Background on British Airways and the UK aircraft market

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Appendix A

Background on British Airways and

the UK aircraft market

The aim of this chapter is to give some background information about British Airways and its role in the United Kingdom aircraft market.

A.1 British Airways

British Airways (BA) is the largest UK based airline company. It operates in the UK air market as well in many other parts of the world. Within the UK, BA is the biggest provider of national flights. Within Europe, BA is the 3rd biggest airline. Its center of operations is based in West London, with numerous flights going out from London Heathrow and London Gatwich every day. To illustrate the company’s enormous size: in 2006 BA carried 36.5 mil-lion passengers and BA’s total revenue was 8.5 bilmil-lion (see British Airways (2006)).

The network British Airways operates on is massive, and people from all parts of the world fly on BA. Being a UK based carrier, the majority of the people flying BA are UK residents. BA is one of the founding members of the Oneworld Airline Alliance, which consists of a few international airlines like American Airlines, Aer Lingus, Iberia and Qantas. This alliance enables the airlines to sell seat tickets on routes they don’t operate on by allocating passengers in alliance-owned aircrafts. This results in a higher passengerload per aircraft and a larger area of operation. This alliance has been quite succesful and BA has been able to sell more tickets effectively at lower rates. In 2006 the Oneworld Alliance operated in approximately 130 countries scheduling around 600 destinations.

A.1.1 Full service carrier

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A.2 The UK air travel market

In the past decade major developments and events have taken place in the air travel market. A few of them will be discussed.

A.2.1 The uprise of low-cost carriers

In 1949 a new aircraft carrier named Pacific Southwest Airlines launched activities as the first low-cost carrier that managed to become profitable. It took a while for the concept to spread around America, but in 1971 Southwest Airlines started operations as a low-cost carrier as well. After that it would take years before low-cost carriers appeared in other parts of the world. When the concept became known though, it quickly became popular. Within the UK, the most succesful low-cost carriers are EasyJet and RyanAir.

Since the launch EasyJet and RyanAir (and others) around 1991-1995, the traditional carriers like Air France - KLM, British Airways, Lufthansa have lost a part of their short haul market share, especially in the UK. In the past decade, the amount of people flying on EasyJet and RyanAir has increased at a fast rate, and even though the demand for low-cost flights is not expected to keep on growing so explosively, it is certain that low-cost carriers will not be a temporary hype in the market.

The main strength of low-cost carriers obviously is their low pricing. By leaving out all the frills on flights, they can cut down on their costs and hence offer tickets at sometimes incredibly low rates. However the price-cutting has to come from somewhere; low-cost carriers operate from airports usually situated a good distance away from the city center and inflight there no complementary drinks or meals, for example.

So far the low-cost carriers have been operating only on short haul routes. It is expected that within a considerable amount of time, no frill flights will be operating on long haul routes as well. Even though there are aspects that make it hard to operate low-cost flights on long haul routes, especially the inflight services, if low-cost carriers would start competing on the long haul market as well, offering flights at low rates, this could have a major impact on the demand for full-service flights.

A.2.2 Major events that affect the market

In the past decades there have been a few major events that have affected the way people think about flying. The most recent event has been in September 2001, when the hostile takeover of a few airplanes, with its well known catastrophical result, caused the aircraft market to hit rockbottom for months. Ever since, people are more aware about safety issues on airplanes. Safety has always been a major issue for people flying on airplanes, but the thought that such a terrible event could happen to anyone, certainly affected the criteria people use when choosing with which carrier to fly.

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Appendix B

Tests for seasonal unit roots

This appendix relies alot on the work done in Hylleberg et al. (1990) and Engle et al. (1990) and Beaulieu and Miron (1993). In these papers tests for seasonal unit roots in monthly data are developed for various frequencies. In this appendix we will discuss the test for monthly data.

We will assume data is monthly, and generated by ϕ(L)xt= et, so every year has p = 12

ob-servations. We are interested in testing seasonal unit roots. In polynomial form the 12th-lag seasonal unit root can be expressed as (L is the lag-operator):

(1 − L12) = (B.1) (1 − L)(1 + L)(1 + L2)(1 + L √ 3 + L2)(1 − L √ 3 + L2)(1 + L + L2)(1 − L + L2) (B.2) (B.3) The twelve roots of this polynomial are shown in table B.1: Each of these roots correspond

No. Root No. Root No. Root

1 1 5 −0.5√3 + 0.5i 9 −0.5 + 0.5√3i

2 −1 6 −0.5√3 − 0.5i 10 −0.5 − 0.5√3i

3 i 7 0.5√3 + 0.5i 11 0.5 + 0.5√3i

4 −i 8 0.5√3 − 0.5i 12 0.5 − 0.5√3i

Table B.1: Roots of polynomial (1 − L12)

to a frequency. The test developed in this chapter, can be used to test a series for whether each of these unit roots are likely to be apparent for particular frequencies.

