ABSTRACT
In the current scenario, handling the cost of transportation has taken into major consideration when deciding the cost structure of an organization. The transportation models have developed during the Second World War & evaluated continuously. Through this study, the student is going to discuss what are the three methods, which can be used to find the initial basic solution. They are namely, “North West Corner, Least Cost & Vogel’s Approximation Method”. Since the cost is higher, initial solution is not the ultimate solution. Therefore, it is compulsory to find the optimal solution. There are two methods for that such as “Stepping Stone Method & Modified Distribution Method”. We assume, the demand of the destination is equal to the supply of the sources. However, the actual scenario is completed difference because there are some cases; the demand is not equal to the supply we call that as “Un Balanced Problem” that can be solved by adding dummy row or column due to the requirement. All the things are clearly described by the support of examples in this study. Each & every business organizations can use these transportation models to reduce the cost of transportation, which directly influence to attain the desired outcomes of the organization such as profit maximization, cost minimization etc.
CHAPTER ONE
INTRODUCTION
1.0. Introduction
This is the chapter which is going to provide a basic introduction about “Transportation Models” with some words about how the transportation models came to world & the major steps which passes with the time. In the latter section, the student aims to discuss narrowly why this concepts & techniques are significance to the world.
1.1. History
Its origins can be traced to “Operations Research” which made its debut during World War II when the Allied forces recruited scientists of various disciplines to assist with military operations. In these early applications, the scientists utilized simple mathematical models to make efficient use of limited technologies & resources. Transportation Modelling is also one of major applications which has evaluated as part of Management Science.
1.2. Definition
This is the technique that uses to distribute the goods & services effectively by reducing the cost of transportation. In other word, the optimal way of procuring the goods from several supply centres to the number of demand centres. Not only that, this is one of major technique which uses to take decisions in management science. This is not only use of minimizing problems but also use for maximizing objectives.
1.3. Significance
- A distribution type problem in which supplies of goods that are held at various locations are to be distributed to other receiving locations.
- The solution of a transportation problem will indicate to a manager the quantities & costs of various routes & the resulting minimum cost.
- Used to compare location alternatives in deciding where to locate factories & warehouses to achieve the minimum cost distribution configurations
CHAPTER TWO
TECHNIQUES
In the second chapter, the student aims to describe the way of solving transportation model by finding final optimal solution depend on the initial basic solution.
2.1. Solution of the Transportation Model
Normally, there are two ways to solve a transportation problem. They are,
- Simplex Method
- Operational Research Method
Simplex Method can be used to transportation problems like how it uses for Linear Programming. In here, very firstly need to build a “Linear Programming Model”. After that, the information of which relate to linear model can be include into simplex table. Then, the problem need to be solved as the way of solving normal liner programming problem.
Example 01
Naman Private Limited is a rice producing company. They have three production factories which locate in three different areas. These daily production of this factories is 100kg, 250kg & 125 kg respectively. This production should be distributed to three whole sellers who live in three different areas. The requirement of them is 175kg, 200 kg & 100kg respectively. The unit cost which need to spend to transport one rice 1kg from one supply location to one whole seller is shown by following table.
Table. 2.1. Unit Cost of Rice Transportation
Whole Sellers
| Factory | Gampaha | Dehiwala | Galle |
| Maradagahamula | 10.00 | 25.00 | 53.00 |
| Polonnaruwa | 35.00 | 48.00 | 75.00 |
| Dambulla | 22.00 | 37.00 | 64.00 |
Source: Developed by the Student
A linear programming model can develop as follows to minimize the transportation cost.
Xij = The quantity of rice which distributes to three whole sellers (j= 1,2,3) by three factories
(i= 1,2,3).
Minimize Z = 10x11 + 25x12 +53x13 + 35x21 + 48x22 + 75x23 + 22x31 + 37x12 + 64x33
Subject to,
X11 + X12 + X13 = 100
Supply Constraint
X21 + X22 + X23 = 250
X11 + X21 + X31 = 175
Demand Constraint
X12 + X22 + X32 = 200
X13 + X23 + X33 = 100
All Xij >= 0
According to above, to take a solution for transport problem through simplex method, very firstly it is required to develop a liner programming model. After that, the solution can be taken in normal way. Even though, there are large number of constraints & decision variables. Therefore, it will take more time to reach for the optimal solution. This draw back of the simplex method has caused to use “Operational Research Solution Method” to take the answer conveniently.
