The Aggie Distribution Company (ADC) has two distribution centers at Dallas and

Question

The Aggie Distribution Company (ADC) has two distribution centers at Dallas and Atlanta (Refer worksheet 'Question5'). They ship their products to distribution centers in Houston, San Jose, Jacksonville, and Memphis. The accounting, production, and marketing departments have provided the information in the data file, which shows the unit cost of shipping between any distribution center and branches, plant capacities over the next planning period, and distribution center demands. ADC's supply chain manager faces the problem of determining how much to ship between each plant and distribution center to minimize the total transportation cost, not exceed available capacity, and meet customer demand. Assume Xamount shipped from plant to distribution center j, where i-1 represents Dallas, i 2 represents Atlanta, j- 1 represents Houston, and so on. Find the optimal transportation cost and quantity. Branch San JoseJacksonvill DC Dallas Atlanta Houston 13 10.75 175 15.25 15.16 325 10.99 9.65 480 Memphis 18.48 18.5 Capacity 1250 750 Demand| 950

Explanation / Answer

-using Hungarian method.

1st iteration is as given in the problem.

Solution:

here given problem is unbalanced row-column.so add 2 new row to convert it into balance.

The number of rows = 5 and columns = 5

Here given problem is balanced.

Step-1: Find out the each row minimum element and subtract it from that row

Step-2: Find out the each column minimum element and subtract it from that column.

Step-3: Make assignment in the opporunity cost table
(1) Rowwise cell (Dallas,J) is assigned, so columnwise cell (atllants,J),(W1,J),(W2,J) crossed off.

(2) Rowwise cell (demand,c) is assigned, so columnwise cell (W1,c),(W2,c) crossed off.

(3) Rowwise cell (W1,m) is assigned, so columnwise cell (W2,m) crossed off.

(4) Rowwise cell (W2,s) is assigned, so columnwise cell (W1,s) crossed off.


Rowwise & columnwise assignment shown in table

Step-4: Number of assignments = 4, number of rows = 5
Which is not equal, so solution is not optimal.

Step-5: Cover the zeros with minimum number of lines
(1) Mark() row atllants since it has no assignment

(2) Mark() column J since row atllants has zero in this column

(3) Mark() row Dallas since column J has an assignment in this row Dallas.

(4) Since no other rows or columns can be marked, therefore draw straight lines through the unmarked rows demand,W1,W2 and marked columns J


Tick mark not allocated rows and allocated columns

Step-6: Develop the new revised table by selecting the smallest element, among the cells not covered by any line (say k = 1.1)
Subtract k = 1.1 from every element in the cell not covered by a line.
Add k = 1.1 to every elment in the intersection cell of two lines.

Repeat stpes 3 to 6 until an optimal solution is obtained.

Step-3: Make assignment in the opportunity cost table
(1) Rowwise cell (Dallas,J) is assigned, so columnwise cell (atllants,J) crossed off.

(2) Rowwise cell (atllants,H) is assigned, so columnwise cell (W1,H),(W2,H) crossed off.

(3) Rowwise cell (demand,c) is assigned, so columnwise cell (W1,c),(W2,c) crossed off.

(4) Rowwise cell (W1,m) is assigned, so columnwise cell (W2,m) crossed off.

(5) Rowwise cell (W2,s) is assigned, so columnwise cell (W1,s) crossed off.


Rowwise & columnwise assignment shown in table

Step-4: Number of assignments = 5, number of rows = 5
Which is equal, so solution is optimal

Optimal assignments are

Optimal solution is

   H s J m c     Dallas 13 15.25 10.99 18.48 1250 atllants 10.75 15.16 9.65 18.5 750 demand 175 325 480 950 0 W1 0 0 0 0 0 W2 0 0 0 0 0    

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