CombatSystemMaintenanceStatusDavid H.OlwellIntroductionandsettingArmyforcesdependuponcombatsystemstomoveandfight.BecauseArmyforcesmoveinverystressfulenvironments,thesecombatsystemseventuallybreak and require repair.Whilea system is awaiting repair,it is considered nonmissioncapableorNMC.Thepercentage ofvehicles awaiting repairis one ofthe key indicators of a unit's ability to fight, and is tracked at all levels ofcommand.The operational readinessrate,or ORRate,isthepercentage ofsystemsinagivenclasswhichareMissionCapable(MC)In this paper,we will discuss three simplemodelswhichdescribetheORrateforthenumberoftanksinatankbattalion inthe1stArmoredDivision.WewillalsodiscusswaysthemodelscouldbeextendedtoimprovetheirusefulnesstothecommanderWeassumethatthereaderisfamiliarwithelementarymatrixalgebra,linearsystemsofdifferenceequations,and optimization using differential calculus.MarkovChainmodelsWebegin with a simple model where a tank can be inone of two states:missioncapable(MC)ornon-mission capable (NMC).Eachday,onaverageacertainpercentageof themission capabletanksbreak,andmovetotheother stateEachday,acertainpercentageoftheNMCtanksarerepairedandbecomeoperational.Graphically,wehave:169
169 Combat System Maintenance Status David H. Olwell Introduction and setting Army forces depend upon combat systems to move and fight. Because Army forces move in very stressful environments, these combat systems eventually break and require repair. While a system is awaiting repair, it is considered nonmission capable or NMC. The percentage of vehicles awaiting repair is one of the key indicators of a unit’s ability to fight, and is tracked at all levels of command. The operational readiness rate, or OR Rate, is the percentage of systems in a given class which are Mission Capable (MC). In this paper, we will discuss three simple models which describe the OR rate for the number of tanks in a tank battalion in the 1st Armored Division. We will also discuss ways the models could be extended to improve their usefulness to the commander. We assume that the reader is familiar with elementary matrix algebra, linear systems of difference equations, and optimization using differential calculus. Markov Chain models We begin with a simple model where a tank can be in one of two states: mission capable (MC) or non-mission capable (NMC). Each day, on average a certain percentage of the mission capable tanks break, and move to the other state. Each day, a certain percentage of the NMC tanks are repaired and become operational. Graphically, we have:
1-p11-p2DMCNMCp2Figure 1Weuse the notation that piis theprobability that a tank stays MC; 1-piis theprobabilitythatatank movesfromMCtoNMC;p2istheprobabilitythatatankmovesfromNMCtoMC,and1-P2istheprobabilitythatatankstaysNMC.Let M,bethenumberoftanks inthebattalion whichare mission capable ondayi, and N,bethe number oftanks non-mission capable.Then we can writetheexpectedtransitionsfromonedaytothenext:MuI= p,M, + p,N,Ni+1I = (1- PI) M, +(1- P2)N,This systemofequationscanbeeasilyexpressed inmatrixnotation:M.IMHPiP2HPnM1-p.1-p.U1'N..3This assumes thatthe p,areknownand constant,and ignores therandomnessinthisproblemby just lookingattheexpectedvalues oftanks operational.(Recall the expected value of an integer-valued variable is not necessarilyaninteger.)Thissystemwill eventuallyreachasteadystate,wheretheexpectedvaluesdonot change fromday to day.The steadystate can befoundby eigenvectoranalysis.A matrix ofprobabilities such as wehave constructed is called aprobability transition matrix, since every entry is non-negative and every column170
170 Figure 1 We use the notation that p1 is the probability that a tank stays MC ; 1-p1 is the probability that a tank moves from MC to NMC; p2 is the probability that a tank moves from NMC to MC, and 1- p2 is the probability that a tank stays NMC. Let Mi be the number of tanks in the battalion which are mission capable on day i, and Ni be the number of tanks non-mission capable. Then we can write the expected transitions from one day to the next: M p M p N N p M p N i i i i i i + + = + = - + - 1 1 2 1 1 1 1 2 ( ) ( ) This system of equations can be easily expressed in matrix notation: M N p p p p M N p p p p M N i i i i i + + + L N M O Q P = - - L N M O Q P L N M O Q P = - - L N M O Q P L N M O Q P 1 1 1 2 1 2 1 2 1 2 1 0 1 1 1 1 0 This assumes that the pi are known and constant, and ignores the randomness in this problem by just looking at the expected values of tanks operational. (Recall the expected value of an integer-valued variable is not necessarily an integer.) This system will eventually reach a steady state, where the expected values do not change from day to day. The steady state can be found by eigenvector analysis. A matrix of probabilities such as we have constructed is called a probability transition matrix, since every entry is non-negative and every column
