NAVAL POSTGRADUATE SCHOOLMonterey, California

AD-A238 336

. • D IC b

ELECTflJUL18S 1991

altD~aB DTHESIS

9I

AVAILABILITY OF AIRCRAFT SUBJECT TO

IMPERFECT PREVENTIVE MAINTENANCE

by

Michael J. Bond

September, 1990

Thesis Advisor: Michael P. Bailey

Approved for public release; distribution is unlimited

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Availability of Aircraft Subject to Imperfect Preventive Maintenance

12 PERSONAL AUTHOR(S)BOND, Michael J.

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16. SUPPLEMENTAL NOTATIONThe views expressed in this thesis are those of tne author and do not reflect the official policy or position of the Departmentof Defense or the U.S. Government.

17. COSATI CODES 1S. SUBJECT TERMS (Continue on reverse If necessary and Identify by block number)FIELD GROUP SUB-GROUP Availability, Imperfect Preventive Maintenance

19. ABSTRACT (Continue on reverse If necessary and Identify by block number)

This thesis studies the Impact that imperfect preventive maintenance has on the availability of aircraft and, as a result, thedecrease In effectiveness. We further consider the practicality of using an imperfect preventive maintenance model fordetermining preventive maintenance schedules. Simulation was used to recreate the operational environment In order tostudy the effects that levels of il nperfect preventive maintenance have on the aircraft effectiveness during execution of anactual fleet tactic.

20 DISTRIBUTION/AVAILABILTIY OF ABSTRACT la. ABSTRACT SECURITY CLASSIFICATION] UNCLASSIFIED/UNLIMITED E] SAME AS RPT.DE DTIC Unclassified

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Approved for public release; distribution is unlimited.

Availability of Aircraft

Subject to

Imperfect Preventive Maintenance

by

Michael J. Bond

Lieutenant Commander, United States Navy

B.E., University of Mississippi, 1979

Submitted in partial fulfillment

of the requirements for the degree of

MASTER OF SCIENCE IN OPERATIONS RESEARCH

from the

NAVAL POSTGRADUATE SCHOOL

September 1990

Author: "ich el J. Bond

Approved by: /4

/M Bailey, Thes Advisor

.Bl BloS n Reader

Peter Purdue, Chairman

Department of Operations Research

ii

ABSTRACT

This thesis studies the impact that imperfect preventive maintenance has on

the availability of aircraft and, as a result, the decrease in effectiveness. We further

consider the practicality of using an imperfect preventive maintenance model for

determining preventive maintenance schedules. Simulation was used to recreate the

operational environment in order to study the effects that levels of imperfect

preventive maintenance have on the aircraft effectiveness during execution of an

actual fleet tactic.

Acoession For 3

NTIS GRA&IDTIC TI 0Unannounced 0I

Justificatio

By -Distri~but ion/

Availability Codes

Avail and/dtDist Special

iii -

THESIS OISCLAIMER

The reader is cautioned that computer programs developed in this research

may not have been exercised for all cases of interest. While every effort has been

made, witbin the time available, to ensure that the programs are free of

computational and logic errors, they cannot be considered validated. Any application

of these programs without additional verification is at the risk of the user.

iv

TABLE OF CONTENTS

I. INTRODUCTION ........................................... 1

II. METHODOLOGY ......................................... 6

A. FLIGHT OPERATIONS ................................. 6

B. IMPERFECT PREVENI VE MAINTENANCE MODEL ........ 7

1. D escription ........................................ 7

2. Model Assumptions .................................. 8

C. THE SIMULATION .................................... 8

1. Prelaunch ......................................... 9

2. Launch .......................................... 10

3. R TB ............................................ 10

4. R ecovery ......................................... 11

5. R T S ............................................ 11

D. MAINTENANCE DATA ................................ 12

III. AN.A. YSIS ............................................. 14

v

IV. CONCLUSIONS ......................................... 19

V. RECOMMENDATIONS .................................. 20

LIST OF REFERENCES ..................................... 21

APPENDIX A. T = 25 HOURS ................................ 22

APPENDIX B. T = 50 HOURS ................................ 28

APPENDIX C. T = 100 HOURS ............................... 34

APPENDIX D. T = 200 HOURS ............................... 40

APPENDIX E. T = 300 HOURS ................................ 46

APPENDIX F. T = 500 HOURS ................................ 52

APPENDIX G. AEWSIM ...................................... 58

A. MAIN PROGRAM .................................... 58

B. W ARM UP ........................................... 64

C. LAU NCH ........................................... 66

vi

D . RTB ............................................... 68

&.RECOVERY ......................................... 69

F. RTS ................................................ 71

G. READIT ..................... 71

H. CONFIRM .............................. 73

1. SCHEDULE . . . . ................................ 75

Jl. PO P . . . ...................................... 77

K. DUMP CALENDAR ............................ 78

L. STAISTICS2 ............................ 79

M. STATISTICS 3 ........................... 80

INITIAL DISTRIBUTION LIST ................................. 81

vii

I. INTRODUCTION

This thesis will study the impact that imperfect preventive maintenance has on

the availability of carrier based aircraft, the potential of using an imperfe,.t

preventive maintenance model for determining an improved preventive maintenance

policy, and the strength of simulating the operating environment when determining

scheduled preN entive maintenance policy. The aircraft used in this thesis is the E-2C

airborne radar surveillance platform.

The E-2C is the primary platform for providing advanced, long range early

warning radar coverage for a carrier battle group (CVBG). Its ability to provide this

support increases the readiness state of the CVBG and the reaction time to a hostile

or unknown surface or airborne threat. To provide this support the aircraft must be

maintained in a mission capable state. This is accomplished in two ways. One, by

increasing the reliability of the systems which comprise the E-2C aircraft. Or, by

maintaining the present systems thereby increasing the availability of the aircraft and

subsequently expanding the amount of radar coverage provided over time.

The E-2C aircraft, as a functioning unit, exists in one of three states; fully

mission capable (FMC), partially mission capable (PMC), or non-mission capable

(NMC). The aircraft is made up of several surveillance, communication and flight

systems, all or part of which are necessary to perform the assigned mission. If all

required systems are operating normally the aircraft is FMC. If one or more systems

fail or malfunction yet the crew is capable of completing the mission in this degraded

state lAej, aircraft is said to PMC. If one or more systems fail or malfunction and

crews' ability to perform the missikr is impacted to such a degree that they can no

longer continue and must abort the flight and return to base, the aircraft is then

declared to be NMC. The systems and ,,hat status an aircraft in the simulation will

assume is defined in Table 1.

2

TABLE 1. SYSTEM DATA

WUC System Description MTITF MTrR Phase FailureStatus

11 Airframe 91.0 11.12 A,B NMC

12 Fuselage 2257.0 7.79 AB NMC

13 Landing Gear 58.0 5.65 AB NMC

14 Flight Control 104.0 11.01 B NMC

22 Turbo Shaft Engines 61.0 5.44 AB NMC

29 Power Plant 61.0 6.97 A,B NMC

32 Hydraulic Propellor 105.0 6.72 A,B NMC

41 Air Cond/Pressurization 164.0 7.72 AB NMC

42 Electrical System 46.0 3.09 NMC

44 Lighting 393.0 2.99 PMC

45 Hyd. Pnu. Power 196.0 7.66 A,B NMC

46 Fuel 120.0 9.29 A NMC

47 Oxygen System 291.0 4.81 PMC

49 Misc. Utility 645.0 5.02 PMC

51 Instrument System 80.0 2.42 NMC

56 Flight Reference 50.0 2.6 NMC

57 Int. Guidance/Fit 311.0 3.86 PMC

58 Internal Fit. Test 1806.0 2.44 PMC

61 HF Communications 322.0 2.14 PMC

62 VHF Communications 1290.0 3.32 FMC

63 UHF Communications 119.0 1.37 PMC

64 ICS 1806.0 1.37 PMC

65 IFF System 4514.0 1.66 NMC

71 Radio Navigation 903.0 2.4 PMC

72 Radar Navigation 2257.0 2.24 NMC

73 Inertial Navigation 127.0 2.14 N'MC

91 Emergency Equipment 3009.0 4.33 NMC

Time of completing Phase "A"/*B" preventive maintenance - 6,5/9.5 hours

3

Since all systems of the E-2C are subject to failure and repair consequently they all

must undergo maintenance. There are two types of maintenance that the systems are

subject to during their operational lifetime, unscheduled and scheduled.

Unscheduled maintenance, or repair, is performed when a system has failed placing

the aircraft in an NMC or PMC status. Maintenance action is then undertaken to

return the aircraft to a fully mission capable status. Scheduled maintenance, or

preventive maintenance, is performed at specified intervals of time for particular

systems. In the case of the E-2C, this will occur after T hours of engine run time.

At the present time, 200 hours is the designated interval, T. The aircraft then enters

what is called a "phased" maintenance period during which selected components

and/or systems are inspected, serviced, or overhauled. By performing this preventive

maintenance the system is kept in a "like new" condition and prevents failures. The

interval between phased maintenance actions is determined by the failure rate of the

system and the reliability standards. The time allotted for the performance of this

preventive maintenance task may also be predetermined by taking into account the

complexity of the system, the difficulty of inspecting, servicing, or overhauling the

system and the average skill level of the maintenance technician. [Ref. l:pp 165-167]

Perfect preventive maintenance is accomplished when the aircraft enters a

preventive maintenance period and all prefailure conditions are discovered and

corrected. Imperfect preventive maintenance occurs when an aircraft enters the

preventive maintenance period and, due to one or more of the causes listed below,

prefailure conditions are not discovered and the system maintains the same prefailure

4

state as before preventive maintenance. There are several reasons why preventive

maintenance might be performed imperfectly:

1. Poor maintenance techniques are used by the maintenance technician

performing the preventive maintenance.

2 Lack of training or skill on the part of the technician.

3. External conditions are less than ideal for the performance of the preventivemaintenance (eg., inclement weather, poor lighting, etc.).

4. Failed systems are repaired with faulty components.

5. Additional damage is done during the accomplishment of preventivemaintenance. [Ref. 1:pp. 165-167]

Thus, assuming that preventive maintenance is, to some degree, imperfect, how

strong an effect will this have on the availability and effectiveness of the aircraft and

what would be a better interval between preventive maintenance periods?

5

II. METHODOLOGY

A. FLIGHT OPERATIONS

In order to fully appreciate the impact that imperfect preventive maintenance

would have on the aircraft it was necessary to place the E-2C in an operationally

stressful environment. Carrier operations conducted in support of a sustained long

range anti-aircraft warfare (LRAAW) scenario was proposed to provide those

conditions. The scenario chosen is the current long range anti-aircraft warfare tactic

and calls for continuous radar coverage at an extended distance from the aircraft

carrier. For maximum coverage a single station had to be maintained with on

station relief, as one E-2C reached the end of its cycle time and departed, another

E-2C entered the station. The time spent on station plus the time spent in transit

to and from the ship to the station required the E-2C to be airborne for a total of

four and a half hours for this mission. The maximum airborne time of the E-2C is

five and a half hours since it is not capable of refueling while airborne. The on

station relief requirement coupled with the inability to refuel dictate that a minimum

of two E-2C's be in FMC or PMC status at all times. This will allow for a third E-

2C to undergo preventive and/or unscheduled maintenance. E-2C squadrons have

four aircraft attached but the fourth E-2C is usually rotated to the hangar deck and

undergoes an extensive (four to six weeks) corrosion inspection and restoration due

to the corrosive effects of the at sea environment.

