Ejercicios del capitulo 4
4.1 La tabla da el número de unidades de sangre tipo A que el hospital
Woodlawn utilizo en las últimas 6 semanas.
Semana de Unidades
Empleados
Agosto 31
360
Septiembre7 389
Septiembre 14 410
Septiembre 21 381
Septiembre 28 3 semanas 368
Octubre 5 374
a. pronostique la demanda para la semana del 12 de octubre con un promedio
móvil de 3 semanas.
Sep 21 = 381 +
Sep28 = 368
Oct 5 = 374
1,123/3 374.33 promedio móvil semana del 12 de octubre.
Utilice un promedio móvil ponderado de tres semanas con ponderaciones de 1.
3. y 6 usando 6 para la semana reciente. Pronostique la demanda para la
semana del 12 de octubre.
Agosto 31
360
Septiembre7 389
Septiembre 14 410
Septiembre 21 381
Septiembre 28 3 semanas 368
Octubre 5 374
SEP 21 410(0.6)+389(0.3)+360(0.1)= 246+116.7+36= 398.7/1= 398.7
SEP 28 381(0.6)+410(0.3)+389(0.1)= 228.6+123+38.9= 390.5/1 =390.5
OCT 5 368(0.6)+381(0.3)+410(0.1)= 220.8+114.3+41= 376.1/1= 376.1
OCT 12 374(0.6)+368(0.3)+381(0.1)= 224.4+110.4+38.1= 372.9/1=372.9
c. Calcule el pronostico para la semana del 12 de octubre aplicando
suavizamiento exponencial con un pronostico de 360 para el 31 de agosto y
a=.2
SEMANAS UNIDADES PRONÓSTICO
Agosto 31 360 360
Septiembre7 389 360 360+0.2
(360-360)=360
Septiembre 14 410 365.08 360+0.2(389-
360)=365.8
Septiembre 21 381 374.64
365.08+0.2(410-365.08)=374.64
Septiembre 28 368 375.91
374.64+0.2(381-374.64)=375.91
Octubre 5 374 374.33
375.91+0.2(368-375.91)=374.33
Octubre 12 374.26
374.33+0.2(374-374.33)=374.26
Pronostico para la demanda de la semana del 12 de octubre = 374.26
4.2 Año 1 2 3 4 5 6 7 8 9 10
11
Demanda 7 9 5 9 13 8 12 13 9 11 7
A) Grafique los datos anteriores observa alguna tendencia ciclos o
variaciones aleatorias?
B) Comenzando el año 4 y hasta el año 12.pronostique la demanda usando
promedios móviles de 3 años. Grafique pronósticos en la misma grafica de
los datos originales.
Año 1 2 3 4 5 6 7 8 9 10 11
Demanda 7 9 5 9 13 8 12 13 9 11 7
9+13+8/3=10
13+8+12/3=11
8+12+13/3= 11
12+13+9/3=11.3
13+9+11/3=11
9+11+7/3=9
El pronóstico para el año 12 tiene una demanda de 9
C) Comenzando en el año 4 hasta el año 12, pronostique la demanda usando
un promedio movil de 3 años, comparaciones de .1 .3 .6 utilizando 6 para el
año reciente GRAFIQUE su pronostico en la misma grafica.
Año demanda Promedio movil ponderado 3 años
1 7
2 9
3 5
4 9 9(6)+5(3)+9(1) /10= 7.8
5 13 13(6)+9(3)+5(1) /10= 11
6 8 8(6)+13(3)+9(1) /10=9.6
7 12 12(6)+8(3)+13(1) /10=10.9
8 13 13(6)+12(3)+8(1)/10= 12.2
9 9 9(6)+13(3)+12(1)/10= 10.5
10 11 11(6)+9(3)+13(1)/10=10.6
11 7 7(6)+11(3)+9(1)/10= 8.4
12
D ) Al observar el pronóstico contra los datos originales, ¿cual considera
que proporciona mejores resultados.
R= Proporciona mejores resultado el año 6 con 9.6 un pronostico promedio,
en comparación con la demanda real 8.
4.3 Regrese al problema 4.2.Desarrolle un pronostico para los años 2 al 12
mediante suavizamiento exponencial con a=.4 y un pronostico para el 1 de 6.
