Example ¹4. Transportation Problem by NorthWest Corner Method (Balanced problem)This solution was made using the calculator presented on the site. Example ¹1. Transportation problem by the Least Cost Method (Balanced problem) Example ¹2. Transportation problem by the Least Cost Method (Unbalanced problem. Fictitious supplier) Example ¹3. Transportation problem by the Least Cost Method (Unbalanced problem. Fictitious consumer) Example ¹5. Transportation problem by the NorthWest Corner Method (Unbalanced problem. Fictitious supplier) Example ¹6. Transportation problem by the NorthWest Corner Method (Unbalanced problem. Fictitious consumer) Problem: The cost of delivery of a unit of production from the supplier to the consumer is located in the lower right corner of the cell.
It is necessary to find a transport plan in which the total cost of delivery will be the lowest. Solution: This is a necessary condition for solving the problem: the total supply of suppliers should be equal to the total needs of consumers. Let's check it. The total supply of suppliers: 10 + 20 + 30 = 60 units. The total needs of consumers: 15 + 20 + 25 = 60 units. The total supply of suppliers equals the total needs of consumers. This is a necessary condition for solving the problem: number of used routes = number of suppliers + number of consumers  1. Therefore, if we have a situation where it is necessary to exclude a column and a row at the same time, we will exclude one thing. We will start filling the table from the upper left corner and gradually move to the lower right corner.
From the NorthWest to SouthEast.
10 = min { 15, 10 }
5 = min { 5, 20 }
15 = min { 20, 15 }
5 = min { 5, 30 }
25 = min { 25, 25 }
Let's calculate the total cost of delivery for the initial solution. 10*5 + 5*3 + 15*2 + 5*1 + 25*2 = 150Is the initial solution optimal? Let's check this using the MODI method (UV method). To each supplier A _{i} we associate a some number U _{i} To each consumer B _{j} we associate a some number V _{j} For the routes used: U + V = cost of delivery. Let's find U _{i} and V _{j} step by step. To do this, we need to enter the value of one of them. Let u_{2} = 0.
There are negative evaluations. Therefore it is possible to reduce the total cost of delivery. STEP ¹1.
The cell A_{1}B_{3} (unused route) is selected ? because its evaluation is negative.
Please, move the mouse cursor to the selected cell A_{1}B_{3}. Only horizontal and vertical cursor movements can be used. Connect the filled cells with a continuous line so that you return to the cell A_{1}B_{3}. The cells located at the vertices of the plotted line form a loop for the selected cell (see highlighted cells in the table below). There is only one loop for cell A_{1}B_{3}.
10 = min { 10, 15, 25 } ?
This transformation will not change the balance.
But this transformation will change the total cost of delivery by: 1 * 10  5 * 10 + 3 * 10  2 * 10 + 1 * 10  2 * 10 = ( 1  5 + 3  2 + 1  2 ) * 10 = 4 * 10 You correctly noted that 4 * 10 = Δ_{13} * 10 ?
We have a new solution. ?
We will calculate the total cost of delivery of the new solution. S = 150 + Δ_{13} * 10 = 150 4 * 10 = 110Is the new solution optimal? Let's check this using the MODI method (UV method). To each supplier A _{i} we associate a some number U _{i} To each consumer B _{j} we associate a some number V _{j} For the routes used: U + V = cost of delivery. Let's find U _{i} and V _{j} step by step. To do this, we need to enter the value of one of them. Let u_{2} = 0.
There are no negative evaluations. Therefore it is not possible to reduce the total cost of delivery. Result:
S_{min} = 110© 20102022 If you have any comments, please write to siteReshmat@yandex.ru 
