Modelingin CalculusIMickA.Sanzotta,DuffCampbellScenario1:CombinedArmsLiveFireWhileparticipating in Cadet Field Training(CFT) in the summer of'g9,you and yourcompany are sent to Ft. Knox to conduct mounted maneuver training. You are placedinchargeoftheair strikeportionof the CombinedArms LiveFireExercise (CALFEX)involving AirForceA10s andArmyAttackHelicopters.Your mission isto ensure theAirForceA1Os and ArmyAttackHelicopters delivertheirordnanceattheappropriatetime and place, and that the air units maintain a minimum safe separation distance of 2nautical miles.At 1400 hours, the OiC provides youwith the following situation:At1430hrsAirForceA10swill be171nauticalmilesfromthetargetareaapproachingat206knots onaheadingof 303degrees.At1430hrstheArmyAttackHelicoptersare88nautical milesfromthetargetarea,approachingat103knotsonaheadingof213degrees.The OiC wants to know howfast the distance between the two air units will be changingat 1430 hrs and if the aircraft units maintain a minimum safe distance throughout themission.Problem1:Determinea reference systemtoexaminetheratesoftheaircraft.Solution 1:Reference System:Positive is moving awayfromthetarget.Thetargetisthe originTimezero (t=0)is now (1430)hrs.TARGETc(t)Figure 1: Sketch of the Air MissionProblem2:Definetheknownvariablesand constants.Solution 2:variables:c(t) is the distance between the 2 air units at time t.(Hypotenuse)a(t) is the distance the army helicopters are from the target.a(t)=-103t+88b(t)isthedistancetheairforceA-lo'sarefromthetargetattimet.b(t)=-206t +171t(0)is 1430 hrs233
233 Modeling in Calculus I Mick A. Sanzotta, Duff Campbell Scenario 1: Combined Arms Live Fire While participating in Cadet Field Training (CFT) in the summer of ’99, you and your company are sent to Ft. Knox to conduct mounted maneuver training. You are placed in charge of the air strike portion of the Combined Arms Live Fire Exercise (CALFEX), involving Air Force A10s and Army Attack Helicopters. Your mission is to ensure the Air Force A10s and Army Attack Helicopters deliver their ordnance at the appropriate time and place, and that the air units maintain a minimum safe separation distance of 2 nautical miles. At 1400 hours, the OIC provides you with the following situation: · At 1430 hrs Air Force A10s will be 171 nautical miles from the target area approaching at 206 knots on a heading of 303 degrees. · At 1430 hrs the Army Attack Helicopters are 88 nautical miles from the target area, approaching at 103 knots on a heading of 213 degrees. The OIC wants to know how fast the distance between the two air units will be changing at 1430 hrs and if the aircraft units maintain a minimum safe distance throughout the mission. Problem 1: Determine a reference system to examine the rates of the aircraft. Solution 1: Reference System: Positive is moving away from the target. The target is the origin. Time zero (t = 0) is now (1430) hrs. Problem 2: Define the known variables and constants. Solution 2: variables: c(t) is the distance between the 2 air units at time t.(Hypotenuse) a(t) is the distance the army helicopters are from the target. a(t)=-103t +88 b(t) is the distance the air force A-10’s are from the target at time t. b(t)=-206t +171 t(0) is 1430 hrs. TARGET a(t) b(t) c(t) Figure 1: Sketch of the Air Mission
Army Attack HelicoptersAirForce A-10sa(0) = 88 nmb(0)= 171 nmanglA = 303 degreesangleB = 213 degreesa'(t)=103 knotsb(0)=206 knotsProblem3:StatesomeassumptionsforthismodelingproblemSolution 3:1.Both air units will maintain their given course and speed2.Bothairunitswill maintainthesamealtitude3.The aircraft units maintain a tight formation and can be considered to be one point in space.Problem4:Determinetheanglebetweenthe2aircraft unitsandtherelationshipbetweenthe units.Solution 4:The angle between the 2 aircraft units, angle =JanglA - anglB. Therefore, angle = 90 degreesTherefore, Pythagorean Theorem applies!Problem5:Determineanexpression relatingall ofthevariables a(t),b(t),andc(t)Solution 5:c(t)? = a(t)? +b(t)2Problem 6: Implicitly differentiate c(t)? = a(t)? +b(t)? with respect to timeSolution 6:.dbdc.da +2 b():= 2 . a(t) . 92.c(t)dtdtdtProblem7:Simplifyand solveforthe change indistancebetweenthetwo aircraftunitsSolution 7:dbda+ b(t)a(t)dcdtdtdtc(t)234
