This preview shows pages 1–20. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Load Factor Analysis The Relationship Between Flight Load
and Passenger Turnaway “mi‘3’ Prepared by: Sales Technology Boeing Commercial Airplane Company (A Division of The Boeing Company)
Seattle, Washington 98124, USA. October 1978 Printed in U.S.A. W7523 What Is the Significance of Load Factor? For every flight over some time period, there is an average
load (factor). 0 When it is "too high,” what are the results? 0 Passengers are turned away—immediate revenue loss. 0 Turnedaway passengers difficult to reattract—potential revenue loss. ' When it is “too low,” what are the results? 0 Excess capacity and operating costs. 0 Profitability less than maximum. 0 When it is “just right,” what are the gains?
0 Passengers are satisfied—they tly when they want. 0 Management is happy—profits are near optimum . The Most Important lrrlormatiorr— Generally Ignored 0 There is some variation around the average load factor. 0 Generally ignored. 0 Harder to track as a management tool. 0 Holds the key to Solving many problems when carefully analyzed. What Can Be Done If This Variation ls Known? 0 it is possible to derive the distribution of passenger demand from load data. 0 From the demand distribution, it is possible to estimate: 0 The total number of passengers dam (“passenger spill").
‘ The total number of flights where at least one passenger was (“flight spill”). 0 A target (desirable) load factor.
0 The exact number of seats required (airplane capacity) for the target load factor. 0 The profit or loss produced by a change in capacity, 0 This information is needed by airlines for efficient operations and for
effective competitive marketing. 0 Techniques for deriving this information are discussed in the
examples that follow. Sequence of Steps for Lead Factor Analysis 1. Start with the right set of data. KNOW YOUR DATA. j
2. Derive the demand distribution. 3. Estimate spill for alternative capacities. 4. Calculate cost/revenue/profit for
alternative capacities. Ill I'L Ill: Ill I't Ill II. III II. Ill III II. Ill ‘II I. 1. Start With the Right 0 There can be seasonal differences.
Variation Average Data W
Combining seasons, days, , , Average Variation directions, and times can
Load ,   . produce errors.
W
Summer Fall
0 There can be daily differences.
'7‘" " T
Mondays Tuesdays Wednesdays Thursdays 0 There can be timeofday differences. a We Eastbound 0800 1300 1700 0 There can be differences in direction. $7 Westbound
0 There can be special problems. 0800 ° . Holidays 1. Start With the Right ma 0 Typical data segmentation. 0 Data in one direction only.
0 Data for one operational season. ° Same day of the week, or similar days, combined. 0 Weekend peak days
' Midweek days 0 Known holiday peaks removed. 0 Time of day. 0 Headtohead flights
0 Timeof—day demand 0 Next—an example set of segmented load data. 1. Start With the Right Data Example of segmented load data:
(Capacity=1 21 ) Similar
days of the week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16, 17 Tue 68 54 59 68 46 97 76 33 53 82 55 75 62 84 41 112 106
Wed 95 45 60 86 100 77 55 73 93 37 39 100 49 69 37 64 78
Thu 36 48 41 70 56 57 80 82 81 88 78 69 66 82 94 51 85
For each day Smallest load Largest load Average 10an
Tue 33 112 68.9
Wed 37 100 68.1
Thu 36 94 69.1 0 In this example: 0 Average loads are approximately equal.
' Variation between smallest and largest loads is similar.
0 No passenger demand is turned away. 0 Because of these similarities:
0 Data can be treated as a single set of 51 loads instead of as three sets oi 17 loads. 0 Increase in sample size gives more confidence in subsequent analysis. 2. Derive the Demand Distribution 0 For a flight with low average load factor: . Ail loads are less than capacity.