6

B. IMPERFECT PREVENTIVE MAINTENANCE MODEL

1. Description

The model used to describe the state of each aircraft in the simulation is

represented in Figure I below:

p

0 - repaired or new state p. probability that preventive maintenance isimperfect

1 - operational state 1-p - probability that preventive maintenance is

2 - failed state perfect

g - repair rate of system i

F(t) - failure distribution of system I

a- transition when system returns to service

Figure 1. Imperfect PM Model

The model represents the three states that a system of the E-2C will transit

during its operation. State 0, the initial state, represents the system in a new

condition or after it has undergone repair or perfect preventive maintenance. It

remains in this state until the system is placed back in operation. State 1 is the

normal operating state, here system i fails at a rate determined by its distribution F(t)

7

or is preventively maintained at time T and returns to state 0. If it fails prior to

performance of preventive maintenance it enters state 2, otherwise it undergoes

preventive maintenance and either remains in state 1 with probability p or returns

to state 0 with probability i-p. State 2 is the failure state. In this state the system

has failed and is subject only to repair and is returned to state 0. This simple model

was embedded in a system simulation which reflected the dynamics of the LRAAW

tactic, preventive maintenance, and contingencies upon return of an NMC or PMC

aircraft.

2. Model Assumptions

The following assumptions pertain to the imperfect preventive maintenance

model used in the simulation. [Ref. 2:p 331]

1. The system undergoes repair at failure or is maintained preventively at

intervals of length T, whichever occurs first.

2. After repair the system is as good as new.

3. The unit after preventive maintenance has the same time of failure as it hadbefore preventive maintenance with probability p (0 < p ( 1), and is as goodas new with probability i-p.

C. THE SIMULATION

As was stated above in Section A, simulation was used to gather the necessary

data for analysis. The LRAAW stationing of the E-2C would provide the basis for

determining if flight operations would have an impact on the preventive maintenance

policy and, combined with the probability of imperfect preventive maintenance, p,

8

ascertain an improved nethod of preventive maintenance scheduling. A primary

motivation for simulating flight operations and studying its impact on preventive

maintenance is due to the goal of maintaining 100 percent radar coverage. During

carrier operations an aircraft might suffer failures which place the aircraft in a PMC

status rather than a NMC status but it could continue its mission. In addition,

preventive maintenance may be delayed until a more advantageous time, ie, until

more than one aircraft is in a FMC or PMC status, and then be taken "off line" and

enter phased maintenance.

The simulation puts the aircraft through the maintenance and flight cycles

repeatively in order that an aircraft may enter the preventive maintenance period

several times. The phases of the operating and maintenance cycles are represented

by five subroutines in the simulation (see Appendix G).

1. Prriaunch

During the pre-launch portion of the flight an available aircraft is readied

for launch. This entails a walk-around inspection of tlie aircraft by the flight crew,

engine startup and system performance checks. The purpose of the inspection and

performance checks is to identify failures which would preclude performing the

mission. If failures are found, the launch is aborted and an alert launch must be

scheduled to fill the station with little lapse in radar coverage. An alert aircraft is

the first aircraft which becomes available for launch. This will include aircraft which

land in an FMC or PMC status and aircraft coming out of a repair or preventive

maintenance.

9

2. Launch

The launch portion encompasses the portion of the flight profile from

launch until the end of on station time when che aircraft departs the station. A

failure during the launch segment has a more degrading effect on radar coverage

than during the warmup portion. During the transit to the station the aircraft

presently on station is preparing to depart leaving the station empty for its relief. If

the relieving aircraft fails, a lapse in coverage exists until a third aircraft is launched.

If this third aircraft is NMC and requires repair or is undergoing preventive

maintenance then the returning aircraft must be "hot spun" back to station. During

a "hot spin", the aircrafts engines are left running, it is refueled and a fresh crew

replaces the crew onboard. The elapsed time for this evolution, from leaving station

until launch and return to station, takes from three and a half to four and a half

hours during which there is no radar coverage.

3. RTB

RTB, or Return To Base, is the final airborne portion of the flight profile.

During this segment of the flight the aircraft has left the station and is handing over

the coverage duties to its relief. While still able to provide a radar picture to the

CVBG until it lands, it provides less long range coverage as it closes the carrier to

land. A failure rendering the departing aircraft NMC during the RTB segment does

not affect the current radar coverage to as great a degree as a failure during the

launch segment since a reFeving aircraft is transiting to station. However, if the

relieving aircraft fails during the warmup portion or while in transit to station in the

10

launch portion, a critical gap in coverage will occur until a third aircraft is made

available or either of the failed aircraft are repaired.

4. Recovery

This portion considers landing and maintenance of the airciaft. A

recovering aircraft is either FMC, PMC or NMC. If NMC it undergoes repair. If

the aircraft returns in an FMC or PMC status one of three actions will occur:

1. If the station is empty (no relieving aircraft in transit) then the returningaircraft is hot spun and launched back to station.

2. If an aircraft is not available as a scheduled relief for the aircraft in transit tostation then the returning aircraft undergoes turnaround maintenance.

3. If all relief requirements are met then the returning aircraft undergoes repairif PMC and/or phase maintenance if T hours of engine run time have lapsedsince last preventive maintenance cycle. It is during this phase maintenanceperiod that the imperfect preventive maintenance model described in SectionB is used.

During turnaround maintenance the aircraft is refueled and the aircrafts' systems are

inspected for any visible damage or failure.

5. RTS

RTS, or Return To Service, is the final portion in the maintenance cycle.

Aircraft leaving preventive maintenance and/or repair must undergo turnaround

maintenance after which they are made available for immediate (alert) launch or else

await a scheduled launch time as a relief.

11

D. MAINTENANCE DATA

The current preventive maintenance schedule calls for the E-2C to enter phased

maintenance every 200 hours of engine run time. At this time all systems subject to

preventive maintenance undergo either phase A or phase B preventive maintenance.

The two phased maintenance cycles, A and B, are independent preventive

maintenance periods. Both are accomplished after 200 hours of accumulated engine

time and it requires 6.5 hours and 9.5 hours, respectively, to accomplish the

preventive maintenance. Therefore, the preventive maintenance cycles are staggered

so that they do not overlap and interfere with the accomplishment of te other phase.

Table 1 lists the systems inspected and serviced during phase A and B preventive

maintenance, respectively.

In order to make the simulation as realistic as possible, it was decided to use

actual failure and repair times as reported by E-2C squadrons operating off carriers

during a deployment period. The data was provided by records maintained at the

Naval Air Development Center, Warminster, Pennsylvania. From the collected data,

five E-2C squadrons where selected. For each squadron a cruise period was selected

and from this period the Maintenance Action Forms (MAF's) were inspected for t ie

failure and repair times. The MAF documents record the aircraft bureau number,

repair and maintenance ac',4vity, the system affected, the type of failure or phase

maintenance action, total engine run time, when the failure was discovered,i.e.,

during flight, before flight (the simulation also allows for a failure to be discovered

inflight, after flight, etc.), what repair action was undertaken and the time that the

12

aircraft spent in maintenance. From the five E-2C squadrons chosen 27,344

maintenance level one (squadron level maintenance) MAFs were investigated which

represented 9,028 total flight hours. Of these 27,344 MAF's, 16,043 were submitted

for performance of unscheduled maintenance and had a "when discovered" code

indicating the failure was found during either the warmup, launch, or RTB portion

of the flight. From analyses of these maintenance action forms a time to failure and

time of repair were derived Zor each system. Limitations on the data provided

constrained failure statistics to a system as a unit as listed in Table 1 and not as

individual components which comprise the system. Due to this restriction

distributions were not available. Therefore is was decided to approximate the failuie

distributions to fit the system. For those systems which where not subjected to

preventive maintenance (electrical and avionic systems) an exponential distribution

was used and the failure rate computed from the data was input to the random

nui jer generator LLRANDOM II [Ref. 3] for a time to failure. For those systems

which underwent preventive maintenance (flight control surfaces, engines, etc.) a

Gamma distribution was used with the computed failure rate and a shape parameter

to produce a time to failure with increasing failure rate (IFR). The author felt that

his fleet experience supported these choices.

13

III. ANALYSIS

The simulation reached steady state at 60 hours and then was run for an

additional time of 6000 hours for each preventive maintenance interval. Selected

phased maintenance intervals varied from 25 to 500 hours. To assess the overall

effects of the imperfect preventive maintenance program the amount of radar

coverage provided during the 6000 hours was divided into 1000 periods of six hours

each. The amount of coverage provided during this period was then calculated as

a percentage and then averaged over the maximum time period to find the mean

coverage. These values were computed for several levels of p (the probability of

imperfect preventive mainterance), 0.00, 0.10, 0.25, 0.50, 0.75, 0.90 and 1.00. This

would cover maintenance accomplishment from perfect (p = 0.00) to imperfect (p

= 1.00). The results of each run are reported in Table 2.

14

TABLE 2. COVERAGE PRECENTAGE

Interval Between Preventive Maintenance

Checks (Hrs)

25 50 100 200 300 500

% .9277 .8829 .8284 .7858 .7822 .7720

0.00 - -std dev .1553 .1750 .1707 .1707 .1662 .1594

% .9150 .8609 .8222 .7883 .7827 .7540

0.10 -.-..-.std dev .1623 .1623 .1608 .1631 .1658 .1643

% .7564 .7687 .7662 .7583 .7505 .7724

0.25 -std dev .1631 .1642 .1614 .1647 .1632 .1588

% .7804 .7537 .7409 .7552 .7650 .7524

0.50std dev .1698 .1610 .1593 .1590 .1626 .1637

% .7660 .7591 .7629 .7616 .7629 .7568

0.75std dev .1604 .1598 .1620 .1629 .1572 .1580

% .7529 .7665 .7595 .7607 .7729 .7672

0.90std dev .1580 .1594 .1569 .1620 .1618 .1624

% .7621 .7699 .7474 .7521 .7541 .7576

.0 td dev [ .1610 .1616 j .1612 .1537 .1528 .1500

15

Since periods of coverag-, were not independent, batch means of the coverage

were computed for epch run and the variance computed from these means.

Coverage provided was then plotted against all intervals between preventive

maintenance periods andp values and is presented in Graph 1. For each preventive

maintenance interval individual values of coverage for 100 time periods were plotted

to study the variance in coverage and length of periods of little or no coverage. For

these graphs, the values of p = 0.10, 0.50 and 0.90 were plotted against p = 0.00 to

better visualize the coverage picture.

0.95- p-0.00 Legend

0.. " P - 0.10 - 0.000 0.9- .pmO- .10P-0.25

0.P -0.50

o 0.8 ... p - 0.75

0.75~~~~~ ~~~ - 0.. 9.... "........