sumsto 1.Itisknownthat everyprobabilitytransition matrixhas at leastoneeigenvalueequal to1,and itsassociated eigenvectoristhesteadystateforthesystem.Wecan solve forthe steady state eigenvector by solving the equationMPiP2p1-pNsinceat the steady statethenumber of vehiclesbreaking exactlybalancesthenumberofvehiclesbeingrepaired.WeobtaintheeigenvectorT p2p2-p +1pMT(-p)PNp,-p,+1bwhereT isthetotal number of tanks in thebattalion(T=M+N).Regardlessofourstartingstate,thetop componentoftheeigenvectortellsushowmanyvehicles,onaverage,wecan expecteventuallytohaveworking,thebottomcomponenttellsushowmanyweexpecttobeawaiting repair.TheMC/NMCratesteadystatescanbefoundsimilarlytobeP2P2-P,+1p(1-p)Np2 - p, +10For example, let's let p, =.95 and p, =.8. This means that only 5% of theworkingvehiclesbreakeachday,and 80%of thebrokenvehicles arerepaired164.58826Weset T=58.Substitutiongives us a steadystate ofandan3.4118eventual steadystateORof 94.1176%.Ofcourse,wecan'thave54.5882tanks.That is theexpected valuefor the number operational, whichis anaverage. Just as the average of 1 and 2 is 1.5, which is not an integer, so canthe averagenumber oftanks operational also notbean integer.We can verify these calculations by constructing a spreadsheet model of thetank maintenance status. Let's start with M = 50 and N =4. Figure 2 is a pictureofhowthesystemonaveragewouldbehave171
171 sums to 1. It is known that every probability transition matrix has at least one eigenvalue equal to 1, and its associated eigenvector is the steady state for the system. We can solve for the steady state eigenvector by solving the equation M N p p p p M N L N M O Q P = - - L N M O Q P L N M O Q P 1 2 1 1 1 2 since at the steady state the number of vehicles breaking exactly balances the number of vehicles being repaired. We obtain the eigenvector T p p p T p p p 2 2 1 1 2 1 1 1 1 - + - - + L N M M M M O Q P P P P ( ) , where T is the total number of tanks in the battalion (T = Mi Ni + ). Regardless of our starting state, the top component of the eigenvector tells us how many vehicles, on average, we can expect eventually to have working, the bottom component tells us how many we expect to be awaiting repair. The MC/NMC rate steady states can be found similarly to be p p p p p p 2 2 1 1 2 1 1 1 1 - + - - + L N M M M M O Q P P P P ( ) . For example, let’s let p1 =.95 and p2 =.8 . This means that only 5% of the working vehicles break each day, and 80% of the broken vehicles are repaired. We set T = 58 . Substitution gives us a steady state of 54 5882 34118 . . L N M O Q P , and an eventual steady state OR of 94.1176%. Of course, we can’t have 54.5882 tanks. That is the expected value for the number operational, which is an average. Just as the average of 1 and 2 is 1.5, which is not an integer, so can the average number of tanks operational also not be an integer. We can verify these calculations by constructing a spreadsheet model of the tank maintenance status. Let’s start with M = 50 and N = 4. Figure 2 is a picture of how the system on average would behave
55545352515004681021214161820Figure 2Let'slook a little closer at the steady state for M, which we said was equal toT p2M..P2-p, +1If wewanttoimprovethe expected steadystatefor M,shouldwetryto increasePr, P2, or some combination of the two? One way to gain insights into thisquestion is to look at the gradient of steady state for M with respect to p1 and p2ThegradientofM isgivenby:LaMaMT(1-p)T p2VMseeadystateNopap,UN(p, - p, + 1)2'(p2 - p + 1)2Weknowfrom calculus that the direction of greatest increase in M is in thedirection of the gradient. That means to obtain the greatest increase in M for afixedsizechangeinthevaluesofpandp2,thechangeinpishouldbeproportionaltop2,andthechangeinp2shouldbeproportionalto1-p1.Doesthismakesense?If we are repairing vehicles quickly already,then itmakes sense to work on notbreakingtheminsteadofworkingon improving ourrepairtime.Thatis whythechangeinpisproportionaltop2.Asp2increases,weworkmoreonp1.172
172 50 51 52 53 54 55 0 2 4 6 8 10 12 14 16 18 20 Figure 2 Let’s look a little closer at the steady state for M, which we said was equal to M T p p p steady state = - + 2 2 1 1 . If we want to improve the expected steady state for M, should we try to increase p1 , p2 , or some combination of the two? One way to gain insights into this question is to look at the gradient of steady state for M with respect to p1 and p2. The gradient of M is given by: Ñ = L N M O Q P = - + - - + L N M O Q M P M p M p T p p p T p p p steady state ¶ ¶ ¶ 1 ¶ 2 2 2 1 2 1 2 1 2 1 1 1 , ( ) , ( ) ( ) . We know from calculus that the direction of greatest increase in M is in the direction of the gradient. That means to obtain the greatest increase in M for a fixed size change in the values of p1 and p2, the change in p1 should be proportional to p2, and the change in p2 should be proportional to 1-p1. Does this make sense? If we are repairing vehicles quickly already, then it makes sense to work on not breaking them instead of working on improving our repair time. That is why the change in p1 is proportional to p2. As p2 increases, we work more on p1