We define θk, k = 1, . . . , 12 to be the k-th root of (1 − L12), where the numbering

corre-sponds to table B.1. In Hylleberg et al. (1990) it is stated that any polynomial ϕ(L) (in this case it is easiest to think of ϕ(L) = (1 − L12)) can be expressed as (under some mild conditions):

p

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Here λk is a set of constants, ϕ∗∗(L) is some polynomial and δk= 1 − 1 θk L, ∆(L) = p Y k=1 δk(L) (B.5) By subtractingPp

k=1λk∆(L) from the first term of B.4, and adding it from the second term

yields: ϕ(L) = p X k=1 λk∆(L) 1 − δk(L) δk(L) + ∆(L)ϕ∗(L) (B.6) Here ϕ∗ = ϕ∗∗+Pp k=1λk.

Straightforward calculation leads to the results shown in table B.2 Because all δk(L), k =

δ1(L) = 1 − L δ5(L) = 1 + (0.5 √ 3 + 0.5i)L δ9(L) = 1 + (0.5 + 0.5 √ 3i)L δ2(L) = 1 + L δ6(L) = 1 + (0.5 √ 3 − 0.5i)L δ10(L) = 1 + (0.5 − 0.5 √ 3i)L δ3(L) = 1 + iL δ7(L) = 1 − (0.5 √ 3 − 0.5i)L δ11(L) = 1 − (0.5 + 0.5 √ 3i)L δ4(L) = 1 − iL δ8(L) = 1 − (0.5 √ 3 + 0.5i)L δ12(L) = 1 − (0.5 − 0.5 √ 3i)L Table B.2: Values of δk(L)

1, . . . , 12 are the components of the polynomial (1 − L12) is obvious that ∆(L) = (1 − L12). A formula for ϕ(L) can now be constructed, but will not be shown due to its incredibly long length. In Beaulieu and Miron (1993) it is stated that for monthly data:

ϕ∗(L)y13t = 12

X

k=1

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Where y1t = (1 + L + L2+ L3+ L4+ L5+ L6+ L7+ L8+ L9+ L10+ L11)xt y2t = −(1 − L + L2− L3+ L4− L5+ L6− L7+ L8− L9+ L10− L11)xt y3t = −(L − L3+ L5− L7+ L9− L11)xt y4t = −(1 − L2+ L4− L6+ L8− L10)xt y5t = − 1 2(1 + L − 2L 2_{+ L}3_{+ L}4_{− 2L}5_{+ L}6_{+ L}7_{− 2L}8_{+ L}9_{+ L}10_{− 2L}11_)x t y6t = √ 3 2 (1 − L + L 3_{− L}4_{+ L}6_{− L}7_{+ L}9_{− L}10_)x t y7t = 1 2(1 − L − 2L 2_{− L}3_{+ L}4_{+ 2L}5_{+ L}6_{− L}7_{− 2L}8_{− L}9_{+ L}11_{+ 2L}11_)x t y8t = − √ 3 2 (1 + L − L 3_{− L}4_{+ L}6_{+ L}7_{− L}9_{− L}10_)x t y9t = − 1 2( √ 3 − L + L3−√3L4+ 2L5−√3L6+ L7− L9+√3L10− 2L11)xt y10t = 1 2(1 − √ 3L + 2L2−√3L3+ L4− L6+√3L7− 2L8+√3L9− L10)xt y11t = 1 2( √ 3 + L − L3−√3L4− 2L5−√3L6− L7+ L9+ √ 3L10+ 2L11)xt y12t = − 1 2(1 + √ 3L + 2L2+ √ 3L3+ L4− L6−√3L7− 2L8−√3L9− L10)xt y13t = (1 − L12)xt

Using OLS parameter estimations for πk, k = 1, . . . , 12 can be obtained. In Beaulieu and

Miron (1993) it is also explained how, based on these estimations, tests can be constructed to test for unit roots at different frequencies. For π1 and π2 one tests whether π1 = 0 versus

