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International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1880
DEVELOPED METHOD FOR OPTIMAL SOLUTION OF TRANSPORTATION
PROBLEM
Poojan Davda1, Jaimin Patel2
1,2Scholar, Department of Mechanical Engineering, L.D. College of Engineering, Ahmedabad, Gujarat.
---------------------------------------------------------------------***----------------------------------------------------------------------
Abstract – Transportation model is a special case of linear
programming problem in which the main objective is to
transport homogeneous commodity from various origins to
different destinations at a total minimum cost. In this paper, a
new developed method is proposed to findanoptimalsolution,
the algorithm for proposed method gives solution in less
iteration compare to Vogel’s approximation method.
Key Words: Transportation Problem, Basic Feasible
Solution, Optimal Solution, Linear Programming
1. INTRODUCTION
Transportation is a logistical problem for organizations
especially for manufacturing and transport companies. This
method is a useful tool in decision-making process which
deals with the distributionof goodsfromseveral points,such
as factories often known as sources, to a number of points of
demand, such as warehouses, often known as destinations.
Transportation cost has significant impact on the cost and
the pricing of raw materials and goods, supplier try to
control the cost of transportation. The way how the
desirable minimum transportation cost can be obtained is
the subject matter of transportation problems in linear
programming. Some conventional methods to find the
minimum transportationcostareNorth WestCornerMethod
(NWCM), Least Cost Method(LCM), Vogel’s Approximation
Method(VAM) and to find the optimality we are using the
MODI method. In this paper, we are introducing a new
proposed method for finding an Initial Basic Feasible
Solution. Two numerical examples are providedtoprove the
claim with stepwise procedure of this new method.
1.1 MATHEMATICAL REPRESENTATION
In a transportation model which involves ‘m’ sources, each
having availability of Ai (i = 1,2,3,….,m) units of product, ‘n’
destinations, each having demand of Bj (j=1,2,3…,n) units of
that product. The cost Cij is the transportation cost. It is
associated with the transportation of on unit of productfrom
ith source to the jth destination for each i and j. The objective
is to determine the number of units to be transformed from
source i to destination j, so that the total transportation cost
is minimum.
The equivalent Linear Programming can be written as
follows: -
Minimize Z = Objective Function
Subject to, Constraints
Where, Non Negativity
Here, Xij is the number of units shipped from source i to
destination j.
2. ALGORITHM FOR PROPOSED METHOD
STEP1: Examine whether the transportation problem is
balanced or not. If it is balanced, then go to next step.
STEP2: Find ‘Second minimum value’ from each row as well
as column.
STEP3: Select the row or column from the matrix which has
maximum value of ‘Second minimum Value’, select the cell
which has minimum cost coefficient. Allocate smallest value
of demand or supply in the cell.
STEP4: Repeat step 2 and step 3 until 2*2 matrix. Apply
VAM’s Penalty approach in 2*2 matrix.
STEP5: Now total minimum cost is calculated as sum of the
product of cost and corresponding allocatevalueofsupplyor
demand.
IMPORTANT POINTS OF ALGORITHM: -
1) Here, ‘Second lowest value’ from row or column means if
2,3,4,5 and 2,2,3,5 given then 3 and 2 is Second lowest value
respectively.
2) In case of tie of maximum value of second minimum
between any two or more rows and columns then find
average of that rows and columns and allocate the cell
according to smallest value of supply or demand having
maximum value of average of row or column.
3) Here for 2*2 matrix apply VAM’spenaltyapproachmeans,
take row difference and column difference called penalty
value of each row and column, allocate cell in row or column
having maximum value of penalty with minimum cost
according to smallest value of supply or demand.
International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1881
2.1 NUMERICAL EXAMPLES
Example 1: - Illustrate
D1 D2 D3 Supply
S1 6 8 10 150
S2 7 11 11 175
S3 4 5 12 275
Demand 200 100 300
Solution: - Since ∑Ai = ∑Bi = 600, the given transportation
problem is balanced; therefore, exist a basic feasiblesolution
to Proposed Method problem.
