Over the weekend I had the opportunity to visit with a friend and colleague at another university. We lamented the fact that students could come out of high school with so little understanding of the math they had seen and so little enthusiasm for the subject.
In discussing this we wondered how a teacher could present material in a way that would give the students a wider perspective on the techniques they were learning. The idea being that thus engaging them on a deeper level should allow them to appreciate and retain the material more successfully.
On a seemingly unrelated note, during my train ride back home I began noodling around with some computations to better understand the economic notion of competitive advantage. In working through the problem, however, I realized that this might work as just the kind of project my friend and I had been discussing.
I present it here for your use, amusement, or enjoyment. Take your pick. I would also be pleased to receive comments, suggestions, and even additional problems or problem ideas.A Factory Problem
There it is. Thinking about this problem gave me some new insights into competitive advantage. I hope you have fun working through these questions, too.
Suppose Alice and Bob both work at a toy factory making plastic dogs and cats. For whatever reason, the factory owners want the number of dogs and the number of cats produced each day to be equal, thus they pay each employee a set amount for each dog and cat pair that the employee produces. (For instance, a dog by itself is not paid for, it must be accompanied by a cat.)
Alice is able to make either 40 dogs per hour or 20 cats per hour. Bob, on the other hand, is only able to make either 30 dogs per hour or 10 cats per hour. We shall assume that Alice can only be making one of the two types of toys at a time, but that she can switch between tasks without any loss of efficiency. The same rules apply to Bob.
After working this way for a while, Alice and Bob begin to wonder if they can make more toys by redistributing the workload between them and submitting their day's production together rather than separately. That is, they wonder whether they can make more toys by having one of them make more dogs than cats and the other make more cats than dogs, but in such a way that the total number of cats matches the total number of dogs.
Suppose Alice and Bob engage in the plan discovered above. Now that they are making more toys, their combined paycheck is larger than before, and the question of how they should split their earnings arises.
Extended Problem
There is a third worker at the factory, Charlene, and she sees that Alice and Bob have increased their paychecks, and so she wants to join in as well. Charlene produces toy dogs at a rate of 50 per hour and cats at a rate of 20 per hour. The same rules apply to Charlene that apply to Alice and Bob.
(Of course, all wage questions could be answered by deciding how many cats and dogs each worker would take from the combined pool of cats and dogs produced, allowing for fractional toys if necessary.)
For Further Exploration
Wednesday, October 22, 2008
Instructive Math Instruction
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4 comments:
Are we invited to post solutions/fairness discussion?
Sure, be my guest.
Although let this be a warning to readers: We may have spoilers below.
Notation convention: A_c is the number of cats produced by Alice, T_Ac is the time in hours that Alice spends producing cats.
•How should Alice divide up her 8 hour work day to maximize her paycheck for the day?
We have four variables: A_c, A_d, T_Ac, T_Ad. We also have four equations of constraint: A_c=20T_Ac and A_d=40T_Ad describe production rates, T_Ac+T_Ad=8 limits the time to one working day, and A_c=A_d stipulates that the number of cats produced must equal the number of dogs produced. Solving this system shows that Alice can produce 106 2/3 pairs of toys, spending 5 h 20 min on cats and 2 h 40 min on dogs. This is also the only division that will produce equal numbers of each toy.
•How should Bob divide up his 8 hour work day to maximize his paycheck for the day?
Solving in the same manner as above, Bob can make 60 pairs of toys, spending 6 h on cats and 2 h on dogs.
•How many total cats (or equivalently, dogs) do Alice and Bob (combined) produce in this way?
Working independently, they produce 166 2/3 pairs of toys each day. I recommend forming the 2/3 of a pair by leaving off the tail and one front leg from the cat and the dog; This pair can be marketed in the “care for a rescued animal” toy line.
•Find an optimal division of labor so that the total number of toys is as large as possible, while still producing equal numbers of cats and dogs. (How many toys are produced this way? Allow non-integer answers.)
