Operations & Logistics Management in Air Transportation - PowerPoint PPT Presentation

Operations & Logistics Management in Air Transportation Professor David Gillen (University of British Columbia ) & Professor Benny Mantin (University of Waterloo) Istanbul Technical University Air Transportation Systems and Infrastructure Air Transportation Management Strategic Planning M.Sc. Program Module 9-10 : 13 June 2014

THE NEWSVENDOR AND APPLICATIONS 2

Additional Slides June 12, 2014 3

Newsvendor Solution: Explanation Marginal Analysis : Suppose you stock S units Marginal Overage Cost with S Marginal Underage Cost with S Probability of S Probability of S P{D ≤ S} 1 - P{D ≤ S} being “over” being “under” Marginal cost of Marginal cost of P{D ≤ S} * C o {1-P{D ≤ S}} * C u over-stocking under-stocking To find the optimal stocking level: Marginal Cost of Marginal Cost of P{D ≤ S} * C o {1-P{D ≤ S}} * C u Over-Stocking Under-Stocking P{D ≤ S} C u / ( C u + C o ) 4

Representing the Decision Problem Mathematically • If D were known, the cost function would be C(S, D) = Cost (S, D) = C o ∙max{0, S -D} + C u ∙ max{0, D -S} = C o ∙ [S -D] + + C u ∙ [D -S] + Had too Had too much stock Little stock Total Cost There exists a fundamental economic tradeoff 0 (S-D) units 5

Impact of Demand Variability Expected Cost Function $300 $275 Expected Cost Curve for Mean = 30, StdDev = 10, CV = 33% Expected Cost Curve for Mean = 30, StdDev = 20, CV = 67% $250 Expected Cost Curve for Mean = 30, StdDev = 30, CV = 100% $225 Deterministic Expected Cost ($) $200 $175 $150 $125 $100 $75 $50 $25 $0 0 10 20 30 40 50 60 70 80 90 100 Stock Level (units) 6

Normal Distribution Tutorial (1) • Let Q be the order quantity, and Start with 0.0180 Center the (  ,  ) the parameters of the  = 100, 0.0160 distribution over 0 normal demand distribution 0.0140  = 25. by subtracting the 0.0120 • Prob {demand is Q or lower} =   0.0100 2 mean D ~ N ( , ) 0.0080 Prob {the outcome of a standard Q = 125 0.0060 normal is z or lower}, where 0.018 0.0040 0.016 0.0020 0.014   P(D ≤ 125) - Q  0.012      0 25 50 75 100 125 150 175 200 z or Q z 0.01  = 0.008 P(Z ≤ 1) 0.006 0.004 • 0.45 Look up Prob {the outcome of a 0.002 Standard normal 0.40 0 standard normal is z or lower} in 0.35 ~ N ( 0 , 1 ) Z -100 -75 -50 -25 0 25 50 75 100 0.30 the Standard Normal Distribution   Q Rescale the vertical  0.25 z  Function Table, or Excel 0.20 and horizontal axes by 0.1 5  125 100 NORMSDIST function.  dividing by the 0.1 0 25 0.05 standard deviation  0.00 1 7 -4 -3 -2 -1 0 1 2 3 4 z-Scale