2.2.2 Operational Research Solution Method
It is very easy to get the solution through this method even there are some steps which need to follow up. The process is shown below by transportation algorithm.
2.2.2.1. Finding the Basic Initial Solution
The basic solution should find after developing of a model for the problem. There is one of compulsory requirement which need to find the basic solution. That is, the demand & supply constraint should be completed. Therefore, before finding the initial basic solution it should be confirm whether the demand & supply quantities are equal. If there is a un-balance in these quantities, it should be added into a dummy column or raw.
Even there are number of methods to find the basic solutions there are three popular techniques usually follow up to reach to the initial basic solution. Such as,
- North West Corner Method
- Least Cost Method
- Vogel’s Approximation Method
- Row Minima Method
- Column Minima Method
2.2.2.1.1. North West Corner Method
In this method, the solution should start from the east west corner cell of the table without considering about the cost. To further understanding, let’s find the basic solution for the very first example as follows,
Table 2.2. The Initial Basic Solution by North West Corner Method
| W/Sellers
Factory |
Gampaha | Dehiwala | Galle | Supply |
| Maradagahamula | 100 | – | – | 100 |
| Polonnaruwa | 75 | 175 | – | 250 |
| Dambulla | – | 25 | 100 | 125 |
| Demand | 175 | 200 | 100 | 475 |
After developing the table by including certain quantities due to North West Corner Method, the total transportation cost can be calculated as follows,
Table 2.3. Total Transportation Cost to North West Corner Method
| Factory | Whole Seller | Transportation Cost (Rs) | Total Cost (Rs) |
| Maradagahamula | Gampaha | 100*10.00 | 1000.00 |
| Polonnaruwa | Gampaha | 75*35.00 | 2625.00 |
| Polonnaruwa | Dehiwala | 175*48.00 | 8400.00 |
| Dambulla | Dehiwala | 25*37.00 | 925.00 |
| Dambulla | Galle | 100*64. | 6400.00 |
| Grand Total | 19350.00 |
Source: Developed by the Student
Even though it’s very convenient to reach the basic solution through North West Corner Method, there are some major weakness that hinder the path as follows,
- Totally ignore the cost which eventually cause for the missing out of the best cost-effective path.
- When the demand & supply are not equal, the problem can be resolved by introducing dummy source or destination where in the aim of transportation model is minimization of cost so introducing a dummy source is not a good solution.
This technique gives a special consideration for reducing the cost. At the beginning, need to identify the lowest cost cell of the table, then a maximum quantity of goods should be distributed to that cell without violating the demand & supply quantities. The maximum amount is the lowest quantity from the demand & supply. After that, the unusable cells should be removed from the solution & the second attention should be given in to the next lowest cost cell by considering remaining unit cost figures. The method can be further understanding by looking at the example in below,
Table 2.4. The Initial Basic Solution by Least Cost Method
| W/Sellers
Factory |
Gampaha | Dehiwala | Galle | Supply |
| Maradagahamula | 100 | – | – | 100 |
| Polonnaruwa | – | 150 | 100 | 250 |
| Dambulla | 75 | 50 | – | 125 |
| Demand | 175 | 200 | 100 | 475 |
Source: Developed by the Student
Table 2.5. Total Transportation Cost to Least Cost Method
| Factory | Whole Seller | Transportation Cost (Rs) | Total Cost (Rs) |
| Maradagahamula | Gampaha | 100*10.00 | 1000.00 |
| Polonnaruwa | Dehiwala | 150*48.00 | 7200.00 |
| Dambulla | Gampaha | 75*22.00 | 1650.00 |
| Dambulla | Dehiwala | 50*37.00 | 1850.00 |
| Grand Total | 11700.00 |
Source: Developed by the Student
When comparing to the total transportation cost of both models, the cost which taken by Least Cost Method is lower by Rs.7650.00 than North West Corner Method. Even Least Cost Method is better than North West Corner Method it is not the best one. Because of following reasons,
- This method does not follow step by step rule for obtaining the optimal solution
- This method is based on the selection through personnel observation when three is tie in the minimum cost it does not follow any systematic rule.