Grafique su nuevo pronostico junto con los datos reales y un pronostico
intuitivo, con base en su inspección visual. Que es pronóstico es mejor?
R=
Año Demanda Pronostico
1 7 6
2 9 6.4 6+0.4 (7-6)=6.4
3 5 7.44 6.4+0.4 (9-
6.4)=7.6
4 9 6.4640 7.44+0.4 (5-
7.44)=6.4640
5 13 7.4784 6.4640+0.4 (9-
6.4640)=7.4784
6 8 9.6870 7.4784+0.4 (13-
7.4784)=9.6870
7 12 9.0122 9.6870+0.4 (8-
9.6870)= 9.0122
8 13 10.2073 9.0122+0.4 (12-
9.0122)=10.2073
9 9 11.3244 10.2073+0.4 (13-
10.2073)=11.3244
10 11 10.3946 11.3244+0.4 (9-
11.3244)=10.3946
11 7 10.6368 10.3946+0.4 (11-
10.3946)= 10.6368
12 9 9.1821 10.6368+0.4 (7-
10.6368)=9.1821
El pronostico mejor es el del año 9 con 11.3244.
4.4 Un centro de procesamiento de cheques usa el suavizamiento exponencial
para pronosticar el número de cheques entrantes por mes. El número de
cheques recibidos en junio fue 40 millones, mientras que el pronóstico era
42 millones. Se empleo un constante de suavizado de .2
R=
a. Cual es el pronóstico para julio?
Nuevo pronostico para julio = 42+.2 (40-42)=41.6 millones de cheques.
b. Si el centro recibió 45 millones enjulio, cual será el pronostico para
agosto?
41.6+.2 (45-41.6)=42.28 millones de cheques.
c. Porque razón podría ser inapropiado este método de pronostico para esta
situación?
R/ Porque podría incrementar el numero de cheques recibidos en el siguiente
mes o mantenerse.
4.5 El hospital Carbondale esta pensando comprar una nueva ambulancia. La
desicion dependerá, en parte del número de millas que habrán de manejar el
próximo año. Las millas recorridas durante los 5 años anteriores son las
siguientes:
Año Millas Promedio
movil 2 años
1 3,000
2 4,000
3 3,400
3,000+4000/2=3500
4 3,800
4,000+3,400/2=3700
5 3,700
3,400+3,800/2=3,600
6
3,800+3,700/2=3,750
a. pronostique el número de millas para el próximo año con un promedio
movil de 2 años.
R= 3,750
b. Encuentre el MAD para su pronostico del inciso a.
Año demanda real pronóstico desviación absoluta
1 3,000
2 4,000
3 3,400 3,500
-100
4 3,800 3,700
-100
5 3,700 3,600
-100
6 3,750
300/3=100
c. Use un promedio movil ponderado de 2 años con ponderaciones de.4 y .6
para pronosticar el numero de millas próximo año (.6 el peso del año mas
reciente) ¿Cuál es el MAD de este pronostico?
Año millas pronostico MAD
1 3,000
2 4000
3 3,400 3,600 200
4000(0.6)+3000(0.4)/1=3,600
4 3,800 3,640 160
3,400(0.6)+4,000(0.4)/1=3,640
5 3,700 3,640 60
3,800(0.6)+3,400(0.4)/1=3640
6 3,740 420/3=140
3700(0.6)+3,800(0.4)/1=3740
d. calcule el pronostico para el año 6 mediante suavizamiento exponencial,
un pronostico inicial para el año 1 de 3,000 millas y a= .5
Año millas pronostico
1 3000 3000
2 4000 3000
3000+.5(3000-3000)=3000
3 3400 3500
3000+.5(4000-3000)=3500
4 3800 3450
3500+.5 (3400-3500)=3450
5 3700 3625 3450+.5 (3800-3450)=3625
6 3662.5 3625+.5 (3700-3625)= 3662.5
4.6 Las ventas en Bettern Inc., fueron como sigue:
Mes Ventas
Enero 20
Febrero 21
Marzo 15
Abril 14
Mayo 13
Junio 16
Julio 17
Agosto 18
Septiembre 20
Octubre 20
Noviembre 21
Diciembre 23
a.