π1 < 0 and π2 = 0 versus π2 < 0 using a t-test. For π4, π6, π8, π10, π12 one tests the

hypoth-esis of the parameter being 0 against a two-sided alternative. If πi, i = 4, 6, 8, 10, 12 is found

equal to zero, it is likely for the series to have a unit root corresponding to the appropriate frequency. If these even parameters are found different from zero, a test can be performed on πi−1= 0 versus πi−1< 0 where, if this parameter is found zero, a unit root is likely for this

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Appendix C

ARIMA model Tables and graphs

Figure C.1: ACF for PSJ series

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Month MAPE October 2005 9.36% November 2005 23.35% December 2005 13.63% January 2006 29.96% February 2006 21.72% March 2006 11.27% April 2006 15.40% May 2006 14.34% June 2006 22.33% July 2006 4.14% August 2006 1.51% September 2006 16.78% MAPE Total 15.32 %

Table C.1: MAPE of the forecast for the evaluation period

AR orders MA orders AICC

0 1 25.59 0 1, 2 25.58 0 1, 12 25.49 0 1, 12, 13 25.28 1 1 25.59 1 1, 12, 13 25.31 1, 12 1, 12, 13 25.52

Table C.2: AICC value for different ARMA models

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Figure C.4: QQ-plot of residuals for the ARIMA model estimation

Figure C.5: Results of test for serial correlation of residuals

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Period 95% LCF 50% LCF 50% UCF 95% UCF Width 50% Width 95% Oct 2005 - Sept 2006 67.82% 92.52% 109.18% 136.66% 16.66% 68.84% Oct 2006 - Sept 2007 62.46% 89.74% 110.93% 141.75% 21.19% 79.29% Oct 2007 - Sept 2008 54.27% 86.85% 113.57% 150.71% 26.72% 96.44% Oct 2008 - Sept 2009 44.46% 83.76% 116.33% 161.03% 32.57% 116.58% Oct 2009 - Sept 2010 32.19% 80.52% 119.46% 169.86% 38.94% 137.67% Oct 2010 - Sept 2011 12.75% 74.87% 124.37% 185.72% 49.50% 172.96% Oct 2011 - Sept 2012 -8.44% 69.18% 128.14% 203.77% 58.96% 212.21% Oct 2012 - Sept 2013 -35.18% 60.04% 136.13% 229.24% 76.09% 264.42% Oct 2013 - Sept 2014 -85.54% 49.01% 148.42% 268.71% 99.41% 354.26% Oct 2014 - Sept 2015 -149.76% 30.84% 161.61% 319.89% 130.77% 469.65%

Table C.3: Accuracy and reliability of the forecasts

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Null Hypothesis Alt. Hypothesis t-value Critical value (95%) π1 = 0 π1 < 0 -0.844912 -1.89 π2 = 0 π2 < 0 -0.569143 -1.87 π4 = 0 π4 6= 0 -0.776210 -1.63 and 1.61 π5 = 0 π5 < 0 -0.168986 -1.88 π6 = 0 π6 6= 0 2.121696* -1.63 and 1.61 π8 = 0 π8 6= 0 -0.650381 -1.63 and 1.61 π10= 0 π106= 0 0.669418 -1.63 and 1.61 π12= 0 π126= 0 -0.399439 -1.63 and 1.61 * Reject at 95% significance

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Appendix D

VAR/VECM Model Tables and

Graphs

Figure D.1: Estimation results of Granger cause test

AR orders AICC VAR

1 102.85

12 101.42

1,12 101.31

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Figure D.2: Results of the Johansen Cointegration test

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Figure D.4: Summary of estimated parameters of the VECM model

AR orders AIC VAR AIC BA only

1 102.62

12 100.93

1,12 101.08

Table D.2: AIC value for different choices of the VECM model

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Figure D.6: Results of the BDS test for the VAR residuals

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Figure D.9: Results of normality test for the residuals of the VAR model

Figure D.10: Results of the BDS test for the VECM residuals

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Month MAPE VECM October 2005 2.46% November 2005 6.05% December 2005 0.44% January 2006 5.62% February 2006 3.48% March 2006 7.24% April 2006 2.32% May 2006 8.63% June 2006 4.36% July 2006 6.76% August 2006 6.07% September 2006 5.34% MAPE total 4.90%

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Variable Estimation Std. error t-value Population size -0.175585 0.14102 -1.24509

