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Linear Programming
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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
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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
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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
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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
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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?
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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
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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
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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
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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?
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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
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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
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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.
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Machine data Product data
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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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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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