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2 Pages

### wbest03_Union_Air

Course: 101 101, Fall 1997
School: Ringling
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Word Count: 164

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4.0 What'sBest! Status Report 4/3/00 10:05 AM Memory Allocated: 8192 Reduction has been turned off CLASSIFICATION STATISTICS Current / Maximum --------------------------------------------------Numeric 106 / 256000 Adjustable 5 Constraints 10 Integers 0 Optimizable 26 Nonlinear 0 Coefficients 55 Model Type: LINEAR The smallest and largest coefficients in the model were: 1.0000000 195.00000 The smallest...

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Coursehero >> Florida >> Ringling >> 101 101

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Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
4.0 What'sBest! Status Report 4/3/00 10:05 AM Memory Allocated: 8192 Reduction has been turned off CLASSIFICATION STATISTICS Current / Maximum --------------------------------------------------Numeric 106 / 256000 Adjustable 5 Constraints 10 Integers 0 Optimizable 26 Nonlinear 0 Coefficients 55 Model Type: LINEAR The smallest and largest coefficients in the model were: 1.0000000 195.00000 The smallest coefficient occurred in constraint cell: Sheet1!H5 on optimizable cell: Sheet1!H5 The largest coefficient occurred in constraint cell: Sheet1!H5 on optimizable cell: Sheet1!F5 Tries: 12 Infeasibility: 0 Objective: 30610 Solution Status: GLOBALLY OPTIMAL. Solution Time: 0 Hours 0 Minutes 2 Seconds Blank cell warning has been turned off. End of report. Union Personnel Airways Scheduling. What'sBest model Shifts Agents hired: Cost/agent: Time Period 6 - 8 AM 8-10 AM 10- Noon Noon-2 PM 2-4 PM 4-6 PM 6-8 PM 8-10 PM 10-Midnite Midnite-6 Total Cost 30610 1 48 170 2 31 160 3 39 175 4 43 180 5 15 195 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 Actually on duty 48 79 79 118 70 82 82 43 58 15 Constraint Objective Labels Dual price Color conventions: Data Decision Formula This sheet is setup to be compatible with the What'sBest solver, see: http://www.lindo.com to download the software. ### ### ### ### ### ### ### ### ### ### Agents needed 48 79 65 87 64 73 82 43 52 15 Dual price #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME? #NAME?
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Ringling - 101 - 101
What'sBest! 7.0 Status Report4/15/04 9:12 PMSolver memory allocated:131072Model before Linearization:CLASSIFICATION STATISTICSOriginal-Numeric23Adjustable2Constraints3Integers0Optimizable9Linearization Enabled:Big M parameter:60000.Del
Ringling - 101 - 101
What'sBest! 10.0.0.0 (Jul 03, 2008) - Library 5.5.1.362 - Status Report DATE GENERATED:Jul 11, 200811:22 AMMODEL INFORMATION:CLASSIFICATION DATACurrentCapacity Limits-Numerics42Variables11Adjustables4UnlimitedConstraints3UnlimitedIntegers
Ringling - 101 - 101
