinations must be satisfied exactly. 9

10 If X ij ( X ij 0 ) is the number of units shipped from source i to destination j, then the equivalent linear programming model will be: Find X ij (i=1,2,..,m; j=1,2,..,n) in order to minimize subject where to : X ij m n Z C ij i 1 j 1 m X a, ij i i 1 n X b, ij j j 1 0. X ij, i 1,2,3,...,m, j 1,2,3...,n, 10

11 The two sets of constraints will be consistent i.e. the system will be in balance if m a i 1 i m b, j 1 j This restriction causes one of the constraints to be redundant ( and hence it can be deleted ) so that the problem will have (m + n 1) constraints and ( mxn) unknowns. 11

12 10.3 Methods for Obtaining Initial Basic Feasible Solution: A feasible solution to a transportation problem is a set of non-negative allocations, X ij that satisfies the rim (row and column ) restrictions. A feasible solution is called a basic feasible solution if it contains no more than m+n-1 non-negative allocations. 12

13 A feasible solution that minimize the transportation cost is called an optimal solution. There are several methods for finding an initial basic feasible solution, out of which three are described here. North-West Corner Rule: This method consists of the following steps: 13

14 1- start with the north west (upper-left ) corner cell of transportation matrix. compare the supply of source 1 (S 1 ) with the demand of destination 1 (D 1 ). (a) if S 1 > D 1, set X 11 = D 1 and proceed horizontally to cell (1,2). (b) if S 1 = D 1, set X 11 = D 1 and proceed diagonally to cell (2,2),and (c) if S 1 < D 1, set X 11 = S 1 and proceed vertically to cell (2,1). 2- Continue procedure, step by step, away from the northwest corner cell till an allocation is made in the south-east corner cell. 14

15 least- Cost Method or Matrix Minima Method. This method consists in allocating as much as possible in the lowest cost cell/cells and then further allocation is done in the cell/cells with second lowest cost and so on. 15

16 3- Vogel s Approximation Method. Vogel s approximation method yields a very good initial solution, which, sometimes may be the optimal solution. It consists of the following steps: (i) enter the difference between the smallest and second smallest elements in each column below the corresponding column and difference between the smallest and second smallest elements in each row to the right of the row. Put these differences in brackets. 16

17 (ii) Select the row or column with the greatest difference and allocate as much as possible, within the constraints of the rim conditions, to the lowest cost cell in that row or column so as to either exhaust the source supply or to satisfy the destination demand. 17

18 (iii) Reduce the supply/ demand units by the amount assigned to the cell and cross out the column/row completely satisfied. (iv) Write down the reduced transportation table omitting rows or columns crossed out in step (iii) Repeat step (i) through (iii) until all allocations have been made. 18

19 Example 10.1 page ( 408) A dairy firm has three plants located through a state. Daily milk production at each plant is as follows: Plant million liters, Plant million liters, and Plant million liters. Each day the firm must fulfill the needs of its four distribution centers. Minimum requirement at each center is as follows: Distribution center 1 7 million liters, Distribution center 2 5 million liters, Distribution center 3. 3 million liters, and Distribution center 4. 2 million liters. 19

20 Cost of shipping one million liters of milk from each plant to each distribution center is given in the following table in hundreds of rupees: Distribution center Plants Formulate the problem as an L.P. problem and find its initial basic feasible solution by : (i) North-west corner rule. (ii) Least Cost Method. (ii) Vogel s approximation method if the object is to minimize the transportation cost. 20

21 Solution: Formulation as L.P. Problem: Step1: Key decision to be made is to find how much quantity of milk from which plant to which distribution center be shipped so as to satisfy the constraints and minimize the cost. Thus the variables are: X 11,X 12, X 13, X 14, X 21, X 22, X 23, X 24, X 31, X 32, X 33, and X

22 These variables presents the quantities of milk to be shipped from different plants to different distribution centers and can be represented in the form of a matrix shown: Distribution Centers X 11 X 12 X 13 X 14 Plant 2 X 21 X 22 X 23 X 24 3 X 31 X 32 X 33 X 34 22

23 In general, key decision is to find the quantity of units from each origin to each destination. Step2: Feasible alternatives are sets of values of X ij 0. Step 3: Objective is to minimize the cost of transportation. Minimize: 2X X X X 14 +X X X 23 + X X X X X 34 23