MES VENTAS
PROMEDIO MOVIL 3 MESES
Enero 20
Febrero 21
Marzo 15
Abril 14
20+21+15=56/3=18.66
Mayo 13
21+15+14=50/3=16.66
Junio 16
15+14+13=42/3=14
Julio 17
14+13+16=43/3=14.33
Agosto 17
13+16+17=46/3=15.33
Septiembre 18
16+17+18=51/3=17
Octubre 20
17+18+20=55/3=18.33
Noviembre 21
18+20+20=58/3=19.33
Diciembre 23
20+20+21=61/20.33
Suavizamiento potencial
a=.3 y un pronostico de septiembre de 18
MES VENTAS
PRONÓSTICOS
Enero 20
Febrero 21
Marzo 15
Abril 14
Mayo 13
Junio 16
Julio 17
Agosto 18
Septiembre 20 18
Octubre 20
18.6 18+0.3 (20-18)=18.6
Noviembre 21
19.02 18.6+0.3 (20-18.6)=19.02
Diciembre 23
19.61 19.02+0.3 (21-19.02)=19.61
ENERO 20.63 19.61+0.3 (23-
19.61)=20.63
c. Conlos datos dados ¿Qué método le permitiría elaborar el pronóstico de
ventas para el próximo mes de marzo? El suavizamiento potencial.
4.7 Doug moodie es el presidente de Garden Product limited. Durante los
últimos 5 años ha perdido a sus vicepresidentes de marketing y de
operaciones que le den pronósticos de ventas. Las ventas reales y los
pronósticos se presentan en la tabla. De acuerdo con MAD, ¿Cuál de los dos
vicepresidentes presento un mejor pronóstico?
AÑO VENTAS VP/MARKETING MAD VP/OPERACIONES MAD
1 167.325 170.000 2.675 160.000
7.33
2 175.362 170.000 5.362
165.000 10.3620
3 172.536 180.000 4.464
170.000 2.54
4 156.732 180.000 23.27
175.000 18.27
5 176.325 165.000 11.33
165.000 11.33
47.1010/5=9.4202
49.832/5=9.9664
El vicepresidente que presento mejor pronostico es el de marketing con un
margen de error de 9.4202
4.8 Las temperaturas diarias altas en la ciudad de Houston durante la
ultima semana fueron las siguientes 93, 94, 9396, 98,90(ayer).
a.
DIAS TEMPERATURA PROMEDIO MOVIL DE 3 DIAS
L 93
M 94
M 93
J 96 a:
93+94+93=280/3=93.33
V 88
94+93+96=283/3=94.33
S 90
93+96+88=277/3=92.33
D 96+88+90=274/3=91.33
B/ pronostique la temperatura alta para hoy usando un promedio móvil de 2
días
Días temperatura promedio móvil de 2 días
L 93
M 94
M 93
93+94/2=93.5
J 95
94+93/2=93.5
V 96
93+95/2=94
S 88
95+98/2=96.5
D 90 98+88/2 = 93
88+90= 89 temperatura
c. calcule la desviación absoluta media con base en un promedio móvil de 2
días.
DÍAS TEMPERATURA PROMEDIO MÓVIL DE 2 DÍAS
MAD
L 93
M 94
M 93 93+94/2=93.5
93-93.5= 0.5
J 95 94+93/2=93.5
96-93.5= 2.5
V 96 93+96/2=94.5
88-94.5= 6.5
S 88 96+88/2=92
90-92= 2
D 90 88+90=89
11.5/4=2.875
d. calcule el error cuadrático medio para un promedio móvil de 2 días.
DÍAS TEMPERATURA PROMEDIO MÓVIL DE 2 DÍAS
ERROR CUADRÁTICO
L 93
M 94
M 93 93.5
0.5*0.5=0.25
J 96 93.5
2.5*2.5=6.25
V 88 94.5
6.5*6.5=42.25
S 90 92
2*2=4
D 89
52.75/4=13.1875
e. calcule el error porcentual absoluto medio para el promedio móvil de 2
días.
DÍAS TEMPERATURA PROMEDIO MÓVIL DE 2 DÍAS
ERROR PORCENTUAL
L 93
M 94
M 93 93.5
100(.5/93)=0.5376
J 96 93.5
100(2.5/96)=2.6042
V 88 94.5
100(6.5/88)=7.3864
S 90 92
100(2/90)= 2.22
D 89
12.7482/4= 3.1871
4.9 P.D. usa un chip x63 en algunas de sus computadoras. Los precios del
chip durante los últimos 12 meses.