What'sBest! 7.0 Status Report4/16/04 11:21 AMSolver memory allocated:131072Model before Linearization:CLASSIFICATION STATISTICSOriginal-Numeric31Adjustable2Constraints3Integers0Optimizable9Linearization Enabled:Big M parameter:60000.De
Ringling - 101 - 101
TITLE Wyndor! ling06b;! Y1 = price! Y2 = price! Y3 = priceGlass Co. Problem.Dual of original model;of plant 1 constraint;of plant 2 constraint;of plant 3 constraint;[Obj] Min= 4*Y1 + 12*Y2 + 18*Y3;[X1][X2]Y12*Y2+ 3*Y3 &gt;= 3;+ 2*Y3 &gt;= 5;!A
Ringling - 101 - 101
! Parametric or sensitivity analysis ofthe RHS of the Wyndor model;SUBMODEL Wyndor:MAX = TotProf;[Profit] TotProf = PCDoor*X1 + 5*X2;! Production time;[Plant1]X1&lt;= 4;[Plant2]2*X2 &lt;= 12;[Plant3] 3*X1 + 2*X2 &lt;= 18;ENDSUBMODELCALC:! Do parametr
Ringling - 101 - 101
! Wyndor Glass Co. Problem. LINDO model;! X1 = batches of product 1 per week;! X2 = batches of product 2 per week;! Profit, in 1000 of dollars;[Profit]MAX= 3 * X1 + 5 * X2;! Production time;[Plant1]X1&lt;= 4;[Plant2]2 * X2 &lt;= 12;[Plant3]3 * X1 +
Ringling - 101 - 101
What'sBest! 7.0 Status Report4/16/04 10:39 AMSolver memory allocated:131072Model before Linearization:CLASSIFICATION STATISTICSOriginal-Numeric21Adjustable3Constraints2Integers0Optimizable8Linearization Enabled:Big M parameter:60000.De
Ringling - 101 - 101
What'sBest! 7.0 Status Report4/16/04 10:48 AMSolver memory allocated:131072Model before Linearization:CLASSIFICATION STATISTICSOriginal-Numeric29Adjustable2Constraints3Integers0Optimizable9Linearization Enabled:Big M parameter:60000.De
Lund - FMA - 021
x = 1 x=0 t=0t &gt; 0 1 (x) = ex 12 2 (x) = 2e2x
Lund - FMA - 021
q b u t u(x, t) x2b x
Lund - FMA - 021
t=0 1sin x10u1t=0 (x )2 x0 u2u(x, t)t=x=0 2 x = &gt; 0 t=0 0t=0 c = 1t &gt; 0
Lund - FMA - 021
t=0 x x = 0 h(x) = 5(x 1.9) (x 2.1) . u(x, 3) 12t &gt; 0u(x, t) 0 &lt; t &lt; 6 x u(3, t)
Lund - FMA - 021
utt uxx = 0u(0, t) = 0u(x, 0) = 0ut (x, 0) = (x 3) (x 6) t = 1 t = 5x &gt; 0, t &gt; 0t&gt;0x&gt;0x&gt;0 t u(x, t) = 0x=1 p0 , p1 p2 0, 1 2 u(x)v (x) ex dx.(u | v ) =0ex p 2
Lund - FMA - 021
p f (x) = x(x) 1/21 [1, 1] f = 1 |f (x)|2 dx utt uxx = 0,u(0, t) = 0, u(x, 0) = 0,ut (x, 0) = (x 1),x &gt; 0, t &gt; 0t&gt;0x&gt;0x&gt;0 x u(r, ) = J0 (02 r),t = 1/2 t = 2
Lund - FMA - 021
0 &lt; x &lt; 1, t &gt; 0 ut auxx = 0,u(0, t) = u(1, t) = 0,t&gt;0u(x, 0) = 2 sin x + 3 sin 4x, 0 &lt; x &lt; 1 , a L2 (I ), I = (1, 1) p 1 (2x 1)(x) p(x)2 dx .1
Lund - FMA - 021
q 2q t = 0 2L
Lund - FMA - 021
TENTAMENSSKRIVNINGKONTINUERLIGA SYSTEM 7.5 hp2009-01-09 kl 14 19LUNDS TEKNISKA HGSKOLAMATEMATIKHJLPMEDEL: Utdelade formelblad (Kontinuerliga system + Funktionsteori), Tefyma ellergymnasietabell och minirknare.Lsningarna ska vara frsedda med ordentl
Lund - FMA - 021
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGKontinuerliga system20090527 kl 8 13HJALPMEDEL: Tefyma eller gymnasietabell, utdelad formelsamling i Kontinuerliga systemsamt minirknare.aLsningarna skall vara frsedda med ordentliga motiveringar.
Lund - FMA - 021
