# Service Qualities and Efficiencies of Different Branches

Pages: 10 (2179 words) Published: March 15, 2013
DCO21020
Operations Research Lecture 2

TAHA Example 2.1-1 (Page 47) : The Reddy Mikks Company The Reddy Mikks Company produces both interior and exterior paints from two raw materials, M1 and M2. Tons of raw material per ton of Maximum daily availability Exterior Paint Interior Paint (tons) Raw material M1 Raw material M2 Profit per ton (\$1000s) 6 1 5 4 2 4 24 6

A market survey indicates that the daily demand for interior paint cannot exceed that for exterior paint by more than 1 ton. Also, the maximum daily demand for interior paint is 2 tons.

Reddy Mikks wants to determine the optimum (i.e. the best) product mix of interior and exterior paints that maximizes the daily profit.

TAHA Example 2.1-1 (Page 47) : The Reddy Mikks Company LP models have three basic components (like all OR models) – 1. 2. 3. Decision variables that we seek to determine. An objective that we need to optimize (minimize or maximize). This involves constructing an objective function. Constraints that the solution must satisfy.

 The variables of the model for solving this problem are : x1 = Tons of exterior paint to be produced daily x2 = Tons of interior paint to be produced daily Total profit from exterior paint = 5x1 thousand dollars Total profit from interior paint = 4x2 thousand dollars  To maximize the daily total profit (in thousands of dollars), the objective function is to maximize f(x1 , x2) where f(x1 , x2) = 5x1 + 4x2

TAHA Example 2.1-1 (Page 47) : The Reddy Mikks Company Usage of raw material M1 by both paints = 6x1 + 4x2 tons / day Usage of raw material M2 by both paints = 1x1 + 2x2 tons / day Usage of a raw material by both types of paint

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Maximum raw material availability

Because the availabilities of raw materials M1 and M2 are limited to 24 and 6 tons daily, therefore the constraints on raw materials are: 6x1 + 4x2  24 (Raw material M1) x1 + 2x2  6 (Raw material M2) The daily production of interior paint should not exceed the daily production of exterior paint by more than 1 ton, so … x2 – x1  1 (This is a limit imposed by the market) The maximum daily demand for interior paint is 2 tons, so …

x2  2

(Another limit imposed by the market)

TAHA Example 2.1-1 (Page 47) : The Reddy Mikks Company Together with the obvious restrictions that the variables x1 and x2 cannot be in negative values, i.e., x1 ≥ 0 and x2 ≥ 0 , a complete model for solving the Reddy Mikks Company problem is: Maximize f(x1 , x2) where f(x1 , x2) = 5x1 + 4x2 Subject to: 6x1 + 4x2  24 x1 + 2x2  6 -x1 + x2  1 x2  2 x1, x2  0

Any value pairs (x1 , x2) that satisfy all these five constraints constitute a feasible solution; otherwise, the solution is infeasible. The goal of the problem is to find the best, or optimal, solution that maximizes the total profit: f(x1 , x2) where f(x1 , x2) = 5x1 + 4x2

Graphical Linear Programming (LP) Solution A two-steps procedure: Step 1: Determination of the ‘feasible solution space’. Step 2: Determination of the optimum solution from among all the feasible points in the solution space.

Graphical LP Solution
Step 1: Determination of the Feasible Solution Space
The non-negativity constraints : x1 > 0 and x2 > 0

mean restricting the solution space to the first quadrant (considering only positive values for both axes) of the Cartesian plane. For each and every inequality constraint, replace the “< ” or “> ” sign by the “ = ” sign. Then graph the boundary equation which is a straight line on the Cartesian plane.

Graphical LP Solution
Step 1: Determination of the Feasible Solution Space
Since (0 , 6) and (4 , 0) satisfy 6x1 + 4x2 = 24 so, the straight line passing through (0 , 6) and (4 , 0) is the graphical representation of 6x1 + 4x2 = 24 This straight line bisects the Cartesian plane into two halves, with only the (x1 , x2) pairs in one of the two halves satisfy 6x1 + 4x2 < 24

Since (0 , 0) satisfies the inequality: 6x1 + 4x2 < 24
together with the non-negativity...

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