234 Army Attack Helicopters Air Force A-10s a(0) = 88 nm b(0) = 171 nm anglA = 303 degrees angleB = 213 degrees a’(t) = 103 knots b’(0) = 206 knots Problem 3: State some assumptions for this modeling problem. Solution 3: 1. Both air units will maintain their given course and speed. 2. Both air units will maintain the same altitude. 3. The aircraft units maintain a tight formation and can be considered to be one point in space. Problem 4: Determine the angle between the 2 aircraft units and the relationship between the units. Solution 4: The angle between the 2 aircraft units, angle = |anglA – anglB|. Therefore, angle = 90 degrees. Therefore, Pythagorean Theorem applies! Problem 5: Determine an expression relating all of the variables a(t), b(t), and c(t). Solution 5: 2 2 2 c(t) = a(t) + b(t) Problem 6: Implicitly differentiate 2 2 2 c(t) = a(t) + b(t) with respect to time. Solution 6: dt db b t dt da a t dt dc 2 ×c(t) × = 2 × ( ) × + 2 × ( ) × Problem 7: Simplify and solve for the change in distance between the two aircraft units. Solution 7: ( ) ( ) ( ) c t dt db b t dt da a t dt dc × + × =
Problem8:Determinethechangeindistancebetweenthe2aircraftunitsat1430hrs(i.e., determine c'(0))Solution 8:dc=230.2994knots or nauticalThedistancebetween themat1430hours is decreasing anddtmiles perhourProblem9:Inordertodetermine if the air units meet the mission,we need to determinehowclosethetwoairunitswill cometoeachother.Determinetheminimumdistancebetweenthe2planes.Dotheyviolatetheminimumsafeseparationdistanceof2nauticalmiles?Solution 9:We need find the minimum of c(t) = a(0)* + b(0). If we minimize what is underneath theradical c(t)? = a(t)? +b(t)*, it follows that we will minimize the function c(t).Figure 2: Plot of c(t)vs. timedbdcdaDifferentiate(t) implicitly:a(t)+b(t)and solve ordtdtdtapply the MCad root finder:guess:t, =0.82 hourst=root(dc(t),t))t=0.8350hoursConversiontoclocktime60 min60secand0.8350hr50.0970min0.0970min5.8200sec1 hr1minTherefore, the air units will be closest to each other at 1520:05 hours.We can now substitutet=0.8350back intoPythagorean's TheoremThus, MinC = a(0.8350) +b(0.8350)2 =2.2361 nmTherefore, they will not violate the 2 nm safety zone235
235 Problem 8: Determine the change in distance between the 2 aircraft units at 1430 hrs (i.e., determine c'(0) ). Solution 8: The distance between them at 1430 hours is decreasing and = dt dc 230.2994 knots or nautical miles per hour. Problem 9: In order to determine if the air units meet the mission, we need to determine how close the two air units will come to each other. Determine the minimum distance between the 2 planes. Do they violate the minimum safe separation distance of 2 nautical miles? Solution 9: We need find the minimum of 2 2 c(t) = a(t) + b(t) . If we minimize what is underneath the radical 2 2 2 c(t) = a(t) + b(t) , it follows that we will minimize the function c(t) . Figure 2: Plot of c 2 (t) vs. time. Differentiate c 2 (t) implicitly: dt db b t dt da a t dt dc = ( ) × + ( ) × and solve or apply the MCad root finder: guess: 0.82 1 t = hours ( ( ), ) 1 1 t = root dc t t t = 0.8350 hours Conversion to clock time: 50.0970 min 1hr 60 min 0.8350 hr × = and 5.8200 sec 1min 60 sec 0.0970 min × = Therefore, the air units will be closest to each other at 1520:05 hours. We can now substitute t = 0.8350 back into Pythagorean's Theorem. Thus, 2 2 MinC = a(0.8350) + b(0.8350) = 2.2361 nm. Therefore, they will not violate the 2 nm safety zone. 0.5 0.6 0.7 0.8 0.9 1 1.1 20 2.5 15 32.5 50 t