MES PRECIO POR CHIP PROMEDIO MÓVIL DE 2 MESES
Enero $1.80
Febrero 1.67
Marzo 1.70 1.80+1.67=3.47/2= 1.735
Abril 1.85 1.67+1.70=3.37/2= 1.69
Mayo 1.90 1.70+1.85=3.55/2= 1.78
Junio 1.87 1.85+1.90=3.75/2= 1.875
Julio 1.80 1.90+1.87=3.77/2= 1.89
Agosto 1.83 1.87+1.80=3.67/2= 1.84
Septiembre 1.70 1.80+1.83=3.63/2= 1.82
Octubre 1.65 1.83+1.70=3.53/2= 1.77
Noviembre 1.70 1.70+1.65=3.35/2= 1.68
Diciembre 1.75 1.65+1.70=3.35/2= 1.68
1.70+1.75=3.45/2= 1.73
b.use un promedio móvil de 3 meses y agréguelo en la grafica creada en el
inicio a.
Mes precio por chip promedio móvil de 3 meses
Enero $1.80
Febrero 1.67
Marzo 1.70
Abril 1.85 1.80+1.67+1.70/3=1.7233
Mayo 1.90 1.67+1.70+1.85/3=1.74
Junio 1.87 1.70+1.85+1.90/3=1.8166
Julio 1.80 1.85+1.90+1.87/3=1.8733
Agosto 1.83 1.90+1.87+1.80/3=1.8566
Septiembre 1.70 1.87+1.80+1.83/3=1.8333
Octubre 1.65 1.80+1.83+1.70/3=1.7766
Noviembre 1.70 1.83+1.70+1.65/3=1.7266
Diciembre 1.75 1.70+1.65+1.70/3=1.6833
1.65+1.70+1.75/3=1.7
c. cual es mejor (usando la desviación absoluta media) el promedio de 2
meses o el promedio de 3 meses?
R= usando la desviación absoluta media, pensaría que es la mejor para
sacar pronósticos.
d. calcule el pronostico para cada mes con suavizamiento exponencial y un
pronostico inicial para enero de $1.80 a=.3 y por ultimo a=5 según MAD que
a es mejor?
MES PRECIO PRONOSTICO
SUAVIZAMIENTO EXPONENCIAL.
Enero 1.80 $1.80
Febrero 1.67 1.84
1.80+.3(1.80-1.80)=1.80
Marzo 1.70 1.79
1.80+.3 (1.80-1.67)=1.84
Abril 1.85 1.763
1.84+0.3 (1.67-1.84)=1.79
Mayo 1.90 1.79
1.79+0.3 (1.70-1.79)=1.763
Junio 1.87 1.823
1.763+0.3 (1.85-1.763)=1.79
Julio 1.80 1.8371
1.79+0.3 (1.90-1.79)=1.823
Agosto 1.83 1.8260
1.823+0.3 (1.87-1.823)1.8371
Septiembre 1.70 1.8272
1.8371+0.3 (1.80-1.8371)=1.8260
Octubre 1.65 1.7890
1.8260+0.3 (1.83-1.8260)=1.8272
Noviembre 1.70 1.7473
1.8272+0.3 (1.70-1.8272)=1.7890
Diciembre 1.75 1.7331
1.7890+0.3 (1.65-1.7890)=1.7473
1.7473+0.3 (1.70-1.7473)=1.7331
4.10 los datos recolectados en las inscripciones anuales para un seminario
de Six Sigma en el Quality, Collage, se muestra en la siguiente tabla.
a.
AÑO INSCRIPCIONES (miles) PROMEDIO MOVIL DE 3
MESES
1 4
2 6
3 4
4 5 4+6+4=14/3= 4.66
5 10
6+4+5=15/3=5
6 8
4+5+10=19/3=6.33
7 7
5+10+8=23/3=7.66
8 9
10+8+7=25/3=8.33
9 12
8+7+9=24/3=8
10 14
7+9+12=28/3=9.33
11 15
9+12+14=35/3=11.66
12
12+14+15=41/3=13.66
b.