D1 D2 D3 S R1 R2
S1 6 8 10(150) 150 8 -
11 11
5 5
S2 7(25) 11 11(150) 175
S3 4(175) 5(100) 12 275
D 200 100 300
C1
C2
6 8 11
7 11 12
The transportation cost is:
Z= 7*25 + 4*175 + 5*100 + 10*150 + 11*150 = 4525
Explanation: -
1) Here, in first step we have maximum value of ‘second
minimum value’ is 11, which appearsinsecondrowandthird
column. So, we will take average of second row and third
column and we got maximum value of average in third
column, so wewill give allocation to cell C13 of 150 unitsand
eliminate row1.
2) In second step there is maximum value of ‘second
minimum’ is 12 in column 3, so we will give allocation to cell
C23 of 150 units and eliminate column3.
3) In final step, for 2*2 matrix, we will take difference of row
and column, we get maximum value in column2, so we will
give allocation to cell C32 of 100 unitsand byfulfillingsupply
and demand constraints we will allocate cell C21 of 25 units
and cell C31 of 175 units.
Example 2: - Illustrate
D1 D2 D3 D4 Supply
S1 9 5 8 10 20
S2 11 17 7 12 15
S3 3 8 10 5 35
S4 13 6 9 2 40
Demand 30 45 20 35
Solution: - Since ∑Ai ≠ ∑Bi, hence adding dummy supply of
20 unitsso that ∑Ai = ∑Bi = 130 and the given transportation
problem is balanced; therefore, exist a basic feasiblesolution
to Proposed Method problem.
D1 D2 D3 D4 S R1 R2 R3 R4 R5
S1 9 5(20) 8 10 20 8 8 9 - -
11 - - - -
5 5 5 5 8
6 6 6 6 13
0 0 0 0 0
S2 11 17 7(15) 12 15
S3 3(30) 8(5) 10 5 35
S4 13 6(5) 9 2(35) 40
S5 0 0(15) 0(5) 0 20
D 30 45 20 35
C1
C2
C3
C4
C5
3 5 7 2
3 5 8 2
3 6 - 2
3 6 - 2
3 6 - -
The transportation cost is:
Z=3*30 + 5*20 + 8*5 + 6*5 + 0*15 + 7*15 + 0*5 + 2*35 =435
Explanation: -
1) Here, in first step we get maximum value of ‘second
minimum value’ of 11, so wewill give allocation to cellC23of
15 units and eliminate row2.
2)In second step, we have maximum value of ‘second
minimum value’ is 8, which appears in first row and third
column. So, wewill take averageoffirstrowandthirdcolumn
and wegot maximum value of average in third column,sowe
will give allocation to cell C53 of 5 units and eliminate
column3.
3)In third step, we get maximum value of ‘second minimum
value’ of 9, so we will give allocation to cell C12 of 20 units
and eliminate row1.
4)In fourth step, we have maximum value of ‘second
minimum value’ is 6, whichappearsinfourthrowandsecond
column. So, we will take average of fourth row and second
column and we got maximum value of average in fourth row,
so wewill give allocation to cell C44 of 35 units andeliminate
column4.
5) In fifth step, we get maximum value of ‘second minimum
value’ of 13, so we will give allocation to cell C42 of 5 units
and eliminate row4.
6) In final step, for 2*2 matrix, we will take difference of row
and column, we get maximum value in column2, so we will
give allocation to cell C52 of 15 units and by fulfilling supply
and demand constraints we will allocate cell C32 of 5 units
and cell C31 of 30 units.
International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072
© 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1882
2.2 COMPARISON OF THE NUMERICAL RESULTS
After proving the results by this proposed method, the
obtained result is compared with solution of other existing
methods.
Method Example 1 Example 2
Proposed Method 4525 435
North-west corner Method 5925 895
Least cost Method 4550 585
VAM 5125 455
MODI-Method 4525 435
3. CONCLUSION
The proposed method considers ‘second minimumvalue’of
each row and each column which is totally new concept. The
solution obtained by the proposed method is optimal with
less iteration compared to other methods. The developed
method is effective for both the large and small size
transportation problem.
REFERENCES
[1] Andersion.R.D, Sweeney.J.D, Williams.A.T & Martin.K.R
(2010): An introduction to Management Science:
Quantitative Approach to Decision.