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGKontinuerliga system20090824 kl 14 19HJALPMEDEL: Tefyma eller gymnasietabell, utdelad formelsamling i Kontinuerliga systemsamt minirknare.aLsningarna skall vara frsedda med ordentliga motiveringar
Lund - FMA - 021
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGKONTINUERLIGA SYSTEM 7.5 hp20100113 kl 8 13HJALPMEDEL: Tefyma eller gymnasietabell, utdelad formelsamling i Kontinuerliga systemsamt minirknare. Lsningarna skall vara frsedda med ordentliga motiveri
Lund - FMA - 021
u(x, t) t=0 xt=0 t 0 u(x, 0) = 100(1 x)0 &lt; x &lt; 12 u u = 0,0 &lt; x &lt; 1, t &gt; 0tx2 u(0, t) = 0, u(1, t) = 100, t &gt; 0u(x, 0) = 100(1 x),0 &lt; x &lt; 1. v (x, t) = u(x, t) 100x 2 v v = 0,0 &lt; x &lt; 1, t &gt; 0tx2 v (0,
Lund - FMA - 021
ut uxx = 0,ux (0, t) = 0,ux (2b, t) = 0,u(x, 0) = q (x) (x b), q 2q u(x, t) = +2 0 &lt; x &lt; 2b, t &gt; 0t&gt;0t&gt;00 &lt; x &lt; 2b . sin k k2 2 t/(4b2 )kx2ecosk2bk =13 1 = ex 2 = e2x ex4 ex e2x 10 a1 = 4 a2 = 3 v = u 1 v
Lund - FMA - 021
x &gt; 0, t &gt; 0 ut Duxx = 0,ux (0, t) = 0,t&gt;01u(x, 0) = (x 1) (x 1 ), x &gt; 0 u(x, t) x t R u 11+11122e(x) /4Dt d +e(x) /4Dt d =4Dt11x+1x1x1x+1+, erf erf + erf x &gt; 0, t &gt; 0.erf 4Dt4Dt4Dt4Dtu(x, t) = =
Lund - FMA - 021
u h 2 u 2u 2 = 0, t2xu(0, t) = 0, u(x, 0) = 0, u (x, 0) = h(x),t u x &gt; 0, t &gt; 0t&gt;0x&gt;0x&gt;0 x R 2 2 uu= 0 , x R, t &gt; 0t2x2xR u (x, 0) = 0, u (x, 0) = h (x), x Rt h u(x, t) = u (x, t) =12x +tx th (y ) dy
Lund - FMA - 021
u x = 0 0 ut (x, 0) 1 u(x,1)1012457xu(x,5)1.51012811x u(1, t) = 0 0 t 2 t 7 p0 = 1 p1 (x) = x 1 p2 (x) = x2 4x + 2. p(x) =(p0 |ex )(p1 |ex )(p2 |ex )711p0 +p1 +p2 = x + x2 .p0 2p1 2p2
Lund - FMA - 021
32 P0 (x) = 1, P1 (x) = x P2 (x) = x2 1 (u|v ) =(P k | f )ak =(P k | P k ) p(x) =u(x)v (x) dx p(x) =121 2ak Pk (x) f p k =0115 a0 = , a1 = a2 =4216531311511+ x + ( x2 ) =+ x + x2 4216 2232 232 t =
Lund - FMA - 021
u(x, t) =d2dx2 uk (t) sin kx k =1(uk (t) + ak 2 2 uk (t) sin kx = 0 .k =1 uk (t) = ck eak2 2 t=0 ck2 sin x + 3 sin 4x = u(x, 0) = ck sin kx,0 &lt; x &lt; 1.k =1 cfw_sinkx k =1 c1 = 2, c4 = 3 ck = 0 u(x, t) = 2ea t
Lund - FMA - 021
ut Duxx = 0,ux (0, t) = ux (2L, t) = 0, u(x, 0) = q ,2q, 3q 2q2k =1 cfw_ k 0 v c2 v v (x) = 0 0u = ( 1 ,0,0) (x), |x| &lt; 12u(x) = 1,|x| = 1 .= u 1ln 2 1 0.01323ck = 2k1 , k = 0, 1, 2,
Lund - FMA - 021
SVAR OCH ANVISNINGARKONTINUERLIGA SYSTEM 7.5 hp20090109LUNDS TEKNISKA HGSKOLAMATEMATIK1. Lgg in ett koordinatsystem s att den 100 gradiga cylindern ligger i intervallet 0 &lt; x &lt; 1och den 0 gradiga i intervallet 1 &lt; x &lt; 2. Med u(x, t) som temperaturen
Lund - FMA - 021
LUNDS TEKNISKA HGSKOLAMATEMATIKLSNINGARKontinuerliga system200905271. Alternativ 1: Vi homogeniserar randvillkoren genom att stta v = u 1. vt vxx = 1v (0, t) = v (1, t) = 0v (x, 0) = 0Sedan anstter vi en sinusserie v (x, t) =k =1 vk (t) sin kx(
Lund - FMA - 021
LUNDS TEKNISKA HGSKOLAMATEMATIKLSNINGARKontinuerliga system200908241. Egenvektorer: k = e5x sin kx,k = 1, 2 , . . .Egenvrden:k = 25 + k 2 2 ,k = 1, 2 , . . .110xSkalrprodukt: (u|v ) = 0 u(x)v (x)e dxVid berkning av (1 |2 ) frsvinner exponenti
