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
Volumetric Analysis tutorial all along with the key concepts of Principle, Requirements, Procedure, Observations and Results, Treatment of Results, Experiment 2B
tutorsglobe.com properties of silver assignment help-homework help by online occurrence and principles of extraction of silver tutors
Zoogeography-Oriental and Australasian Regions tutorial all along with the key concepts of Fauna of Oriental Region, Fauna of Australasian Region and Fauna of whole Australasian Region
tutorsglobe.com proteins assignment help-homework help by online nutrition tutors
tutorsglobe.com choices and preferences of consumer assignment help-homework help by online demand and supply tutors
tutorsglobe.com properties of fluorine assignment help-homework help by online halogen family tutors
tutorsglobe.com properties of copper sulphate assignment help-homework help by online copper sulphate tutors
tutorsglobe.com chelates assignment help-homework help by online terminologies tutors
Theory and lecture notes of Simplified Database System all along with the key concepts of Simplified Database System, Illustration of Database with Conceptual Data Model, Example of a simple Database. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Simplified Database System.
www.tutorsglobe.com offers production possibility curve (ppc) homework help, assignment help, and economics solutions by online economics tutors.
tutorsglobe.com bacterial diseases assignment help-homework help by online bacteria-structure tutors
theory and lecture notes of static characteristics iii, all along with the key concepts of inverter design, operation and output characteristic, transfer characteristic and logic voltages. tutorsglobe offers homework help, assignment help and tutor’s assistance on static characteristics iii.
Work-in-progress cost audit program - the work-in-progress has been actually verified and that it agrees along with the balance in the incomplete cost cards.
tutorsglobe.com direct and indirect taxes assignment help-homework help by online types of tax tutors
Brief Survey of Microbes as Friends and Foes tutorial all along with the key concepts of Microorganisms as Friends, Microorganisms and Food Production, Production of Pharmaceuticals, Production of Organic Acids, Microorganisms and Agriculture and Se0wage Treatment
1949054
Questions Asked
3689
Tutors
1448284
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!