[2] Hay. A (1977) Linear Programming: elementary
geographical applications ofthetransportationproblem
[3] Erlander S.B (2010) Cost-MinimizingChoiceBehaviorin
Transportation Planning: A Theoretical. Page 8-10
[4] Kumar. T, and Schilling S.M (2010). : Comparison of
optimization Technique in large scale Transportation
Problem
[5] Aguaviva.E(1997).: Quantitative TechniquesinDecision
Making 97 Ed. Pages 57-91
[6] Adlakha.V, Kowalski.k, Lev.B (2006).Solving
transportation problems with mixed constraints.
International journal of management Science and
Engineering management vol.1, pp.47-52
[7] Utdallas (2009). : Transportation and Related
Problems.http://www.utdallas.edu/”Scniu/OPRE-
6201/documents/A_transportation_Problems.xls
[8] Abdual Quddoos, Shakeel Javaid and M.M.Khalid, A New
Method for Finding an Optimal Solution for
Transportation Problems, International Journal on
Computer Science and Engineering, 4(7)(2012), 1271-
1274.
[9] G.B.Dantzig, Application of the simplex method to a
transportation problem, Activity Analysis of production
and allocation, (T.C. Koopmans ed.), Wiley, New York,
(1951), 359-373.
[10] V.J. Sudhakar, N. Arunnsankar, T. Karpagam, ‘A new
approach for find an Optimal Solution for Trasportation
Problems’, European Journal of Scientific Research 68
254-257, 2012.
[11] Neetu M.Sharma and Ashok P.Bhadane, An alternative
method to north west corner method for solving
transportation problem, International Journal for
Research in Engineering Application & Management,
1(12)(2016), 1-3
[12] Reena G.Patel and P.H.Bhathawala, An Innovative
Approach to Optimum Solution of a Transportation
Problem, International Journal ofInnovativeResearchin
Science, Engineering Technology, 5(4)(2016), 5695-
5700.
[13] H.S. Kasana and K.D. Kumar, ‘Introductory Operation
Research: Theory and Applications’, Springer PP,2004.
BIOGRAPHIES
Poojan Davda student at L.D.
college of Engineering,
Ahmedabad in B.E. final year.
Jaimin Patel student at L.D.
college of Engineering,
Ahmedabad in B.E. final year.

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IRJET- Developed Method for Optimal Solution of Transportation Problem

  • 1. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072 © 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1880 DEVELOPED METHOD FOR OPTIMAL SOLUTION OF TRANSPORTATION PROBLEM Poojan Davda1, Jaimin Patel2 1,2Scholar, Department of Mechanical Engineering, L.D. College of Engineering, Ahmedabad, Gujarat. ---------------------------------------------------------------------***---------------------------------------------------------------------- Abstract – Transportation model is a special case of linear programming problem in which the main objective is to transport homogeneous commodity from various origins to different destinations at a total minimum cost. In this paper, a new developed method is proposed to findanoptimalsolution, the algorithm for proposed method gives solution in less iteration compare to Vogel’s approximation method. Key Words: Transportation Problem, Basic Feasible Solution, Optimal Solution, Linear Programming 1. INTRODUCTION Transportation is a logistical problem for organizations especially for manufacturing and transport companies. This method is a useful tool in decision-making process which deals with the distributionof goodsfromseveral points,such as factories often known as sources, to a number of points of demand, such as warehouses, often known as destinations. Transportation cost has significant impact on the cost and the pricing of raw materials and goods, supplier try to control the cost of transportation. The way how the desirable minimum transportation cost can be obtained is the subject matter of transportation problems in linear programming. Some conventional methods to find the minimum transportationcostareNorth WestCornerMethod (NWCM), Least Cost Method(LCM), Vogel’s Approximation Method(VAM) and to find the optimality we are using the MODI method. In this paper, we are introducing a new proposed method for finding an Initial Basic Feasible Solution. Two numerical examples are providedtoprove the claim with stepwise procedure of this new method. 