- 0.75- -r ........ ... -----. p- 1.00

0L 0.7-

0.65 ....26 so 10 300 S0

T (hrs)

Graph 1. T .vs. Percentage Coverage

16

Analysis was first concentrated on the impact that imperfect preventive

maintenance would have on the mean radar coverage. From Table 2 and Graph 1

it may be seen that for short preventive maintenance cycles, 25, 50 and 100 hours,

the coverage decreases as p increases until the probability of imperfect preventive

maintenance reaches 0.25. After this value of p is reached coverage does not differ

appreciably and greater values have no noticeable effect. As the interval between

preventive maintenance increases to 200, 300 and 500 hours, it may be seen that p

has no appreciable affect at any value. Next, investigation of Graphs 2 thru 37 in

Appendices A thru F was conducted to evaluate the variance in coverage provided

from time period to time period.

At an interval of 25 hours between preventive maintenance actions, Graph 2

of Appendix A shows little variance in coverage at p = 0.00. Coverage is maintained

at a consistent level and intermittently drops below 80 percent coverage during any

one time period. At values of p larger than 0.00, Graphs 3, 4, 5, 6 and 7 there is a

greater variability in levels of coverage obtained in a time period and asp increases

above 0.50 gaps in coverage become apparent as aircraft fail and replacements are

not available.

At a 50 hour interval between preventive maintenance actions,as seen in

Appendix B, the effect of a less than optimum preventive maintenance interval

becomes apparent. At p = 0.00 (perfect preventive maintenance) there are time

periods of less than 50 percent coverage at approximately evenly spaced intervals.

At less than perfect preventive maintenance, p = 0.10 and greater, this same effect

17

is observed untilp exceeds 0.50. At this value and greater the failure rate overcomes

any positive effects of preventive maintenance and coverage fluctuates drastically

from periods of high coverage to periods of little or no coverage.

At a 100 hour interval (Appendix C) it is again apparent that there is al effect

induced by the amount of time between preventive maintenance actions. It is also

apparent, even at p = 0.00 and 0.10, that there is a greater variability of coverage

provided from period to period and fewer periods of 100 percent coverage and

increased number of periods with less than 80 percent coverage. As p increases to

0.25 and beyond the coverage becomes more erratic and vacillates between periods

of low (less than 60 percent) coverage to periods of moderate coverage (70 to 80

percent).

At intervals of 200 hours or greater (Appendices D, E, and i) between

preventive maintenance actions consistent high o: moderate coverage is no longer

attainable and though coverage in some periods reaches 100 percent this level is not

continued and drops during subsequent periods to little or no coverage. At intervals

of 300 and 500 hours it is seen that there is virtually no difference in coverage over

all values of p.

Periods of highly variable or moderate (70 percent or less) coverage are

unsatisfactory when conducting long range surveillance operations. With the E-2C

platform providing the primary means of detecting, tracking and reporting of

unknown air contacts these gaps in coverage that resulted from the simulation "blind"

the CVBG commander and allow for undetected aircraft to close the formation.

18

IV. CONCLUSIONS

This thesis studies the impact that imperfect preventive maintenance has on the

availability of aircraft and as a result the decrease in effectiveness. It further looks

at the feasibility of using an imperfect preventive maintenance model for determining

an improved preventive maintenance schedule and the use of simulation as a driver

for the model. For those systems that are subject to preventive maintenance, the

model assumes that maintenance is performed imperfectly with some probability p

and perfectly with probability i-p. Various levels of p were studied at several

intervals T between successive preventive maintenance actions.

The following conclusions are drawn from the results of the study:

1. There is a level of p such that above this value preventive maintenance isineffective and the failure rate of the system overwhelms any advantage ofpreventive maintenance.

2. As used in the model, there is a time interval T, the time betwee.i preventivemaintenance checks, above which preventive maintenance is notadvantageous. Availability of the aircraft and radar coverage does not changefor any value of p.

3. The use of simulation for estimating the impact of imperfect preventivemaintenance on aircraft systems in an operating environment is practical andshould be investigated for use in determining optimal preventive maintenanceschedules.

19

V. RECOMMENDATIONS

As new aircraft enter the fleet to replace the present systems, these aircraft will

have increased capabilities and longer lifetimes. Along with these better capabilities

will be more complex systems that will entail greater and possibly different

maintenance requirements. As the system changes, the maintenance technicians'

knowledge mu--t grow to the level of the system. Therefore the following

recommendations are made:

1. Based upon the output of the simulation and the possible causes of imperfectpreventive maintenance listed in Chapter I, a model of imperfect preventivemaintenance should be investigated for use in determining optimal preventivemaintenance schedules for present and future carrier based aircraft.

2. That the training of the maintenance technician concentrate as much onproper preventive maintenance techniques as it does on repair. That suchtraining include what a prefailure condition looks like and other indicatorsnecessitating repair or replacement of the system prior to failure.

3. That the intensity of carrier operations and environmental conditions underwhich preventive maintenance is performed be considered as factors inmodels used to determine optimal preventive maintenance schedules.

20

LIST OF REFERENCES

1. Bazovsky, I., Reliabililty Theory and Practice, pp. 165-167, Prentice-Hall, Inc.,1961.

2. Nakagawa, T., "Optimum Policies When Preventive Maintenance is Imperfect,"IEEE Transactions On Reliability, v. R-28:4, pp. 331-332, October 1979.

3. Naval Postgraduate School Technical Report, NPS 55-81-005, The New NavalPostgraduate School Random Number Package-LLRANDOM II, by Lewis,P.A.W. and Uribe, L., Monterey, California, 1986.

4. Bailey, M. P., Naval Postgraduate School, Department of Operations Research,1990.

21

APPENDIX A. T = 25 HOURS

- p = 0. 0025 Hr PM I nterval1- p = 0. 10

I:I IJ a IIluI I , if 1 I 1I 1 f 4 A11 1I ~~ i I I l i Hi j l 1 .1 ( i t I I I lil

n jj~ I I III II I ItI Ii I I I I I

011 1 I I 1 I V 11I l 111. 1111 f I I I I I I II 1 1I F I 'i II I I 1 9lil t]l~ II I 1 I II

-) I I I i I 1111' 1 11 1111 Il I I 4 .. ..I

It I I I I I I 1 1 it I I I itI... I I I f I lI I 1 1 I I I I I I I I I

1)) u:I 1 11 11 11 1 I IIII ll 11111 Ii I IIit I

0 2 ..... ......... . I~. L I ....... I ... 4 .............. .......... 4 ... ..... ....

0- . . ............. .. i . .............. ...........1. .............. i............... . . . ..............

0 20 40 w0 logi

Time Period

Graph 2. (P=0.O .vs. P=.10) : 25 Hr

22

-p = 0. 0025SHr PM I ntervalI-p =o0.25

r r't T

1 1 MI I A l I 1 11 11111111 1 1 I I I I If I1111 itI( I (I

Iii:fl( j .. 1 -1 +... il ... ... I iM. 4 A I i i

ifg R~l f I ~ 1.1j? 11 1 1(1 1. .1w 11 1I 1 111 11 f IJ M II I Ii I I ilI Ii I

II1) V111 II I I1 1 1

UI Il 1 :~I1' li

Li II I Lii I I :I I 1iii Ii 1 1 1 ! I f ii

lii II I I lii II : v I II I i*4-.J 0,4 -i-Zl ............ ..... LA I .. 1........ I I ... ..... .....I.

It) I I! 111 Ill I 1 1 1 le I I)()I jIf II III 1 9 1 11 1

40 '100I

0.ph (P 00.v. =.2)2U

23

-p = 0. 0025 Hr PM Interval-p = 0. 50

..... ... .....

iili 1 1.1 it I~ L I N ,uJI

I it L I I M 1 11. lit11 $1 1 1 1 (iI II I VI IIII A n! i t I I Ill I I

....... .......... ii tJD' . I iI I IIIl i I I i I ~V 1 It 1 1

I II II I I I I iI It :I I I I 11 )I

I ltI I1111 I1I1I 11 11o ... .... 4.1 .... 444 W ..... 4 1..4 ........ .1 .- -4 .. 11....

C)II II I II I I I 1 1I: I t IIt 1 11 I I 1 1I 1

II II IIIt I Ii) I II II 1I 11

Q.. 0.2..... . ........

.. ...... . .. U ......... ...... . ...

C20 40 go so 100

T jme Peri!od

Graph 4. (P .00 .vs. P=0.50) : 25 Hr

24

-p = 0,0025 Hr PM Interval -- p = o,75

-', ] ] m r -],, " '" ' '" ' / l ' i "] , .... , ... ... ....... .. ,,-- '•~~~Ml rII/Il

III .... I 'r' ""'1 I I I I I I I I I

1111 1 11 1 1 I I I I I I il1 1I Iii I I lI II I I ' i I I i1'I I I 1

If II I I II I II I I I I I I it

*II II II II f II I I III . I I I I II

1 I 1 II 1 1 1 I II I I I I I i t

I il t I I I I I II I I I I* 04 -I ..I .. ..................

4'' ' .......................... |'""T ...... ..... .wr "ll .............. fI1 it . .. ......... ... 14 4 1...... . 1...t....

I' ill II I lii III III ICI Ill II I1! Il II II I .) I 11 1 I I Ill II IISiI I I II II

1 H11 I1. f:II II II I

1 I II II I i IICLill I i It ... 11 .........

I fi I V II:[I I i I iI.... .... .... .... .... ... ..i .................. ......I T ............................... .-.. ....... ....g ....... .. .. ..: ..................., -

0 20 40 so SO W00

Time Period

Graph 5. (P = 0.00 .vs. P = 0.75) : 25 Hr

25

-p = 0,0025 Hr PM Interval -p = o.9o

..... . ... .... ....... .. - "L 1 Il IR @ . .... .. .... .. .. ..... ......I IjA III I 11 k I j ... .IJV l u

" ! I II ti l~ II ' I .. I | .. . .1 I Ill I I l II I I.J I I II. II IV II ii .1 I II II V II I I II I

if i I if I If iI I I I I I I 1 )

C) .I ;..I.II , , .I. I I II .i.. .....I..

1 Ii I I 1 l 11 1 1,SII II I I I I I li l LI I I I I I I II I'

a, 4 ......... I .... ...... . ... # 1. ..... .l..... b.l. ................... I....IV :... .............. .. . ..... ., ......i,0I II 'II IIUIi II I !I LI I II

. I II 111 II II! III I(DII I'I L '1I i II

II II II I I LI V III I, II 1 I I I l I I

I II I I I I I

- A- " T....1........................ rl......... It............. ....................... I.... ' ............. ........... l............. il

-!II I I I 1.1 LII II I I II II II

2) I 'II l I I If II iIiIfi I I It

(:1) II I Ii""!............................2... ........................................................... ..... ....... .

iI I 11I IIi liit !l iI

I 1111 IV I11l t i

-.i............. .. ....... ... .................... j L.......... L... ... __ _ _ _ _ _ _ __ _ _ _ _ _ .0 20 40 OD So 100

Time Period

Graph 6. (P = 0.00 .vs. P = 0.90) : 25 Hr

26

-p =0. 0025 Hr PM Interval -p =1.00~

I ~ I I All H iii 1 1) A

- N 1 1 i l I I I 1:111 I I III I I III I I I 'I f--. . ... IJ.. .. 1 1:011 1 1 .I l. IIJ? ..I.. . . .1 ,

1, 1 1111111 1 I N 1 1 I i I I I I VI IH t filt ifIM I 1 11:1 1 11 1 U lf 11 i t I fl

16i :11 If1 11 I1 I 1 1 :lII I It I il I H 1 1 1 If it

C..) : if Ul 11 I 1 1 M ill I I 1 I 1l11 ,i i f ' I III) Ill it 1 1 1 iii I I til II V I I 11 I1 I~i

IIfII 144 ,14.I. ................. W .I

Ii,4 1 1. 1 Ill III I I I IIIt VI II I I Ii I I 1 1 l

It. I....TI

it 1 V 1 1

........... I..........I......... ................. U............... j.........