Oil price -319.6234 330.913 -0.96588

Disp. Income 0.717319 12.3604 0.05803

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Appendix E

PMS Model Table and Graphs

Constant No. of carriers Oil price Pop. size Sept. 2001 AICC

0 1 1 1 1 -4.147 1 1 1 1 1 -4.231 0 1 1 1 0 -4.205 1 1 1 1 0 -4.261 0 0 1 1 0 -4.091 1 0 1 1 0 -4.323

Table E.1: AICC for various regression model choices

Constant No. of carriers Oil price Pop. size Sept. 2001 AR(4) MA(4) AICC

1 0 1 1 0 1 0 -5.095

1 0 1 1 0 0 1 -5.052

1 0 1 1 0 1 1 -5.545

Table E.2: AICC for best regression model, with added AR and MA component

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Figure E.2: QQ-plot of the PMS series

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Appendix F

EViews source code for simulations

F.1 Simulation of forecasts and confidence limits for an arima

model

’COMMENTS INCLUDED IN CAPS

’---REQUIRES INPUT---%series = "ba" ’NAME OF SERIES

!constant = 0 ’INCLUDE (1) OR EXCLUDE (0) CONSTANT IN REGRESSION !aantevaluatie = 12 ’SIZE OF EVALUATION PERIOD

!forecastsize = 120 ’HOW MANY MONTHS TO FORECAST?

%modelspec = "MA(1) SMA(12)" ’SET AR, SAR, MA, SMA ORDERS TO BE ESTIMATED !simsize = 500 ’HOW MANY SIMULATIONS?

!samplesize = 117 ’HOW MANY MONTHS OF DATA IS THERE? !simulate = 1 ’PERFORM SIMULATION YES (1) OR NO (0) !sort = 0 ’APPLY SORTING ALGORYTHM YES (1) OR NO (0) !lost = 13 ’HOW MUCH DATA IS LOST BY DIFFERENCING OF DATA

’--- CALCULATION BEGINS HERE ---matrix(!forecastsize,!simsize) prefore ’DECLARE MATRIX VARIABLES

matrix(!forecastsize,!simsize) foreresid

matrix(!forecastsize+!samplesize,!simsize) foredifseries matrix(!forecastsize+!samplesize,!simsize) foreseries matrix(!forecastsize+!samplesize,!simsize) foreseriessorted matrix(!simsize,!forecastsize+!samplesize) fsstrans !effsamplesize = !samplesize - !lost

IF !constant = 1 then %constant = "c " else

%constant = "" endif

smpl @first @last ’APPLY DIFFERENCING TO DATA series d0 = %series

series d1 = d0 - d0(-1) series d121 = d1 - d1(-12) %model = %constant + %modelspec

smpl @first @last-!forecastsize ’USE ONLY DATA UP TO EVALUATION PERIOD WHILE ESTIMATING LS d121 %model ’ESTIMATE MODEL PARAMETERS

series resid01 = resid ’STORE RESIDUALS equation dy1.ls d121 %model ’PRODUCE FORECASTS smpl @last-!forecastsize @last

dy1.forecast(d,m=n, n=6) df1

!c1 = dy1.@coefs(1) ’STORE FIRST PARAMETER ESTIMATED !c2 = dy1.@coefs(2) ’STORE 2ND PARAMETERS ESTIMATED smpl @first @last

series dfor = d121 series residfor = resid01 if !constant = 1 then

%modelstring = @str(!c1)+"+ " + @str(!c2)+"*residfor(-1)" + @str(!c3)+"*residfor(-12)" + @str(!c2) + "*" + @str(!c3) + "*residfor(-13) + residfor" else

%modelstring = @str(!c1)+"*residfor(-1)" + @str(!c2)+"*residfor(-12)" + @str(!c1) + "*" + @str(!c2) + "*residfor(-13) + residfor" endif

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dummy(!i,3) = dummy(!i,2) - dummy(!i-1,2) next

for !j = 1 to !simsize for !i = 1 to !forecastsize

prefore(!i,!j) = @round(@runif(1,!effsamplesize)) ’FILL MATRIX WITH RANDOM NUMBERS TO USE FOR EMPIRICAL DISTRIBUTION SAMPLING foreresid(!i,!j) = dummy(prefore(!i,!j),1) ’FILL MATRIX WITH SAMPLED RESIDUALS

smpl @last-!forecastsize+!i @last-!forecastsize+!i ’SET CURRENT FORECAST INDEX series residfor = foreresid(!i, !j) ’ADD ONE RESIDUAL TO CURRENT RESIDUAL SERIES series dfor = %modelstring ’ADD ONE FORECASTED VALUE TO FORECASTED SERIES foredifseries(!i+!samplesize,!j) = dfor(!samplesize+!i) ’PUT VALUE INTO MATRIX next

foredifseries(!i,!j) = d121(!i) ’FILL TOP PART OF MATRIX WITH REAL SERIES VALUES foreseries(!i,!j) = d0(!i) ’MAKE SURE THE FIRST VALUES ARE THE REAL VALUES next

for !i = 1+!samplesize to !samplesize+!forecastsize

foreseries(!i,!j) = foredifseries(!i,!j) + foreseries(!i-1,!j) + foreseries(!i-12,!j) - foreseries(!i-13,!j) next

