Example ¹1. Transportation Problem by the Least Cost Method (Balanced problem)This solution was made using the calculator presented on the site. 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 ¹4. Transportation problem by the North-West Corner Method (Balanced problem) Example ¹5. Transportation problem by the North-West Corner Method (Unbalanced problem. Fictitious supplier) Example ¹6. Transportation problem by the North-West 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. First of all we will use the routes with the lowest cost of delivery.
20 = min { 20, 30 }
10 = min { 25, 10 }
10 = min { 15, 10 }
15 = min { 15, 20 }
5 = min { 5, 5 }
Let's calculate the total cost of delivery for the initial solution. 10*1 + 15*3 + 5*4 + 20*1 + 10*2 = 115Is 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 is negative evaluation. Therefore it is possible to reduce the total cost of delivery. STEP ¹1.
The cell A_{2}B_{2} (unused route) is selected because its evaluation is negative.
Please, move the mouse cursor to the selected cell A_{2}B_{2}. 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_{2}B_{2}. 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_{2}B_{2}.
5 = min { 5, 20 } ?
This transformation will not change the balance.
But this transformation will change the total cost of delivery by: 2 * 5 - 4 * 5 + 2 * 5 - 1 * 5 = ( 2 - 4 + 2 - 1 ) * 5 = -1 * 5 You correctly noted that -1 * 5 = Δ_{22} * 5 ?
We have a new solution. ?
We will calculate the total cost of delivery of the new solution. S = 115 + Δ_{22} * 5 = 115 -1 * 5 = 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© 2010-2024 If you have any comments, please write to matematika1974@yandex.ru |