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Operations Research

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Operations Research (MTH601)
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Formulate this problem as a linear programming model by defining each component of the model separately and
then combining the components into a single model.
Decision Variables
The decision confronting management in this problem is how many plates and mugs to produce. As such,
there are two decision variables that represent the number of plates and mugs to be produced on a daily basis. The
quantities to be produced can be represented symbolically as,
X1 = the number of plates to produce
X2 = the number of mugs to produce
The Objective Function
The objective of the company is to Maximize total profit. The company's profit is the sum of the individual
profits gained from each plate and mug. As such, profits from plates is determine by multiplying the unit profit for
each plate, Rs. 4, by the number of plates produced, X1. Likewise, profit derived from mugs is the unit profit of a
mug, Rs. 5, multiplied by the number of mugs produced, X2. Thus, total profit, Z, can be expressed mathematically
as
Maximize Z = 4X1 + 5X2
where
Z = total profit per day
Rs 4X1 = profit from plates
Rs 5X2 = profit from mugs
By placing the term Maximize in front of the profit function, the relationship expresses the objective of the
firm to Maximize total profit.
Model Constraints
This problem has two resources used for production, which are limited, labor and clay. Production of plates
and mugs require both labor and clay. For each plate produce, one hour of labor is required. Therefore, the labor
used for the production of plates is 1X1 hours. Similarly, each mug requires two hours of labor; the labor used for
the production of mugs is 2X2 hours. Thus, the labor used by the company is the sum of the individual amounts of
labor used for each product.
1X1 + 2X2
However, the amount of labor represented "1X1 + 2X2" is limited to 40 hrs per day, thus, the complete labor
constraint is
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Operations Research (MTH601)
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1X1 + 2X2 < 40 hours
The "less than or equal to ( < )" inequality is employed instead of an equality ( = ) because the forty hours
of labor is a maximum limitation that can be used, but not an amount that must be used, but not an amount that must
be used. This allows the company more flexibility in that it is not restricted to use the 40 hours exactly, but whatever
amount necessary to Maximize profit up to and including forty hours. This means that the possibility of "idle or
excess capacity" (i.e., the amount under forty hours not used) exists.
The constraint for pottery clay is formulated in the same way as the labor constraint. Since each plate
requires four pounds of clay, the amount of clay used daily for the production of plates is 4X1 pounds, and since
each mug requires three pounds of clay, the amount of clay used for mugs daily is 3X2. Given that amount of clay
available for production each day is 120 pounds, the material constraint can be formulated as
4X1 + 3X2 < 120 pounds
A final restriction is that the number of plates and mugs produced be either zero or a positive value, since it
would be impossible to produce negative items. These restrictions are referred to as nonnegative constraints and are
expressed mathematically as
X1 > 0, X2 > 0
The complete linear programming model for this problem can now be summarized as
Maximize
Z = Rs. 4X1 + 5X2
Subject to
1X1 + 2X2 < 40
4X1 + 3X2 < 120
X1, X2 > 0
The solution of this model will result in numerical values for X1 and X2, which will maximize total profit,
Z. As one possible solution, consider X1 = 5 plates and X2 = 10 mugs. First we will substitute this hypothetical
solution into each of the constraints in order to make sure that the solution does not require more resources than the
constraints show are available.
1(5) + 2(10) < 40
25 < 40
4(5) + 3(10) < 120
50 < 120
and
4(5) + 3(10) < 120
50 < 120
Thus, neither one of the constraints is violated by this hypothetical solution. As such, we say the solution is
feasible (i.e., it is possible). Substituting these solution values in the objective function gives Z = 4(5) + 5(10) = Rs.
70. However, the maximum profit.
Now consider a solution of X1 = 10 plates and X2 = 20 mugs, This would result in a profit of
Z = Rs. 4(10) + 5 (20) = 40 + 100 = Rs. 140
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Operations Research (MTH601)
84
While this is certainly a better solution in terms of profit, it is also infeasible (i.e., not possible) because it
violates the resource constraint or labor:
1(10) + 2(20) < 40
50 < 40
Thus, the solution to this problem must both Maximize profit and not violate the constraints. The actual
solution to this model which achieves this objective is X1 = 24 plates and X2 = 8 mugs, with a corresponding profit
of Rs. 136.
