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Linear Programming.  Linear Programming provides methods for allocating limited resources among competing activities in an optimal way.  Any problem.

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Presentation on theme: "Linear Programming.  Linear Programming provides methods for allocating limited resources among competing activities in an optimal way.  Any problem."— Presentation transcript:

1 Linear Programming

2  Linear Programming provides methods for allocating limited resources among competing activities in an optimal way.  Any problem whose model fits the format for the linear programming model is a linear programming problem.  Wyndor Glass Co. example  Two variables – Graphical method  Maximize profit

3  Mary diagnosed with cancer of the bladder → needs radiation therapy  Radiation therapy  Involves using an external beam to pass radiation through the patient’s body  Damages both cancerous and healthy tissue  Goal of therapy design is to select the number, direction and intensity of beams to generate best possible dose distribution  Doctors have already selected the number (2) and direction of the beams to be used  Goal: Optimize intensity (measured in kilorads) referred to as the dose

4 Area Fraction of Entry Dose Absorbed by Area (Average) Restriction on Total Average Dosage, Kilorads Beam1Beam2 Healthy Anatomy0.40.5minimize Critical Tissues0.30.1≤ 2.7 Tumor Region0.5 = 6.0 Tumor Center0.60.4≥ 6.0

5 Graph the equations to determine relationships Minimize Z = 0.4x 1 + 0.5x 2 Subject to: 0.3x 1 + 0.1x 2 ≤ 2.7 0.5x 1 + 0.5x 2 = 6 0.6x 1 + 0.4x 2 ≥ 18 x 1 ≥ 0, x 2 ≥ 0

6  In order to ensure optimal health (and thus accurate test results), a lab technician needs to feed the rabbits a daily diet containing a minimum of 24 grams (g) of fat, 36 g of carbohydrates, and 4 g of protein. But the rabbits should be fed no more than five ounces of food a day.  Rather than order rabbit food that is custom-blended, it is cheaper to order Food X and Food Y, and blend them for an optimal mix.  Food X contains 8 g of fat, 12 g of carbohydrates, and 2 g of protein per ounce, and costs $0.20 per ounce.  Food Y contains 12 g of fat, 12 g of carbohydrates, and 1 g of protein per ounce, at a cost of $0.30 per ounce.  What is the optimal blend?

7 Daily Amount Food Type Daily Requirements (grams) XY Fat812≥ 24 Carbohydrates12 ≥ 36 Protein21≥ 4 maximum weight of the food is five ounces: X + Y ≤ 5 Minimize the cost: Z = 0.2X + 0.3Y

8 Graph the equations to determine relationships Minimize Z = 0.2x + 0.3y Subject to: fat: 8x + 12y ≥ 24 carbs: 12x + 12y ≥ 36 protein: 2x + 1y ≥ 4 weight:x + y ≤ 5 x ≥ 0, y ≥ 0

9 When you test the corners at: (0, 4), (0, 5), (3, 0), (5, 0), and (1, 2) you get a minimum cost of sixty cents per daily serving, using three ounces of Food X only. Only need to buy Food X

10 You have $12,000 to invest, and three different funds from which to choose. Municipal bond:7% return CDs:8% return High-risk acct:12% return (expected)  To minimize risk, you decide not to invest any more than $2,000 in the high-risk account.  For tax reasons, you need to invest at least three times as much in the municipal bonds as in the bank CDs.  Assuming the year-end yields are as expected, what are the optimal investment amounts?

11 Bonds (in thousands):x CDs (in thousands):y High Risk:z  Um... now what? I have three variables for a two-dimensional linear plot  Use the "how much is left" concept  Since $12,000 is invested, then the high risk account can be represented as  z = 12 – x – y

12 Constraints: Amounts are non-negative: x ≥ 0 y ≥ 0 z ≥ 0  12 – x – y ≥ 0  y ≤ –x + 12 High risk has upper limit z ≤ 2  12 – x – y ≤ 2  y ≤ –x + 10 Taxes: 3y ≤ x  y ≤ 1/3 x Objective to maximize the return: Z = 0.07x + 0.08y + 0.12z  Z = 1.44 - 0.05x – 0.04y

13 When you test the corner points at (9, 3), (12, 0), (10, 0), and (7.5, 2.5), you should get an optimal return of $965 when you invest $7,500 in municipal bonds, $2,500 in CDs, and the remaining $2,000 in the high-risk account.

14 Machine data Product data

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16 max 45 x P + 60 xQxQ Objective Function s.t. 20 x P +  1800 12 x P + 28 x Q  1440 15 x P + 6 x Q  2040 10 x P + 15 x Q  2400 demand Are we done? nonnegativity Are the LP assumptions valid for this problem? Optimal solution x * P = 81.82 x * Q = 16.36 Structural constraints x P ≥ 0, x Q ≥ 0 x P  100, x Q  40 10 x Q

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19  Optimal objective value is $4664 but when we subtract the weekly operating expenses of $3000 we obtain a weekly profit of $1664.  Machines A & B are being used at maximum level and are bottlenecks.  There is slack production capacity in Machines C & D.

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