Decision step strategy: Applying the general method to a specific problem
Taken from: J. Fleet, F. Goodchild, R. Zajchowski, “Learning for Success”, 2006. See R. Zajchowski for a completed example.
To help learners focus on the process of solving problems, rather than on the mechanics of formula and calculations.
The focus is on correct application of concepts to specific situations. This strategy helps you to increase your awareness of the mental steps you make in problem solving, by “forcing” you to articulate your inner dialogue regarding procedure.
Identify the key decisions that determine what calculations to perform. In lecture, try to record the decision steps the professor uses but may not write down or post.
- Analyze solved examples, using brief statements focusing on steps you find difficult:
- What was done in this step?
- How was it done; what formula or guideline was followed?
- Why was it done?
- Any spots or traps to watch out for?
- Test run the decision steps on a similar problem, and revise until the steps are complete and accurate.
Example: Decision steps in Calculus for max/min word problems
Problem: A peanuts manufacturer wishes to design a can to hold dry-roasted peanuts. The volume of the cylindrical can is 250 cm3, and the circular top of the can is made from aluminum while the sides and the bottom are made from stainless steel. If aluminum is twice as expensive as stainless steel, what are the most economical dimensions of the can?
|1. Identify Quantity to be maximized/ minimized (Q)||C=Cost per can|
|2. Diagram when possible (including variables)
a. Shapes (perimeter, area, volume)
b. Equations to be graphed (axes, levels, distances)
|3. Make equation for questions using terms from formulas
a. Perimeter (P), surface area (S.A.), volume (V)
b. Pythagorean relationship
c. Sums, differences
d. Cost of steel (k), overall cost (C)
e. Distance between points
– Define variables
– Often need to combine equations
|4. Substitute the given volume value into equation (1) to get h(r) and substitute into equation (3) to get C(r).|
|5. Set the 1st derivative of overall cost (C) with respect to radius to 0 to find the radius that gives the optimum overall cost||Cross out “k” in both terms since it is common in both, rearrange equation, and solve for r:
r=2.98 cm and from (1): h=8.947cm
|6. Check 2nd derivative to verify the values found for “r” and “h” indeed give a minimum cost. (if 2nd derive >0, min. cost is found; if 2nd derive <0, max cost is found)||Since the right hand side of the equation can never be negative, r=2.98cm gives the minimum cost.|
|7. State answer; watch significant figures||The most economical dimensions for the can are r=3.0 cm and h=8.9 cm.|
The ‘what’ and the ‘how’
Note that these decision steps try to capture WHAT and especially HOW each step is carried out – including possible alternatives that can be tweaked so that the student is not left wondering how to make the decision needed. Most textbook steps tend to give the WHAT only. For example, these are steps from a calculus textbook:
- Determine the quantity Q to be maximized or minimized
- If possible, draw a figure illustrating the problem
- Write an equation for Q in terms of another variable of the problem
- Take the derivative of the function in step 3 … etc.
From Washington A.J. (2000). Basic Technical Mathematics with Calculus (7th ed.), Addison Wesley Longman.
Decision steps for rational expressions
Math 172. Used with permission.
- Read question.
- Make table:
- Identify cases (include a third case if total or difference of both cases)
- Put equation at top of table
- W = r x tor
- Total Cost= Cost/person x #of people
- Fill in columns of table with knowns and unknowns:
- Use letters for formulas above for unknowns
- If two columns are filled, then do third by algebra
- Watch! Do previous step carefully!
- Set up equation:
- Sum? Then add rates
- Difference? Then subtract rates
- Watch! Which rate is bigger? Then add to smaller
- Solve resulting equation for one of the cases
- Find answer for ‘other’ case
- Check by substituting answer into its respective case
- Write answer in appropriate format
- Carefully following these steps should allow you to solve any problem of this kind. If these steps don’t quite ‘work’ adjust them so that they do.
- As you can see, good decision steps often explain HOW to do a complicated or new step quite carefully. They are much more than just a general approach e.g. “Read question, create table, set up and solve equations”
Good decision steps also can – and should – include some ‘watch’ steps to remind you to be careful in spots where it is easy to make careless errors.
Quantitative concept summary strategy
Taken from: Fleet, J., Goodchild, F. and Zajchowski, R., “Learning for Success”, 2006
To provide a structure for organizing fundamental, general ideas. The mental work involved in constructing the summary helps clarify the basic ideas and shift the information from working memory to long-term memory. This is an excellent study tool, for quick review.
The organizational elements are:
- You can identify key ideas by referring to the course outline, chapter headings in the text, lecture outline. Sometimes concepts are thought of individually, other times they are meaningfully grouped for better recall. Eg. Depreciation, Capital Cost Allowance, and Half-Year Rule; acid, base and PH.
Use general categories to organize material, and then add specific details as appropriate. Sample general categories may include:
- Allowable key formula- check summary page of text or ask professor
- Definitions- define every term, unit and symbol
- Additional important information- sign conventions, reference values, meaning of zero values, situations in which formula do not work, etc
- Simple examples or explanations- use your own words, diagrams, or analogies to deepen your thinking and check your understanding
- List of relevant knowns and unknowns: to help you know which concepts are associated with which problems, use crucial knowns to help distinguish among problems.
