Transportation Algorithm for Minimization Problem (MODI Method)
Step 1
Make the transportation table entering the origin capacities ai, the cost cij and destination requirement bj
Step 2
Determine an initial basic feasible solution by vogel's technique or by any of the known method.
Step 3
For all the basic variables xij, solve the system of equations ui + vj = cij, for all i, j for which cell (i, j) is in the basis, starting at first with some ui = 0, compute the values of ui and vj on the transportation table
Step 4
Calculate the cost differences dij = cij - ( ui + vj ) for all the non-basic cells
Step 5
Apply optimality test by testing the sign of each dij
Step 6
Assume the variable xrs enter the basis. Assign an unknown quantity ? to the cell (r, s). Then make a loop that begins and ends at the cell (r, s) and links some of the basic cells. The amount ? is added to and subtracted from the transition cells of the loop in such a way that the availabilities and necessities remain fulfilled.
Step 7
Allocate the largest possible value to the ? in such a manner that the value of at least one basic variable comes out to be zero and the other basic variables remain non-negative. The basic cell whose allotment has been made zero will leave the basis.
Step 8
Now, go back to step 3 and repeat the process until an optimal solution is achieved.
Worked Examples
Illustration 1
Determine an optimal solution
W1
W2
W3
W4
Availability
F1
19
30
50
10
7
F2
70
40
60
9
F3
8
20
18
Requirement
5
14
Answer
1. Put vogel's approximation method for determining the initial basic feasible solution
Penalty
5(19)
(30)
(50)
2(10)
X
(70)
7(40)
2(60)
(40)
8(8)
10(20)
Minimum transportation cost comes out to be 5 (19) + 2 (10) + 7 (40) + 2 (60) + 8 (8) + 10 (20) = Rs. 779
2. Test for Non-degeneracy
The initial fundamental feasible solution has m + n - 1 i.e. 3 + 4 - 1 = 6 allocations in independent positions. Therefore optimality test is fulfilled.
3. Computation of ui and vj : - ui + vj = cij
ui
u1= -10
u2 = 40
u3 = 0
vj
v1 = 29
v2 = 8
v3 = 0
v4 = 20
Allocate a 'u' value to zero. (Convenient rule is to choose the ui, which has the maximum number of allocations in its row)
Assume u3 = 0, then
u3 + v4= 20 which means 0 + v4 = 20, so v4 = 20
u2 + v4= 60 which means u2 + 20 = 60, so u2 = 40
u1 + v4= 10 which means u1 + 20 = 10, so u1 = -10
u2 + v3= 40 which means 40 + v3 = 40, so v3 = 0
u3 + v2= 8 which means 0 + v2 = 8, so v2 = 8
u1 + v1= 19 which means -10 + v1= 19, so v1 = 29
4. Computation of cost differences for non basic cells dij = cij - ( ui + vj )
cij
ui + vj
-2
-10
69
48
29
0
dij = cij - ( ui + vj )
32
1
-18
11
5. Doing Optimality test
dij < 0 i.e. d22 = -18
so x22 is entering the basis
6. Creation of loop and allotment of unknown quantity ?
We assign ? to the cell (2, 2). Reallocation is done by transferring the maximum possible amount ? in the marked cell. The value of ? is achieved by equating to zero to the corners of the closed loop. That is min (8-?, 2-?) = 0 which gives ? = 2. Hence x24 is outgoing as it turns out to be zero.
5 (19)
2 (10)
2 (30)
7 (40)
6 (8)
12 (20)
Minimum transportation cost comes out to be 5 (19) + 2 (10) + 2 (30) + 7 (40) + 6 (8) + 12 (20) = Rs. 743
7. Enhanced Solution
u2 = 22
v3 = 18
(60)
51
42
52
As dij > 0, an optimal solution is achieved with minimal cost of Rs.743
Cell Growth tutorial all along with the key concepts of Cellular Differentiation, Cell Turnover, Renewing Cells, Cell Number, Cell Populations, Cell Size Regulation in Mammals, Types of Cell Division, Cellular Growth Disorders
Theory and lecture notes of Building Models all along with the key concepts of building models, essence of model-building, homework help, assignment help, understanding macroeconomics. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Building Models.
The term overhead involves factory overheads, office overheads and selling and distribution overheads.
Tissue system is a group of tissues that performing an identical function irrespective of its position in the plant body.
tutorsglobe.com somatic embryogenesis assignment help-homework help by online micropropagation tutors
theory and lecture notes of static characteristics i, all along with the key concepts of source amplifier, operation as an amplifier, operation as a switch and logic voltages. tutorsglobe offers homework help, assignment help and tutor’s assistance on static characteristics i of mos transistor.
Chemical properties of Carbohydrates tutorial all along with the key concepts of Reactions involving Carbohydrates, Osazone Formation, Chain Shortening and Lengthening and Formation of Glycosides
Proper Care should be taken when checking the biasing voltage of old CRT that should be appropriate to the new CRT.
www.tutorsglobe.com offers addition reactions of dienes homework help, addition reactions of dienes assignment help, online tutoring assistance, organic chemistry solutions by online qualified tutor's help.
tutorsglobe.com anatomical adaptations assignment help-homework help by online xeric habitats characterization tutors
Common-sized financial statements are general financial statements (like the income statement, statement of financial position and statement of cash flows) that are expressed in terms of a number of base figures.
tutorsglobe.com interphase assignment help-homework help by online mitotic cell cycle tutors
tutorsglobe.com hmp shunt pathway assignment help-homework help by online carbohydrate metabolism tutors
www.tutorsglobe.com offers entropy and free energy homework help, entropy and free energy assignment help, online tutoring assistance, physical chemistry solutions by online qualified tutor's help.
in the dark ages earlier than cell phones, people who actually required mobile- communications capability installed radio telephones in their cars.
1948055
Questions Asked
3689
Tutors
1447820
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!