Lund - FMA - 021
LUNDS TEKNISKA HGSKOLAMATEMATIK1. Modell fr 0 &lt; x &lt; 1,Lsning:u(x, t) = 100x +k =1LSNINGARKontinuerliga system 7.5 hp20100113t &gt; 0: ut uxx = 0u(0, t) = 0, u(1, t) = 100u(x, 0) = 100(1 x)20022(1 + (1)k )ek t sin kxk2. Temperaturen r tidsobe
Lund - FMA - 430
LUNDS TEKNISKA HGSKOLAMATEMATIKTENTAMENSSKRIVNINGFlerdimensionell analys2008-05-23, kl 8-13INGA HJLPMEDEL. Motivera lsningarna vl.1. Lt D vara det begrnsade omrde som begrnsas av linjen y = x och parabeln y = x 2 .Berkna dubbelintegralen20x 2 dx d
Lund - FMA - 430
Q(h, k) f (x, y ) = 3xy x3 + y 3 . ff+2= 25yxyu = x + 2yv = 2x + y f (x, 0) = x2 + 2x f (x, y ) = e4x 2 xy 2, x &gt; 0 f (x, y ) = 1 grad f (1, 2) f P : (1, 2) (4,
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGFlerdimensionell analys20081023, kl. 813INGA HJALPMEDEL. Lsningarna skall vara frsedda med ordentliga motiveooringar.1.a. Denera begreppet stationr punkt fr en funktion av era variabler.ao(0.2
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGFLERDIMENSIONELL ANALYS20081216 kl 813HJALPMEDEL: Inga.Lsningarna skall vara frsedda med ordentliga motiveringar. Rita grna gurer i frekomooaomande fall.1. Berkna dubbelintegralena(x + y ) dx
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGFlerdimensionell analys20090112, kl. 813.INGA HJALPMEDEL. Lsningarna skall vara frsedda med ordentliga motiveooringar.1.a) Vad menas med att en kvadratisk form r negativt denit?ab) Finn alla lo
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGFLERDIMENSIONELL ANALYS2009-03-10 kl 8-13INGA HJALPMEDEL. Lsningarna skall vara frsedda med ordentliga motiveringar.oo1. a) Bestm en ekvation fr tangentplanet till ytan x + y 2 + z 3 = 3 i punkten
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGFlerdimensionell analys20090414, kl. 1419.INGA HJALPMEDEL. Lsningarna skall vara frsedda med ordentliga motiveooringar.1.a) Denera begreppen stationr punkt och lokal extrempunkt.ab) Finn alla l
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGFlerdimensionell analys2009-05-26 kl 813INGA HJALPMEDEL.Lsningarna skall vara frsedda med ordentliga motiveringar. Skriv fullstndiga meningarooaoch frklara dina beteckningar. Ge tydliga och enkl
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGFlerdimensionell analys2009-08-26 kl 813INGA HJALPMEDEL.Lsningarna skall vara frsedda med ordentliga motiveringar. Skriv fullstndiga meningarooaoch frklara dina beteckningar. Ge tydliga och enkl
Lund - FMA - 430
TENTAMENSSKRIVNINGFLERDIMENSIONELL ANALYS2009-10-24 kl 813LUNDS TEKNISKA HGSKOLAMATEMATIKINGA HJLPMEDEL.Lsningarna skall vara frsedda med ordentliga motiveringar.1. Bestm strsta och minsta vrde till funktionenf (x, y ) = x2 + (y 1)2p omrdet D = c
Lund - FMA - 430
TENTAMENSSKRIVNINGFLERDIMENSIONELL ANALYS2009-12-15 kl 813LUNDS TEKNISKA HGSKOLAMATEMATIKINGA HJLPMEDEL.Lsningarna skall vara frsedda med ordentliga motiveringar.1. Berknax2 dxdy,Ddr D r omrdet som begrnsas av olikheterna y x2 , y 1 och x 0.2.