Ifwearebreakingvehiclesfrequently,itmakessensetoimproveourrepaircapability.(1-p1)is thefraction of vehicles expected to“break"each timeperiod.Asthat increases,so shouldp2,accordingtothegradient.Noticethat unlessp2is zero, orp,isone,we should work on improving bothpercentages. The gradient tells us how to allocate our effort between them.There are at least two shortcomings of this model.First, it is deterministic:weonlyworkwiththeexpectednumberofvehiclesineachcategory,andthisignores therandomness of the true behavior,as well as the integer nature of theobjects.Wecan'thave54.5882tanksmissioncapable:wecanonlyhave54or55.Wedon'tknowfromthismodelhowmuchvariationtoexpectaroundtheaverage.Thesecondshortcomingisthatit isverysimple.Itdoesn'trecognizethatvehiclesbreakatdifferentratesdependinguponwhetheroneisdrivingtheminthe field or not. Vehicles are repaired at different rates depending upon theavailabilityofsparepartsandmechanics,anduponhowmanyothervehiclesarewaitingtoberepaired.Theavailabilityofpartsdependsontheprioritycodeofthe unit.If the unit level maintenance can not repair the vehicle,it is sent tohigherlevelmaintenance,whichhasdifferentrepairrates.Abettermodelwould incorporatethese additional featuresHowever, this simple model is useful.It allows us to gain understandingaboutoursystem,whichisthehallmarkofaneffectivemodel.Binomial EquationItispossibleto improvethemodel intheprevioussectionbyconsideringastochasticmodel.Thiswillallowustonotonlyunderstandtheaveragebehaviorof our model, butalso howmuch thenumber of Mctanksvaries around thataverage.There isa significantdifferencebetween54.5882(plus orminus.1),and 54.5882(plus orminus 10).Wewill usethebinomial distribution forourmodels.Let's considerthe number oftanksthat are mission capable each day.Thatnumberwillbethesumofthosetanksthatweremissioncapablethepreviousdayandstayed inthat state,plusthe number oftanks that were repaired theprevious day.Thiscanbemodeled asthesumoftwobinomiallydistributedrandomvariables.LetM,bethenumberofmissioncapabletankstheprecedingperiod,andTbethetotal number oftanks.ThenM+=Bin(pr, M)+Bin(p2,T-M)173
173 If we are breaking vehicles frequently, it makes sense to improve our repair capability. (1-p1 ) is the fraction of vehicles expected to “break” each time period. As that increases, so should p2, according to the gradient. Notice that unless p2 is zero, or p1 is one, we should work on improving both percentages. The gradient tells us how to allocate our effort between them. There are at least two shortcomings of this model. First, it is deterministic: we only work with the expected number of vehicles in each category, and this ignores the randomness of the true behavior, as well as the integer nature of the objects. We can’t have 54.5882 tanks mission capable: we can only have 54 or 55. We don’t know from this model how much variation to expect around the average. The second shortcoming is that it is very simple. It doesn’t recognize that vehicles break at different rates depending upon whether one is driving them in the field or not. Vehicles are repaired at different rates depending upon the availability of spare parts and mechanics, and upon how many other vehicles are waiting to be repaired. The availability of parts depends on the priority code of the unit. If the unit level maintenance can not repair the vehicle, it is sent to higher level maintenance, which has different repair rates. A better model would incorporate these additional features. However, this simple model is useful. It allows us to gain understanding about our system, which is the hallmark of an effective model. Binomial Equation It is possible to improve the model in the previous section by considering a stochastic model. This will allow us to not only understand the average behavior of our model, but also how much the number of MC tanks varies around that average. There is a significant difference between 54.5882 (plus or minus .1), and 54.5882 (plus or minus 10). We will use the binomial distribution for our models. Let’s consider the number of tanks that are mission capable each day. That number will be the sum of those tanks that were mission capable the previous day and stayed in that state, plus the number of tanks that were repaired the previous day. This can be modeled as the sum of two binomially distributed random variables. Let Mi be the number of mission capable tanks the preceding period, and T be the total number of tanks. Then Mi+ = Bin p Mi Bin p T Mi + - 1 1 2 ( , ) ( , )