’TRANSFORM THE MATRIX AND STORE TO EXCEL FILE foreseriessorted = foreseries for !j = 1 to !samplesize+!forecastsize for !i = 1 to !simsize fsstrans(!i,!j) = foreseriessorted(!j,!i) next next fsstrans.write c:\forecasts.xls ’---SORTING (BUBBLESORT)---IF !sort=1 then

for !k = 1+!samplesize to !samplesize+!forecastsize for !j = 1 to !simsize-1

for !i = 1 to !simsize-1

if foreseriessorted(!k,!i) > foreseriessorted(!k,!i+1) then !temp = foreseriessorted(!k,!i+1) foreseriessorted(!k,!i+1) = foreseriessorted(!k,!i) foreseriessorted(!k,!i) = !temp endif next next next

’endif ’SORTING ENDS

’USING THE SORTED MATRIX, OBTAIN SERIES OF CONFIDENCE LIMITS !n25 = @round(0.25*!simsize)

!n75 = @round(0.75*!simsize) !n01 = @round(0.01*!simsize) !n99 = @round(0.99*!simsize) !n50 = @round(0.50*!simsize)

for !i = !samplesize to !samplesize+!forecastsize-1 smpl @first+!i @first+!i

series conflow50 = foreseriessorted(!i+1, !n25) series confhigh50 = foreseriessorted(!i+1, !n75) series conflow99 = foreseriessorted(!i+1, !n01) series confhigh99 = foreseriessorted(!i+1, !n99) series confmid = foreseriessorted(!i+1,!n50) next

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F.2 Simulation of forecasts and confidence limits for a VECM

model

’COMMENTS INCLUDED IN CAPS

’---REQUIRES INPUT---!aantevaluatie = 12 ’SIZE OF EVALUATION PERIOD

!forecastsize = 120 ’HOW MANY MONTHS TO FORECAST? !simsize = 500 ’HOW MANY SIMULATIONS?

!samplesize = 81 ’HOW MANY MONTHS OF DATA IS THERE? !simulate = 0 ’PERFORM SIMULATION YES (1) OR NO (0) !sort = 0 ’APPLY SORTING ALGORYTHM YES (1) OR NO (0)

!lost = 13 HOW MUCH DATA IS LOST BY DIFFERENCING OF DATA AND USING AR COEFFICIENTS if !simulate=1 then

’--- CALCULATION BEGINS HERE ---smpl @first @last

series bab = ba series bmib = bmi series eab = ea series ryb = ry matrix(!forecastsize,!simsize) prefore1 matrix(!forecastsize,!simsize) foreresid1 matrix(!forecastsize,!simsize) prefore2 matrix(!forecastsize,!simsize) foreresid2 matrix(!forecastsize,!simsize) prefore3 matrix(!forecastsize,!simsize) foreresid3 matrix(!forecastsize,!simsize) prefore4 matrix(!forecastsize,!simsize) foreresid4 matrix(!forecastsize+!samplesize,!simsize) foredifseries1 matrix(!forecastsize+!samplesize,!simsize) foredifseries2 matrix(!forecastsize+!samplesize,!simsize) foredifseries3 matrix(!forecastsize+!samplesize,!simsize) foredifseries4 matrix(!forecastsize+!samplesize,!simsize) foreseries matrix(!forecastsize+!samplesize,!simsize) foreseriessorted matrix(!simsize,!forecastsize+!samplesize) fsstrans !effsamplesize = !samplesize - !lost