Problem1o:Aretheassumptionsyoumadereasonable?Solution 10:It is reasonable to assume the aircraft will fly straight and level at a constant velocity in atightformation.This model is not based on many simplifying assumptions.The OIC should bevery confident of the data and insights provided by this analysis. Initially, at 1430 hrs theaircraft units are closing in on each other at 230.3 knots per hour and during the exercise willmaintainatleast2.236nmdistancefromeachother.IftheaircraftdeviatefromtheflightplanstheOICmustbenotifiedimmediately.Scenario2:TanksarenotDiscreteConsideratankbattlebetweencountryXandCountryY.Analystshavereasontobelievethatifx(t)representsthenumberoftankscountryXhasafterthoursofcombat.and y(t) is the number of tanks Country Y has, then the differential equations governingapurearmorbattlemighttaketheformdx-A.x-ydtdy=-B-x-ydtProblem1a:Doesthisseemlikea reasonablemodel?ExplainSolution 1a:A way to interpret this model is that we have a rate of change that is proportional to what ispresent,buttheconstant"ofproportionalitydependsonhowmanyenemyarepresent:dx-(A·y)-x.Another way is to note that the factor x·y is a good measure of thedtinteractions between the two populations: the higher each factor, the higher the product, and viceversa. A further consideration is that the rate of change is zero when there are no enemy present.While this may not accord with real life (see Problem 2), it is a reasonable model.Problem1b:WhatdotheparametersAandBrepresent?Solution 1b:The parameter A is a measure of how fast countryXloses tanks. As such, it is a combination ofthe probability of two opposing tanks meeting (and thus is dependent on time of day, weather,236
236 Problem 10: Are the assumptions you made reasonable? Solution 10: It is reasonable to assume the aircraft will fly straight and level at a constant velocity in a tight formation. This model is not based on many simplifying assumptions. The OIC should be very confident of the data and insights provided by this analysis. Initially, at 1430 hrs the aircraft units are closing in on each other at 230.3 knots per hour and during the exercise will maintain at least 2.236 nm distance from each other. If the aircraft deviate from the flight plans the OIC must be notified immediately. Scenario 2: Tanks are not Discrete Consider a tank battle between country X and Country Y. Analysts have reason to believe that if x(t) represents the number of tanks country X has after t hours of combat, and y(t) is the number of tanks Country Y has, then the differential equations governing a pure armor battle might take the form ï ï î ï ï í ì = - × × = - × × B x y dt dy A x y dt dx Problem 1a: Does this seem like a reasonable model? Explain. Solution 1a: A way to interpret this model is that we have a rate of change that is proportional to what is present, but the “constant” of proportionality depends on how many enemy are present: A y x dt dx = - ( × )× . Another way is to note that the factor x ×y is a good measure of the interactions between the two populations: the higher each factor, the higher the product, and vice versa. A further consideration is that the rate of change is zero when there are no enemy present. While this may not accord with real life (see Problem 2), it is a reasonable model. Problem 1b: What do the parameters A and B represent? Solution 1b: The parameter A is a measure of how fast country X loses tanks. As such, it is a combination of the probability of two opposing tanks meeting (and thus is dependent on time of day, weather
terrain,as well as tactical doctrine and target acquisition capabilities)as well as the probabilityof a“kil"once an encounter is made (and thus dependent on weapon characteristics, armor, etc.)Problem1c:Whatdoyouexpectthelong-termbehaviorofthesystemtobe?Solution1c:Given that there are no reinforcements, one expects that both forces will attrit (at rates whichdepend on the relative sizes of A and B)until one is down to zero, at which point the opposingforcewill remainataconstantstrengthfromthenon.Theactual result(whowins")woulseem todepend on therelative sizes of AandB,but also on the initial conditions.Numericalexperiments with different parametervalues and different initial conditions can help here.Problem1d:ClassifythissystemofdifferentialequationsSolution1dThis is a second-order, non-linear, homogeneous, autonomous (independent of time) system.As such, one expects not to be able to solve it analytically, however, as the next two parts show,this particular system can in fact be solved explicitly.11asolutionofthissystem?Problem1e:Is x(t)J(t):Bt'AtSolution 1e:dxdyd(1d(1Wehaveand similarlyAlso,dt(BBdtdtBtdtA111-A.x.and-B-X.JSo this particular solution isBtAtAt?Btverified.11Problem1f:Is x(t):thegeneralsolutionofthissystem?Whyy(t)B(t+C)A(t+C)NOT?Solution 1f:dxddy11+Cand(t + C)-2WehavedtBdtdt(B(t+C)dtBA1andwhile -A-x-Bt+cB(t+C)A(t+C)237
