problem solving: practice & approaches
DESCRIPTION
Problem Solving: Practice & Approaches. Practice solving a variety of problems Strategies for solving problems More Practice. General Idea of This Lesson. Programming is like learning a language - PowerPoint PPT PresentationTRANSCRIPT
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Problem Solving: Practice & Approaches
1. Practice solving a variety of problems2. Strategies for solving problems3. More Practice
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General Idea of This Lesson
• Programming is like learning a language– You need to learn the vocabulary (keywords), grammar (syntax), and
how to use punctuation (symbols)
• Problem solving is like learning to cook– A novice chef has a recipe – An master chef can create their own recipe
Both tasks require practice!
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Review: Scientific Problem-Solving Method
1. Problem Statement2. Diagram3. Theory4. Assumptions5. Solution Steps6. Identify Results & Verify Accuracy7. Computerize the solution
a. Deduce the algorithm from step 5b. Translate the algorithm to lines of codec. Verify Results
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Example #1: Balancing a fulcrum
A 30-kg child and a 20-kg child sit on a 5.00-m long teeter-totter. Where should the fulcrum be placed so the two children balance? (Note: an object is in static equilibrium when all moments balance.)
Using the supplied worksheet, solve the problem work on the first couple of steps:
On your ownWith your neighborsWhat did you get?
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Example #1: Balancing a fulcrum
1. Problem Statement:a) Givens:
m1 = 30 kgm2 = 20 kgL = 5m
b) Find: Location of the fulcrum
2. Diagram
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Example #1: Balancing a fulcrum
3. TheoryForce = Mass * Acceleration (F = m*a)
Moment = Force * Distance (M = F*d)
MF = 0, The sum of moments about the fulcrum equals zero at static equilibrium
4. AssumptionsMass of teeter-totter is negligibleEarth’s gravitational constant
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2
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Example #1: Balancing a fulcrum
5. Solution Steps1. Forces: F1 = m1 * g F2 = m2 * g
2. Moments: M1 = F1 * L1 M2 = -F2 * L2
3. Equilibrium: MF = 0 = M1 + M2
0 = F1 * L1 -F2 * L2
0 = m1 * g * L1 -m2 * g* L2
thus: m1 *L1 = m2 *L2
4. Get rid of L2 using: L2 = L – L1
m1 *L1 = m2 (L-L1) thus: L1 = m2 * L / (m1 + m2)
5. L1 = 20*5/(20+30)=100/50=2.00 m
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Example #1: Balancing a fulcrum
6. Identify results and verify
L1 = 2.00 m
Does this make sense?– Units?– Overall Dimension?– Easy to imagine!– Can you rerun the analyses with other givens using Step 5?
This is the key to Computer Programming!!
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Problem Solving Strategies
• The trouble with Step 5: “Solution Steps”
There can be many approaches to solving the same problem
• Creativity is an important component on how we view and approach problems:
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Creativity
• Connect the following 9 dots with four continuous lines without lifting your pencil
Sometimes you will need to think outside the box
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Problem Solving Strategies (Polya, 1945)
• Utilize analogies– Flow through a piping system can be modeled with electronics
Resistors – Fluid FrictionCapacitors – Holdup tanksBatteries – Pumps
• Work Auxiliary Problems– Remove some constraints
• Generalize the problem Ex: L1 = m2 * L / (m1 + m2)
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Problem Solving Strategies (Polya, 1945)
• Decompose & Recombine problems– Break the problem into individual components
Calculate Cost of Area(𝑝2+1 ) (𝑞2+1 ) (𝑟2+1 ) (𝑠2+1 )
𝑝𝑞𝑟𝑠 ≥16
Prove the following equation
2 x 2 x 2 x 2 = 16
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Problem Solving Strategies (Polya, 1945)
• Work backwards from the solution
Ex: Measure exactly 7 oz. of liquid from an infinitelylarge container using only a 5 oz. container and an 8 oz. container
Solution:1. Fill 5 oz container and empty into 8oz 2. Fill 5 oz container again, then pour to top-off 8oz container
(2 oz remaining in 5 oz)3. Empty 8 oz and fill with the remaining 2oz from 5oz container4. Fill 5 oz container and add it to the 8oz container
8 5 7?
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Example #2: Fuel tank design
A fuel tank is to be constructed that will hold 5 x 105 L. The shape is cylindrical with a hemisphere top and a cylindrical midsection. Costs to construct the cylindrical portion will be $300/m2 of surface area and $400/m2 of surface area of the hemispheres. What is the tank dimension that will result in the lowest dollar cost?
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1. Problem Statement:Givens:
CostHemisphere = $300/m2
CostCylinder = $400/m2
VolumeTank = 500,000 L
Find: Size for minimum cost
2. Diagram
Example #2: Fuel tank design
R
H
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Example #2: Fuel tank design
3. TheoryVolume Cylinder : Volume Hemisphere: Surface Area
Cylinder: Surface Area Hemisphere:
4. AssumptionsNo dead air spaceConstruction cost independent of sizeOther costs do not change with tank dimensionsThickness of walls is negligibleBottom portion of tank is free.
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Example #2: Fuel tank design
5. Solution Steps1.
Substitute equations
2.
Solve for H with respect to R
Substitute equations
Solve for Cost with respect to radius
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Example #2: Fuel tank design
5. Solution StepsCalculate minimum cost with respect to radius
MATLAB can be used to calculate minimum cost • Plot the data - plot(R,Cost)• Identify minimum for an array of costs - min(Cost)
• Numerical Methods (Iterative solutions)
0 1 2 3 4 5 6 7 8 9 100.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6x 10
5
Radius (R), m
Cos
t (C
), D
olla
rsCost $155,027 $111,310 $95,106 $91,416 $95,239 $104,432 $117,925
Radius 2 3 4 5 6 7 8
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Example #2: Fuel tank design
5. Solution StepsSolve for H with identified R
where
R
H
H = 3.03m
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Example #2: Fuel tank design
6. Identify results and VerifyTake the derivative of the cost function
Does this make sense?– Units?– Overall Dimension?– Can you rerun the analyses with other givens using Step 5?
5m
3m
5VTank = 500,000L
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Wrapping Up
• Utilize the 7 step process before you begin programming• Be clear about your approach• Think creatively• Use a couple of strategies when understanding a problem• Practice!• Use MATLAB to make your life easier
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Try it yourself
• What if the fuel tank had two hemispheres?
R
H
A fuel tank is to be constructed that will hold 5 x 105 L. The shape is cylindrical with a hemisphere top, a hemisphere base and, and a cylindrical midsection. Costs to construct the cylindrical portion will be $250/m2 of surface area and $300/m2 of surface area of the hemispheres. What is the tank dimension that will result in the lowest dollar cost?