Mi+1isthe sumof twobinomial random variables,but it itself isnotbinomiallydistributedunlessp,=p2.How does Mi+,behave? Let's imagine2o battalions all startingat with the samenumber of MC and NMC tanks (54 and 4), each with p1 = 0.95 and p2 = 0.80, andeachfollowing ourmodel.We cangraph their number of Mc tanks using aspreadsheet, and obtainFigure3:Numberofoperationalvehiclesa545Y45020406080100DayFigure3This imageof20differenthistories(oneforeachofthebattalions)givesusafeelforthevariabilitywecanexpect inthenumberofoperationaltanks.Wecanexpecttohavebetween51and58tanksMC usually,andonlyabout1inevery2000 days would we have as few as 47 tanks operational.We can alsousethe spreadsheetto do somesimple“what--if"analysis.Wecanputpiandp2 in cells,andhaveall the calculationsrefertothosecells.Thenaswevarythevaluesofthosecells,wecandynamicallyseetheeffect inoperationalreadinessrate.AtFigure4,weseeanexample,withp,=.85,p,=.5.:174
174 Mi+1 is the sum of two binomial random variables, but it itself is not binomially distributed unless p1 = p2. How does Mi+1 behave? Let’s imagine 20 battalions all starting at with the same number of MC and NMC tanks (54 and 4), each with p1 = 0.95 and p2 = 0.80, and each following our model. We can graph their number of MC tanks using a spreadsheet, and obtain Figure 3: 45 48 51 54 57 0 20 40 60 80 100 Day Number of operational vehicles Figure 3 This image of 20 different histories (one for each of the battalions) gives us a feel for the variability we can expect in the number of operational tanks. We can expect to have between 51 and 58 tanks MC usually, and only about 1 in every 2000 days would we have as few as 47 tanks operational. We can also use the spreadsheet to do some simple “what-if” analysis. We can put p1 and p2 in cells, and have all the calculations refer to those cells. Then as we vary the values of those cells, we can dynamically see the effect in operational readiness rate. At Figure 4, we see an example, with p p 1 2 =.85, =.5.:
BDapin0.850.52Numberof5506740830o10201110121301402040608010015Day1617DABCDEFGHIKKLIDGFigure 4Noticethatthisdata is seriallycorrelated.Iftheprevious daywas low,the nextday tends to be low, as well. if the previous day was high, the next day tends tobe high. It is obvious from the graph that the day to day levels of M; are notindependent.Of course,weknewthat since inourmodelthedistribution ofMi+1was afunctionof M,Thistypeofpictorial representationofastochasticdiscretesystemisveryusefulto demonstrate the properties of a maintenance systemto a commander.Pictures tell stories that mere numbers often fail to convey. The picture of theMarkov chainmodel shows onlythelong-termtrend oftheaveragenumberoftanks.Thesepicturesfromthesimulationshowboththelong-termaverageandthevariability.WhiletheaveragenumberoftanksMCmayreachanequilibriumvalue, the variability of the actual number of tanks up at any given time does notgotozero!Therandomnessremains.EstimatingtheparametersHowdowedeterminep,andp2?Wemayneverknowthemexactly.Thebestwecanhopetodois to estimatethemfromhistorical data.Wehave assumed inour earlier models that p1and p2are constant.This is a very strong assumption.If itholds,wecan estimatep1andp2frompriorrecords.Inthesectionafterthis,wediscussanapproachforrelaxingtheassumptionof constanttransitionprobabilities.Fornow,we continueto assumethatthe transition probabilities are constantHowmayweestimatethem?175
175 Figure 4 Notice that this data is serially correlated. If the previous day was low, the next day tends to be low, as well. If the previous day was high, the next day tends to be high. It is obvious from the graph that the day to day levels of Mi are not independent. Of course, we knew that since in our model the distribution of Mi+1 was a function of Mi. This type of pictorial representation of a stochastic discrete system is very useful to demonstrate the properties of a maintenance system to a commander. Pictures tell stories that mere numbers often fail to convey. The picture of the Markov chain model shows only the long-term trend of the average number of tanks. These pictures from the simulation show both the long-term average and the variability. While the average number of tanks MC may reach an equilibrium value, the variability of the actual number of tanks up at any given time does not go to zero! The randomness remains. Estimating the parameters How do we determine p1 and p2? We may never know them exactly. The best we can hope to do is to estimate them from historical data. We have assumed in our earlier models that p1 and p2 are constant. This is a very strong assumption. If it holds, we can estimate p1 and p2 from prior records. In the section after this, we discuss an approach for relaxing the assumption of constant transition probabilities. For now, we continue to assume that the transition probabilities are constant. How may we estimate them?