237 terrain, as well as tactical doctrine and target acquisition capabilities) as well as the probability of a “kill” once an encounter is made (and thus dependent on weapon characteristics, armor, etc.) Problem 1c: What do you expect the long-term behavior of the system to be? Solution 1c: Given that there are no reinforcements, one expects that both forces will attrit (at rates which depend on the relative sizes of A and B) until one is down to zero, at which point the opposing force will remain at a constant strength from then on. The actual result (who “wins”) would seem to depend on the relative sizes of A and B, but also on the initial conditions. Numerical experiments with different parameter values and different initial conditions can help here. Problem 1d: Classify this system of differential equations. Solution 1d: This is a second-order, non-linear, homogeneous, autonomous (independent of time) system. As such, one expects not to be able to solve it analytically, however, as the next two parts show, this particular system can in fact be solved explicitly. Problem 1e: Is At y t Bt x t 1 , ( ) 1 ( ) = = a solution of this system? Solution 1e: We have 1 1 - 1 1 - 2 ÷= - ø ö ç è æ ÷= ø ö ç è æ = t B t dt B d dt Bt d dt dx and similarly 1 - 2 = - t dt A dy . Also, 2 1 1 1 Bt At Bt A x y A ÷= - ø ö ç è æ ÷× ø ö ç è æ - × × = - × and 2 1 At - B ×x ×y = - . So this particular solution is verified. Problem 1f: Is ( ) 1 , ( ) ( ) 1 ( ) A t C y t B t C x t + = + = the general solution of this system? Why NOT? Solution 1f: We have 1 2 ( ) 1 ( ) 1 ( ) 1 - - ÷= - + ø ö ç è æ = + ÷ ÷ ø ö ç ç è æ + = t C B t C dt B d dt B t C d dt dx and 2 ( ) 1 - = - t + C dt A dy , while 2 ( ) 1 ( ) 1 ( ) 1 B t C A t C B t C A x y A + = - ÷ ÷ ø ö ç ç è æ + ×÷ ÷ ø ö ç ç è æ + - × × = - × and
1-B.x-VThuswehaveaninfinitenumberofsolutions,asCisanarbitraryA(t +C)?constant.However,this is NOTa general solution,asweneed twoarbitraryconstantstodealwith all possible initial conditions.This solution will only work for initial conditions that havex(0)AtheratioJ(O)BBDFe-BDTDAExtraCredit:1a.Showthatx(t)=t)=isthegeneral solution.1-Fe-BDr,1-Fe-BDrSolutionExtraCredit1a:BBDFe-BDID"Fe-BDtdx-BD°Fe-BDtDAAFirst note that x.Also,1-Fe-BDr(1- Fe-BDr)21-Fe-BDt(1-Fe-BDr)2dtand this is in fact - A.x y.Furthermore,B"D"FB-RD(BDFe-BDr)(1- FeDFeAAdydt(1 Fe-BDr)2BD"F2e-2BDtB"D"F2B2D"FB2D°F2BDtDAAAA(1- Fe-BDr )?(1-Fe-BDtand this is the same as -B-x y.As there are two arbitrary constants in this formula, this is in fact thegeneral solution tothe systemof equationsExtra Credit: 1b. If A=0.0009, B= 0.00075, x(0)=110 and y(0) = 100,findthe particular solution. What is the long-term result of an armor battle?Solution Extra Credit 1b:DSubstituting in 1 = 0 , we get 120 = x(0)andBDFBF (BFD0.00075F100 = y(0) =(120),and so(120)A(1-FA0.0009A(1-F)110(0.0009)=1.1. Therefore D=(1-F)120=-0.1(120)=-12. Thus the particular120(0.00075)11e 0.009-11e 0.009t12-12solution is x(t):The resultand y(t) :11.1e 0.00911.1e 0.009111.1e0.009r1.1e0.0091is that x(t) →0 and y(t) 10 . So although Country Y started off with ten less tanks than238