24 minimize m i 1 n j 1 C ij X ij n j 1 m i 1 C X ij ij. Step 4: Constraints are (availability or supply) X 11 +X 12 +X 13 +X 14 = 6, (for milk plant1) X 21 +X 22 +X 23 +X 24 =1, (for milk plant2) X 312 +X 32 +X 33 +X 34 =10, (formilk plant3) Which express as : n X j 1 ij S, i i 1,2,3,......,m 24

25 Requirement or demand: X 11 +X 21 +X 31 =7, (for distribution center1) X 12 +X 22 +X 32 =5, (for distribution center2) X 13 +X 23 +X 33 =3, (for distribution center3) X 14 +X 24 +X 34 =2, (for distribution center4) In general m X i 1 ij D, j j 1,2,3,......,n In general, such a situation will involve (mxn) variables and (m+n) constraints. 25

26 (i) Initial Basic Feasible Solution by North-West Corner Rule: supply /0 (6) /0 (1) /5/2/0 (5) (3) (2) requirement 7/1/0 5/0 3/0 2/

27 (i) set X 11 equal to 6, namely the smaller of the amount available at S 1 (6) and that needed at D 1 (7) and (ii) proceed to cell (2,1) (rule c ). Compare the number of units available at S 2 ( namely 1) with the amount required at D 1 (1) and accordingly set X 21 = 1. (iii) proceed to cell(3,2) (rule b). Now supply from plant S 3 is 10 units while the demand for D 2 is 5 units; accordingly set X 32 equal 5. Proceed cell (3,3) (rule a ) and allocate 3 there. 27

28 (v) proceed to cell (3,4) (rule a ) and allocate 2 there. It can be seen that the proposed solution is feasible solution since all supply and requirement constraints are fully satisfied. The transportation cost associated with the solution is: Z = Rs.(2x6+1x1+8x5+15x3+ 9x2)x100 = Rs. 11,600. Note: A cell that gets an allocation is called a basic cell. 28

29 (ii) Initial Basic Feasible Solution by least Cost Method. Her, the lowest cost cell is (2,2) and maximum possible allocation ( meeting supply and requirement positions) is made here. maximum feasible allocation in call (2,2) is (1). This the supply position of plant 2. Therefore, row 2 is crossed out, indicating that no allocations are to be made in cells (2,1), (2,3), and (2,4). 29

30 The next lowest cost cell is (1,1), max. possible allocation of (6) is made and row 1 is crossed out. Next lowest cost in row 3 is (3,1) and allocation of (1) is done.and so on supply 1 2 (6) / (1) 6 1 1/0 3 5 (1) 8 (4) 15 (3) 9 (2) 10/9/5/2/0 requirement 7/1/0 5/4/0 3/0 2/

31 Z =Rs.(2x6+0x1+5x1+8x4+15x3+9x2)x100 =RS. 11,00 Which is less than the cost associated with the solution obtained by North-West Corner Method. (iii) Initial Basic Feasible Solution by Vogel s approximation Method. This method consists of the following steps: 31

32 Step 1 : Write down the cost matrix as shown below: enter the difference between the smallest and second smallest elements in each row and column as shown supply (1) 6 (1) 1/0 (1) (3) Requirement (1) (3) (5) (6) 32

33 Substep2: Select the row or column with the greatest difference and allocate as much as possible within the restrictions of the rim conditions to the lowest cost cell in the row or column selected. In case a tie occurs, allocate to the cell associated with the lowest cost. Since (6) is the largest number, we choose column 4 and allocate as much as possible to the cell (2,4) as it has lowest cost 1 in column 4. Since supply is 1 while the requirement is 2, maximum possible allocation is (1). 33

34 Substep3: Cross out the row or column completely satisfied by the allocation just made. Then the table reduced to: and repeat last steps Supply / 1 (5) (3) requirement 7 5/0 3 1 (3) (5) (4) (2) 34

35 Substep4: (a) Repeat steps 1 to 3 until all assignments have been made column (2) exhibit the greatest difference of (5) units to cell (1,2), since it has the smallest transportation cost in column 2 and this column is crossed out and we get the following table: supply /0 (5) (1) (4) requirement 7/6 3 1 (3) (4) (2) 35

36 (b) differences are recalculated. The maximum difference is (5). Therefore, we allocate (1) to the cell (1,1) since it has the lowest cost in row 1. Then we get the following table: (C) As cell (3,1) has the lowest cost, maximum possible allocation of (6) is made. Likewise, next location of (1) is made in cell (3,4) and (3) in cell (3,3) as shown supply requirement /0 (6) (3) (1) 6/0 3/0 1/0 36