0 20 40 so So iDD

Time Per iod

Graph 7. (P= 0.00 .vs. P =1.00) : 25 Hr

27

APPENDIX B. T = 50 HOURS

-p = 0. 00

50 Hr PM I ntervalI p = 0.10o

.............. l.,* -VT-T

AI Al I A A I tI I I I

I ''.1,A i 1u I IV I I I I II~ R h I I I iU Il l) LI 11 1 1 1 I

Ill 11 111 if II I I I I

ill I I I I I I Ill i I I Il I I LIIll I I ul 1 I1 1 ifi I I I I ifI I I I I I

>2 Ill I I UI l I I IIIIII I N till I 1 1 1

oi aS ... 14 I ...... 4 . . ....... . 1 ........ 1 141

I1 1, 1 1 1 1I 1 1111111 1 1 1 l iII 1 1 1 1I I 1) fill I I I I I ifI I I I II I 1111 L I if if

I. I I f I Iilli l iI iI I I I I I f11 I I I I If i

7 ** ...:.......... It T......... .. . ..... I TI1 4cI II 1 II Ill If I

'a) I IIIf Ill II

I 111 T:

If

0 20 40 OD so 100

Time Period

Graph 8. (P=0.00 .vs. P=0.10) : 50 Hr

28

-p = 0. 0050 Hr PM IntervalI -p = 0.25

.~~ ~ ~ .. ... ........ . ... .. ..."... .. .. ..... ..... ....I~~ t ~ I~'n Ill

II I I IV 1 1 1l 1i111

ii iI I 1k . I II I. iI Ii I I f 1 1 l i iI I I I

. t I I 41 Lt .4 .. Ii . ...... iI ii I I ItI I i t I I i t IIII I

i i I i I Ii I lii III I Ii

4 -- ..... .......... I ....... .. 4 4 1,1 4 -4. . .... I ........I.-fill 1 0 1i 1 11 1 Il I I I Ii

T',i i I gii Ililly1 i il ti

II Ii il I

ii I........ V ...... . ................ ....... .... ...

0 0 40 so so 100

Time Period

Graph 9. (P =0.00 .vs. P =0.25) : 50 Hr

29

-p= 0. 005D Hr PM I ntervaI-p= 0. 50

...... .... .... ... . ..

A31 1 A 111 I I ~ Il 1 I I It A11 IIl I Ii t i I I I I Itl It I I I I I IIi III I I I f 111 11 1 1I iI 1 I

li iiI I I ~ I t I I I I

. .. l....J..... ..I..11............II ~ ) i IIIiI I I I (I

11I i I it Ill 1iI I I11) 1ii I iii I IlI if V i 1 1 1 1 I II

I I I lii I I 1 111 1 1 14..' ~Il it I I I I

--. V................................. .... ........

I. it i

if I

0 20 40 OD so 100

Time Period

Graph 10. (P 0.00 .vs. P 0.50) : 50 Hr

30

-p = 0.0050 Hr PM Interval -- p = 0.,75

I 1 1 '. I ' ' '... ....... .. I ...

I. I I, 1JAIvi Ai Ii Ii t~lliltl IiiI Vii: ii ,ii

i 11 11 i I lil .iJ* .....

IlI I II : 1 1 ii 1 1 1 1I I i I I I I li lt W 11I 1 i I I I I t i if itV "..

II I i i I I 0III II I l I I1 II II i"

o a.-. Li 1-i4.i 1 4Ui41 14ii .................0 II 1 i V I1l I I ltl V i I I I I I i I I Ii

i II I liii I I uii I ill II I I I ii IIi i i i ii ii lii i i ii I Vil I I I l 1 i ii i l

i I I I ill I I ill i I i I I III l II Iii li II i ii U II IIi I 1 !i II I i ifII II I l l III il I It II I II 1 l 1 11 IIIII IN I I I I :u II Il I 1 .1 1 11

... .... ~ 1 ~ ~ rr i .. .... .:....... 1 .....SI II Iii ii I il II IIII ii IIa.) ii I ! ' I I I I I i l IIII) ii .. I - . i ill UII I) ' I iii I

G.) . . . I If . I I I . .I r I ii i

ii II !

..... ................ .. .... ............. ............. ................... ............................. ............................... -I i I i , I.......I.,.j...t , ..L I

20 40 sd so 100

Time Period

Graph 11. (P=0.00 .vs. P-0.75) : 50 Hr

31

-p = 0. 0050 Hr PM I ntervalI p = 0.90o

I I AAI I I 1 IfAA I Ii I ( I II I t It if I

A I L I I I t II I I I 1 1 1 11 it I 1 1111

(I 1 I I Ill I I li I 1 11 .11 1 II I11 I 111 111 L

0 s 1.- 4I J.... ~. I 14 1.1 1 .. . .. ..44...14 ..... 44 ... .....4...Ii I I ifI I I I I I I I 1 111 Il I

It) 11 11 11 11 1 1 1 11 I

UI II II

) U 11

Q. 0.2 .......... 1

Time Period

Graph 12. (P= 0.00 .vs. P 0.90) : 50 Hr

32

-p = O.,OO05D Hr PM Interval -p = i.00cl

..... ... A 1 ...... -- .. .. ...... ... .. ...

I i I L~ I iI

11 1 1 I 1 1 1fL :1 1 i t I 1 1I I I I I I I I I I ii I Iii

ltIII ( I I:I II I U:II I I I I I I Iii I 1 11111 1 1ii 1>0 IMi I II ; I I Ii I l II I 1 :1 1 1!.. I Ii liI I I~ I

L)II ::111 1i 1i U I I I) II 1 111111 II' i ii I

i Ii III ii I Ii I A iiI I I I I II I Ii I I 1 ill 1

I 1 11 1i I A I I1 I I il Ii I 111 ii l f .i .I i.. III- I ..... -1..... ... ....1 .V .... V I " -A11 1I 1 1 lilt I II 11 11 1i

04~~I 11 fir r 111 ,I 1

cv III I

0.2 WI 1:u................. ............. ..I . ...............

0 20 40 $0 s 100

Time Period

Graph 13. (P= 0.00 .vs. P= 1.00) : 50 Hr

33

APPENDIX C. T = 100 HOURS

-p = 0. 00100 Hr PM I ntervaI-- p = 0. 10

... .... I* 17 ... .....11 I ...... ...I~~ ~ I lI I Ili I i" f T..

0) 1 1 f I1 i I I iL

vI Il I I I 11 11 1 Il I I I it

4-oJ ...... ... r.1 ~ .. I 1. ..... ......................... ....... . .I U ...... ITi II I ........ 11

I I I I 1 11

(a) I IIl

U 1 II11 2

IfI

0 20 40 IO w Ic

Time Period

Graph 14. (P 0.00 .vs. P 0.10) : 100 Hr

34

-p = 0.00100 Hr PM I nterval-- p = 0.25

I3 1, l, -J i l l

Ii l ' I I I 1 I I I

I l t II I II II |1

I I I f I I I I I I 1 i lf I II .0 , 0 ".6 '" "'-4 ...414 ... ........ 1. '" ..I ... V 1 ........ ................ 4..... 1-""'" ............ ' " I I .. .1 .1. ... .."..I .-Ill II | I t~II II I l III I I

I I I I f I II III I II

( II IIIJ i II~' 11 II I

4 J 0 a. ... ........ T ...... ... .............. ... 1 1 t- T-IT. 4 . .

C fI

I I I I I II

I i V II II I III i II II ! ISI

itil IIII i ItI" . I II III II I II IIII II IIII11 I II II

O - o 2 .. ."1...... ............ ..... ......................... .... .......................... ....... ..' .................. .. ............ ................. . "

44 04 Ir IrII I.II li II I II.lii I II

0 0 40 410 soa 10

Time Period

Graph 15. (P=0.00 .vs. P=025) : 100 Hr

35

-p = 0. 00100 Hr PM I nt-erval-- p = 0.50

I I II A It I l~tRI Il Aj IIit it I I 1 I

I I I i I I~ L I I I

I1111 1i 0 Iti11 1 1:l 1i I i 1 11 1 1i 11 1i 1 1: 1 1 1 1 1 1~

ii i I i I l I ii I I Ili I I I I 1i1:

it i.. ..... ..i .... ......v.... .... .. .. ..... i i

i1 11 li IJ I I it I i I I ii I l iii ii i~i I tll i I I il li

ti.) li IJ 1 11 H I I li

T4 .i............................ . TiU I

tI I i f 1U i

0. ..... ... .... ... ...........................

I ..... ... ... L I I ........................ L ........L......

040 so SO10

Time Period

Graph 16. (P =0.00 .vs. P = 0.50) : 100 Hr

36

-p = 0.00100O Hr PM Interval-p = 0.75

Iil Aa il

... ...... I .... I .I .. (IA .. ........ .I I I I I H 11 11 1 1 1 1 11 IV 1 I I ( I I I

I I Ill I I I / I I I I I i 11 1I

1) :111 Ir it U 1I I Itil v 1 1 Ii itiI ll liiI ii I it 1 1 I i fI i

I0 t ill 1 1 11 I ifI I I I I

~IFIT ....... int ................ r

Cl) I ii LI ltI

H I till I

................ Li ... ............ .............

0 20 40 GD so01

Time Period

Graph 17. (P 0.00 vs. P 0.75) :100 Hr

37

-p = 0.00100 Hr PM I nterva I-- p = o. 90

-+1.,-~~~~~~~~~~~~~~ ..... ........,.'7',,,+'",,..... ....,.-.it I I'

" l I I / II Illi l ! / l Ji l ' 1 I "

i I I l I I I I I I I I I I I I I

I I I II I I tl I 1 1

II I I 1 f I III I iI I I>I I I I 11I1111]111I

.i I = 4 " 1" ' '"....... .... " -I ....I.............. .. . .... I1 ................... I ... ...... I .......... I ... .1 14 ........ 41.... .

u3 II HA ('1I (I I~1P 1 1 -1

I I I I III i I I ll II

.. I I II I I I I I ,1t I Il 1 II i I

Ii" 1.1 II 1 1 I itS 4 - "" ... .. .............. 11 ................. . ....1 ................... 11 ....... I .... I .......... 3 -.........i ili , ,Ulii 1:1II IIfi Il V: 11 I I I 111I Ii R: lii Itl II I

Q.2 ........ ................... I I l ilt 1i................ Ii .... -- -' - * .......

