We have to choose between two ways to produces watches. Watches produced by process A give after one month an average time of 12:01 with a standard deviation of 1 minute, while watches produced by process B show 12:00 on average, but with a standard deviation of 2 minutes. Which process is best?
- Set up the quality loss function.
QLF: L(X,theta) = k*(X-theta)^2 as the loss is $10 when the deviation >= 3 minutes, we can determine k: 10 = k*3^2 => k = 10/9 10 2 L(X,theta) = ---- ( X - theta ) 9 - Use the expected loss
to choose which process is best.
E[L(X,theta)] = k*sigma^2 + k*(mu-theta)^2, where mu and sigma are mean and standard deviation Process A: mu = 1201, sigma = 1 10 2 10 2 20 E[L(X,theta)] = ---- 1 + ---- 1 = ---- 9 9 9 Process B: mu = 1200, sigma = 2 10 2 10 2 40 E[L(X,theta)] = ---- 2 + ---- 0 = ---- 9 9 9 The process A is best because it gives a lower expected loss than process B.
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- As marathon organizers, how would we estimate
the probability of breaking the world record at our event?
Assume our budget allows to pay five top athletes.
Athlete a runs a marathon on average in time mu(a) and with standard deviation sigma(a), a = 1,2,..,5. Let t = sample(mu(a),sigma(a)) samples running time for athlete a. We simulate the marathon N times and record each time the minimum time, which is the time of the winner.
- Describe the computational procedures we would apply,
using pseudo code.
run_min := 24 hours x 60 minutes x 60 seconds; for i from 1 to N do for a from 1 to 5 do t(a) := sample(mu(a),sigma(a)); end for; m = minimum(t(1),t(2),t(3),t(4),t(5)); if m < run_min then run_min := m; end if; end for; report run_min.
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- Set up the recursion relation for the balance B(n)
at month n.
B(0) = 1000 ( 0.2 ) B(n) = B(n-1)*( 1 + ----- ) - 50 ( 12 ) ( 61 ) B(n) = B(n-1)*(----) - 50 ( 60 ) - Solve the recursion for B(n).
( 61 ) B(1) = 1000*(----) - 50 ( 60 ) ( 61 ) B(2) = B(1)*(----) - 50 ( 60 ) ( ( 61 ) ) ( 61 ) = ( 1000*(----) - 50 )*(----) - 50 ( ( 60 ) ) ( 60 ) ( 61 )2 ( 61 ) = 1000*(----) - 50*(----) - 50 ( 60 ) ( 60 ) ( 61 )3 ( 61 )2 ( 61 ) B(3) = 1000*(----) - 50*(----) - 50*(----) - 50 ( 60 ) ( 60 ) ( 60 ) n-1 ( 61 )n --- ( 61 )k B(n) = 1000*(----) - 50 > (----) ( 60 ) --- ( 60 ) k=0 ( 61 )n (----) - 1 ( 61 )n ( 60 ) = 1000*(----) - 50 * -------------- ( 60 ) 61 ---- - 1 60 ( 61 )n B(n) = 3000 - 2000 * (----) ( 60 ) - Determine for which n, B(n) = 0.
Evaluate B(n) using a calculator: n = 24 : B(24) = 26.17 n = 25 : B(25) = -23.39 So the last payment at n=25 is 50-23.39.
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- Using * for convolution and xx for the DFT of x,
write the property
x * y -> xx . yy
down for frames of four elements.x = (x(0) x(1) x(2) x(3)) (x(0) x(1) x(2) x(3)) ... y = (y(0) y(1) y(2) y(3)) (y(0) y(1) y(2) y(3)) ... z = x*y = (z(0) z(1) z(2) z(3)) (z(0) z(1) z(2) z(3)) ... zz = xx . yy, zz(k) = xx(k).yy(k), for k = 1,2,3,4. - Show the property holds for frames of four elements.