2.2.2.1.3. Vogel’s Approximation Method (VAM)
Penalty Cost is taken into consideration in this method that means, the cost which has to incur because of not transporting to the lowest cost destination. Very firstly, need to calculate the penalty cost which relates to each raw & column. The formula of the calculation of penalty cost as follows,
Penalty Cost = Second Lowest Unit Cost – The Lowest Unit Cost
After all, the first distribution need to be done to the location which has the highest penalty cost. Then the same method can be followed up to the remain rows & columns. The method can be further understood by paying the attention to solve Naman Trading example as follows,
Table 2.6. The Initial Basic Solution by Vogel’s Approximation Method
| Penalty Cost Table | |||||
| Step | Penalty Cost | Largest Penalty Cost | Allocation | Elimination | |
| Row
1 2 3 |
Column
1 2 3 |
||||
| 1 | 15 13 15 | 12 12 11 | 15, col 1 | X11 =100 | Row 1 |
| 2 | 0 13 15 | 13 11 11 | 15, col 2 | X31 = 75 | Col 1 |
| 3 | 0 27 27 | 0 11 11 | 27, col 2 | X22 = 200 | Col 2 |
| 4 | 0 75 64 | 0 0 11 | 75, col 3 | X23 = 50 | Row 2 |
| 5 | 0 0 64 | 0 0 64 | 64, col 3 | X33 = 50 | Col 3 |
Source: Developed by the Student
Consideration of the penalty cost & the steps involved in applying Vogel’s Approximation Method are shown Table 2.6. In the table, step 1 row shows the penalty cost for the 3 rows & 3 columns & the largest penalty cost is 15 for column 1. Therefore, the first allocation is for the least cost cell in column 1, namely for cell (1,1). X11=min (S1 =100, D1 =175) =100. This eliminates row 1 & the demand in column 1 is reduced to 75. According to this way, remaining steps have followed till satisfy the both demand & supply constraint together. The final allocation for each destination are graphically represented by Table 2.8 as follows,
Table 2.7. The Initial Basic Solution by Vogel’s Approximation Method
| W/Sellers
Factory |
Gampaha | Dehiwala | Galle | Supply |
| Maradagahamula | 100 | – | – | 100 |
| Polonnaruwa | – | 200 | 50 | 250 |
| Dambulla | 75 | 50 | – | 125 |
| Demand | 175 | 200 | 100 | 475 |
Source: Developed by the Student
Table 2.8. Total Transportation Cost to Vogel’s Approximation Method
| Factory | Whole Seller | Transportation Cost (Rs) | Total Cost (Rs) |
| Maradagahamula | Gampaha | 100*10.00 | 1000.00 |
| Polonnaruwa | Dehiwala | 200*48.00 | 9600.00 |
| Polonnaruwa | Galle | 50*75.00 | 3750.00 |
| Dambulla | Gampaha | 75*22.00 | 1650.00 |
| Dambulla | Galle | 50*64.00 | 3200.00 |
| Grand Total | 19200.00 |
Source: Developed by the Student
Even though, Vogel’s Approximation Method is very systematic & absorb a lesser time in solving transportation problem with less computation there are some major draw backs which hinder the path,
- Provide an approximation solution to the given problem
- The method is tedious when the given matrix is large one.
2.3.1. Find the Optimal Solution
Finding an initial basic solution is not the end result for transportation problems because it needs optimal solution. The objective of finding final solution is trying to reduce the transport cost further more by changing the way of distribution. Firstly, need to clarify that whether the basic requirement is fulfilled or not which is M+N-1. In other words, the value of subtraction one from the sum of rows & columns of a transportation table need to be equal to the number of occupied cells. The word “Occupied “means that, the cells which includes in to the distribution process where as “Un Occupied” means the cells which not include into a certain distribution process. There are two basic methods which can be used to find the optimal solution such as,
- Stepping Stone Method
- Modified Distribution Method
2.3.1.1. Stepping Stone Method
This is a developed method to reduce the cost furthermore by distribution to the unoccupied cells. There are some certain steps which need to follow up as follows,
- The distribution can be done to unoccupied cell which locate in any side from an occupied cell.