1.1 MATHEMATICAL REPRESENTATION In a transportation model which involves ‘m’ sources, each having availability of Ai (i = 1,2,3,….,m) units of product, ‘n’ destinations, each having demand of Bj (j=1,2,3…,n) units of that product. The cost Cij is the transportation cost. It is associated with the transportation of on unit of productfrom ith source to the jth destination for each i and j. The objective is to determine the number of units to be transformed from source i to destination j, so that the total transportation cost is minimum. The equivalent Linear Programming can be written as follows: - Minimize Z = Objective Function Subject to, Constraints Where, Non Negativity Here, Xij is the number of units shipped from source i to destination j. 2. ALGORITHM FOR PROPOSED METHOD STEP1: Examine whether the transportation problem is balanced or not. If it is balanced, then go to next step. STEP2: Find ‘Second minimum value’ from each row as well as column. STEP3: Select the row or column from the matrix which has maximum value of ‘Second minimum Value’, select the cell which has minimum cost coefficient. Allocate smallest value of demand or supply in the cell. STEP4: Repeat step 2 and step 3 until 2*2 matrix. Apply VAM’s Penalty approach in 2*2 matrix. STEP5: Now total minimum cost is calculated as sum of the product of cost and corresponding allocatevalueofsupplyor demand. IMPORTANT POINTS OF ALGORITHM: - 1) Here, ‘Second lowest value’ from row or column means if 2,3,4,5 and 2,2,3,5 given then 3 and 2 is Second lowest value respectively. 2) In case of tie of maximum value of second minimum between any two or more rows and columns then find average of that rows and columns and allocate the cell according to smallest value of supply or demand having maximum value of average of row or column. 3) Here for 2*2 matrix apply VAM’spenaltyapproachmeans, take row difference and column difference called penalty value of each row and column, allocate cell in row or column having maximum value of penalty with minimum cost according to smallest value of supply or demand.
  • 2. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072 © 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1881 2.1 NUMERICAL EXAMPLES Example 1: - Illustrate D1 D2 D3 Supply S1 6 8 10 150 S2 7 11 11 175 S3 4 5 12 275 Demand 200 100 300 Solution: - Since ∑Ai = ∑Bi = 600, the given transportation problem is balanced; therefore, exist a basic feasiblesolution to Proposed Method problem. D1 D2 D3 S R1 R2 S1 6 8 10(150) 150 8 - 11 11 5 5 S2 7(25) 11 11(150) 175 S3 4(175) 5(100) 12 275 D 200 100 300 C1 C2 6 8 11 7 11 12 The transportation cost is: Z= 7*25 + 4*175 + 5*100 + 10*150 + 11*150 = 4525 Explanation: - 1) Here, in first step we have maximum value of ‘second minimum value’ is 11, which appearsinsecondrowandthird column. So, we will take average of second row and third column and we got maximum value of average in third column, so wewill give allocation to cell C13 of 150 unitsand eliminate row1. 2) In second step there is maximum value of ‘second minimum’ is 12 in column 3, so we will give allocation to cell C23 of 150 units and eliminate column3. 3) In final step, for 2*2 matrix, we will take difference of row and column, we get maximum value in column2, so we will give allocation to cell C32 of 100 unitsand byfulfillingsupply and demand constraints we will allocate cell C21 of 25 units and cell C31 of 175 units. Example 2: - Illustrate D1 D2 D3 D4 Supply S1 9 5 8 10 20 S2 11 17 7 12 15 S3 3 8 10 5 35 S4 13 6 9 2 40 Demand 30 45 20 35 Solution: - Since ∑Ai ≠ ∑Bi, hence adding dummy supply of 20 unitsso that ∑Ai = ∑Bi = 130 and the given transportation problem is balanced; therefore, exist a basic feasiblesolution to Proposed Method problem. D1 D2 D3 D4 S R1 R2 R3 R4 R5 S1 9 5(20) 8 10 20 8 8 9 - - 11 - - - - 5 5 5 5 8 6 6 6 6 13 0 0 0 0 0 S2 11 17 7(15) 12 15 S3 3(30) 8(5) 10 5 35 S4 13 6(5) 9 2(35) 40 S5 0 0(15) 0(5) 0 20 D 30 45 20 35 C1 C2 C3 C4 C5 3 5 7 2 3 5 8 2 3 6 - 2 3 6 - 2 3 6 - - The transportation cost is: Z=3*30 + 5*20 + 8*5 + 6*5 + 0*15 + 7*15 + 0*5 + 2*35 =435 Explanation: - 1) Here, in first step we get maximum value of ‘second minimum value’ of 11, so wewill give allocation to cellC23of 15 units and eliminate row2. 2)In second step, we have maximum value of ‘second minimum value’ is 8, which appears in first row and third column. So, wewill take averageoffirstrowandthirdcolumn and wegot maximum value of average in third column,sowe will give allocation to cell C53 of 5 units and eliminate column3. 3)In third step, we get maximum value of ‘second minimum value’ of 9, so we will give allocation to cell C12 of 20 units and eliminate row1. 4)In fourth step, we have maximum value of ‘second minimum value’ is 6, whichappearsinfourthrowandsecond column. So, we will take average of fourth row and second column and we got maximum value of average in fourth row, so wewill give allocation to cell C44 of 35 units andeliminate column4. 5) In fifth step, we get maximum value of ‘second minimum value’ of 13, so we will give allocation to cell C42 of 5 units and eliminate row4. 6) In final step, for 2*2 matrix, we will take difference of row and column, we get maximum value in column2, so we will give allocation to cell C52 of 15 units and by fulfilling supply and demand constraints we will allocate cell C32 of 5 units and cell C31 of 30 units.