We have eight variables: A_c, A_d,, T_Ac, T_Ad, B_c, B_d,, T_Bc, T_Bd. We also have seven equations of constraint: A_c=20T_Ac, A_d=40T_Ad, B_c=10T_Bc and B_d=30T_Bd describe production rates, T_Ac+T_Ad=8 and T_Bc+T_Bd=8 limit the time to one working day, and A_c+B_c=A_d+B_d stipulates that the number of cats produced must equal the number of dogs produced. We can introduce P as the number of pairs produced and since the number of cats must equal the number of dogs, it is equal to A_c+B_c and A_d+B_d. Solving the above system to describe A_d and B_d in terms of some time gives A_d=320-40T_Ac and B_d=-180+45TA_c. Summing these gives P=140+5T_Ac. This says that the number of pairs is directly related to the time Alice spends making cats, with the most pair possible when she spends her whole day making cats. The law of competitive advantage would concur. Setting T_Ac=8 results in A_c=160, A_d=0, B_c=20, and B_d=180 (i.e. Bob makes dogs for 6 h and cats for 2 h), for a total of 180 pairs of toys. The “care for a rescued animal” toy line will have to be discontinued…for the time being. There are two interesting notes: In order for it to be possible to have equal numbers of cats and dogs, Alice must spend at least 4 h making cats, otherwise she produces too many dogs for she, in the rest of her day, and Bob, in his whole day, to match with cats; their teamed production will surpass total individual production if Alice spends more than 5 h 20 min producing cats. If she spends less time than this, all teaming efforts will have…gone to the dogs.
There discussion of fairness revolves around the compensation rates from producing a dog and from producing a cat. If cats and dogs were equally compensated, Alice would take a 33% pay cut to team up (as she would be producing only 2/3 as many total toys), and unless she and Bob were married, his 50% raise would not be a great incentive to do so. Seeing as both workers make dogs at a fast pace than cats, it would make sense to suggest that cats are more difficult to make and might deserve more compensation. We will say that a pair of toys is paid a unit amount and that each toy is worth a share of that unit (i.e. if dogs paid nothing they would have a 0 share and cats would have a 1 share; if equal, they both would have a 1/2 share). Consistent with her production rates, Alice needs cats to pay at least a 2/3 share for her to make more as part of the team (106 2/3 c+106 2/3 d=160 c, where c is share of cat and d is share of dog). Since Bob is spending most of his day making dogs, he would do better with dogs receiving a greater share, in fact Bob needs dogs to pay at least a 1/4 share for him to make more as part of the team (60 c+60 d=180 d+20 c). So, both agreeing to team up would not be likely unless cats received between a 2/3 and 3/4 share. For both to experience the same percent pay increase, cats would need to pay a 18/25 (72%) share; they would each get an 8% pay raise by joining the team at this rate ((180(1-c)+20 c)/(60)=(160 c)/(106 2/3)). Suppose Bob is a communist and he wants the toys compensated so that they would each get paid the same, cats would need a 9/16 share (≈56%) (160 c=20 c+180 d), far too low for Alice to join for reasons other than charity. Suppose Alice is a libertarian and wants a pay increase commensurate with her faster rate of production (i.e. individually she makes 106 2/3 pairs for every 60 pairs of Bob, 7/9 or 78% more, so she wants her final pay to still be 78% higher than Bob’s). This would require that cats get a 288/337 share (≈85%) (((160 c)/(106 2/3))/( (180(1-c)+20 c)/(60))), much to high for Bob to join for reasons other than stupidity. Another option would be to compensate the toys so each worker gets the same absolute increase in pay (e.g. both would earn 20 units more). This would occur at a 17/24 share (≈71%) (106 2/3-160 c=60-20 c-180(1-c)). Note that the equal absolute and equal relative increases are very similar.
•How many cat and dog pairs can Charlene make in a regular 8 hour work day, working by herself?
Solving in the same manner as above, Charlene can make 114 2/7 pairs of toys, spending 5 5/7 h on cats and 2 2/7 h on dogs.
•What is the maximum number of cat and dog pairs that Alice, Bob, and Charlene can make all together?
Expanding our above system to include Charlene’s variables and constraints, as well as adding her production to the totals, we find that total production is P=260+5T_Ac-5T_Cd/2. The absence of critical points over the 0-8 h region leaves the boundary points as candidates for optimization. The optimum solution has Alice and Charlene making only cats. This presents a problem because given the other constraints, Bob must now make cats for negative 2 h and use the time he gained by working backwards in time to produce an additional two hours worth of dogs, for 10 h in total. The troublesome result is T_Bc=7T_Cd/4-2. In order for this to be positive, Charlene must spend at least 8/7 h making dogs. As any more would move away from the optimum solution, this can be taken to be the practical optimum solution; it gives a total production of 297 1/7 pairs of toys. Bob is making only dogs and Charlene is making 57 1/7 dogs and 137 1/7 cats, enabling her to contribute to the “mend a rescued animal” toy line.