37 The above repetition can be made in a single matrix as follows: Supply 1 Plants 2 3 (1) 2 3 (5) (1) (6) (3) 9 (1) (1) 6/1/0 (1) (3) 1/0 10/4/3/0 (1) (3) (5) (4) Requirement (1) (3) (5) (6) (3) (5) (4) (3) (4) (2) 2 37

38 The cost of transportation associated with the above solution is: Z =Rs.( 2x1 + 3x5 + 1x1 + 5x6 + 15x3 + 9x1 ) x 100. = Rs. 10,200. Which is evidently the least of all the values of transportation cost found by different methods. Since Vogel s approximation method results in the most economical initial feasible solution, we will use this method for finding such a solution, unless otherwise asked. 38

39 Example 10.2 page (415): Solve the last example ( 10.1 ) by the transportation technique. Solution: The transportation technique involves representing the problem in the form of a matrix, Finding initial basic feasible solution, Testing the solution for optimality, and If required making iteration to arrive at an optimal solution. 39

40 Step I: Set up Transportation table: Distribution Centers ( Destinations) Supply Plants (origins) Requirement total Step II : Find an initial Basic Feasible Solution. (like last example) 40

41 The associate transportation cost is: Rs. 10, Supply 1 Plants (1) 3(5) (1) 5(6) 8 15(3) 9 (1) (1) 6/1/0 (1) 1/0 (3) 10/4/3/0 (1) (3) (5) (4) Requirement (1) (3) (5) (6) (3) (5) (4) (3) (4) (2) 2 41

42 Step III: Perform Optimality Test: An optimality test can be performed only on that feasible solution in which: (a) number of allocations is m+n-1, where m is the number of rows and n is the number of columns. In the given situation, m=3 and n=4 and, therefore the required number of allocations is 6. Since the actual number of allocations as seen from table is also 6, this conditions is satisfied. 42

43 These m+n-1 allocations should be in independent positions. Allocations are in independent positions if it is not possible to increase or decrease any allocation without either changing the position of the allocations or violating the row and column restrictions. A simple rule for allocations to be in independent positions is that it is impossible to travel from any allocation, back to itself, by a series of horizontal and vertical jumps from one occupied cell to another. 43

44 The u-v method or Modified distribution ( MODI) Method or, unit coast-penalty method used for performing optimality test consists of the following steps: Substep1: Set up a cost matrix containing the costs associated with the cells for allocations have been made. This matrix for the given example : 44

45 Cost matrix for allocated cells. Substep2: Enter a set of numbers v j across the top of the matrix and a set of numbers u i across the left side so that their sums equal the costs entered in substep1.. 45

46 Thus u 1 + v 1 =2, u 3 + v 1 =5 u 1 + v 2 =3, u 3 + v 3 =15, u 2 + v 4 =1, and u 3 + v 4 =9. Let v 1 =0, then u 1 =2, u 3 = 5, v 2 = 3-2 =1 v 3 = 15-5=10, v 4 = 9-5= 4, and u 2 =1-4=-3 Therefore, the matrix may be written as: 46

47 u j v j Substep3: Fill the vacant cells with the sums of u i and v j. As shown above. Substep4: Subtract the cell values of the matrix of substep3 from the original matrix. 47

48 The resulting matrix is called cell evaluation matrix Cell evaluation matrix 48

49 Substep5: If any of the cell evaluations are negative, the basic feasible solution is not optimal. In the present example, since two cell evaluations are negative, step should be taken to obtain an optimal solution. 49

50 Step IV : This involves the following substeps: SubstepI: From the cell evaluation matrix, identify the cell with the most negative entry. Since, in our case, two cells have the same negative entry (-1), We can identify either of them. Let us choose the cell(1,3). 50

51 Substep2: Write down again the initial feasible solution Substep3: Trace a path in this matrix consisting of a series of alternate horizontal and vertical lines. 51

52 The path begins and terminates in the identified cell. All corners of the path lie in the cells for which allocations have been made. The path may skip over number of occupied or unoccupied cells. Substep4: Mark the identified cell as positive and each occupied cell at the corners of the path alternately ve, +ve, -ve, and so. 52

53 Substep5: Make a new allocation in the identified cell by entering the smallest allocation on the path that has been assigned a ve sign. Add and subtract this new allocation from the cells at the corners of the path, maintaining the row and column requirements. 53

55 This causes one basic variable (cell) to become zero and other variables ( cells) remain nonnegative. The basic cell whose allocation has been made zero, leaves the solution. The total cost of transportation for 2 nd feasible solution is: Z= Rs.(3x5+11x1+1x1+7x5+2x15+1x9)x100 = 10,100. Which is less than for the first (starting) feasible solution. 55