Ouranswerwilldependonwhattypeofdatawehaveavailable.IfwehavedatathattellsusforeachdaythenumberofvehiclesthatstayedMC,thatwentfromMCtoNMC,thatwentfromNMCtoMC,andstayedNMC,wecanestimatethetransitionprobabilitiesdirectly.Unfortunately,thedatawehavefromthe1stArmored Divisiononlyhas thenumber of MCvehiclesforeachreportingperiod.We canusethat datato estimatep1and p2 in our binomial model by constructingadiscretedynamicalsystemofexpectedvaluesunderourbinomialmodelwhereweallowthetransitionprobabilitiestobevariables.WehavethatEDM.= P,M, + P,N,,where E(M)means theexpectedvalue of M.We then calculate residuals,whicharethedifferencebetweentheexpectedvalueandtheobserved values.Wesquare the residuals,and choose the two values of p,andp2that minimize thesumofthesquared errors.UsingtheSolvemacrofromExcel,wefindthevalues ofp,andp2thatminimizethe squaredresiduals.Theoutputis shown inFigure5.Thespreadsheetissearchingthroughallpossiblevaluesofp,andp2tofindthevaluesthatminimizethesumofsquaredresiduals.Thesevaluesarealllistedonthefirst lineoftheoutput.WeseethatthebestestimatefromthisdataisthatP1=96.66%andp2=32.11%.176
176 Our answer will depend on what type of data we have available. If we have data that tells us for each day the number of vehicles that stayed MC, that went from MC to NMC, that went from NMC to MC, and stayed NMC, we can estimate the transition probabilities directly. Unfortunately, the data we have from the 1st Armored Division only has the number of MC vehicles for each reporting period. We can use that data to estimate p1 and p2 in our binomial model by constructing a discrete dynamical system of expected values under our binomial model where we allow the transition probabilities to be variables. We have that E Mi+ = p Mi p Ni + 1 1 2 b g , where E(M) means the expected value of M. We then calculate residuals, which are the difference between the expected value and the observed values. We square the residuals, and choose the two values of p1 and p2 that minimize the sum of the squared errors. Using the Solve macro from Excel, we find the values of p1 and p2 that minimize the squared residuals. The output is shown in Figure 5. The spreadsheet is searching through all possible values of p1 and p2 to find the values that minimize the sum of squared residuals. These values are all listed on the first line of the output. We see that the best estimate from this data is that p1 = 96.66% and p2 = 32.11%
p1p2sum0.321071113.54520.966653MNDayEMr2336554543554.129130.0166743454453.483550.2667255263353.483552.200907632520.0370152.192385263152.192380.037015263052.192380.037015295352.192380.652249528530.02625652.837965275352.837960.0262565265352.837960.0262562553552.837960.0262565245352.837960.0262565235352.837960.0262568225052.837968.0540355215350.901224.4048935205352.837960.02625619104852.8379623.405896185249.610055.7118531753552.192380.6522495165352.837960.0262565155352.837960.02625614104852.8379623.405899134949.610050.3721639124950.255631.5766179114950.255631.57661710481050.255635.087885535949.6100511.49175585352.837960.026256755352.837960.026256654452.837961.350329535553.483552.299634045854.1291314.98365035856.065873.74084255356.065871.136089315554.129130.758418Figure5177