Concept Summary: Example
Allowable Key Formula:
Henderson-Hasselbalch: pH = pKa + log([A–]/[HA])
pH = -log(H+)
pOH + pH = 14
x = 10pH-pKa / 1+10pH-pKa
Definitions of each symbol, and its units:
pKa: acid dissociation constant
[A–]: conjugate base concentration
x: unprotonated fraction
Additional important information: (e.g. sign conventions, special characteristics, when concept doesn’t work, special cases, etc.)
Physiologic pH = 7.4
pH > pKa –> deprotonated
pH < pKa –> protonated
pH = pKa –> 50/50
One pH unit above the pKa –> 90% deprotonated, 10% remains protonated (and vice versa)
This only applies to weak acids and bases, hence buffers.
*** In questions regarding clinical acid-base imbalances, dysentery/diarrhea typically suggest acidosis (endogenous alkali is not absorbed, travels too fast through the intestines), and vomiting/emesis suggest alkalosis (all endogenous acid is excreated, not absorbed).
Simple examples, explanations, cases:
What fraction of substance X (pKa = 6.0) remains protonated at pH = 7.3?
x = 107.3-6.0/1+107.3-6.0 = 0.95 (deprotonated)
1=0.95 = 0.05 protonated
What is the concentration of acid in a buffer with pH = 5.5, pKa = 5.5, and salt concentration = 0.5M?
pH = pKa + log([A–]/[HA])
5.5 = 5.5 + log(0.5/HA)
log(0.5/HA) = 0 –> [HA] = 1
Relevant knowns, and unknowns: (and words/phrases from word problems that signal these)
Anything mentioning buffers, including blood/plasma.
“protonated / deprotonated fraction”
“weak acid / base pH”
By permission from website of R. Zajchowki.
Example: Concept summary for Ordinary Simple Annuities
Concept title: Ordinary Simple Annuities
- Key allowable formula(s):
- Ordinary annuity à payment at end of each payment period
- Simple annuity à interest period = payment period
- Definition of each new symbol and its units:
- PV is the present value of the annuity in $
- FV is the future value of the annuity in $
- PMT is the regular payment per period in $
- n is the number of payments made
- I is the interest rate per payment period
- Additional important information: (sign conventions, special characteristics, reference & zero values, when concept does not work, special cases, etc.)
- In problems check interest period – payment period & that payment is at the end of the period; otherwise, formulas need changing!
- Be careful to find “i” per period
- Watch signs on “n” value ! FV>PV and PMT is small in comparison
- You cannot find “i” with formulas: a calculator is needed
- Otherwise you can find any of the 4 symbols above if you know 3 others (e.g., find FV knowing PMT, i, n).
- Simple examples of explanations:
These formulas can “compress” an annuity into a value – either PV or FV. Mortgage is a simple example: I am loaned some money (PV) to buy a house. I pay “PMT” per month at i% for n payments (usually 25 years, and n = 300). Then I can find FV and I end up paying (knowing) PV, i, n, and I can find PMT.
- Relevant knowns and unknowns: (words or phrases from word problems that signal these)
- To be an annuity, problem must say “annuity” or “series of payments” or “regular payments”, etc.
- To be an ordinary or a simple annuity, you need to check “pay at the end of the month or quarter”, etc. and that the pay period is the same as the interest period.
- PV is often “loan of…”; “price of…”
- FV is often “accumulates to …” or “how much after…”
Range of problems strategy: Common types of difficult problems
Taken from: J. Fleet, F. Goodchild, R. Zajchowski, Learning for Success, 2006
Expand your thinking in preparation for exams, where problems are not exactly the same as you have previously solved. Work from an existing problem, and make it more challenging by adding or changing:
Hidden knowns: needed information is hidden in a phrase or diagram Eg. “at rest” means initial v = 0 in physics.
Multipart-same concept: a problem may comprise 2 or more sub-problems, each involving the same concept. This type of problem can be solved only by identifying the given information in light of these sub-problems
Mulitpart-different concepts: same idea as above, but the sub-problems involve the use of different concepts
Multipart-simultaneous equations: same idea as above, but no single sub-problem can be solved by itself. You may have 2 unknowns and 2 equations or 3 unknowns and 3 equations, and you will need to solve them simultaneously, e.g. using substitution, comparison, addition and subtraction, matrices, etc.
Work backwards: some problems look different because to solve them you have to work in reverse order from problems you have previously solved
Letters only: when known quantities are expressed in letters, problems can look different. If you follow the decision steps, they are not usually as difficult.
“Dummy variables”: sometimes a quantity that you think should be a known is not specified because it is not really needed – that is, it cancels out. E.g. mass in work-energy problems, temperature in gas-law problems.
Red herrings, unnecessary information: a problem may give you more information than is needed, which is confusing if you think you should use everything provided.
McMaster University’s academic resources website. There are 3 videos on Problem Solving illustrating general ideas (Problem Solver I), differences in applying concepts vs. formula chasing (Problem Solver II), and applying the Decision Steps strategy (Problem Solver III).
Richard Zajchowski’s Resources web page, with examples of completed Concept Summaries, Decision Steps and other strategies.
Fleet, J, Goodchild, F, Zajchowski, R Learning for Success: Effective strategies for students, Thomson Nelson, 4th ed, 2006
Whimbey, A, Lockhead, J, Problem Solving & Comprehension, New Jersey: Lawrence Erlaum Associates, 5th ed., 1991
Woods, DR, Problem-based Learning: How to gain the most from PBL, Waterdown, ON: DR Woods, 1994