56 Step V: Check for Optimality: Required number of allocations = m+n-1= 3+4-1=6, and actual number of allocations is also 6. Further these 6 allocations are in independent positions. Above conditions being satisfied, an optimality test can be performed as follows: SubstepI: Set up the cost matrix containing the costs associated with cell for allocations have been made. 56

57 u i \v j Cost matrix for allocated cells. Substep2: Enter a set of numbers v j across the top of the matrix and a set of numbers u i across the left side so that their sums equal the costs entered in substep1.. 57

58 Thus u 1 + v 2 =3, u 3 + v 1 =5 u 1 + v 3 =11, u 3 + v 3 =15, u 2 + v 4 =1, and u 3 + v 4 =9. Let v 1 =0, then u 3 =5, u 2 = -3, v 3 = 15-5=10 u 1 = 11-10=1, v 4 = 9-5= 4, and v 2 =3-1=2 Therefore, the matrix may be written as: 58

59 u j v j Substep3: Fill the vacant cells with the sums of u i and v j. As shown above. Substep4: Subtract the cell values of the matrix of substep3 from the original matrix. 59

61 substep5: Since one cell value is negative, the second feasible solution is not optimal. Step VI: Iterative Towards an Optimal Solution: SubstepI: In the cell evaluation matrix, identify the cell with negative entry. It is the cell (2,3). 61

62 Substep2: Write down again the feasible solution Substep3: Trace a path in this matrix consisting of a series of alternate horizontal and vertical lines. 62

63 substep4: Make the identified cell as +ve and other alternatively ve and +ve. Substep5: Make the new allocation in the identified cell by entering the smallest allocation on the path which has been assigned a negative sign. Subtract and add this amount from other cells. We get the following tables: 63

64 Third feasible solution 64

65 From this allocation matrix the transportation cost is: Z=Rs.(5x3+1x11+1x6+1x15+2x9+7x5)x100 = 10,000. Thus it is better solution. 65

66 Step VII: Test for optimality (a) Number of allocations is m+n-1 i.e 6. (b) These m+n-1 allocations are in independent positions. Hence repeat the following steps: Substep1: Set up the cost matrix containing costs associated with cells for which allocations have been made. This is in the following table: 66

67 u i \v j Cost matrix for allocated cells. Substep2: Enter a set of numbers v j across the top of the matrix and a set of numbers u i across the left side so that their sums equal the costs entered in substep1. 67

68 Thus u 1 + v 2 =3, u 3 + v 1 =5 u 1 + v 3 =11, u 3 + v 3 =15, u 2 + v 3 =6, and u 3 + v 4 =9. Let v 1 =0, then u 3 =5, u 2 = -4, v 3 = 15-5=10 u 1 = 11-10=1, v 4 = 9-5= 4, and v 2 =3-1=2 Therefore, the matrix may be written as: 68

69 u j v j Substep3: Fill the vacant cells with the sums of u i and v j. As shown above. Substep4: Subtract the cell values of the matrix of substep3 from the original matrix. 69

71 Substep5: Since all cell values are positive, the third feasible solution given by the following matrix is optimal solution. 71

72 Distribution Centers supply (5) (1) Plants (1) (7) (1) (2) 10 Requirement Z=Rs.(5x3+1x11+1x6+1x15+2x9+7x5)x100 = 10,000. And the optimal ( minimum) transportation cost= Rs. 10,

73 Example 10.7 page(438): Find the optimum solution to the following transportation problem in which the cells contain the transportation cost in rupees. W 1 W 2 W 3 W 4 W 5 available F F F F Required

74 Find a basic feasible solution: We shall use Vogel s approximation method to find basic feasible solution. The basic feasible solution is given by the following table: 74

75 W 1 W 2 W 3 W 4 W 5 Available F 1 F 2 F 3 F 4 Required 7 (5) 8 6 (15) 5 (10) 30/25/ (30) (15) (20) (5) 40/25/5 (1)(1)(2) (2) 30(1)(2) 20/15 (1)(1)(1)(1) 10 (1)(1)(1)(1) (1) (1) (2) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) 75

76 السالم عليكم و رحمة هللا و بركاته 76

4. Linear Programming

4. Linear Programming /9/08 Systems Analysis in Construction CB Construction & Building Engineering Department- AASTMT by A h m e d E l h a k e e m & M o h a m e d S a i e d. Linear Programming Optimization Network Models -