II: III V

0l i0 40 1

i

Time Period

Graph 18. (P =0.00 .vs. P = 0.90) : 100 Hr

38

-P = 0. 00100 Hr PM I ntervaI-- p = 1.00o

A I I 1 :1 1 II . is I II I I I it

I1 A 1U11 1I 1 1 1I 1 1 1

I I if I II1 I I I I IIifIII I1 I II I I Ii I I . II I I I I til 1

>f I I 1 1 t I I I I 1 II 1o1 1 ..... ... ....... 11 ...... .. .. I ..... -4 4...I 4 . fl I II H

it II 1 11 1 f f I I I Ii ll) IVi Il I I I I I I i t I i f I 1 111VI tI I II I I I I I I I 11 1I .1 1 11 I

.4 ............... ...... .I ......... T11 T I ..It:

iI I I II lu I

I I HI 111.. .. ... . .. ........... ... .. ... .... ........ ...... .... .. ...............

it

0 040 sow 100

Time Period

Graph 19. (P= 0.00 vs. P 1.00) : 100 Hr

39

APPENDIX D. T = 200 HOURS

- p = 0. 00200 Hr PM I nterva 1-p = 0.10o

I II Ii IA %VI It Ii1: 1 1 1 ill 1 ~1

*.lL ~It-4 ..... ..4 II1 I f Ii I I~ I11 114 .I

III 111~11 1 1 1I I i I I I I i

Id Itf 11.1 II IUJ i I 1 1 I I

V I I ii::I[ Ii it

IT I II I 111 .9 II

ci.) It ;i III Ii I 11 1

1 II ti 1.1 UI

.... .... ... .... .... ............ I ........ ............ I.. .... .. .... .. .. ...). .i

o 4D wi w i0

aTim PerIiod

Gaph2.(=.0v.P01):20H

II Ii 40

-p = 0.00200 Hr PM IntervalI-p = 0.25

................ rr

I I F I I I A II A 2 I 1 1 I IiI 11

a)1. 4.1 -4I 4 1

L IfI Ii fi I I I I ~ I I

>' if u 11 1 I I

II if t I I I I II II l I

4- 0 ................ L ........... II ... ..I ............ .i 'I..i ... ..........

Tim Pe i l

0t41

-p = 0,00200 Hr PM I nterval-- p = 0.50

-rM, I.I ii 1 •f , 1 •11I I AA

SI II II I A1 ". 1 .... I .... I ...

ll I III N I I 1 '....

Ifl JJUYII . l I I III(L. " ULU It I i"

SI I II 1 l I i I.(3 ,,I,.,. 1 IIII II | 1 I Ii Iii_...!, !

.r ... ......... ... ....... ..-. . ................... .... .4 .............. ......... l.. .,

I IU i ~i illlI IIIi|

i III |1TI I I l "ii i! 1 II ii Ii I1

(a) I II II 11 I

(a) I 1i If III II II I ~

.. .. .. . .. ... ... ... ... ....... ....... .. ............

................. .......... . ...T ......................... ..... ... . . . . . ..T "

20I I0 so so io

Grp 2I ( 1 v)

i 2Ii I

, , , I 1 . , , , I

0 )40 1 30 100

Time Period

Graph 22. (P = 0.00 .vs. P = 0.50) : 200 Hr

42

-F? =0. 00200 Hr- PM I ntervaI-- p =0. 75

.......-.... .. ...... ... . T i ..... -.....II I IAIIA a

I!,,I p: (III I II Iiill I I I I 1 11 f1 III 1 1 11

0I 1 F I I III:1 I I I III II I I 1 0 11 1l11 11 I

a a . I I 1 9 H 1 11 1 *i I *..I I ifI III I I 1i 1 ~ 11 1 I I(1 i. . .: lI

I II1 11 I ts 1 1 1 IIIII

LV I U II f I 1 II 11 11 1i 1uli uI Ii it 11111 11 11 Ul I 111111

i f I 11 lII I1 1 if I I I Iji 'II I io o.............I4 4......~. !U 4

.)11 11 1 ft11 111 1:iII I I11 1 1 111 1I ':1 11u I 1 1( 1 :1 iI I

0i 20 40 ID00I

Grap 23 (P 0.0v.P0.5 0 HrI

* 0.)I Il ~ i 43

-p = 0.00200 Hr- PM I nterval-- p = 0.90

-" " "A nJ r' .......... .. "1" v...... F-c - --r . .. r... ....... .. -S 1I II I I i 'i A,' I I I I 1 .

i i I I t N I I I i I I I I

. .. .. . .. .. . I ; .t' I"I I ,1! -(a) T- T It 111

. I Ir ii ~Ir 1 1 f U II li HI . - I i i I I I ll I I ' '"I i I I !

> l I .I. . . . .1 ... ....... 4444. .. -. '-ilij -il ....... ........... Il .. 44. . .. ....... 4 ............ 4I . . .... . ..... ,!.

I I I I I I I 1i I 1 1 I 11 I II 111 fl II Ii ii II l i 'l .05II I IIII If II II I 1 I1

,iiith 1 1 1i i i iI

0 - ..' ............ ............. ................ T " I" . .. .............................. .. ........ ..... .............. .....

¢ "II II I I i I1i I( it ii II i I 'I

4'T iti r. . . . . . . . .............. .............

Ii II Ill, I(. I ii ii il

r] I iiI V

1: _ . - .:.. .. . .......... . ....... ..................... ............ .............................. . . ....... .. ..... ... ...

0 20 40 sod so 410

Time Period

Graph 24. (PO.00 .vs. P I0.90) : 200 Hr

44

-p = 0. 00200 Hr PM Interval--p = 1.00

ft*U A A Ik I 1 1

ii I I I ii I It II Ii I I

1 1 I I .1 ) 1 1 11 1 i II I I I I I

> 1I I if 1) 11 11ii 1 M

-1. -.4 ... .... ...... I .... V- 1 1- 4 - .i t 1 .... .... ... ..II

Ib. A I I I I II II i I I I ll I ii 111I1 I If I f i 1 9 11 1 1 M l I i

o 4 ....I.J ......II A~.U .. . . .. I.

lii IfI~II~Ill iiL l I U I I ilI Ii

I I 111T I me Per j II

G.aph 25. (P 0.0 vs P 1.00 : 200 Hr I

45 T....

APPENDIX E. T = 300 HOURS

-p = 0.0030D Hr PM I nterval-- p = o. lo

..... ..........I...... 1..........11... ....... ..........2. wI Il ' i i IA l111 I 11 * I I I i :

I I. I I I I I I I I I I I

a, a. I I '1 I ' i IA I l H I ~ 1 I |1il ! ":11 1 I Iii IAJI I I I II t I I I

.1.f'F IT.

0 . ............ .....

i.I I I I tl11 II IiIt i l I 1t.... ... i. I.. IHI)t . I.ll I V[ I .

S II II I ll 1- 1a, ! ' I I!II I II~ II II 11 11 IllI

o., "' .... ... .....T- "l. .".--........... ..... ............... .r .... .. rr 'r~ l ' " ''

! Il 1 tl; 111!I 112 U t IAIV l I III i, 1, 111 II I Ii ' II iII I I 111 I II II II

0.41T~... ..... ....... i 1

i~ if Q I l II I I

C U II tif Ili I I I II I

(1)I 1 1 21I lII I I I I i HC) l . I I t I 1 I! II 1

SI tI i I I Ii ii 1 11 U I .

!iIl I 11 I I t i

a -.......... ........ .1. .................. ... I........ ..

0 I0 40 00 so 100

Time Period

Graph 26. (P=0.00 .vs. P=0.10) : 300 Hr

46

-p = 0. 003OO Hr- PM I ntervalI-- p = 0.-25

til' TI I ~ Ll '~

I t

I i I ! ~ I I II It I I

I) I I I I I I I TI I I I 1 1.

o ..... .... 1. .... .... . ....... . ... ...

I~il I~I 1 111 I II I H il

L.. ............ I.0,0 40 so so1 IiD

V im PeriodI

Grp 27. (P1 0.0 vs PI.530H

I II Il I 47

p = 0.00300 Hr PM Interval-- p = 0 50

I .. . ..... . . . . . . . . . . . I .A I I I I I I A. I I I

I I I l i 1 l II I i I I

" I III I ~1 II

• I liI lI 1Iii

- iI I V I II Ii lj I..I I I II III ,

0.4 -'"" .... "r i'", ...... ,t ....... ............ ri W1'' i;-

,II III II ,~iIIII i I H l

II

( ,O alUIl

1................... ........ ..

SII II

I I II 1

0 - 4-. 1 ..................... 4 -..................... ... 11 ............. .......... ... .. ... .

Graph 28. (P=0.00 .vs. P=0.50) : 300 Hr

48

-p = 0.00300 Hr PM Interval-- p = 0.751I rrTIi l i ...... "" . . ... .. ." . ' ....... ... ..... ... .. ...... ............. T.1 ....... ..... "rP "T"r .......... r"

I I II I A I I I A Al I i I

I I l11I t I H ! I ifIi t I 1 l11 1 Aill 1 11 M 111 1 1I I 1t11

I I ' II"'. I" "1 " I I 'lf'lil : I J j/l~ tI ilt l !I I II I II I II 11

I t l III "l f i Is•~ ,"I- i Il 1 1I II I ! liL.ii i I 9 Ilii IlI li l! I .

(3)I i Ii III ! I I

0.4 ... . . . ..... ........ .... 1........." ... "

II I II III i Ii

I i I I "I II i I ii i I I I I i II I ici -'.i ii itH - -0.2 "". . .. . ............. .."1""". ....................... ...... .......... .... ................

/)I i II '1 li i

I I I I I

I . . I , , , I , ,

0 20 40 so so 1 "0

Time Period

Graph 29. (P=0.00 .vs. P=0.75) : 300 Hr

49

P=0. 00300 Hr PM I ntervalI-- p = 0.90o

..... ..... .... .... ..... ......... I ..... I. . II ii

N 11i 1~t tI 1 1 ill 11 F

.. .... ... I I I I i I I I I I a...... ...* ~~j

If I IIU.Ill .... tll

It l I I i I i I I Ii V

0r r. .... . ~ ...... I

.. .... .... ... ....i. ... .... .... ..... .... . .. .. .. .... .... ..i. .... .... .... ...

11 1 1 il0 40Il so N

II ~ Tm Peri II 1iodI

Grap 30 (P 0.0 vs P 0.0 : 300 Hr

II tll llt Iii50

-p = 0. 00300 Hr PM I nterva I--p = 1.00o

II I I I (I I I I It I itI till I II il III i 1 I1

I I iI Vi ~ i

1 1 A II

ifi 11 1 11i 1 1 1 11 1 1iii if 11I3 1 i I i

.1 Vi I i 11 ) 111 1i 1 1

0~ ~~~~ .... ........' U4..V. . .~ . I

l0) i I Y V I ki~ i ! I i

.. ... .... .... .. .. .......... ..........