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGFlerdimensionell analys2010-01-11 kl 813INGA HJALPMEDEL.Lsningarna skall vara frsedda med ordentliga motiveringar. Skriv fullstndiga meningarooaoch frklara dina beteckningar. Ge tydliga och enkl
Lund - FMA - 430
LUNDS TEKNISKA HGSKOLAMATEMATIKLSNINGARFlerdimensionell analys2008-05-231. Linjen y = x och parabeln y = x 2 skr varandra fr x sdanaatt x = x 2 vilket kan skrivas x (x 1) = 0. Skrningspunkterna blir (0, 0) och (1, 1). Dubbelintegralen kan berknasme
Lund - FMA - 430
LUNDS TEKNISKA HGSKOLAMATEMATIKSVAR OCH ANVISNINGARFlerdimensionell analys2008-08-25a) Svar: Att Q(h, k) &gt; 0 fr alla (h, k) = (0, 0).1.b) Svar: (0,0) och (1, 1) r stationra punkter. (1, 1) r en lokal minimipunkt.2. Svar: f (x, y) = 6y2 x2 + xy + g
Lund - FMA - 430
444332102123100012x34012x34012x34524y31-3-2-10-2112-1-1x00-1123-2-32-200123
Lund - FMA - 430
x y x 0 x 2 y x22(x + y ) dxdy =(x + y )Dy =x02x2 dx = 20 x332=0y22xdxy =x16.3163 xy +02=dx =u =x+yv =xy f (x, y ) g(u, v) fxfy f (x, y ) = g(u(x, y ), v(x, y ) =g= u2vg(u, v ) = C 1 f (x, y
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKLOSNINGSFORSLAGFlerdimensionell analys20090112, kl. 813.1.a. Att Q(h, k ) &lt; 0 fr alla (h, k ) = (0, 0).ob. Fr att nna stationra punkter deriverar vi partiellt och f drmed ekvaoaar ationsystemetfx = 2xy 4y = 2y (
Lund - FMA - 430
LUNDS TEKNISKA HGSKOLAMATEMATIKLSNINGARFLERDIMENSIONELL ANALYS2009-03-10 kl 08-131. a) Ytan r nivyta till f (x, y, z ) = x + y 2 + z 3 , allts gradf r normalvektor till tangentplanet. grad f = (1, 2y, 3z ) grad f (1, 1, 1) = (1, 2, 3) tangentekvation
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIK1.LOSNINGSFORSLAGFlerdimensionell analys20090414, kl. 1419.a) Se sidorna 99 och 100 i boken.b) Fr att nna stationra punkter deriverar vi partiellt och f ekvationssysteoaarmetfx = 2x 4 + y 2 = 0fy = 2xy = 0Fr d
Lund - FMA - 430
LUNDS TEKNISKA HGSKOLAMATEMATIKLSNINGARFlerdimensionell analys2009-05-261.D=0=123xy 23x123dx =x2 x2 2 7 x2327dx =xy 2 dy01y11x y dxdy =501= . =011(x x 2 ) dx =171x1Svar: Dubbelintegralens vrde r .7Anm.: Det g
Lund - FMA - 430
LUNDS TEKNISKA HOGSKOLAMATEMATIKTENTAMENSSKRIVNINGFlerdimensionell analys2009-08-26 kl 813LOSNINGSFORSLAGx11.1xy dy dx =xy dxdy =0D2x033 x3 d x = .282. a) Se lroboken.ab) Se lroboken.ac) L g (x, y, z ) = x2 y 2 1 z 3 . Vi f att g
Lund - FMA - 430
LUNDS TEKNISKA HGSKOLAMATEMATIKLSNINGARFLERDIMENSIONELL ANALYS2009-10-241. Vi hittar stationra punkter genom att lsa fx = 2x = 0, fy = 2(y 1) = 0 vilketger (x, y ) = (0, 1) som ligger p randen. Fr att underska randen anstter vi x(t) =cos(t), y (t)
Lund - FMA - 430
LUNDS TEKNISKA HGSKOLAMATEMATIKLSNINGARFLERDIMENSIONELL ANALYS2009-12-151. Vi brjar med att integrera med avseende p y .110x201x2 [y ]1 2 dx =x0x2 (1 x2 )dx =1x2x2 dy dx =D=11x2 dxdy =03xx5351dy dx =x21=0211=.35152.
Lund - FMA - 430
LUNDS TEKNISKA HGSKOLAMATEMATIKSVAR OCH ANVISNINGARFlerdimensionell analys2010-01-111. De stationra punkterna visar sig vara (0, 0), (1, 1) och (1, 1). Underskning av demger svaret.Funktionen har inget strsta vrde, ty f (x, 0) = x4 3x2 = x4 (1 x32
Ashford University - ABS - psy 304
KOHLBERG'S MORAL STAGESKolberg's theory specifies six stages of moral development, arranged in three levels.Level I: Preconventional/PremoralMoral values reside in external events, or in bad acts. The child is responsive to rules, but viewsthem in ter
Ashford University - ABS - psy 304
Portfolio ProjectJennifer Davis8/22/11COM321: Communication TheoryDuvan ArsolaMy motivation for studying communication is to learn how to better communicate withpeople. People sometimes have problems getting their point across to other people. I fee
Ashford University - ABS - psy 304
Different cultures have many different ways of doing things including how people raise theirchildren. Because of how people are able to communicate faster and easier with the moreadvances in technology we have today the way people are doing things are c
Ashford University - ABS - psy 304
Different cultures have many different ways of doing things including how people raise theirchildren. Because of how people are able to communicate faster and easier with the moreadvances in technology we have today the way people are doing things are c
Ashford University - ABS - psy 304
During the first two years of life, children quadruple their weight and increasetheir height by two-thirds. This rate slows down between 2 and 3 years when childrengrow only about 3.5 inches and gain only about 4 pounds. Between the ages 4 and 6, thein