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Table of Contents:
  1. Introduction:OR APPROACH TO PROBLEM SOLVING, Observation
  2. Introduction:Model Solution, Implementation of Results
  3. Introduction:USES OF OPERATIONS RESEARCH, Marketing, Personnel
  4. PERT / CPM:CONCEPT OF NETWORK, RULES FOR CONSTRUCTION OF NETWORK
  5. PERT / CPM:DUMMY ACTIVITIES, TO FIND THE CRITICAL PATH
  6. PERT / CPM:ALGORITHM FOR CRITICAL PATH, Free Slack
  7. PERT / CPM:Expected length of a critical path, Expected time and Critical path
  8. PERT / CPM:Expected time and Critical path
  9. PERT / CPM:RESOURCE SCHEDULING IN NETWORK
  10. PERT / CPM:Exercises
  11. Inventory Control:INVENTORY COSTS, INVENTORY MODELS (E.O.Q. MODELS)
  12. Inventory Control:Purchasing model with shortages
  13. Inventory Control:Manufacturing model with no shortages
  14. Inventory Control:Manufacturing model with shortages
  15. Inventory Control:ORDER QUANTITY WITH PRICE-BREAK
  16. Inventory Control:SOME DEFINITIONS, Computation of Safety Stock
  17. Linear Programming:Formulation of the Linear Programming Problem
  18. Linear Programming:Formulation of the Linear Programming Problem, Decision Variables
  19. Linear Programming:Model Constraints, Ingredients Mixing
  20. Linear Programming:VITAMIN CONTRIBUTION, Decision Variables
  21. Linear Programming:LINEAR PROGRAMMING PROBLEM
  22. Linear Programming:LIMITATIONS OF LINEAR PROGRAMMING
  23. Linear Programming:SOLUTION TO LINEAR PROGRAMMING PROBLEMS
  24. Linear Programming:SIMPLEX METHOD, Simplex Procedure
  25. Linear Programming:PRESENTATION IN TABULAR FORM - (SIMPLEX TABLE)
  26. Linear Programming:ARTIFICIAL VARIABLE TECHNIQUE
  27. Linear Programming:The Two Phase Method, First Iteration
  28. Linear Programming:VARIANTS OF THE SIMPLEX METHOD
  29. Linear Programming:Tie for the Leaving Basic Variable (Degeneracy)
  30. Linear Programming:Multiple or Alternative optimal Solutions
  31. Transportation Problems:TRANSPORTATION MODEL, Distribution centers
  32. Transportation Problems:FINDING AN INITIAL BASIC FEASIBLE SOLUTION
  33. Transportation Problems:MOVING TOWARDS OPTIMALITY
  34. Transportation Problems:DEGENERACY, Destination
  35. Transportation Problems:REVIEW QUESTIONS
  36. Assignment Problems:MATHEMATICAL FORMULATION OF THE PROBLEM
  37. Assignment Problems:SOLUTION OF AN ASSIGNMENT PROBLEM
  38. Queuing Theory:DEFINITION OF TERMS IN QUEUEING MODEL
  39. Queuing Theory:SINGLE-CHANNEL INFINITE-POPULATION MODEL
  40. Replacement Models:REPLACEMENT OF ITEMS WITH GRADUAL DETERIORATION
  41. Replacement Models:ITEMS DETERIORATING WITH TIME VALUE OF MONEY
  42. Dynamic Programming:FEATURES CHARECTERIZING DYNAMIC PROGRAMMING PROBLEMS
  43. Dynamic Programming:Analysis of the Result, One Stage Problem
  44. Miscellaneous:SEQUENCING, PROCESSING n JOBS THROUGH TWO MACHINES
  45. Miscellaneous:METHODS OF INTEGER PROGRAMMING SOLUTION