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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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Monterey, CA 93943-5000 Monterey, California 93943-5000
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11. TITLE (Including Security Classificatlon)
Availability of Aircraft Subject to Imperfect Preventive Maintenance
12 PERSONAL AUTHOR(S)BOND, Michael J.
13 TYPE OF REPORT 13b. TIME COVERED 14. DATE OF REPORT (Year, Month, Day) 15. Page CountMaster's Thesis FROM TO 1990, September 90
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
22a. NAME OF RESPONSIBLE INDIVIDUAL 22b. TELEPHONE (Include Area Code) 22c. OFFICE SYMBOLMichael P. Bailey .. (408)646-2085 OR/Ba
DD Form 1473, JUN 86 Previous editions are obselete. SECURITY CLASSIFICATION OF THIS PAGES/N 01 02-LF-01 4-6603 Unclassified
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
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0.4 ... . . . ..... ........ .... 1........." ... "
II I II III i Ii
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/)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
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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
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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
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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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.. .. .... .. .... .. ... .. .. ...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
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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..... 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
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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
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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
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