Estime la demanda de nuevo para los años del 4 al 12 con un promedio móvil
ponderado donde la inscripción del año mas reciente tiene una peso de 2 en
los otros dos años un peso de 1.
AÑO INSCRIPCIONES (miles) PROMEDIO PONDERADO
1 4
2 6
3 4
4 5 4(2)+6(1)+4(1)=18/4=4.5
5 10
5(2)+4(1)+6(1)=20/4=5
6 8
10(2)+5(1)+4(1)=29/4=7.25
7 7
8(2)+10(1)+5(1)=31/4=7.75
8 9
7(2)+8(1)+10(1)=32/4=8
9 12
9(2)+7(1)+8(1)=33/4=8.25
10 14
12(2)+9(1)+7(1)=40/4=10
11 15
14(2)+12(1)+9(1)=49/4=12.25
12
15(2)+14(1)+12(1)=56/4=14
c. Grafique los datos originales y los pronósticos ?Cual de los métodos
parece mejor?
El promedio móvil de 3 meses es el que esta mas cerca de las inscripciones.
4.11 Use suavizamiento exponencial con constantes de suavizado de 0.3 para
pronosticar las inscripciones al seminario del problema 4.10 para comenzar
el procedimiento suponga que el pronostico para el año 1 fue una
inscripción de 5.000 personas
Año inscripciones (miles) pronostico
suavizamiento exponencial
1 4 5.0
2 6 4.7 5.0+0.3 (4- 5.0)=4.7
3 4 5.09 4.7+0.3 (6-4.7)=5.09
4 5 4.763 5.09+0.3
(4-5.09)=4.763
5 10 4.8341
4.763+0.3 (5-4.763)=4.8341
6 8 6.3839
4.8341+0.3 (10-4.8341)=6.3839
7 7 6.8687
6.3839+0.3 (8-6.3839)=6.8687
8 9 6.9091
6.8687+0.3 (7-6.8687)=6.9091
9 12 7.5364
6.9091+0.3 (9-6.9091)=7.5364
10 14 8.8755
7.5364+0.3 (12-7.5364)=8.8755
11 15 10.4129
8.8755+0.3 (14-8.8755)=10.4129
12 11.7890
10.4129+0.3 (15-10.4129)=11.7890
4.12 En los problemas 4.10 y 4.11 se desarrollaron tres pronósticos de las
inscripciones al seminario que son pronóstico móvil de 3 meses, ponderado y
suavizamiento exponencial. Con MAD como criterio ¿Cuál de los tres métodos
de pronósticos es mejor explique su respuesta?
" "
" "
"Pronostic" " " " "
"o " " " " "
"Año "Corazón "Suavizante =.6 "Suavizante " "
" " " "=.9 " "
"1 "45 "41 "41 " "
"2 "50 "43.=41+.6(45-41) "45 " "
"3 "52 "47=43.4+.6(50-43.4)"49 " "
"4 "56 "50=47.36+6(52-47-36"52 " "
" " ") " " "
"5 "58 "54=50.14+.6(56.50.1"56 " "
" " "4) " " "
"6 "? "56=53.65+.6(58-53-6"58 " "
" " "5) " " "
" " " " " "
" " " " " "
" " " " " "
"B. "Trasplante " " " "
" "de " " " "
"Año "Corazón "Promedio Móvil " " "
" " "3años " " "
"1 "45 " " " "
"2 "50 " " " "
"3 "52 " " " "
"4 "56 "(45+50+52)/3=49 " " "
"5 "58 "(50+52+56)/3=53 " " "
"6 "? "(52+56+58)/3=55 " " "
" " " " " "
" " " " " "
" " " " " "
"C. "Trasplante " " "Proyección de "
" "de " " " "
"Año X "Corazón Y "X2 "XY "Tendencia "
"1 "45 "1 "45 "2.8 + 13.23 "
" " " " "(1)=16.04 "
"2 "50 "4 "100 "2.8 + 13.23 "
" " " " "(2)=29.27 "
"3 "52 "9 "156 "2.8 + 13.23 "
" " " " "(3)=36.88 "
"4 "56 "16 "224 "2.8 + 13.23 "
" " " " "(4)=50.39 "
"5 "58 "25 "290 "2.8 + 13.23 "
" " " " "(5)=68.96 "
"6 "55 "36 "330 "2.8 + 13.23 "
" " " " "(6)=82.19 "
"Ex= 21 "Ey=261 "Ex2=91 "Exy=1,145 " "
" " " " " "
" " " " " "
"X= 21/6 = 3.5 " " " "
" " " " " "
"Y= 261/6= 43.5 " " " "
" " " " " "
" " " " " "
"b= 1,145- (6) (3.5) "231.5 "13.23 " "
"(43.5) " " " "
" 91-(6) (3.5"2) "17.5 " " "
" " " " " "
"a= 43.5 - 13.23 (3.5) "2.18 " " "
" " " " " "
"La Ecuación de mínimos cuadrados para la tendencia es Y= " "
"2.81 mas13.23x " "
" " " " " "
" " " " " "
" " " " " "
4.14 regrese al problema4.13 con base en el criterio de MAD.Cual de los 4
pronósticos es mejor?
