SOLUCIONARIO
Enviado por joavy • 2 de Diciembre de 2014 • 3.867 Palabras (16 Páginas) • 176 Visitas
Chapter 19
More on Variation and Decision Making Under Risk
Solutions to Problems
19.1 (a) Continuous (assumed) and uncertain – no chance statements made.
(b) Discrete and risk – plot units vs. chance as a continuous straight line between 50 and 55 units.
(c) 2 variables: first is discrete and certain at $400; second is continuous for $400, but uncertain (at this point). More data needed to assign any probabilities.
(d) Discrete variable with risk; rain at 20%, snow at 30%, other at 50%.
19.2 Needed or assumed information to be able to calculate an expected value:
1. Treat output as discrete or continuous variable .
2. If discrete, center points on cells, e.g., 800, 1500, and 2200 units per week.
3. Probability estimates for < 1000 and /or > 2000 units per week.
19.3 (a) N is discrete since only specific values are mentioned; i is continuous from 0 to 12.
(b) The P(N), F(N), P(i) and F(i) are calculated below.
N 0 1 2 3 4
P(N) .12 .56 .26 .03 .03
F(N) .12 .68 .94 .97 1.00
i 0-2 2-4 4-6 6-8 8-10 10-12
P(i) .22 .10 .12 .42 .08 .06
F(i) .22 .32 .44 .86 .94 1.00
(c) P(N = 1 or 2) = P(N = 1) + P(N = 2)
= 0.56 + 0.26 = 0.82
or
F(N 2) – F(N 0) = 0.94 – 0.12 = 0.82
P(N 3) = P(N = 3) + P(N 4) = 0.06
(d) P(7% i 11%) = P(6.01 i 12)
= 0.42 + 0.08 + 0.06 = 0.56
or
F(i 12%) – F(i 6%) = 1.00 – 0.44
= 0.56
19.4 (a) $ 0 2 5 10 100
F($) .91 .955 .98 .993 1.000
The variable $ is discrete, so plot $ versus F($).
(b) E($) = $P($) = 0.91(0) + ... + 0.007(100)
= 0 + 0.09 + 0.125 + 0.13 + 0.7
= $1.045
(c) 2.000 – 1.045 = 0.955
Long-term income is 95.5cents per ticket
19.5 (a) P(N) = (0.5)N N = 1,2,3,...
N 1 2 3 4 5 etc.
P(N) 0.5 0.25 0.125 0.0625 0.03125
F(N) 0.5 0.75 0.875 0.9375 0.96875
Plot P(N) and F(N); N is discrete.
P(L) is triangular like the distribution in Figure 19-5 with the mode at 5.
f(mode) = f(M) = 2 = 2
5-2 3
F(mode) = F(M) = 5-2 = 1
5-2
(b) P(N = 1, 2 or 3) = F(N 3) = 0.875
19.6 First cost, P
PP = first cost to purchase
PL = first cost to lease
Use the uniform distribution relations in Equation [19.3] and plot.
f(PP) = 1/(25,000–20,000) = 0.0002
f(PL) = 1/(2000–1800) = 0.005
Salvage value, S
SP is triangular with mode at $2500.
The f(SP) is symmetric around $2500.
f(M) = f(2500) = 2/(1000) = 0.002 is the probability at $2500.
There is no SL distribution
AOC
f(AOCP) = 1/(9000–5000) = 0.00025
f(AOCL) is triangular with:
f(7000) = 2/(9000–5000) = 0.0005
Life, L
f(LP) is triangular with mode at 6.
f(6) = 2/(8-4) = 0.5
The value LL is certain at 2 years.
19.7 (a) Determine several values of DM and DY and plot.
DM or DY f(DM) f(DY)
0.0 3.00 0.0
0.2 1.92 0.4
0.4 1.08 0.8
0.6 0.48 1.2
0.8 0.12 1.6
1.0 0.00 2.0
f(DM) is a decreasing power curve and f(DY) is linear.
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