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This is a procedure of determining the potential if any for improving each of the non basic variables in terms of the objective function. To determine this potential each of the non basic variables ( empty cells) is considered one by one. For each such cell well find out as to what effect on the total cost would be if one unit is assigned to this cell. Within this information then we come to known whether the solution is optimal or not. If not we improve that solution. To understand it we refer to the example containing the initial basic feasible solution obtained by least cost rule.
In a business, statistics is used to study the demand and market characteristics of the product or service being sold. In fact, market research
LCD Projectors LCD projectors are several steps ahead of conventional OHPs. These projectors are more compact and more powerful and can be directly linked to a
Graphic Methods a. Scatter Diagram: Scatter diagram is a special type of dot chart. Under this methods the given data are plotted in a graph paper in the form of d
#q.2 Q.2 Six Operators are to be assigned to five jobs with the cost of assignment in Rs. given in the matrix below. Determine the optimal assignment. Which operator will have no a
Question 1) What are the advantages of retailing. Write a brief note on retailing in India Question 2) Define franchising. What are the advantages and challenges of franch
CRITICAL EXAMINATION OF THE APPLICATION OF DECISION THEORY TECHNIQUES OR MODEL IN BUSINESS DECISION MAKING IN NIGERIA
Uses of Mean Deviation Mean Deviation is rarely being used as a measure of dispersion. But due to easy and simplicity in calculation. It is rather useful in busin
User Groups: The next step would be to identify more specifically the potential user groups. This may be done on the basis of interviews, questionnaires, study of organisation
A paper mill produces two grades of paper viz., X & Y. Because of raw material restrictions, it cannot produce more 400 tons of grade X paper & 300 tons of grade Y paper in a week.
Maximize Z= 3x1 + 2X2 Subject to the constraints: X1+ X2 = 4 X1 - X2 = 2 X1, X2 = 0
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