- The distribution can’t be done from un occupied cell to another un occupied cell
- When distributing, it is okay to escape one cell location
- There are should be only one path for distribution to one location
Example 02
Sanjeewa Private Limited Company has three stores & there are three factories to provide goods for these stores. The initial basic solution has already found by using North West Corner Method as follows,
Step 01- Finding Initial Basic Solution
Table 2.9. The Initial Basic Solution by North West Corner Method
| Stores
Factory |
1 | 2 | 3 | Supply |
| A | 150 | – | – | 150 |
| B | 50 | 50 | – | 100 |
| C | – | 70 | 180 | 250 |
| Demand | 200 | 120 | 180 | 500 |
Source: Developed by the Student
Table 2.10. Total Transportation Cost to North West Corner Method
| Factory | Stores | Transportation Cost (Rs) | Total Cost (Rs) |
| A | 1 | 150*20.00 | 3000.00 |
| B | 1 | 50*6.00 | 300.00 |
| B | 2 | 50*10.00 | 500.00 |
| C | 2 | 70*15.00 | 1050.00 |
| C | 3 | 180*12.00 | 2160.00 |
| Grand Total | 7010.00 |
Source: Developed by the Student
Step 02- Verification of the Rule of M+N-1
Number of Rows + Number of Columns -1 =Number of Occupied Cells
3 + 3 -1 = 5
After the fulfilment of basic requirement, then need to identify the un occupied cells. And also need to find the ways how to distribute for these un occupied cells from the occupied cells. If the net transportation cost takes a negative value that’s the path which need to change the previous distribution path. If there are more than one negative values it is good to choose the highest value. In contrast if there is a positive value that means the cost will be increased if change the previous path. That means the existing way is good & should not to change. Therefore, we need to find optional critical paths until the values take positive or zero value.
Table 2.11. Finding Critical Path, I
| Un Occupied Cells | Paths | Values |
| A to 2 | -20+9-10+6 | -15 |
| A to 3 | -20+5-12+15-10+6 | -16 |
| B to 3 | -10+18-12+15 | +11 |
| C to 1 | -15+2-6+10 | -9 |
Source: Developed by the Student
The cost values of the paths which leads to un occupied cells are shown above. Normally (-) & (+) values can be seen as one after one. From among the four paths the critical path is the way which owns the highest negative value.
Critical Path
A to 3 -20+5-12+15-10+6 = -16
It means that, the total transportation cost can be reduced by Rs.16.00 as a result of distribution one unit to A3 cell. After deciding the critical path, it needs to clarify how much units need to be transferred to this location. It is done by comparing the quantity values which relate to the points of the critical path as follows,
Critical Path -20+5-12+15-10+6
Quantities 150 180 50
According to that, 50 units of quantity should be transferred through this critical path. Not only the distribution style but also transportation cost is completely changed than before.
Table 2.12. Total Transportation Cost of the First Critical Path
| Quantity | Unit Cost (Rs) | Total Cost (Rs) |
| 100 | 20.00 | 2000.00 |
| 50 | 5.00 | 250.00 |
| 100 | 6.00 | 600.00 |
| 120 | 15.00 | 1800.00 |
| 130 | 12.00 | 1560.00 |
| Grand Total | 6210.00 |
Source: Developed by the Student
But this can’t be the optimal solution. Therefore, it is compulsory to examine again whether this is the final solution or not like below,
Table 2.13. Finding Critical Path II
| Un Occupied Cells | Paths | Values |
| A to 2 | -5+9-15+12 | +1 |
| B to 2 | -6+0-15+12-5+20 | +16 |
| B to 3 | -6+18-5+20 | +27 |
| C to 1 | -12+2-20+5 | -25 |
Source: Developed by the Student
Critical Path
C to 1 -12+2-20+5 =-25
Quantity 130 100
Table 2.14. Total Transportation Cost of the Second Critical Path
| Quantity | Unit Cost (Rs) | Total Cost (Rs) |
| 100 | 20.00 | 200.00 |
| 150 | 5.00 | 750.00 |
| 100 | 6.00 | 600.00 |
| 100 | 2.00 | 200.00 |
| 120 | 15.00 | 1800.00 |
| 30 | 12.00 | 360.00 |
| Grand Total | 3710.00 |
Source: Developed by the Student