’VECM PARAMETERS WERE ESTIMATED OUTSIDE THIS PROGRAM

%modelstring1 = "- 0.2556051861*( bab(-1) - 0.9886779704*bmib(-1) + 0.8649801996*eab(-1) - 1.556080634*ryb(-1) - 114780.7673 ) + 0.5282564328*d(bab(-12)) + 0.8214402712*d(bmib(-12)) + 0.298879077*d(eab(-12)) + 0.133688532*d(ryb(-12)) - 603.7782796" %modelstring2 = "0.01369383986*( bab(-1) - 0.9886779704*bmib(-1) + 0.8649801996*eab(-1) - 1.556080634*ryb(-1) - 114780.7673 ) + 0.09923645571*d(bab(-12)) + 0.3236762101*d(bmib(-12)) - 0.01870869858*d(eab(-12)) + 0.01303020487*d(ryb(-12)) - 260.9234507" %modelstring3 = "0.04026690849*( bab(-1) - 0.9886779704*bmib(-1) + 0.8649801996*eab(-1) - 1.556080634*ryb(-1) - 114780.7673 ) + 0.1849987031*d(bab(-12)) + 0.06646652068*d(bmib(-12)) + 0.4959754388*d(eab(-12)) - 0.06031674275*d(ryb(-12)) + 10170.7363" %modelstring4 = "0.3289418209*( bab(-1) - 0.9886779704*bmib(-1) + 0.8649801996*eab(-1) - 1.556080634*ryb(-1) - 114780.7673 ) + 0.436308588*d(bab(-12)) - 0.04815624783*d(bmib(-12)) + 0.4787866778*d(eab(-12)) - 0.3323894514*d(ryb(-12)) + 10669.15731"

’---SIMULATION---if !simulate = 1 then

smpl @first+!lost @last-!forecastsize ’

resid01.cdfplot(o=dummy1) ’CREATE EMPIRICAL DISTRIBUTION OF RESIDUALS dummy1(1,3) = dummy1(1,2) ’CALCULATE PROBABILITIES FROM DENSITY for !i = 2 to !effsamplesize

dummy1(!i,3) = dummy1(!i,2) - dummy1(!i-1,2) next

for !j = 1 to !simsize for !i = 1 to !forecastsize

prefore1(!i,!j) = @round(@runif(1,!effsamplesize)) ’FILL MATRIX WITH RANDOM NUMBERS TO USE FOR EMPIRICAL DISTRIBUTION SAMPLING foreresid1(!i,!j) = dummy1(prefore1(!i,!j),1) ’FILL MATRIX WITH SAMPLED RESIDUALS

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smpl @last-!forecastsize+!i @last-!forecastsize+!i ’SET CURRENT FORECAST INDEX series residfor1 = foreresid1(!i, !j) ’ADD ONE RESIDUAL TO CURRENT RESIDUAL SERIES series residfor2 = foreresid2(!i, !j)

series residfor3 = foreresid3(!i, !j) series residfor4 = foreresid4(!i, !j)

series dfor1 = %modelstring1 + residfor1 ’ADD ONE FORECASTED VALUE TO FORECASTED SERIES series bab = dfor1 + bab(-1)

series dfor2 = %modelstring2 + residfor2 series bmib = dfor2 + bmib(-1) series dfor3 = %modelstring3 + residfor3 series eab = dfor3 + eab(-1)

series dfor4 = %modelstring4 + residfor4 series ryb = dfor4 + ryb(-1)

foreseries(!i+!samplesize,!j) = bab(!samplesize+!i) ’PUT VALUE INTO MATRIX next

TRANSFORM THE MATRIX AND STORE TO EXCEL FILE ’foreseriessorted = foreseries ’for !j = 1 to !samplesize+!forecastsize ’for !i = 1 to !simsize ’fsstrans(!i,!j) = foreseriessorted(!j,!i) ’next ’next ’fsstrans.write c:\text.xls ---SORTING (BUBBLESORT)---IF !sort=1 then

for !k = 1+!samplesize to !samplesize+!forecastsize for !j = 1 to !simsize-1

for !i = 1 to !simsize-1

if foreseriessorted(!k,!i) > foreseriessorted(!k,!i+1) then !temp = foreseriessorted(!k,!i+1) foreseriessorted(!k,!i+1) = foreseriessorted(!k,!i) foreseriessorted(!k,!i) = !temp endif next next next

endif ’SORTING ENDS

USING THE SORTED MATRIX, OBTAIN SERIES OF CONFIDENCE LIMITS !n25 = @round(0.25*!simsize)

!n75 = @round(0.75*!simsize) !n01 = @round(0.01*!simsize) !n99 = @round(0.99*!simsize) !n50 = @round(0.50*!simsize)

for !i = !samplesize to !samplesize+!forecastsize-1 smpl @first+!i @first+!i