Let P(t) be the performance level of someone learning a skill as a function of t
ID: 1949220 • Letter: L
Question
Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dP/dt represents the rate at which performance improves. If M is the maximum level of performance of which the learner is capable, then a model for learning is given by the differential equation dP/dt = k(M-P(t)) where k is a positive constant.Two new workers, Andy and John, were hired for an assembly line. Andy could process 11 units per minute after one hour and 13 units per minute after two hours. John could process 10 units per minute after one hour and 16 units per minute after two hours. Using the above model and assuming that P(0)=0, estimate the maximum number of units per minute that each worker is capable of processing.
Andy: ?
John: ?
Explanation / Answer
You have: dP(t)/dt=k(M-P(t)) This is a separable equation: dP/(P - M) = -k dt Integrate both sides: ln(P - M) - ln(Po - M) = -k*t where Po is the value of P at t = 0 (i.e., P(0)). ln((P-M)/(Po - M)) = -k*t P(t) = M + (Po-M)*exp(-k*t) We are told that Po = 0, so: P(t) = M*(1 - exp(-k*t)) This equation holds for all workers, but each worker will have different values M, which is their limiting (maximum possible) performance, and each individual will have a different learning rate constant (k). These are the two unknown parameters remaining in the solution. Now use the two data points given for each of the workers to solve for the values of k and M for each worker. We know that Mark processed 12 unit/min after 1 hour, and 15 units/min after 2 hrs. This gives us 2 equations in 2 unknowns: 12 unit/min = M*(1 - exp(-k*1hr)) 15 units/min = M*(1 - exp(-k*2hr)) (12 units/min)/(1 - exp(-k*hr)) = M (15 units/min)/(1 - exp(-2k*hr)) = M Subtract the second equation from the first : (12 units/min)/(1 - exp(-k*hr)) = (15 units/min)/(1 - exp(-2k*hr)) (1 - exp(-k*hr))/(12 units/min) = (1 - exp(-2k*hr))/(15 units/min) 1/12 - exp(-k*hr)/12 = 1/15 - exp(-2k*hr)/15 1 - 5exp(-k*hr) + 4*exp(-2k*hr) = 0 Remember that exp(2x) = exp(x)^2, so, letting z = exp(-k*hr) we have: 4z^2 - 5z + 1 = 0 Using the quadratic formula, we find that z = 1 and z = 1/4, so, exp(-k*hr) = 1 and exp(-k*hr) = 1/4 The first solution corresponds to the solution k = 0, which is the "trivial" solution (no learning takes place). The second solution is the "interesting" one, then: exp(-k*hr) = 1/4 Take the log of both sides: -k*hr = ln(1/4) = -ln(4) k = ln(4)/hr Now use this value of k in one of the equations above to solve for M: (12 unit/min)/(1 - exp(-k*hr)) = M (12 unit/min)/(1 - exp(-ln(4))) = M (12 unit/min)/(1 - exp(ln(1/4))) = M (12 unit/min)/(1 - 1/4) = M M = 16 unit/min. This is the maximum number of units per minute that Mark will be capable of processing. Furthermore, the learning law for the Mark is: P(t) = (16 unit/min)*(1 - exp(ln(1/4)*t/hr)) You can follow the same procedure to solve for k and M for the second worker.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.