TRANSPLANTE PROMEDIO SUAVISANTE
SAV.
AÑO . CORAZON MOVIL MAD = 6
MAD = 9 MAD
1 45 41 4.00 41 4.00
2 50 43 7.00 45 5.00
3 52 47 5.00 49 3.00
4 56 49 7.00 50 6.00 52 4.00
5 58 52.67 5.00 54 4.00 56 2.00
MAD 6.2 5.20 3.60
"Proyección de Tendencia 0.64 MAD es la mejor porque es la"
"mas pequeña "
"que los otros " " " "
"pronósticos " " " "
" " " " " "
"Suavizante Exponencial" " " "
".6 MAD= 5.2 " " " "
" " " " " "
"Suavizante Exponencial" " " "
".9 MAD= 3.6 " " " "
" " " " " "
"Promedio Móvil " " " "
"MAD=6.2 " " " "
AÑO TRAN. DE CORAZON X2 XY TENDENCIA
MAD
1 45 1 45 45.8 0.80
2 50 4 100 49 1.00
3 52 9 156 52.2 0.20
4 56 16 224.00 55.4 0.60
5 58 25 290.00 58.6 0.60
/5=0.64
"X= 15/5 " " " " " "
"= 3 " " " " " "
" " " " " " "
"Y= 261/5= 52.2 " " " " "
" " " " " " "
" " " " " " "
"b= 815- (5) (3) (52.2) "32 "3.2 " " "
" 55-(5) (3"2) "10 " " " "
" " " " " " "
"a=52.2 - 3.2 (3) "42.6 " " " "
" " " " " " "
" " " " " " "
"La Ecuación de mínimos cuadrados para la tendencia es Y= 42.6 + "
"3.2x "
" " " " " " "
4.15 Retome el problema resuelto 4.1 déla pagina136. Use un móvil de 3 años
para pronosticar las ventas volkwagen beetle en nevada durante 2003.
Año ventas promedio movil de 3 años
1998 450
1999 495
2000 518
2001 563
450+495+518/3=487.66
2002 584
495+518+563/3=525.33
2003 ¿
518+563+584/3=555 se pronostican las ventas
Las ventas del 2003 según el pronostico del promedio móvil de 3 años
seria 555
4.16 Retome el problema resuelto de 4.1 usando el metodo de proyección de
tendencia pronostique las ventas de volkswgen beetle en nevada 2003.
AÑO X VENTAS
X2 XY
1998 1 450 1 450
1999 2 495 4 990
2000 3 518 9 1554
2001 4 563 16 2252
2002 5 584 25 2920
EX= 15 EY = 2,610 EX2 = 55 EXY= 8,166
X = 15/5 = 3
Y =2,610/5=522
B= 8,166-(5) (3) (522) 336 33.6
55-(5) (3 '' 2) 10
A= 522-33.6 (3) 421.2
La ecuación de mínimos cuadrados para la tendencia es Y = 421.2+33.6
4.17 vuelva al problema resuelto 4.1 con constantes de suavizado de .6 .9 y
pronostique las ventas de volkwagen.Que efecto tiene la constante de
suavizado? use MAD para determinar cual de las tres constantes de suavizado
(3.6.9) da el pronostico mas acertado.