Ii II............... ................. .. ........ ..... ..... ..

I 20 40 s

Tim Period

Grap 3 1.( .0is .0):30H

II I 51

APPENDIX F. T SO50 HOURS

-p = 0.00500 Hr PM I ntervalI-- P = 0.10

I - . .~ rin.....................I4

I I I I i I At I B A IIIJ I I It I AAil it A

U. D m I 1 13 Vl f-i 7 'l T IT-i

I i il I I i Ii W U.1 I I I It I 111 1i i 11 11 lNIl I I I I

0 .... ... .... .. ....... .. .31 1 ....... iii 4 ........ 1 11 ..

1 1: 1 1i ii I i i : I 1 1)

11 It 11 I I it: l i

................ .ia . .. ..... 1 ....... I .......... ..... ............Ii II II I r I

, I --. ) I I I I I . I I . I F0 20 40 so

ii Time Period

Grap 32. (P =i 0.0 vs .0):SOH

II II1 I52

-p = U.OO500~ Hr- PM I ntervaI-- p = 0. 25

..r I .. I.. - -... rr, ,,*

I 11 iV I I I

rV A II t ! I II I I liI .'i I I III~ AI I I I I A I I P 1 1

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ci I it~ I I LIlj C IL IIh(D iv I U11 LI I 1 II 1 1 1

1* 7 '* I - I- ....... -'TI* -11 .....11 ItI 1 111 11 111 11I

iir 117

.. ..... .. .. .... ..... .... .. ... ... ..... .... .. .. ..... . .... .... ....

(I. 20i I0 OD II1 IIL

(..) LII Liim PeiodI II

Grap 33 (P 0.0 vs. PI 0.25 : 50H

ci53

- p = 0.00500 Hr PM Interval1-- p = 0.50

.. ..... .... , ... . .. ....I11 t il ii 1 11 1 L IL I L i

1 11 11 11 1 flit Lit I ILiINI i ll11 I A I I IA I II I I I

111 lt11 .1 1 1 tI lIl 1111 I

1 It: I M11 I I 1 1: 1 It: IIcoI 1 111I 1 11: [1 I 1 Iit

I- 1111 tIll/LI 1 l11i I Iltil I fI I I II I; 1 Il 1 U lf

~~j) i fL . Sl.I V III1.1 if I 11

0.6 ... 1.. 44 t. .. IIIJJ. .4 ......... J.4 .... 44. II t ........ .I....I III I I 11 I

04 1 TI Hi 1 1:1

I I 1 I f IfI

S 0.2 .. ... ........ ......

.... ..... .

.......... .... ... ........................... ........... .................. ....... ..... ...........

0 20 40 so iD

Time Per iod

Graph 34. (P 0.00 .vs. P =0.50) : 500 Hr

54

-p = 0.00500 Hr- PM I ntervalI-- p = 0.75

A It .A I 1.IL 11A A AI i I I II H IC~ I I I (JI It

I I I I II I lt I 1 l I I l ill I I

I I I I I Iii It I I~ I t l I It

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..... .... IL IIL. ................ LI.V It I

.. .. .... .. .... .. ... .. .. ...I .. ..... .. .. .. ... .. ... ... .. .

............. ..............

0 20 40 so w 100

T ime Perijod

Graph 35. (P =0.00 .vs. P =0.75) : 500 Hr

55

-p = 0. 00500 Hr- PM I ntervaI-- p = 0,.90

trig11", .... A" i.

itM l I I I I k I I)A I AfI 1I 0h 1i1IN il 1 1 1 1 1 1 11 1~l 11l 1i1

rV i l hIt I I I1 I I I I I I I I1 I

0) 1 11 ' t111Ii 1:2 I !I II I 1 h 1i

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U Il 111 i I II I i 11....... T1 ...... f ... I..... I

0..... .... ...... .......V .......

...... ... ..... . ............ ............................

0 20 40 60 0

Time Period

Graph 36. (P .00 .vs. P =0.90) : 500 Hr

56

-p = 0,00500 Hr- PM I nterva I-- p = 1.00o

I till II AI I I t I HIIJIt I fl Iit t it I Il .I t it Ml l 'I I [, It I t I f I ii I 1 11

I I I I I I V I I AI i I 1 1I l I Ii I i I I 1~ 11 l

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Il I1 1 i I I tlI It I I i I I I I I I I I

......il..... I . ....

0 4 -... 4 - .. ... ... .... ... ......

I 1 Ii I I

1II

iit

t I I U 1

0 20 40 OD so 10

Time Period

Graph 37. (P 0.00 .vs. P~ 1.00) : 500 Hr

57

APPENDIX G. AEWSIMA. MAIN PROGRAM

CCC AIRBORNE EARLY WARNING PLATFORM STATIONING SIMULATIONC BYC LCDR MICHAEL J. BONDCC THIS PROGRAM SIMULATES THE STATIONING AND RECOVERY OF

AEW PLATFORMS IN A DEFENSIVE AAW POSTURE AROUND THECARRIER BATTLE GROUP.

C IT FURTHER SIMULATES THE REPAIR AND MAINTENANCEPERFORMED ON THE PLATFORM DURING THE TURNAROUNDPHASE AND PERIODIC MAINTENANCESCHEDULE. ASSUMPTIONS MADE IN THE MODELING OF THEMAINTENANCEPORTION ARE:

A) REPAIR IS PERFECT AND RETURNS THE SYSTEM TO A GOOD ASNEW STATE.

B) PREVENTIVE MAINTENANCE IS IMPERFECT WITH PROBABILITYP AND PERFECT WITH PROBABILITY 1-P. IF THE PM IS IMPER-FECT THEN IT MAINTAINS ITS ORIGINAL FAILURE TIME. IF ITIS PERFECT THEN THE SYSTEM IS RETURNED TO AS GOOD ASNEW STATE.

C) THE PLATFORM UNDERGOES PREVENTIVE MAINTENANCEACCORDING TO A PLANNED PREVENTIVE MAINTENANCE CYCLE.

CC PARAM(1) = A/C NUMBERC (2) = ENGINE START TIMEC (3) = STATION ASSIGNEDC (4) = STATE (FMC, PMC, NMC)C (5) = TOTAL TIME ON ENGINEC (6) = TIME TO FAILUREC (.)=CC STANUM Station number a/c is assigned to

ABORT Aircraft fails on deck or on stationQUEUE Aircraft in FMC or PMC status for assignment

58

EMPTY Flags station as empty, no replacementFLAG Flags program to fill empty stationMAXAC Maximum aircraft on flight deckHANGAR Storage space for PARAM0MINFAL Minimum failure time

FALTIM MTrF for system iRFAIL Random time to fail for system iMAXPRM Maximum number of parameters in PARAM(ENDURE Mission flight timeTS Time to stationTOS Time on stationNOW Current program timeSTATUS Aircraft status (up or down, for stats)COVER Station is filled (for stats)DTIME Maximum down timeFLYTIM Total time on engine from startup to recoveryHSTIM Hotspin timeTATIM Turnaround timeREPTIM MTTR for system iRTIME Random assigned time for repair of systemHTPM(1) Hours till next pm phase A for aircraft #HTPM(2) " BINWORK(#) Denote" whether system i undergoes phase A/B pmPERFPM Random probability of perfect pmIMPERF Probability of imperfect pmMAXTIM Maximum running time of program

C*8 88 8 88 8 8 *ss* 8** t *88t~~8s888*i88*8*8*s*88*88**~ssst$ 8*$s88

C THIS MODULE CONTAINS:C -MAIN PROGRAMC -DOITC

PROGRAM AEWSIM

C MAIN PROGRAM

CALL READITCALL CONFRMN=1

10 CONTINUEIF(N .EQ. 1)IMPERF = 0.0IF(N .EQ. 2)IMPERF = .10

59

IF(N .EQ. 3)IMPERF = .25IF(N .EQ. 4)IMPERF = .50IF(N .EQ. 5)IMPERF = .75IF(N .EQ. 6)IMPERF = .90IF(N .EQ. 7)IMPERF = 1.0WRITE(12,*)'RUN NUMBER ',NCALL INITZECALL DOITN=N+1IF(N .LE. 7)GOTO 10STOPEND

C***s.sssssssssssg*.S*..sss*.*.**sgtslsssss**.gssss*.*.*s s**.**

SUBROUTINE DOIT [Ref 6]

C THIS SUBROUTINEC - HANDLES THE PASSAGE OF TIME AND THE EXECUTION OF

EVENTSC - STOPS AS SOON AS MAXTIM TIME UNITS HAVE TRANSPIREDCC EVENT 1 IS THE PRELAUNCH/WARMUP PHASE OF THE LAUNCH

CYCLEC EVENT 2 IS THE LAUNCH OF THE AIRCRAFT TO STATIONC EVENT 3 IS RETURN TO BASE COMMAND FOR EACH AIRCRAFTC EVENT 4 IS THE RECOVERY OF THE AIRBORNE AIRCRAFTC EVENT 5 IS THE RETURN TO SERVICE OF THE AIRCRAFT AFTER

MAINT.C EVENT 6 IS A STATISTICAL EVENT

C1000 CONTINUE

CC .............. PULL AN EVENT OFF THE CALANDARC

CALL POP(TIMEVTNUMBER,PARAM)CC .............. STOPPING CONDITIONS CHECKEDC

IF((NUMBER .NE. 0).AND. (NOW .LE. REAL(MAXTIM)))THENIF(TIMEVT .LT. NOW)THEN

WRITE(*,*)'TIME IS GOING BACKWARD, TIME=',NOWCALL DMPCALSTOP

60

ENDIFIF(NUMBER .NE. 7)THEN

NOW = TIMEVTENDIFGOTO (10,20,30,40,50,60,70), NUMBER

c10 CONTINUE

CALL WARMUP(PARAM)GOTO 99

C ............... LAUNCH EVENTC20 CONTINUE

CALL LAUNCH(PARAM)GOTO 99

CC ............... RETURN TO BASE (RTB) EVENTC30 CONTINUE

CALL RTB(PARAM)GOTO 99

CC ............... RECOVERY EVENTC40 CONTINUE

CALL RECOV(PARAM)GOTO 99

CCC ............... RETURN TO SERVICE (RTS) EVENTC50 CONTINUE

CALL RTS(PARAM)GOTO 99

CCC ............... A STATISTICAL ROUTINEC60 CONTINUE

CALL STAT2(PARAM)GOTO 99

CC .............. ANOTHER STATISTICAL ROUTINE

61

70 CONTINUECALL STAT3(PARAM)GOTO 99

C99 CONTINUE

CGOTO 1000

ELSEC WRITE(*,*)'SIMULATION COMPLETED AT TIME',NOWC CALL DMPCAL

IF(NOW .LT. REAL(MAXTIM))THENWRIT(*,*)'SIMULATION ENDED DUE TO EMPTY CALANDAR'

ENDIFENDIFRETURNEND

,*** **** ,ll .************.*******tll 8 S **** ***sss*****s***SS$ ~S SSS SSISSSSS8***

SUBROUTINE INITZE

C THIS SUBROUTINE WILLC - INITIALIZE ALL STATE VALUESC - SCHEDULE INITIAL SORTIES FOR LAUNCHC - INTIALIZE THE EVENT CALANDER

CEPS = 1.OE-6L = 1.0DO 5 1 = 1,MAXAC

STATUS(1) = 1.0COVER(I) = 0.0

5 CONTINUEQUEUE = MAXACDO 20 1 = 1,LISTL-1

TIME(I) = 0.0EVTNUM(I) = 0

DO 10 J = 1,MAXPRMPASSED(I,J) = 0.0

10 CONTINUENXTEVT(I) I + 1

20 CONTINUENXTEVT(LISTL) = 0FSTEVT = 0AVAIL = 1

62

NOW = 0.0DO 100 1 = 1,MAXAC

HANGAR(I,1) = REAL(I)DO 90 J = 2,MAXPRM

HANGAR(Ij) = EPS90 CONTINUE

100 CONTINUEDO 110 1 = 1,MAXAC

EMPTY(I) = .FALSE.FLAG(I) = .FALSE.