At the second path, total transportation cost has reduced by Rs.2500.00 than first critical path. But again, need to clarify is it the optimal solution or not like below,
Table 2.15. Finding Critical Path II
| Un Occupied Cells | Paths | Values |
| A to 1 | -5+20-2+12 | +25 |
| A to 2 | -5+9-15+12 | +1 |
| B to 2 | -6+10-15+2 | -9 |
| B to 3 | -6+18-12+2 | +2 |
Source: Developed by the Student
According to that, there is a negative value in B to 2 paths. Therefore, again it requires a modification,
Critical Path
B to 2 -6+10-15+2 = -9
Quantity 100 120
Table 2.16. Total Transportation Cost of the Third Critical Path
| Quantity | Unit Cost (Rs) | Total Cost (Rs) |
| 150 | 5.00 | 750.00 |
| 100 | 10.00 | 1000.00 |
| 200 | 2.00 | 400.00 |
| 20 | 15.00 | 300.00 |
| 30 | 12.00 | 360.00 |
| Grand Total | 2810.00 |
Source: Developed by the Student
At the third path, total transportation cost has reduced by Rs.900.00 & Rs.3400 than second & first critical paths respectively. But again, need to clarify is it the optimal solution or not like below
Table 2.17. Finding Critical Path III
| Un Occupied Cells | Paths | Values |
| A to 1 | -5+20-2+12 | +25 |
| A to 2 | -5+9-15+12 | +1 |
| B to 1 | -10+6-2+15 | +9 |
| B to 3 | -10+18-12+15 | +11 |
Source: Developed by the Student
According to Table 2.17 all the values are positive which means no cost reduction furthermore. This is the optimal solution.
2.3.1.2. Modified Distribution Method
This is an easier method to find the optimal solution than Stepping Stone Method. In here no need to follow up several steps instead of that the solution can be taken by comparing the cost of several paths. M+N-1 formula verification is a compulsory requirement in this method also. After that, two variables are substituted to the transportation table. “U” uses to introduce rows & “V” for column. Then the below mentioned formula should be used,
Cij = Ui + Vj
Ui = Row Values
Vj = Column Values
Cij = The Cost Value of the Point of Intersect both Row & Column/ Cost of the Occupied Cell
It is required to consider the value of U or V as zero when finding the certain values for them. After deciding all the values of the occupied cells then can find how much cost can be reduced by distributing to the unoccupied cells. It can be calculated by using the Index of Cost Improvement.
CII = Cij – Ui – Vj
Cij = Cost of Un Occupied Cell
CII = Cost Improvement Index
If the value of CII is negative it means that the cost can be reduced furthermore in other hand the positive value reflects the cost is going to be increased. When there are number of negative values, need to distribute to the highest cost path. According to that, the calculations need to be done till the values are being positive or zero.
Example 03
The Production of Sanduni Private Limited should be distributed into three persons who are in three several locations. The Initial Basic Solution has calculated by North West Corner Method as follows,
Table 2.18 Initial Basic Solution by North West Corner Method
| Stores
Factory |
1 | 2 | 3 | Supply |
| A | 150 | 100 | – | 250 |
| B | 100 | 50 | 150 | |
| C | – | 100 | 100 | |
| Demand | 150 | 200 | 150 | 500 |
Source: Developed by the Student
Table 2.19. Total Transportation Cost to the North-West Corner Method
| Quantity | Unit Cost (Rs) | Total Cost (Rs) |
| 150 | 8.00 | 250.00 |
| 100 | 8.00 | 800.00 |
| 100 | 12.00 | 1200.00 |
| 50 | 14.00 | 700.00 |
| 100 | 10.00 | 1000.00 |
| Grand Total | 4900.00 |
Source: Developed by the Student
Before all, need to verify the main formula as below,
M + N – 1 = No of Occupied Cells
3 + 3 -1 = 5
The first rule is already satisfied. Therefore, it is time to find the values for Cij = Ui + Vj. Only the occupied cells should use for this,
Occupied Cells Cij = Ui + Vj
U1 V1 8 = 0 + V1 U1 = 0 V1 = 8
U1 V2 8 = 0 + V2 U2 = 4 V2 = 8
U2 V2 12 = U2 + V8 U3 = 0 V3 = 10
U2 V3 14 = 4 + V3
U3 V3 10 = U3 + 10
The value of CII should be calculated after finding the values for U & V figures.
