IEOR 4000: Production Management - Columbia Universitygmg2/4000/assol/sol5.pdf · IEOR 4000:...
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IEOR 4000: Production Management Prof. Gallego Solution to HW 5, Fall 2004, Prepared by Lin Li 1 Note that X is lognormal with mean μ and standard deviation σ. We have μ = E(X) = exp(ν + τ 2 2 ) (1) σ 2 = E(X 2 ) - (E(X)) 2 = exp(2ν +2τ 2 ) - exp(2ν + τ 2 )= μ 2 (exp(τ 2 ) - 1) From these, we get, cv 2 =( σ μ ) 2 = exp(τ 2 - 1) which implies that τ 2 = ln(cv 2 + 1), thus τ = ln(1 + cv 2 ) Plug the result of τ into equation (1), we can find ν which is ν = ln μ - ln √ 1+ cv 2 2 Since D is lognormal with mean μ and variance σ 2 , ln(D) is normal with mean ν and variance τ 2 . Pr(D ≤ x)= Pr(ln D ≤ ln x)= Pr( ln D - ν τ ≤ ln x - ν τ ) = Φ( ln x - ν τ ) Use the formula of ν and τ in problem 1, Pr(D ≤ x) = Φ( ln( x μ ) + ln( √ 1+ cv 2 ) ln(1 + cv 2 ) ) With μ fixed, as σ 2 →∞, the ln(1 + cv 2 ) →∞ and ln( x μ ) + ln( √ 1+ cv 2 ) ln(1 + cv 2 ) →∞ Thus, Pr(D ≤ x) → 1. 3 In the lognormal model, denote the base stock level asS, then ln( x μ ) + ln( √ 1+ cv 2 ) ln(1 + cv 2 ) = z β Solve for S, we have S = μ exp( ln(1 + cv 2 )z β - ln( 1+ cv 2 )) In this case, as σ 2 →∞, the ln(1 + cv 2 ) →∞ ln(1 + cv 2 )z β - ln( √ 1+ cv 2 )= ln(1 + cv 2 )(z β - 1 2 ln(1 + cv 2 )) → -∞, thus, S → 0. 4 For normal model, S = μ + z β σ, so S μ =1+ z β cv For lognormal model, S = μ exp( ln(1 + cv 2 )z β - ln( √ 1+ cv 2 )), so S μ = exp( ln(1 + cv 2 )z β - ln( 1+ cv 2 )) See the excel file for solution. 1
Transcript of IEOR 4000: Production Management - Columbia Universitygmg2/4000/assol/sol5.pdf · IEOR 4000:...
IEOR 4000: Production Management
Prof. GallegoSolution to HW 5, Fall 2004, Prepared by Lin Li
1 Note that X is lognormal with mean µ and standard deviation σ. We have
µ = E(X) = exp(ν +τ2
2) (1)
σ2 = E(X2)− (E(X))2 = exp(2ν + 2τ2)− exp(2ν + τ2) = µ2(exp(τ2)− 1)
From these, we get,cv2 = (
σ
µ)2 = exp(τ2 − 1)
which implies that τ2 = ln(cv2 + 1), thus τ =√
ln(1 + cv2)Plug the result of τ into equation (1), we can find ν which is ν = lnµ− ln
√1 + cv2
2 Since D is lognormal with mean µ and variance σ2, ln(D) is normal with mean ν and varianceτ2.
Pr(D ≤ x) = Pr(lnD ≤ lnx) = Pr(lnD − ν
τ≤ lnx− ν
τ) = Φ(
lnx− ν
τ)
Use the formula of ν and τ in problem 1,
Pr(D ≤ x) = Φ(ln( x
µ ) + ln(√
1 + cv2)√ln(1 + cv2)
)
With µ fixed, as σ2 →∞, the√
ln(1 + cv2)→∞ and
ln( xµ ) + ln(
√1 + cv2)√
ln(1 + cv2)→∞
Thus, Pr(D ≤ x)→ 1.
3 In the lognormal model, denote the base stock level asS, then
ln( xµ ) + ln(
√1 + cv2)√
ln(1 + cv2)= zβ
Solve for S, we haveS = µ exp(
√ln(1 + cv2)zβ − ln(
√1 + cv2))
In this case, as σ2 →∞, the√
ln(1 + cv2)→∞√ln(1 + cv2)zβ − ln(
√1 + cv2) =
√ln(1 + cv2)(zβ − 1
2
√ln(1 + cv2))→ −∞, thus, S → 0.
4 For normal model, S = µ + zβσ, soS
µ= 1 + zβcv
For lognormal model, S = µ exp(√
ln(1 + cv2)zβ − ln(√
1 + cv2)), so
S
µ= exp(
√ln(1 + cv2)zβ − ln(
√1 + cv2))
See the excel file for solution.
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