110 CONTINUESTAGAR =0.0

STANUM =STATNS

DO 1301I 1,STATNSDO 120 J = 1,MAXPRM

PARAM(J) = EPS120 CONTINUE

PARAM(3) = REAL(STANUM)STANUM = STANUM - 1CALL SCHDLE(NOW+ STAGAR,1,PARAM)STAGAR = STAGAR + STAG

130 CONTINUECALL SCHDLE(6.0,7,PARAM)CALL SCHDLE(0. 1,6,PAR-AM)

C CALL DMPCALRETURNEND

63

B. WARMUP

SUBROUTINE WARMUP(PARAM)

C THIS SUBROUTINE WILLC -FLAG A STATION EMPTY IF NO AIRCRAFT AVAILABLEC -SAMPLES EACH SYSTEM FOR FAILURE DURING WARMUP

PERIODC -SCHEDULESC :A LAUNCH EVENT

CSTANUM = INT(PARAM(3))ABORT = .FALSE.IF(QUEUE .EQ. 0)THEN

EMPTY(STANUM) = .TRUE.FLAG(STANUM) = .TRUE.

C WRITE(11,5)STANUM,NOW5 FORMAT(1X,'STATION ',I2,' IS EMPTY AT TIME ',F7.1)

RETURNELSE

DO 201 = 1,MAXACIF(HANGAR(I,1) .GE. 0.5) THEN

DO 10 J = 1,MAXPRMPARAM(J) = HANGAR(Ij)HANGAR(Ij) = EPS

10 CONTINUEQUEUE = QUEUE -1PARAM(2) = NOWPARAM(3) = REAL(STANUM)GOTO 40

ENDIF20 CONTINUE

ENDIF40 CONTINUE

MINFAL = 100000.0DO 50 K = 6,MAXPRM

IF(PARAM(K) .EQ. EPS)THENCALL RANNUM(5,SEED(K-5),FALTIM(K-5),4.0,0,RFAIL)PARAM(K) = RFAIL

ENDIF50 CONTINUE

64

DO 60 K = 6,MAXPRMIF(PARAM(K) .LT. 0.5)THEN

ABORT = .TRUE.MINFAL = MIN(MINFALPARAM(K))IF(MINFAL .LT. 0.1)MINFAL = 0.1TF(K .GE. NUM)THEN

PARAM(4) = 1.0ELSE

PARAM(4) = 2.0ENDIF

ENDIF60 CONTINUE

IF(ABORT)THENC WRITE(11,70)PARAM(l),NOW+ MINFAL70 FORMAT(1XF4.1,'FAILED ON DECK AT TIMEP,F7.1)

CALL SCHDLE(NOW +MINFAb,4,PARAM)CALL SCHDLE(NOW+ MINFA,1l,PARAM)

ELSECALL SCHDLE(NOW + 0.5,2,PARAM)

ENDIFSTATUS(INT(PARAM(l))) =1.0

COVER(INT(PARAM(1))) =0.0

RETURNEND

65

C. LAUNCH

SUBROUTINE LAUNCH(PARAM)

C THIS SUBROUTINE WiLLC -LAUNCH AN AIRCRAFT TO STATIONC -SAMPLES EACH SYSTEM FOR AN INFLIGHT FAILUREC -SCHEDULESC :A RELIEF AIRCRAFTC :A RETURN TO BASE TIME BASED UPON FAILURE STATUS

ABORT = .FALSE.DO 10 K = 6,MAXPRM

PARAM(K) =PARAM(K) - 0.510 CONTINUE

COVER(INT(PARAM(l))) = 1.0MINFAL = 100000.0DO 20 K = 6,MAXPRM

IF(PARAM(K) .LT. ENDURE)THENIF(PARAM(K) .GE. NUM)T HEN

PARAM(4) = 1.0ELSE

ABORT = .TRUE.MINFAL = MIN(MINFALPARAM(K))PARAM(4) = 2.0

ENDIFENDIF

20 CONTINUEIF((ABORT) .AND. (MINFAL .LT. TrS+TOS))THEN

CALL SCHDLE(NOW + MINFAL,3,PARAM)ELSE

CALL SCHDLE(NOW + iTS + TOS,3,PARAM)ENDIFIF((ABORT) .AND. (MINFAL .LT. TTS))THEN

CALL SCHDLE(NOW, 1,PARAM)ELSE

CALL SCHDLE(NOW + TTS + TOS-O.5, 1,PARAM)ENDIFIF(MINFAL .LT. ENDURE)THEN

C WR1TE(11,50)PARAM(1),MINFAL+ NOW50 FORMAT(1X,F4.1,'NWILL FAIL AT TIME ',F7.1)

ENDIF

66

STATUS(INT(PARAM(l))) =1.0

COVER(INT(PARAM(1))) =1.0

RETURNEND

67

D. RTB

SUBROUTINE RTB(PARAM)

C THIS SUBROUTINE WILLC -SCHEDULE A RECOVERY TIME ON THE CALENDAR BASED

UPONC A) IF IT IS STILL IN TRANSIT TO STATION ORC B) ON STATION

IF((NOW - PARAM(2)+0.5) .LT. 1TS)THENCALL SCHDLE(NOW + (NOW-(PARAM(2) + 0.5)),4,PARAM)

ELSECALL SCHDLE((NOW + TTS),4,PARAM)

ENDIFIF(PARAM(4) .GT. 1.5)STATUS(INT(PARAM(1))) = 0.0IF(PARAM(4) .GT. 1.5)COVER(INT(PARAM(1))) = 0.0RETURNEND

68

E. RECOVERY

SUBROUTINE RECOV(PARAM)

C THIS SUBROUTINE WILLC -HOT SPIN AN AIRCRAFT' TO AN EMPTY STATION IF NO OTHER

AIRCRAFTC ARE AVAILABLE AND WILL NOT UNDERGO PREVENTiVE

MAINTENANCEC -REPAIR RECOVERING AIRCRAFT' IF REQUIRED AND PERFORM

PREVENTIVEC MAINTENANCE OF SYSTEMS NOT SUBJECT TO REPAIR

CDTIME = 0.0COVER(INT(PARAM(1))) = 0.0IF(PARAM(4) .GT. 1.5)STATUS(INT(PARAM(1))) = 0.0FLYTIM(INT(PARAM(l))) = NOW - PARAM(2)PARAM(5) = PARAM(5) + FLYTIM(INT(PARAM(l)))DO 51 = 6,MAXPRM

PARAM(I) = PARAM(I) - FLYTIM(INT(PARAM(1)))5 CONTINUE

DO 10 1 = 1,STATNSIF(EMPTY(I) .AND. (PARAM(4) .LT. 1.5))THEN

EMPTY(I) = .FALSE.PARAM(3) = REAL(I)CALL SCHDLE((NOW +HSTIM),5,PARAM)

C WRITE(11,6)PARAM(l),l6 FORMAT(] X,'HOTSPIN ',F4.1,' TO STATION %,12)

RETURNENDIF

10 CONTINUEIF((QUEUE .LT. 1) .AND. (PARAM(4) .LT. 1.5))THEN

CALL SCHDLE(NOW + TATIM,5,PARAM)RETURN

ENDIFDO 20 1 = 6,MAXPRM

IF(PARAM(I) .LT. 0.0)THENPARAM(I) =EPS

PARAM(4) =0.0

CALL RANNUM(3,SEED(I-5),REPTIM(I-5),0.0,0,RTIME)DTIME = MAX(RTIME,DTIME)

69

ENDIF20 CONTINUE

IF(HTPM(INT(PARAM(1)))/PARAM(5) .LT. 1.0)THENDO 30 K = 6,MAXPRM

IF(INWORK(K-5) .GT. 0.5)THENCALL RANNUM(1,SEED(K-5),0.0, 1.0,0,PERFPM)IF(PERFPM .GT. IMFERF)PARAM(K) = EPSDTIME = DTIME + HTPM(6)

ENDIF30 CONTINUE

HTPM(INT(PARAM(1))) = HTPM(INT(PARAM(1))) + HTPM(7)ENDIFCALL SCHDLE((NOW +DTIME +TATIM),5,PARAM)

C WRJITE(11,60)PARAM(1),NOW+ DTIME+TATIM60 FORMAT(1X.,F4.1,' IS IN FOR MAINTENANCE UNTIL TIME ',F7.1)

RETURNEND

70

F. RTS

SUBROUTINE RTS(PARAM)

C THIS SUBROUTINE WILLC -RETURN AIRCRAFT TO THE QUEUEC -SCHEDULE AN AIRCRAFT FOR IMMEDIATE LAUNCH IF A

STATION ISC EMPTY

DO 10 1 = l,MAXPRMHANGAR(INT(PARAM(l)),I) =PARAM(I)

10 CONTINUESTATUS(INT(PARAM(1))) = 1.0QUEUE = QUEUE + 1DO 20 J = 1,STATNS

IF(FLAG(J))THENPARAM(3) = REAL(J)CALL SCHDLE(NOW,1,PARAM)FLAG(J) = .FALSE.