Unoccupied Cells CII = Cij – Ui – Vj
U1 V3 CII = 8 – 0 – 10 = -2
U2 V1 CII = 18 – 4 – 8 = 6
U3 V1 CII = 8 – 0 – 8 = 0
U3 V2 CII = 12 – 0 – 8 = 4
In the place of U1 V3 has a negative value. Therefore, the transport cost can be reduced by distributing this place.
Place Path
U1 V3 -8+8-14+12
Quantity 100 50
Like above, the quantity which need to transfer is 50. After distributing 50units to new cell the transport cost is decreased as follows,
Table 2.20 The Total Transportation Cost
| Quantity | Unit Cost (Rs) | Total Cost (Rs) |
| 150 | 8.00 | 1200.00 |
| 50 | 8.00 | 400.00 |
| 50 | 8.00 | 400.00 |
| 150 | 12.00 | 1800.00 |
| 100 | 10.00 | 1000.00 |
| Grand Total | 4800.00 |
Source: Developed by the Student
The cost is reduced by Rs.100 than initial basic solution cost. But this might not be the optimal solution. Therefore, again it is necessary to follow up the same process.
Occupied Cells Cij = Ui + Vj
U1 V1 8 = 0 + V1 U1 = 0 V1 = 8
U1 V2 8 = 0 + V2 U2 = 4 V2 = 8
U1 V3 8 = 0 + V3 U3 = 2 V3 = 8
U2 V2 12 = U2 + 8
U3 V3 10 = U3 + 8
The value of CII should be calculated after finding the values for U & V figures.
Unoccupied Cells CII = Cij – Ui – Vj
U2 V1 CII = 18 – 4 – 8 = 6
U2 V3 CII = 14 – 4 – 8 = 2
U3 V1 CII = 8 – 2 – 8 = -2
U3 V2 CII = 12 – 2 – 8 = 2
In here also U3V1 unoccupied cell has a negative value. Therefore, this is not the optimal solution.
Place Path
U3 V1 -10+8-8+8
Quantity 100 150
Table 2.21 The Total Transportation Cost
| Quantity | Unit Cost (Rs) | Total Cost (Rs) |
| 50 | 8.00 | 400.00 |
| 50 | 8.00 | 400.00 |
| 150 | 8.00 | 1200.00 |
| 150 | 12.00 | 1800.00 |
| 100 | 8.00 | 800.00 |
| Grand Total | 4600.00 |
Source: Developed by the Student
The cost is reduced by Rs.200 than the first turn & by Rs.300 than initial basic solution cost. But this might not be the optimal solution. Therefore, again it is necessary to follow up the same process,
Occupied Cells Cij = Ui + Vj
U1 V1 8 = 0 + V1 U1 = 0 V1 = 8
U1 V2 8 = 0 + V2 U2 = 4 V2 = 8
U1 V3 8 = 0 + V3 U3 = 0 V3 = 8
U2 V2 12 = U2 + 8
U3 V1 8 = U3 + 8
The value of CII should be calculated after finding the values for U & V figures.
Unoccupied Cells CII = Cij – Ui – Vj
U2 V1 CII = 18 – 4 – 8 = 6
U2 V3 CII = 14 – 4 – 8 = 2
U3 V2 CII = 12 – 0 – 8 = 4
U3 V3 CII = 10 – 0 – 8 = 2
There are no more negative values which means that this is the optimal solution.
| Stores
Factory |
1 | 2 | 3 | Supply |
| A | 50 | 50 | 150 | 250 |
| B | 150 | 50 | 150 | |
| C | 100 | 100 | 100 | |
| Demand | 150 | 200 | 150 | 500 |
Source: Developed by the Student
2.3.3. Special Cases of Transportation Problem
2.3.3.1. Unbalanced Supply & Demand
Still the chapters discuss about the problems which demand & supply equal. But there are some special cases which the demand & supply are not equal. It can be further understood by below example,
Example 04
| S1 | S2 | S3 | S4 | |
| F1 | 4 | 6 | 8 | 13 |
| F2 | 13 | 11 | 10 | 8 |
| F3 | 14 | 4 | 10 | 13 |
| F4 | 9 | 11 | 13 | 8 |
Source: Developed by the Student
The capacity of the organization per week is F1=50, F2=70, F3=30, F4=50. The weekly requirement of the stores is S1=25, S2= 35, S3=105, S4=20. The initial basic solution has taken by VAM as follows,
Making equal the demand & supply together.