w^(-9)*( z(3) = y(3)x(0) + y(2)x(1) + y(1)x(2) + y(0)x(3) ) w^(-0)*( z(0) = y(3)x(1) + y(2)x(2) + y(1)x(3) + y(0)x(0) ) w^(-3)*( z(1) = y(3)x(2) + y(2)x(3) + y(1)x(0) + y(0)x(1) ) + w^(-6)*( z(2) = y(3)x(3) + y(2)x(0) + y(1)x(1) + y(0)x(2) ) ---------------------------------------------------------------- z(0)*w^(-0) + z(1)*w^(-3) + z(2)*w^(-6) + z(3)*w^(-9) = zz(3) Collecting the terms at the right of the sum: x(0)*( y(0)w^(-0) + y(1)w^(-3) + y(2)w^(-6) + y(3)w^(-9) = yy(3) ) + x(1)*( y(0)w^(-3) + y(1)w^(-6) + y(2)w^(-9) + y(3)w^(-0) = yy(3)w^(-3) ) + x(2)*( y(0)w^(-6) + y(1)w^(-9) + y(2)w^(-0) + y(3)w^(-3) = yy(3)w^(-6) ) + x(3)*( y(0)w^(-9) + y(1)w^(-0) + y(2)w^(-3) + y(3)w^(-6) = yy(3)w^(-9) ) which equals zz(3) = x(0)yy(3)w^(-0) + x(1)yy(3)w^(-3) + x(2)yy(3)w^(-6) + x(3)yy(3)w^(-9) = xx(3).yy(3) The same holds for zz(0), zz(1), and zz(2).
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We are obliged to produce at least 20 units of A and to offer at least 10 units of B. Our storage capacity is limited to 250 units. We have only 600 man hours available, one unit of A requires two man hours, while one unit of B takes three man hours.
- Set up the linear programming model for this problem.
max 5 x1 + 7 x2 subject to -x1 <= -20 - x2 <= -10 x1 + x2 <= 250 2x1 + 3x2 <= 600 - Graph the feasibility region.
- From the graph, compute the optimal solution.
value at ( 20,186): 5 20 + 7 186 = 1402 at (150,100): 5 150 + 7 100 = 1450 at (240, 10): 5 240 + 7 10 = 1270 The optimum occurs when 150 units of A, and 100 units of B are produced.
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year || 1950 | 1960 | 1970 | 1980 | 1990
------++------+------+------+------+------
size || 151 | 179 | 203 | 226 | 249
We use P(t) = exp(a t + b) to model the size of
the population (in millions) at year t.
- Set up the system of linear equations we need to solve
in order to find a and b.
ln(P(t)) = a t + b [ 1950 1 ] [ ln(151) ] [ 1960 1 ] [ a ] [ ln(179) ] [ 1970 1 ] [ ] = [ ln(203) ] [ 1980 1 ] [ b ] [ ln(226) ] [ 1990 1 ] [ ln(249) ] X [a b]^T = P - Explain how to solve this overconstrained linear system,
but do not solve it.
We can either reduce the system to a 2-by-2 linear system, as X^T X [a b]^T = X^T P or apply QR to X, X = QR, Q^T Q = I, R upper triangular and solve a triangular system in a and b: R [a b]^T = Q^T P.
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- Suppose the equipment will save the company
each year one hundred thousand dollars.
Is this investment worth doing?
The present value of the savings at r = 0.05 are 100,000*(exp(-r) + exp(-2r) + .. + exp(-12r)) 1 - exp(-13r) = 100,000 --------------- 1 - exp(-r) = 980,005.30 < 1,000,000 So the investment is not worth doing. - Compute the break even annual savings, i.e.:
how large should the annual savings be in order
to gain from this investment?
1,000,000 savings = ----------------- ( 1 - exp(-13r) ) (---------------) ( 1 - exp(-r) ) = 102,040.26 At this savings, the investment breaks even, no profit and no loss.
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- Compute the elasticity of this demand function.
D'(p) D(p) = 8 + 20*p^(-2) e = - p ------ D(p) D'(p) = -40*p^(-3) -40/p^3 = - p ------------ 8 + 20/p^2 40 = ----------------- 2 ( 20 ) p ( 8 + ---- ) ( p^2 ) 40 = ------------ 8 p^2 + 20 - Explain what this means for the price the producer
will charge for the product.
The price where the revenue does not change occurs at e=1. Thus the price the producer will charge for the product satisfies the equation 40 1 = ------------ <=> 8 p^2 + 20 = 40 8 p^2 + 20 <=> 8 p^2 = 20 <=> 4 p^2 = 10 <=> p^2 = 10/4 => p = 10/sqrt(2) is the price the producer will charge
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