0 Loads are the same as demand.
0 There is zero spill. 0 Excess capacity and cost may exist. Histogram of Frequency 0 Next—use the segmented load data from the previous page to
examine a realistic situation with adequate capacity. 2. Derive the Demand Distribution Demand distribution and spill for lowload—factor example:
(Capacity :121) Load Recorded load
interval (passengers) Histogram of Load Data 3135 33 3640 36,137, 37, 39 I
4145 41, 41, 45 
4650 46, 48, 49 C I
5155 53, 54, 55, 55 apacity: 121 Seats —\ 5660 56, 57, 59, 60
6165 61, 62, 64 Number of
6670 66, 68, 68, 69, 69, 70 Occurrences 7175 73, 75
7680 76, 77, 78, 78, 80
8185 81, 82, 82, 82, 84, 85
8690 86, 88
9195 93, 94, 95
96100 97, 100,100
101105
106110 106
111115 112 . Loa CL interval Total passengers carried 73,502 , , _ ,
Total capacity (121) x (51) 6,171 0 There is adequate capac1ty In this case— Load factor 56.7%
(pass. carried/total capacity) Demand factor 56.7%
(3,502/total capacity) ° What would happen if capacity were reduced? Average pass. spilled per flight 0 no passengers are turned away. .:11i “Mme ..l.~.....«.m..l ....‘ WWW 3. Estimate Spill fer Alternative Capacities 0 If capacity is reduced: ' Load factors become higher.
' Some loads will be at capacity. 0 Some loads will be less than demand. 9 SP Histogram oi Frequency Capacity Flights where demand
is less than capacity Flights where some demand
turned away 0 Next—use the same segmented load data but reduce capacity. ‘ . I _ l .      3. Estimate Spill for Alternative Capacities Demand distribution and spill for highloadfaotor example:
(Capacity = 90) Load Recorded load 
interval (passengers) Histogram of Load Data
3135 33 3640 36, 37, 37, 39 41—45 41, 41, 45 4650 46, 48, 49 5155 53, 54, 55, 55 5660 56, 57, 59, 60 6165 61, 6'2, 64 6670 66, 68, 68, 69, 69, 70
7175 73, 75 76—80 76, 77, 78, 78, 80
8185 81, 82, 82, 82, 84, 85
8690 86, 88 (and eight loads of 90) Flights where demand I Capacity:
is less than capacity Flights where some demand
turned away Frequency of
Occurrence 3 Total pass. carried 3,425 ' ‘v‘
Total capacity (90) x (51) 4,590 g
/ 7 Load factor 746% a
(pass. carried/total capacity) m S a 8 g 8 o m 8 8 g a 8
Demand factor 76.3% d m ,5 (D m 0? “' 'T 'T '1' F; '— r ‘9 F‘ to I" (D I— (D
(3,502/total capacity) " "' N N 00 an an o o .— .—
A era e ass s illed er fli ht 12 = 151 '— F '— '— v g p . p p g  51  Load interval $\ Eadw [.7 ./O 0 There is insufficient capacity in this case. a 77 passengers are turned away. 12 3. Estimate Spill for Alternative Capacities
Spill Estimation for Several HighLoadFactor Flights Original capacity: Alternative capacities 121 seats 110 seats 100 seats 90 seats 80 seats 70 seats
Total passengers carried 3,502 3,500 3,484 3,425 3,315 3,103 (121)x(51)=6,171 (110)xl51)=5,610 l100)xl51)=5,100 l90)x(51)=4,590 l80lxl51l=4,080 (70)x(51)=3,570 Total capacity Load factor (%)
(Pass. carried/ total capacity) 56.7 62.4 68.3 74.6 81.3 87_1
Demand factor l%l
(3,502/t0tal capacity) 56.7 62.4 68.7 76.3 85.8 98.1
Spilled passengers (3,502)—(3,500l=2 l3,502l—(3,484l=18 l3,502)—l3,425l=77 l3,502l—(3,315l=187 (3,502)(3,108)=394
Averagepassengms 2 +8 77 +37 394 , . _=' —=.35 ——=1.51 _= _ __=
spilled per flight 0 51 O4 51 51 51 3 67 51 7.73 0 From this it is possible to quickly assess the spill for any alternative capacity offered for this flight. 0 Such information is valuable whenever it can be developed, because it permits a profit/ loss
comparison for alternative capacities. It is assumed that:
0 The revenue loss from one spilled passenger is $200. 0 The operating cost of one added seat is $50. Capacity 70 80 90 100
Cost of capacity (Assume 70 is base) Base (0) $500 $1,000 $1,500 A Cost $500 $500 $500 $500
Spilled passengers 7.73 3.67 1.51 .35 A Spill 4.06 2.16 1.16 .31 A Bevenuegain $812 $432 $232 $62
Revenue gain minus cost +$312 —$68 —$268 —$438 0 An increase in capacity from 70 to 80 (and the corresponding reduction in spill) ~ can produce an increase in profit of $812 — $500= $312. 110 $2,000 .04 $550 ._..— .04
$8
$542 0 Most profitable operation occurs when the last seat that is addedstill produces a positive increase in profit (between 80 and 90 seats). 0 Other levels of capacity are less profitable. 4. Calculate Cost/Revenue/Proﬁt for Alternative Capacities 121 $2,550 13 14 What Next? o So far we have seen that: 0 Load data must be screened to remove variations other than
those that are random. 0 These loads can then be used to estimate the demand
that generated the loads. 0 The resulting demands can be used to examine the economics of
. . . i
various capacrty alternatives. 0 To relate demands to loads so that one can be estimated from the other requires generalization of this technique
mathematically. ‘ . This process is discussed on the following pages. Describing the Distribution cl Demand Mathematically 0 It is assumed that the distribution of demand is a normal distribution. 0 A normal distribution is defined by two parameters:
 The mean or average/.1 , a measure of central tendency.