  • 3. International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072 © 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1882 2.2 COMPARISON OF THE NUMERICAL RESULTS After proving the results by this proposed method, the obtained result is compared with solution of other existing methods. Method Example 1 Example 2 Proposed Method 4525 435 North-west corner Method 5925 895 Least cost Method 4550 585 VAM 5125 455 MODI-Method 4525 435 3. CONCLUSION The proposed method considers ‘second minimumvalue’of each row and each column which is totally new concept. The solution obtained by the proposed method is optimal with less iteration compared to other methods. The developed method is effective for both the large and small size transportation problem. REFERENCES [1] Andersion.R.D, Sweeney.J.D, Williams.A.T & Martin.K.R (2010): An introduction to Management Science: Quantitative Approach to Decision. [2] Hay. A (1977) Linear Programming: elementary geographical applications ofthetransportationproblem [3] Erlander S.B (2010) Cost-MinimizingChoiceBehaviorin Transportation Planning: A Theoretical. Page 8-10 [4] Kumar. T, and Schilling S.M (2010). : Comparison of optimization Technique in large scale Transportation Problem [5] Aguaviva.E(1997).: Quantitative TechniquesinDecision Making 97 Ed. Pages 57-91 [6] Adlakha.V, Kowalski.k, Lev.B (2006).Solving transportation problems with mixed constraints. International journal of management Science and Engineering management vol.1, pp.47-52 [7] Utdallas (2009). : Transportation and Related Problems.http://www.utdallas.edu/”Scniu/OPRE- 6201/documents/A_transportation_Problems.xls [8] Abdual Quddoos, Shakeel Javaid and M.M.Khalid, A New Method for Finding an Optimal Solution for Transportation Problems, International Journal on Computer Science and Engineering, 4(7)(2012), 1271- 1274. [9] G.B.Dantzig, Application of the simplex method to a transportation problem, Activity Analysis of production and allocation, (T.C. Koopmans ed.), Wiley, New York, (1951), 359-373. [10] V.J. Sudhakar, N. Arunnsankar, T. Karpagam, ‘A new approach for find an Optimal Solution for Trasportation Problems’, European Journal of Scientific Research 68 254-257, 2012. [11] Neetu M.Sharma and Ashok P.Bhadane, An alternative method to north west corner method for solving transportation problem, International Journal for Research in Engineering Application & Management, 1(12)(2016), 1-3 [12] Reena G.Patel and P.H.Bhathawala, An Innovative Approach to Optimum Solution of a Transportation Problem, International Journal ofInnovativeResearchin Science, Engineering Technology, 5(4)(2016), 5695- 5700. [13] H.S. Kasana and K.D. Kumar, ‘Introductory Operation Research: Theory and Applications’, Springer PP,2004. BIOGRAPHIES Poojan Davda student at L.D. college of Engineering, Ahmedabad in B.E. final year. Jaimin Patel student at L.D. college of Engineering, Ahmedabad in B.E. final year.