Total Supply = 200
Total Demand = 185
Additional Supply Units = 15
According to that, the deficiency is in the demand. Therefore, it is required to add a dummy row to the table. The cost of the dummy column needs to consider as zero.
Table 2.24 Finding Initial Basic Solution
| S1 | S2 | S3 | S4 | Dummy | Supply | |
| F1 | 4 | 6 | 8 | 13 | 0 | 50 |
| F2 | 13 | 11 | 10 | 8 | 0 | 70 |
| F3 | 14 | 4 | 10 | 13 | 0 | 30 |
| F4 | 9 | 11 | 13 | 8 | 0 | 50 |
| Demand | 25 | 35 | 105 | 20 | 15 | 200 |
Source: Developed by the Student
After developing the transportation table, the initial basic solution need to be find,
Table 2.25 Initial Basic Solution
| Stores
Factory |
S1 | S2 | S3 | S4 | Dummy | Supply |
| F1 | 25 | 55
5 |
20 | 0 | 0 | 50 |
| F2 | 0 | 0 | 70 | 0 | 0 | 70 |
| F3 | 0 | 30 | 50 | |||
| F4 | 0 | 0 | 15 | 20 | 15 | 50 |
| Demand | 25 | 35 | 105 | 20 | 15 | 200 |
Source: Developed by the Student
Row Penalty Cost
| 4 | 2 | 2 | 2 | 5 | 0 |
| 8 | 2 | 2 | 2 | 2 | 2 |
| 4 | 6 | 0 | 0 | 0 | 0 |
| 8 | 1 | 1 | 3 | 5 | 5 |
Source: Developed by the Student
Column Penalty Cost
Table 2.26 Column Penalty Cost
| 5 | 2 | 2 | 0 | 0 |
| 5 | 2 | 2 | 0 | 0 |
| 5 | 5 | 2 | 0 | 0 |
| 0 | 5 | 2 | 0 | 0 |
| 0 | 0 | 3 | 0 | 0 |
Source: Developed by the Student
Total Transportation Cost
Table 2.27 Total Transportation Cost
| Factory | Stores | Quantity | Unit Cost (Rs) | Total Cost (Rs) |
| F1 | S1 | 25 | 4.00 | 100.00 |
| F1 | S2 | 5 | 6.00 | 30.00 |
| F1 | S3 | 20 | 8.00 | 160.00 |
| F2 | S3 | 70 | 10.00 | 700.00 |
| F3 | S2 | 30 | 4.00 | 120.00 |
| F4 | S3 | 15 | 13.00 | 195.00 |
| F4 | S4 | 20 | 8.00 | 160.00 |
| F4 | Dummy | 15 | 0.00 | 0.00 |
| Grand Total | 1465.00 |
Source: Developed by the Student
CHAPTER THREE
CONCLUSION
This is the chapter which is going to discuss about essence of the entire study that has described all the way.
When we are living as a person in the society & a member of an organization always have to take the decisions. Even the time is limited we have to take the most corrected decisions. If a person has a broad knowledge & experience that the decision making is not a difficult task. Transportation is a major criterion of every organization which requires a higher level of attention because it consumes huge amount of cost.
The first chapter of the study describes what is the transportation modelling, how this technique has evaluated over the time & why the transportation techniques are that much important to an organization. At the second chapter, the student discusses three major techniques which normally use to find the initial basic solution including the draw backs of the techniques separately. At the latter part, two specific ways are explained which use to find the optimal solution to decrease the cost of transportation furthermore. Finally, the student describes, the way of solving “Unbalanced Problem” when the demand & supply are not equal together.
Transportation cost is a vital element of the total cost structure for any business. In this modern era of hypercompetitive competition, the key criteria for any business is to grab high profits & success by reducing its cost. Hence transportation cost being an important & major element of the cost structure need to be reduced (taken care off) to the extent possible.
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