 The standard deviation, tr , a measure of dispersion of the distribution about the mean. 0 The ratio, 6 lil, is the coefficient of variation, K, and describes the shape of the demand distribution.
 Typical values of K for the airline industry range between .20 and .40. 0 Unique properties of the normal distribution:
v Approxhnate’ry 68% of all thedemand data'will fall within one standard deviation of the mean
(it i U l.
o Approximately 95% of all the demand data will fall within two standard deviations of the mean
(p i 2 rr ). O Substantial experience with actual airline data confirms that the assumption of a normal distribution
is valid and realistic. 15 Quick But Approximate Estimates of 0‘, u, and K o 95% of the time, the demand values will fall within two standard deviations of the mean. 0 For example, if it is thought that 95% of the time the demands would fall between 50 passengers and
100  50
————~——— = 25. 150 passengers, the mean would be 100 and the standard deviation would be 2 nqAnd the K (=% would be .25. 0 A word of caution is necessary when considering this method, because quite often the range of data is
estimated to be too small, resulting in a low estimate for K. This will cause a low estimate for spill. a Precise techniques for estimating ,u, a, and K with low and high load factors are in the Appendix. Example: Estimate ,u, a, and K
for Segmented Load Data Used Previously  The 51 load values range from 33 to 112.  49 values lie between 36 to 106. o The 49 values contain 5591—: 96% of all loads. . A 95% interval of where thedemand vaers fall“ could be between 34.5 and 109.
34.5 + 109  Therefore, the mean would be 2 = 71.8
 The standard deviation would be = 18.7
18.7
0 A d :———: _
n K 71.8 26 17 r Mathematical Formulation The expression for calculating the mean (first moment) of Example
a truncated normal distribution (the case when spill occurs) can be Written in terms of load factor (L) and demand cons'der the case Where: factor (D), as follows: H = 72
o = 19
1 1 1 1 C = 80
= _ — _ ____. _ __ _ __ +
L (D 1)F0(KD K) KDfO<KD K) 1
Then: K = 19/72=.264
0 Demand standard deviation D = 72/30 = .900
where: K: =—______....
u Demand mean KD = (.264)(.900)=.238
7777 ,, ,, ﬂ, 1 ,, WHH, W, , ,
— 2m2 — _ —— = 42
folxl=eXp(X/) KD K '1
V 2 1r
L = (.900 — 1.000) F0 (.421)  .238 10 (.421) + 1.000
F0(x) = 1" f0(t)dt = (~.100)(.6632) —(.238)(.3651)+1.000
_°° = .847
__ Average load _
L " Capacity ' LOad ‘acmr Spill = C(D— L) = 80(.900—.847) d d 4.240 passengers per flight
D = W = Demand factor
Capacrty This expression can be evaluated using normal probability
tables for various values of K and D. SDill is calculated by C(D — L), where C is aircraft capacity. Tables of spill as a function of L and D for various
values of K have been calculated from this expression
and appear in the Appendix of this document. 18 What Did the Mathematical Formulation Accomplish ? Provided the necessary data base to start detailed analysis: 0 Base for finding demand distribution. 0 Demand mean
0 Demand standard deviation
0 K factor 0 Base for estimating spilled passengers. ~ ,BQEDQQ1039993930”?an sway passengers
0 Revenue gained for alternate choices of increased capacity Provides first step in estimating carrier’s market share. I9 ...
View
Full
Document
 Fall '06
 hansman

Click to edit the document details