C WRITE(1 1,15)J,PARAM(1)K 15 FORMAT(1X,'STATION ',I2,' WAS EMPTY ',F4.1,' TO FILL')

RETURNENDIF

20 CONTINUERETURNEND

G. READIT

SUBROUTINE READIT

C THIS SUBROUTINEC -READS PARAMETERS AND VARIABLE FROM THE DATA FILE

READ(9,*)HSTIMREAD(9,*)MANTIMREAD(9,*)TATIMREAD(9,*)ALTIM

RAD,)ENDUREREAD(9, 2)STAG

RAD(,)SECTOR

71

READ(9,*)SENSORREAD(9,* )DETECTREAD(9,')SPEEDREAD(9,*)TrSREAD(9,*)TOSREAD(9,*)STATNSDO 100 1= 1,MAXIRV

AEAD(9,*)FALTIM(l),REPTIM(I)100 CONTINUE

DO 1011I = 1,MAXRVFALTIM(I) = (1/FALTIM(I))*4.0REPTIM(I) = (1/REPTIM(I))

101 CONTINUEDO 105 I = 1,MAXRV

READ(9,*)INWORK(I)105 CONTINUE

READ(9,)HTPM(),HTPM(2),HTPM(3),HTPM(4),HTPM(5)&,HTPM(6),H-TPM(7)

107 CONTINUEDO 110 1 = 1,MAX.RV

READ(9,*)SEED(I)110 CONTINUE

RETURNEND

/2

H. CONFIRM

SUBROUTINE CONFRM

C THIS SUBROUTINEC -READS PARAMETERS AND VARIABLES TO AN OUTPUT DATA

FILE

C..........READ THE DATA INTO THE FILENVRITE(10,')' INPUT DATA *****s~WRiTE(10,*)'PWRITE(10,5)MAXTIM

5 FORMAT(1X.,'SIMULATION RUN TIME (MIN) 0,17)WRITE( 10, 10)HSTIM

10 FORMAT(1X'HOTSPIN TIME (MIN) 7,4.1)WRITE(1O,15)MANTIM

15 FORMAT(1X'MAN UP TIME (MIN) 7,4.1)WRITE(10,20)TATIM

20 FORMAT(1)('TUTRNAROUND TIME (MIN) W,4.1)WRITE( 10,25)ALTIM

25 FORMAT(1Xt'ALERT LAUNCH TIME (MIN) W,4.1)WRITE(10,35)ENDURE

35 FORMAT(1X,'CREW FATIGUE TIME (HRS) 7,4.1)WRITE(10,40)STAG

40 FORMAT(lX,'STAGAR INITIAL SORTIES TO STATION (MIN) ',F4.1)WRITE(10,45)SECTOR

45 FORMAT(lX'SECrOR. OF AEW COVERAGE (DEG) 7,5.1)WRITE( 10,50)SENSOR

50 FORMAT(1X,'SENSOR COVERAGE (NM) 7,5.1)WRITE( 10,55)DETECr

55 FORMAT(1X,'REQUIRED RAID DETECTION RANGE (NM) ',F5.1)WRITE( 10,60)SPEED

60 FORMAT(1X,'PLATFORM SPEED TO STATION (KTS) ',F5.1)WRITE( 10,65)TrS

65 FORMAT(1X,,'TIME TO STATION (HR) 7,5.1)WRITE( 10,70)TOS

70 FORMAT(1X.,'TIME ON STATION (HR) 9,F5.1)WR1TE( 10,75)STATNS

75 FORMAT(lX'NUMBER OF STATIONS tIIi)WRITE(10,*)"

"VITE(10,)' ***** 8888****8*88888**

WRITE(10, 8)''1

73

WRITE(10,*)' MTTF MTTRWRrrE(10,)'DO 105 1 = 1,MAXRV

WRIT(10,100)FALTIM(l),REPTIM(I)100 FORMAT(l1X,F5.2,3XF5.2)105 CONTINUE

'WRITE(1,)o'WRITE(10,*)yDO 1l1 I = 1,MAXRV

WRITE(10,1 10)INWORK(I)110 FORMAT(1XF3.1)111 CONTINUE

WRITE(10,*)'WRITE(10,*)'WRITE(10,*)) HOURS TO PREVENTIVE MAINTENANCEWRITE(10,2*)'WRITE(10,*)' A/C NUMBER TIME TO TIMEWRITE(10,*)' 1 2 3 4 5 PERFORM BETWEEN'WRITE(10,1 12)HTPM( 1),HTPM(2),HTPM(3),HTPM(4),HTPM(5),

&HTPM(6),HTPM(7)112 FORMAT(1X,F5.1,3X ,F5.1,3XF5.1,3X..F5. 1,3X,F5.1,3XF3.1,3X.,F5.1)

VWRITE(10,*)' SEEDSDO 120 I = 1,MAXRV

WRrTE(10,*)SEED(I)120 CONTINUE

RETURNEND

74

I. SCHEDULE

SUBROUTINE SCHDLE(EVTTIM,NEWNUM,PARAM) [Ref. 6]

C THIS SUBROUTINE SCHEDULES AN EVENTC -AT THE TIME EVENTIMEC -WITH THE PROPER EVENT NUMBER EVTNUMC -WITH THE PASSED PARAMETERS PARAM GIVENCC EVTTIM BE REALC NEWNUM MUST BE INTEGERC PARAM MUNST BE AN ARRAY OF MAXPRM REALS

IF(AVAIL .LE. 0.5) THENWRITE(*) 'CALANDAR FULL'WRITE(10,*) 'CALANDAR FULL'STOP

ENDIFIF(FSTEVT .EQ. 0)THEN

EVENT = AVAILAVAIL = NXTEVT(AVAIL)NXTEVT(EVENT) = 0FSTEVT = EVENT

ELSEIF(TIME(FSTEVT) .GT. EV'ITIM)THEN

C ........ INSERT THE EVENT ON THE TOP OF THE CALANDAREVENT = AVAILAVAIL = NXTEVT(AVAIL)NXTEVT(EVENT) = FSTEVTFSTEVT = EVENT

ELSEC ........ THE EVENT IS NOT THE FIRST EVENT IN THE CALANDAR

LSTEVT = FSTEVTCUREVT = NXTEVT(FSTEVT)COUNT = 0

200 CONTINUEIF(CUREVT .EQ. 0)THEN

C ........ WE HAVE RUN OUT OF CALANDAR, THE EVENT IS LASTE'ENT = AVAILAVAIL = NXTEVT(AVAIL)NXTEVT(EVENT) = 0NXTEVT(LSTEVT) = EVENT

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ELSEIF(TIME(CUREVT) .GT. EVTTIM)THEN

C.. ........ INSERT THE EVENT BETWv EEN LSTEVT AN'D CUREVTEVENT =AVAIL

AVAIL =NXTEVT(AVAIL)

NXTEVT(EVBNT) =CUREVT

NXTE VT(LSTE VT) EVENTELSE

C.. ........MOVE DOWN THE CALANDARLSTEVT =CURE VTCUREVT = NXTEVT(CUREVT)COUNT = COUNT + 1IF(COUNT .GT. USTL)THEN

WRpJTE(*,*)'INFTIflTE LOOP IN THE EVENT LIST'STOP

ENDIFGOTO 200

ENDIFENDIF

ENDIFENDIF

C ..... ASSIGN THIE OTHER VALUES AT THE ARRAY LOCATION EVENTTIME(E VENT) = EVTTIMEVTNUM(EVENT) = NEWNUMDO 2011I = 1,MAXPRM

PASSED(EVENT,I) =PARAM(I)

201 CONTINUERETURNEND

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J. POP

SUBROUTINE POP(EVITIM,NUMEVTPARAM) [Ref. 6]

C THIS SUBROUTINE TAKES THE FIRTS EVENT OFF THE CALANDARAND

C RETURNS THE PERTINENT DATACC EVTHM IS THE TIME OF THE EVENTC NUMEVT IS THE EVENT NUMBER OF THE EVENTC PARAM IS AN ARRAY OF FIVE REAL PASSED PARAMETERSCC EVTrIM AND NUMEVT ARE RETURNED AS 0 IF THE CALANDER IS

EMPTY

CIF(FSTEVT .EQ. 0)THEN

NUMEVT = 0EVTTIM = 0.0RETURN

ELSEEVENT = FSTEVTNUMEVT = EVINUM(EVENT)EV'TIM = TIME(EVENT)DO 501 I= 1,MAXPRM

PARAM(I) = PASSED(EVENTI)501 CONTINUE

FSTEVT = NXTEVT(EVENT)NXTEVT(EVENT) = AVAILAVAIL = EVENTRETURN

ENDIFEND

77

L DUMP CALENDAR

SUBROUTINE DMPCAL [Ref. 6]

CC DUMPS THE CURRENT CALANDAR TO THE OUTPUT FILE UNIT 10C

C SIM06250WRITE(10, S),WRITE(10, 405) SIM06270WRITE(10, 401) FSTEVT, AVAIL SIM06280WRITE(10, 406) SIM06290EVENT = FSTEVT SIM06300COUNT = 0 SIM06310

404 CONTINUE SIM06320IF (EVENT.NE.0.AND.COUNT.LT.LISTL + 1) THEN

WRITE(10, 402) EVENT, TIME(EVENT), EVTNUM(EVENT),& NXTEVT(EVENT),(PASSED(EVENT, J),J = 1, 5)

EVENT = NXTEVT(EVENT)COUNT = COUNT + 1 SIM06370GOTO 404 SIM06380

ENDIF SIM06390C SIM06400405 FORMAT(' OUTPUT FROM THE ROUTINE DMPCAL, THE CALANDARDUMPER',/)401 FORMAT(' FIRST EVENT (FSTEVT) = ',I10,/, SIM06420

& 'FIRST AVAILABLE SLOT (AVAIL) = ', /10,/) SIM06430406 FORMAT(' EVENT TIME NUMBER NEXT EVNT')SIM06440402 FORMAT(I4, 4X, F8.3, 4X, 14, 4X, 14,/15X, 5(2X, F8.3)) SIM06450

RETURN SIM06460END

78

L STATISTICS 2

SUBROUTINE STAT2(PARAM)

CCALL SCHDLE(NOW +0. 1,6,PARAM)IF(NOW .LE. .15)THEN

AEW = 0.0DO 10 1 = 1,MAXAC

RVAIL(I) =0.0

DTIME(I) =0.0

UTIME(I) =0.0

PDOWN(I) =0.0

10 CONTINUEENDIFPAEW = 0.0DO 201I = 1,MAXAC

IF(STATLJS(I) .GT. 0.5)THENUTIME(I) = UTIME(I) + 0.1

ELSEDTIME(I) = DTIME(l) + 0.1

ENDIFRVAIL(I) = UTIME(1)/(UTIME(I) +DTIME(I))IF(COVER(I) .GT. 0.5)PAEW = 0.1

20 CONTINUEAEW = AEW + PAEWRETURNEND

79

M. STATISTICS 3

SUBROUTINE STAT-3(PARAM)

CIF(K .LE. 999)CALL SCHDLE(NOW+ 6.O,7,PARAM)TAEW = AEW/6.0IF((K .GT. 199.5) .AND. (K .LT. 300.0))THEN

WRITE( 11,20)TAEW20 FORMAT(1XF7.4)

ENDIFC

TITAEW = TTAEW + TAEWAEW =0.0RETURNEND

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INITIAL DISTRIBUTION LIST

No. of Copies

1. Defense Technical Information Center 2Cameron StationAlexandria, Virginia 22304-6145

2. Library, Code 52 2Naval Postgraduate SchoolMonterey, California 93943-5002

3. Navy Tactical Support Activity 1Washington Navy YardWashington, D.C. 20374

4. Dr. Michael P. Bailey, Code OR/BANaval Postgraduate SchoolMonterey, California 93943-5002

5. CAPT Paul S. Bloch, USN, Code OR/BLNaval Postgraduate SchoolMonterey, California 93943-5002

6. LCDR Michael J. Bond, USNCarrier Airborne Early Warning Squadron One Hundred FifteenFPO San Francisco, California 96601

81