238 2 ( ) 1 A t C B x y + - × × = - . Thus we have an infinite number of solutions, as C is an arbitrary constant. However, this is NOT a general solution, as we need two arbitrary constants to deal with all possible initial conditions. This solution will only work for initial conditions that have the ratio B A y x = (0) (0) . Extra Credit: 1a. Show that BDt BDT BDt Fe DFe A B y t Fe D x t - - - - = - = 1 , ( ) 1 ( ) is the general solution. Solution Extra Credit 1a: First note that 2 2 1 1 (1 ) BDt BDt BDt BDt BDt Fe D Fe A B Fe DFe A B Fe D x y - - - - - - = - × - × = . Also, 2 2 (1 ) BDt BDt Fe BD Fe dt dx - - - - = , and this is in fact - A×x ×y . Furthermore, ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 (1 ) (1 ) (1 ) (1 ) BDt BDt BDt BDt BDt BDt BDt BDt BDt BDt BDt Fe e A B D F Fe A B D F e e A B D F e A B D F Fe DFe BDFe A B e A B D F Fe dt dy - - - - - - - - - - - - - = - - + - = - ÷ ø ö ç è æ - ÷ ÷ ø ö ç ç è æ - - = and this is the same as - B×x ×y . As there are two arbitrary constants in this formula, this is in fact the general solution to the system of equations. Extra Credit: 1b. If A = 0.0009, B = 0.00075, x(0) = 110 and y ( 0 ) = 100 , find the particular solution. What is the long-term result of an armor battle? Solution Extra Credit 1b: Substituting in t = 0 , we get F D x - = = 1 120 (0) and (120) 0.0009 0.00075 (120) (1 ) 1 100 (0) F A BF F D A BF A F BDF y ÷= = ø ö ç è æ - = - = = , and so 1.1 120(0.00075) 110(0.0009) F = = . Therefore D = (1- F)120 = - 0.1(120) = - 12 . Thus the particular solution is 1.1 1 12 1 1.1 12 ( ) 0.009 0.009 - = - - = t t e e x t and 1.1 1 11 1 1.1 11 ( ) 0.009 0.009 0.009 0.009 - = - - = t t t t e e e e y t . The result is that x(t) ® 0 and y(t) ® 10 . So although Country Y started off with ten less tanks than
Country X, the greater survival rate meant that Country Y's armor force survived and CountryX's didn't.Problem2:Supposethedifferential equationsareinsteaddx=-0.0009·x·ydtdy=-0.00065.x·y-0.01.ydtWhat change in the situation does the new system above suggest?Solution 2:Notice that the parameter B has decreased from 0.00075 to 0.00065, but an additional termO.Ol-y,hasbeensubtracted.Thiswould indicatethatCountryY'stankshavebecomesomewhat harder to kill (an improved design?) but they are lost at a steady 1% rate, regardless ofthe number of enemy present.One might speculate that the improved tank design is subject tobreakdown,orperhapsthepresenceofarandomelementlikeaminefieldiscausingthesteadydecline in Country Y's tanksProblem 3:Suppose thedifferential equations are insteaddx0.0009.x.yl.05dtdy-0.00065.x.y-0.01.y(dtWhat change in the situation does the new system above suggest?Solution3:Now the exponent on the y term has slightly increased, so that large numbers of country Y'stanks will kill significantly more of CountryX's tanks.Perhaps some new group-huntingdoctrineorcapabilityhasbeenaddedtotheYsideoftheequation.Problem 4: Suppose the differential equations are insteaddx-0.0009.x.y1.05 + 3dtdy-0.00065.x.y-0.01.JdtWhat change in the situation does the new system above suggest?239
239 Country X, the greater survival rate meant that Country Y’s armor force survived and Country X’s didn’t. Problem 2: Suppose the differential equations are instead ï ï î ï ï í ì = - × × - × = - × × x y y dt dy x y dt dx 0.00065 0.01 0.0009 What change in the situation does the new system above suggest? Solution 2: Notice that the parameter B has decreased from 0.00075 to 0.00065, but an additional term, 0.01×y , has been subtracted. This would indicate that Country Y’s tanks have become somewhat harder to kill (an improved design?) but they are lost at a steady 1% rate, regardless of the number of enemy present. One might speculate that the improved tank design is subject to breakdown, or perhaps the presence of a random element like a minefield is causing the steady decline in Country Y’s tanks. Problem 3: Suppose the differential equations are instead ï ï î ï ï í ì = - × × - × = - × × x y y dt dy x y dt dx 0.00065 0.01 0.0009 1.05 What change in the situation does the new system above suggest? Solution 3: Now the exponent on the y term has slightly increased, so that large numbers of country Y’s tanks will kill significantly more of Country X’s tanks. Perhaps some new group-hunting doctrine or capability has been added to the Y side of the equation. Problem 4: Suppose the differential equations are instead ï ï î ï ï í ì = - × × - × = - × × + x y y dt dy x y dt dx 0.00065 0.01 0.0009 3 1.05 What change in the situation does the new system above suggest?