177 p1 p2 sum 0.966653 0.321071 113.5452 Day M N EM r2 36 55 3 35 54 4 54.12913 0.016674 34 54 4 53.48355 0.266725 33 52 6 53.48355 2.200907 32 52 6 52.19238 0.03701 31 52 6 52.19238 0.03701 30 52 6 52.19238 0.03701 29 53 5 52.19238 0.652249 28 53 5 52.83796 0.026256 27 53 5 52.83796 0.026256 26 53 5 52.83796 0.026256 25 53 5 52.83796 0.026256 24 53 5 52.83796 0.026256 23 53 5 52.83796 0.026256 22 50 8 52.83796 8.054035 21 53 5 50.90122 4.404893 20 53 5 52.83796 0.026256 19 48 10 52.83796 23.40589 18 52 6 49.61005 5.711853 17 53 5 52.19238 0.652249 16 53 5 52.83796 0.026256 15 53 5 52.83796 0.026256 14 48 10 52.83796 23.40589 13 49 9 49.61005 0.372163 12 49 9 50.25563 1.576617 11 49 9 50.25563 1.576617 10 48 10 50.25563 5.087885 9 53 5 49.61005 11.49175 8 53 5 52.83796 0.026256 7 53 5 52.83796 0.026256 6 54 4 52.83796 1.350329 5 55 3 53.48355 2.299634 4 58 0 54.12913 14.98365 3 58 0 56.06587 3.74084 2 55 3 56.06587 1.136089 1 55 3 54.12913 0.758418 Figure 5
Modelingthetransition probabilitiesas functionsofothervariablesLet'slookataplotof theresidualsfromtheprevioussection,givenatFigure6:Residuals2Residuals2520-3Figure6We see that thereare some days where the residuals are much bigger than theothers are,indicating that our estimated model does notfit toowell on thosedays.Can we find informationto explain whythe model doespoorly on thosedays?Oneapproachistoallowpandp2tobefunctionsof someothervariables.Forexample,p1mightverywellbeaffectedbywhetherornotthebattaliontanksweredriventhatday.p2mightbeaffectedbythenumberofmechanicsavailable,thenumberofsparepartsavailable,and eventhenumberofNmcvehicles:themorebroken,thelesslikelyanyoneofthemmightbefixedWe were not able to collect data on these predictors, but we can find the daysthe battalion was inthe field using its vehicles.To illustrate the method,we adda predictortoour data set withvalue1if thebattaliondrovethevehicles inthefield that day,and valueO if not.Itis still possibletohavea vehiclebecomeNMC if itisnotdrivenbecauseofthepreventivemaintenancechecksandservices (PMCS),but itismuchless likely.Wedefinep3as theamount that p,changes whenthebattalion is inthefield.Wethen have EM,=(p,+p,/,)M,+p,N,,where l, is an indicator variable equalto 1 if the battalion is in the field that day. Just as before, we use Excel to solveforall three pvalues,and obtain the spreadsheet output at Figure7,below.178
178 Modeling the transition probabilities as functions of other variables Let’s look at a plot of the residuals from the previous section, given at Figure 6: Residuals -5 -4 -3 -2 -1 0 1 2 3 4 1 4 7 10 13 16 19 22 25 28 31 34 Residuals Figure 6 We see that there are some days where the residuals are much bigger than the others are, indicating that our estimated model does not fit too well on those days. Can we find information to explain why the model does poorly on those days? One approach is to allow p1 and p2 to be functions of some other variables. For example, p1 might very well be affected by whether or not the battalion tanks were driven that day. p2 might be affected by the number of mechanics available, the number of spare parts available, and even the number of NMC vehicles: the more broken, the less likely any one of them might be fixed. We were not able to collect data on these predictors, but we can find the days the battalion was in the field using its vehicles. To illustrate the method, we add a predictor to our data set with value 1 if the battalion drove the vehicles in the field that day, and value 0 if not. It is still possible to have a vehicle become NMC if it is not driven because of the preventive maintenance checks and services (PMCS), but it is much less likely. We define p3 as the amount that p1 changes when the battalion is in the field. We then have EMi+ = p p I i Mi p Ni + + 1 1 3 2 ( ) , where Ii is an indicator variable equal to 1 if the battalion is in the field that day. Just as before, we use Excel to solve for all three p values, and obtain the spreadsheet output at Figure 7, below: