Transportation Problem

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1 Transportation Problem. Production costs at factories F, F, F and F 4 are Rs.,, and respectively. The production capacities are 0, 70, 40 and 0 units respectively. Four stores S, S, S and S 4 have requirements of,, 0 and 0 units respectively. Using transportation cost matrix find the transportation plan that is optimal considering the production costs also. Solution: Factory Factory Stores S S S S 4 F 4 6 F F 9 F Stores S S S S 4 Supply F F F F Demand 0 0 Total demand = = 8 Total supply = = 0 Total demand Total supply. Therefore the problem is unbalanced. Since the total supply is more than the total demand add an extra column with zero entries and demand as the difference between their sums to make the problem balanced one Vogel s approximation method:

2 //0 (4) /()/(4 ) /() 70 /4 (8)/()/()/() 40 / (4)/(6)/()/() 0/0 (8)/()/()/() /0 /0 0/80 0/0 /0 () () ()/(0) (0)/(0) (0) Minimum cost = = 490 The number of basic cells = 8 = number of rows +number of columns. Therefore the solution is non degeneracy. Modi method: For basic cells u i + v j c ij = ε u = 0 u = u = u 4 = v = 4 v = v = 8 v 4 = v = For non basic cells

3 u + v c = = 4 < 0 u + v 4 c 4 = 0 + = 0 < 0 u + v c = 0 0 = < 0 u + v c = + 4 = 7 < 0 u + v c = + = 7 < 0 u + v 4 c 4 = + 8 = < 0 u + v c = = 8 < 0 u + v 4 c 4 = + = 8 < 0 u + v c = 0 = 0 u 4 + v c 4 = = 0 u 4 + v c 4 = + = 4 < 0 u 4 + v c 4 = 0 = > ε ε ε 0 ε u = 0 u = u = u 4 = v = 4 v = v = 8 v 4 = v = Put ε =, we get

4 ε u = 0 u = u = u 4 = v = 4 v = v = 8 v 4 = v = For non basic cells u + v c = = 4 < 0 u + v 4 c 4 = 0 + = 0 < 0 u + v c = 0 0 = < 0 u + v c = + 4 = 7 < 0 u + v c = + = 7 < 0 u + v 4 c 4 = + 8 = < 0 u + v c = 0 = < 0 u + v c = = 8 < 0 u + v 4 c 4 = + = 8 < 0 u + v c = 0 = 0 u 4 + v c 4 = = 0 u 4 + v c 4 = + = 4 < 0 Since u i + v j c ij 0, i, j. Optimum solution is reached. Minimum cost = = 4. Find the minimum cost distribution plan to satisfy demand for cement at three construction sites from available capacities at the three cement plants given the following transportation costs (in Rs) per ton of cement moved from plants to sites.

5 From Stores Capacity (tons/month) P P P Demand (tons/month) Solution: Total demand = = 700 Total capacity = = 900 Total demand Total capacity. Therefore the problem is unbalanced. Since the total capacity is more than the total demand add an extra column with zero entries and demand as the difference between their sums to make the problem balanced one Vogel s approximation method: /00/0 (00) (60)(6) 00/00 (0)(0)(0) 000/700 ()(0) (0) (4) (4) (4) () () (0)

6 Minimum cost = = The number of basic cells = 6 = number of rows +number of columns. Therefore the solution is non degeneracy. Modi method: For basic cells u i + v j c ij = u = 0 u = 0 u = 6 v = 00 v 4 = 0 v = 40 v = 60 For non basic cells u + v c = = 8 < 0 u + v 4 c 4 = = 0 > 0 u + v c = = 0 < 0 u + v c = = 0 < 0 u + v c = = 0 < 0 u + v 4 c 4 = = < ε ε ε 00 ε ε 700 ε 9 7 0

7 Put ε = 00, we get u = 0 u = 0 u = 6 v = 00 v 4 = 0 v = 40 v = 60 For non basic cells u + v c = = 8 < 0 u + v c = = 0 < 0 u + v c = = 0 < 0 u + v 4 c 4 = = 0 < 0 u + v c = = 0 < 0 u + v 4 c 4 = = 6 < 0 Since u i + v j c ij 0, i, j. Optimum solution is reached. Minimum cost = = A dairy firm has three plants located in a state. The daily production at each plant is as follow: Plant : 6 million litres, Plant : million litre and plant : 0 million litres. Each day, the firm must fulfill the needs of its four distribution centers. Minimum requirement at each center is as follows. Distribution center : 7 million litres, Distribution center : million litres, Distribution center : million litres, and Distribution center 4: million litres. Cost in hundreds of rupees of shipping one million litre from each plant to each distribution centre is given in the following table:

8 Distribution Center D D D D 4 Plant P 7 P 0 6 P 8 9 Determine an optimal solution to minimize the total cost. Solution: Distribution Center Capacity D D D D 4 Plant P 7 6 P 0 6 P Requirement 7 Since the sum of capacities is equal to the sum of requirements then the problem is balanced. Vogel s approximation method: //0 () () /0 ()(0) 0 ()(4) 7 6 () () 0 () () () (4) ()

9 Minimum cost = = 0 The number of basic cells = 6 = number of rows +number of columns. Therefore the solution is non degeneracy. Modi method: For basic cells u i + v j c ij = u = 0 u = u = v = v = v = v 4 = 6 For non basic cells u + v c = 0 + = > 0 u + v 4 c 4 = = < 0 u + v c = + = 4 < 0 u + v c = + 0 = < 0 u + v c = + 6 = > 0 u + v c = + 8 = < 0

10 ε ε ε ε 8 9 Put ε =, we get u = 0 u = 4 u = 4 v = v = v = v 4 = For non basic cells u + v c = 0 + = < 0 u + v 4 c 4 = = < 0 u + v c = 4 + = 4 < 0 u + v c = = < 0 u + v c = = > 0 u + v c = = < 0

11 7 ε ε ε + ε 8 9 Put ε =, we get u = 0 u = u = 4 v = v = v = v 4 = For non basic cells u + v c = 0 + = < 0 u + v 4 c 4 = = < 0 u + v c = + = < 0 u + v c = + 0 = < 0 u + v 4 c 4 = + = < 0 u + v c = = < 0 Since u i + v j c ij 0, i, j. Optimum solution is reached. Minimum cost = = 00

12 4. A company has factories at F, F and F which supply to warehouses at W, W and W. Weekly factory capacities are 00, 60 and 90 units, respectively. Weekly warehouse requirement are 80, 0 and 0 units, respectively. Unit shipping costs (in rupees) are as follows: W W W Factories F 6 0 F F Determine optimal distribution for this company to minimize total shipping cost. Solution: W W W Capacities F F F Requirement Since the sum of capacities is equal to the sum of requirements then the problem is balanced. Vogel s approximation method: /40 (4) (4) 60 /40 (6) (4) 90/0 (8) (0) 80 ()/() 0/ 0 () 0/ 60/0 (4)/ (6)

13 Minimum cost = = 90 The number of basic cells = = number of rows +number of columns Therefore the solution is non degeneracy. Modi method: For basic cells u i + v j c ij = u = 0 u = u = 4 v = 6 v = 0 v = For non basic cells u + v c = = 0 < 0 u + v c = + 8 = 8 < 0 u + v c = = 6 < 0 u + v c = = 0 < 0 Since u i + v j c ij 0, i, j. Optimum solution is reached. Minimum cost = = 90. Solve the transportation problem using Vogel s approximation method and check its optimality using MODI method. From To Availability A B C I II III Requirements 4 Solution: Since the sum of Availability is equal to the sum of requirements the problem is balanced.

14 Vogel s approximation method: (0) (0) (4) (4) 4//0 (0) (0) (40) () (0) Minimum cost = = 80 The number of basic cells = 4 < number of rows +number of columns Therefore degeneracy occurs. To make the solution has non degeneracy select the minimum value in the non basic cell and make it as basic cell with 0 value entered in the box The number of basic cells = =number of rows +number of columns Therefore the solution is non degeneracy. Modi method:

15 For basic cells u i + v j c ij = u = 0 u = 40 u = 70 v = 0 v = 0 v = 0 For non basic cells u + v c = = 40 < 0 u + v c = = > 0 u + v c = = 0 < 0 u + v c = = 0 < 0 + ε 0 ε ε ε Put ε = 0, we get

16 0 0 0 u = u = 40 u = 9 v = 0 v = v = 4 For non basic cells u + v c = = > 0 u + v c = = 6 < 0 u + v c = = 7 < 0 u + v c = = < 0 Since u i + v j c ij 0, i, j. Optimum solution is reached. Minimum cost = = Consider the following transportation problem: Factory Godowns 4 6 Stock Available A B C D Demand It is not possible to transport any quantity from factory B to godown. Determine: (i) Initial solution by Vogel s Approximation method. (ii) Optimal basic feasible solution. (iii) Is the optimal solution unique? If not find the alternative optimum basic feasible solution. Solution: Since the total stock availability is equal to the total demand, the problem is balanced.

17 Vogel s approximation method: /40 /0 () /()/() /(4) 0 /0 ()/()/()/()/() 90 /70 /0 /0 (0)/(0)/(4)/()/() ()/()/(0)/(0)/(0) 60 0/0 0/0 40/0 40/0 ()/(0) () 40/0/0 (4) () () (0) (0)/(0) () The number of basic cells = 8 < number of rows +number of columns Therefore degeneracy occurs. To make the solution has non degeneracy select the minimum value in the non basic cell and make it as basic cell with 0 value entered in the box

18 Minimum cost = Modi method = u = 0 u = u = u 4 = v = v = v = 9 v 4 = v = v 6 = For non basic cells u + v c = = > 0 u + v c = = > 0 u + v 4 c 4 = = < 0 u + v c = + = 9 < 0 u + v 4 c 4 = + = 9 < 0 u + v c = + = < 0 u + v 6 c 6 = + = < 0 u + v c = + = < 0 u + v c = + 0 = 8 < 0 u + v 6 c 6 = + 8 = 8 < 0 u 4 + v c 4 = + 0 = 8 < 0 u 4 + v c 4 = = < 0 u 4 + v 4 c 44 = + 6 = 4 < 0 u 4 + v c 4 = + 9 = 7 < 0 u 4 + v 6 c 46 = + = < 0

19 ε 0 0 ε ε 0 + ε ε ε Put ε = u = 0 u = u = u 4 = v = 7 v = v = 4 v 4 = 0 v = 0 v 6 = For non basic cells u + v c = = < 0 u + v 4 c 4 = = 7 < 0 u + v c = = < 0 u + v c = + = 4 < 0 u + v 4 c 4 = + 0 = 9 < 0 u + v c = + 0 = < 0 u + v 6 c 6 = + = 0

20 u + v c = + 7 = < 0 u + v c = + 0 = < 0 u + v 6 c 6 = + 8 = < 0 u 4 + v c 4 = + 0 = < 0 u 4 + v c 4 = = < 0 u 4 + v 4 c 44 = = 4 < 0 u 4 + v c 4 = = 7 < 0 u 4 + v 6 c 46 = + = 7 < 0 Since u i + v j c ij 0, i, j. Optimum solution is reached. Minimum cost = = 0 7. The Burgarf Co. distributes its products from ware houses to major metropolitan areas. Next months anticipated demand exceeds the warehouse supplies and the management of Burgraf would like to know how to distribute their product in order to maximize the revenue. Their distribution costs per unit are shown in the table. To be competitive Burgraf must charge a different price in different cities. Selling prices are Rs. 9.9, Rs. 0., Rs. 9., Rs.., Rs. 0.9 in cities,,, 4 and. From Warehouse 4 Supply Demand Find the optimal allocation to maximize the revenue. Solution: From Warehouse 4 Supply Demand Total demand = = 400 Total supply = = 800 Total demand Total supply. Therefore the problem is unbalanced.

21 Since the total supply is less than the total demand add an extra row with zero entries and demand as the difference between their sums to make the problem balanced one Convert the maximization problem into a minimization problem by subtracting all the values from the highest value (0.9)/(0.0) 000 /400/00/0 (. ) /(0.0) /(0.0)/(0.6) 600 (0) /(0) /(0) 00/ /0 800 (0.) (8.) (0.4) (8.4) (0.) (7.7) (0.) (0.6)

22 u = 0 u = 0. u = 8.6 v =. v =. v =. v 4 = 0 v = 0.9 For non basic cells u + v c = = 0. < 0 u + v c = = 0. > 0 u + v c = =. < 0 u + v c = = 0.6 < 0 u + v c = = 0.4 < 0 u + v c = = 0.0 < 0 u + v 4 c 4 = =. < 0 u + v 6 c 6 = = 0.4 < 0 ε 0 ε ε 600 +ε Put ε = 0

23 u = 0 u = 0.4 u = 8.8 v = 0.9 v = 0.9 v = 0.9 v 4 = 0. v = 0.9 For non basic cells u + v c = = 0. < 0 u + v c = =. < 0 u + v 4 c 4 = = 0. < 0 u + v c = = 0.6 < 0 u + v c = = 0. < 0 u + v c = = 0.0 < 0 u + v 4 c 4 = =. < 0 u + v 6 c 6 = = 0.04 < 0 Since u i + v j c ij 0, i, j. Optimum solution is reached Maximum cost = = 6

24 Assignment Problem. At the end of a cycle schedule a trucking firm has a surplus of one variable in each of the cities,,, 4 and and a deficit of one vehicle on each of the cities A, B, C, D, E and F. The costs in rupees of transportation and handling between the cities are given below. Find the assignment of surplus, vehicles to deficit cities which results in minimum total cost. Also find out which city will not receive a vehicle? A B C D E F Solution: Since the number of rows is less than the number of column, the problem is unbalanced. To make it balance (i.e., number of rows =number of columns) add a row with zero entries. A B C D E F Row reduction: A B C D E F Number of lines drawn Number of rows. Optimality condition is not satisfied.

25 A B C D E F Number of lines drawn = Number of rows. Therefore Optimality condition is satisfied. B, F, C, 4 A, D, 6 E Minimum cost = = 448. A company is faced with the problem of assigning 4 machines to different jobs (one machine to one job only). The profits are estimated as follows. Solve the problem to maximize the total profits. Job Machine A B C D Solution: Since the number of rows is more than the number of columns, the problem is unbalanced. To make it balance (i.e., number of rows =number of columns) add two columns with zero entries. 4 6 A B C D E F Converting maximization problem into a minimization problem by subtracting each value of the row from its highest element in the row.

26 4 6 A B C D E F Column reduction: 4 6 A B C D E F Number of lines drawn Number of rows. Optimality condition is not satisfied. 4 6 A B C D E F Number of lines drawn = Number of rows. Therefore Optimality condition is satisfied. E, A, B, 4 D, F, 6 C Maximum cost = = 8. A computer centre has three expert programmers. The head of the centre wants to allocate projects to programmers. The development time in hours is given in the table below. Allocate the projects to the programmers, in-order to minimize the development time. Projects Programmers A B C Solution:

27 A B C Since the number of rows is equal to the number of columns, the problem is balanced. Row reduction: A B C Column reduction: A B C Number of lines drawn = Number of rows. Therefore Optimality condition is satisfied. A, B, C A Maximum cost = = A department has five employees with five jobs to be performed. The time (in hours) each mean will take to perform each job is given in the effectiveness matrix. Employees I II III IV V A 0 6 Jobs B C 0 7 D E Solution: Since the number of rows is equal to the number of columns, the problem is balanced. Row reduction:

28 A B C D E I II III IV V Number of lines drawn Number of rows. Optimality condition is not satisfied. A B C D E I II III IV V Number of lines drawn = Number of rows. Therefore Optimality condition is satisfied. A II, B I, C V, D III, E IV Minimum cost = =. Solve the assignment problem. Machine P Q R S A B C D Solution: Since the number of rows is equal to the number of columns, the problem is balanced. Row reduction:

29 A B C D P Q R S Column reduction: A B C D P Q R S Number of lines drawn Number of rows. Optimality condition is not satisfied. A B C D P Q R S Number of lines drawn Number of rows. Optimality condition is not satisfied. A B C D P Q R S Number of lines drawn = Number of rows. Therefore Optimality condition is satisfied. A R, B P, C Q, D S Minimum cost = = 9

30 6. A company has five jobs to be done. The following matrix shows the return in rupees of assigning i th machine i =,,, 4, to the j th job j = A, B, C, D, E. Assign the five jobs to the five machines to maximize profit. A B C D E Solution: Since the number of rows is equal to the number of columns, the problem is balanced. Row reduction: 4 A B C D E Column reduction: 4 A B C D E Number of lines drawn Number of rows. Optimality condition is not satisfied. 4 A B C D E Number of lines drawn = Number of rows. Therefore Optimality condition is satisfied. E, D, A, 4 C, B

31 Minimum cost = = Travelling Salesman Problem. A delivery truck must leave a warehouse and visit four customer delivery points and return to the warehouse. Find out the best route where ( represents the warehouse). Distance (Miles) To I From Solution: Row reduction: Number of lines drawn = Number of rows. Therefore Optimality condition is satisfied. 4, Minimum cost = = 9 and 4 = + = 8 and 4 = + = 4 4 and = + 4 = 6 4 and = = 8

32 and = + = 8 and = + 7 = Since and 4 give the minimum extra amount we select this route. 4 Minimum cost = =. A salesman has to travel five cities in a month without revisiting any city. Determine an optimal travel plan for the time matrix (in hours) given. To City A B C D E A From City B C D 7-7 E A B C D E A B C D E Row reduction: A B C D E A B C D E Column reduction:

33 A B C D E A B C D E Number of lines drawn Number of rows. Optimality condition is not satisfied. A B C D E A B C D E Number of lines drawn = Number of rows. Therefore Optimality condition is satisfied. A B D A, C E C Minimum cost = = 0 AC and EB = + 0 = AE and CB = + 0 = BC and ED = + 4 = BE and CD = + = DC and EA = + 0 = DE and CA = + = 6 Since AE and CB give the minimum extra amount we select this route. A E C B D A Minimum cost = =

The Transportation Problem

CHAPTER 12 The Transportation Problem Basic Concepts 1. Transportation Problem: BASIC CONCEPTS AND FORMULA This type of problem deals with optimization of transportation cost in a distribution scenario