To judge fairness, we will determine the range in which the share from producing a cat will give each worker at least as much as when working individually. Since Alice is making all cats, as above, she needs it to be at least a 2/3 share to have an increase. and Bob needs it no more than a 3/4 share. For Charlene cats must be at least a 5/7 share (≈71%) ((114+2/7)*c+(114+2/7)*d=(57+1/7)*d+(137+1/7)*c). The range that is acceptable to all three workers is now 5/7 to 3/4. There is no value that results in a uniform pay increase. At 8/11 Alice and Bob receive a 9.1% increase, but Charlene only gets a 0.9% increase. As the share is increased, Alice and Charlene benefit more and Bob benefits less. Bob and Charlene both get a 2.1% pay increase when cats have a 35/47 share (≈74%); Alice would get a 12% increase. Considering absolute increases, Alice and Bob would have equal increases at a 43/60 share (≈72%) and Bob and Charlene would have equal increases at a 82/112 share (≈74%). Note that the equal absolute and equal relative increases are very similar. It is not possible for Alice and Charlene to experience an equal increase.
•Suppose one person in the group is absent one day. How should the remaining two individuals divide up their work duties? (We have already asked this question when it is Charlene who is absent.)
Solving in the same manner as above, if Alice and Charlene are working, Alice should make all cats and Charlene should spend 3 3/7 h making cats; they would make 228 4/7 pairs of toys. If Bob and Charlene are working, Charlene should make only cats and Bob should make cats for 2 h; they would make 180 pairs of toys.
•In the above scenarios, how should the remaining two workers (fairly) divide up the wages?
For Alice and Charlene, an equal relative increase of 3.4% occurs with cats getting a 69% share and an equal absolute increase occurs at a 69% share. Note that in the case of Alice and Charlene, Charlene is playing to Alice’s competitive advantage so 71% is now the upper bound for her to receive a benefit. For Bob and Charlene an equal relative increase of 2.1% occurs with cats getting a 74% share and an equal absolute increase occurs at a 73% share.
•Allowing the three workers to work together or not, in any way they choose, and considering the total output, which grouping of workers produces the largest number of toys?
All possible divisions of labor are considered in the problem of optimizing the group of three; therefore, the situation described above creates the most toys.
Since payment is only delivered to a worker if a pair of toys is made, the division that produces the most toys will enable the most earnings. We have shown above that the total pay is maximized when the total number of toys is maximized and that under such conditions, each worker can receive at least the same amount of pay as when working individually; this occurs when cats are paid at a share between 5/7 and 3/4. The benefit to each worker differs as the rate varies within that range.
One thing to consider is that, depending on the rate of a cat, a worker may experience a greater increase in pay from a paring rather than the group of three. If we consider relative wage increases for instance, Bob always benefits from lower rates on cats and Alice always benefits from higher rates on cats. Alice gets the most benefit from the group of three, whereas Bob gets the most benefit from being paired with Alice. Charlene also gets the most benefits from being paired with Alice. If one worker is to work alone, the most toys will be made if Charlene is working alone (294 2/7 pairs of toys).
Chances are that since Bob wants to work with Alice and Alice wants to work with everyone, some verbal agreement will be made regarding a group of three. This will lead Bob to be enthused about the increase in pay he will be getting in the near future. He can buy a house. But since he doesn’t yet have this pay increase he’s going to look for a low introductory rate, perhaps even sub-prime. He’ll say he got a really good mortgage and when asked if he stands by that statement, he’ll say you can take it to the bank. His constant puns will likely get on the nerves of Alice so much so that she’ll consider taking the smaller benefit of working just with Charlene, who’s thrilled because then she’ll get have more potential for benefit. It won’t be long before the ladies think they’re the cat’s meow and make good on their threat, leaving Bob without his raise, just as his adjustable rate mortgage has an interest rate hike. His house will now be foreclosed, leaving him in the doghouse, as it were. He’ll start begging them to take him back on, only to be told he’s barking up the wrong tree. Bob is now disgruntled so he goes postal and spills merchandise all over the production floor, causing it to rain cats and dogs. It was at this time that the author realized that he could not wrap this up with a clever spoonerism on a well known cliché and that he better get back to grad school....
Purdue has a Problem of the Week that they publish in the student daily, The Exponent. They're generally some type of analysis or geometry, but mostly they remind me of the Putnam: http://www.math.purdue.edu/pow/
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