Solution 4:To combat this, Country X is now reinforcing the battle at a steady rate of three tanks per hourProblem 5:Suppose the differential equations are insteaddx=0.0009.x-y1.05+3dtC0.00065.x·y0.01.y+0.005.xdtWhat change in the situation does the new system above suggest?Solution5:Country Y is now reinforcing their forces also, but not at a steady rate. The rate of reinforcementdepends on how many enemy tanks are still "alive"Problem 6:Suppose thedifferential equations are instead[ = -009 ( (0) 0 3dtdy=-0.00065·x(t)· y(t)-0.01·y(t)+0.005·x(t-4)(dtWhat change in the situation does the new system above suggest?Solution 6:There is a four-hour (realistic?) time delay in the reinforcements for Country Y- informationfrom the battle and logistics are delaying thereinforcement process.Problem7:Howwouldyoutake intoaccountfriendlyfire?Solution 7:Friendly fire would add probably negative x’ and y’ terms to the system, just as we use xyterms to model interactions between forces, we use xand y?terms to model interactionsbetween friendly forces.One hopes the coefficients on these terms are very small240
240 Solution 4: To combat this, Country X is now reinforcing the battle at a steady rate of three tanks per hour. Problem 5: Suppose the differential equations are instead ï ï î ï ï í ì = - × × - × + × = - × × + x y y x dt dy x y dt dx 0.00065 0.01 0.005 0.0009 3 1.05 What change in the situation does the new system above suggest? Solution 5: Country Y is now reinforcing their forces also, but not at a steady rate. The rate of reinforcement depends on how many enemy tanks are still “alive”. Problem 6: Suppose the differential equations are instead ï ï î ï ï í ì = - × × - × + × - = - × × + 0.00065 ( ) ( ) 0.01 ( ) 0.005 ( 4) 0.0009 ( ) ( ) 3 1.05 x t y t y t x t dt dy x t y t dt dx What change in the situation does the new system above suggest? Solution 6: There is a four-hour (realistic?) time delay in the reinforcements for Country Y – information from the battle and logistics are delaying the reinforcement process. Problem 7: How would you take into account friendly fire? Solution 7: Friendly fire would add probably negative 2 x and 2 y terms to the system, just as we use x ×y terms to model interactions between forces, we use 2 x and 2 y terms to model interactions between friendly forces. One hopes the coefficients on these terms are very small
Problem8:Whatwouldthesystemlooklikeifthereisathree-waybattlebetweenCountry X, CountryY,and Country Z? (This has occurred duringWorld War Il in China[theJapanese,theChineseCommunists,andtheKuomintang];and intheformerYugoslavia duringthebreakup [the Serbs,theBosnians,andthe Croats].)Solution 9:One might expect, absent other factors as in #2-#7 above, to get a system of the formdx-A-X-y.zdtdy=-B.x.y-2dtdz=-C.x.y-z[dtHowever, this is incorrect as it only seems to model three-way interactions, which should be rare.It also does not reduce to the two-way case when z=O.A more realistic model might look likedx-A.X.y-D.X-zdtdy=-B·x·y-E.y·zdtdz-C·X·z-F·y·z[dtwhere each parameter now measures interactions and kill rates for each possible two-wayencounter.Thecomplications introduced in#2-#7couldalsobemodeled241
241 Problem 8: What would the system look like if there is a three-way battle between Country X, Country Y, and Country Z? (This has occurred during World War II in China [the Japanese, the Chinese Communists, and the Kuomintang]; and in the former Yugoslavia during the breakup [the Serbs, the Bosnians, and the Croats].) Solution 9: One might expect, absent other factors as in #2-#7 above, to get a system of the form ï ï ï î ï ï ï í ì = - × × × = - × × × = - × × × C x y z dt dz B x y z dt dy A x y z dt dx However, this is incorrect as it only seems to model three-way interactions, which should be rare. It also does not reduce to the two-way case when z = 0. A more realistic model might look like ï ï ï î ï ï ï í ì = - × × - × × = - × × - × × = - × × - × × C x z F y z dt dz B x y E y z dt dy A x y D x z dt dx where each parameter now measures interactions and kill rates for each possible two-way encounter. The complications introduced in #2-#7 could also be modeled