Año unidades pronostico 0.3 MAD 0.6 MAD 0.9
MAD
1998 450 410 40 410
40 410 40
1999 495 422 73 434
61 446 49
2000 518 443.9 74.1 470.6
47.4 490.1 27.9
2001 563 466.13 96.87 499.04
63.96 515.21 47.79
2002 584 495.19 88.80 537.41
46.59 558.22 25.77
2003 521.83
565.36 581.43
372.77/5= 74.55 258.85/5=57.79
190.4690/5= 38.09
4.18 Regrese problema resuelto 4.1 y a los problemas de 4.15 y 4.16 con
MAD como criterio? Utilizaría suavizamiento exponencial con constante de
suavizado de .3 como se muestra en el problema 4.1 un promedio móvil de 3
años o usaría la tendencia para predecir las ventas de volkwagen beetle?
Explique su respuesta?
"Año "Ventas "Suavisante "MAD "Promedio Móvil 3años"MAD "
" " "Exponencial .3 " " " "
"1998 "450 "410 "40 " " "
"1999 "495 "422 "73 " " "
"2000 "518 "444 "74 " " "
"2001 "563 "466 "97 "488 "75 "
"2002 "584 "495 "89 "525 "59 "
"MAD.. "373/5= "74.6 "134/2= "67 "
" " " " " " "
" " " " " " "
" " " " " " "
"MAD Promedio Móvil de 3 años es de 67 " " " "
"MAD Suavizante exponencial con constante de .3 es " " "
"de 74.6 " " "
"MAD Tendencia es de 5.6 por lo tanto es " " " "
"la mejor " " " "
" " " " " " "
" " " " " " "
"Año "X "Ventas "X2 "XY "Tendenci"
" " " " " "a "
"1998 "1 "450 "1 "450 "4.8 "
"1999 "2 "495 "4 "990 "6.6 "
"2000 "3 "518 "9 "1,554 "4 "
"2001 "4 "563 "16 "2,252 "7.4 "
"2002 "5 "584 "25 "2,920 "5.2 "
" "EX= 15 "Ey=2,610 "Ex2=55 "Exy=8,166 "28 "
" " " " " "/ 5 = "
" " " " " "5.6 "
" " " " " " "
" " " " " " "
" " " " " " "
"X= 15/5 " " " " " "
"= 3 " " " " " "
" " " " " " "
"Y=2,610/5= 522 " " " " "
" " " " " " "
" " " " " " "
"b= 8,166- (5) (3) "336 "33.6 " " "
"(522) " " " " "
" 55-(5) (3"2)"10 " " " "
" " " " " " "
"a=522 - 33.6 (3) "421.2 " " " "
" " " " " " "
" " " " " " "
"La Ecuación de mínimos cuadrados para la tendencia es Y= 421.2 + 33.6x " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
4.19 los ingresos en el despachote abogados de wesson para el periodo de
febrero a julio han sido?
" " "F+ "T+ "FIT+ "
"Mes "Ingreso "Pronostico x mes "Tendencia "Pronostico "
" "Anual " "suavizada "incluyendo "
" " " " "Tendencia "
" " " " " "
"Feb "70 "65 "0 "65 "
"Mar "68.5 "70+.1(70-65)=70.5 "1.9 "72.4 "
"Abr "64.8 "70.5+.1(68.5-70.8)=70"1.56 "71.86 "
" " ".3 " " "
"May "71.7 "70.3+.1(64.8-70.3)=69"1.36 "71.11 "
" " ".75 " " "
"Jun "71.3 "69.75+.1(71.7-69.75)="1.13 "71.08 "
" " "69.95 " " "
"Jul "72.8 "69.95+.1(71.3-69.95)="0.93 "71.02 "
" " "70.09 " " "
"Ago " "70.09+.1(72.8-70.09)="0.8 "69.56 "
" " "70.36 " " "
" " " " " "
" " " " " "
" " " " " "
" " " " " "
"Marzo... = .2(70.5-65)+(1-.2)" " " "
"(0) " " " "
" =" " " " "
".2(5.5)+0.8 " " " " "
" =" " " " "
"1.9 " " " " "
" " " " " "
"Abril... = " " " "
".2(70.3-70.5)+(1-.2) (1.9) " " " "
" =" " " " "
".2(0.2)+1.52 " " " " "
" =" " " " "
"1.56 " " " " "
" " " " " "
"Mayo... = " " " "
".2(69.75-70.3)+(1-.2) (1.56) " " " "
" = " " " "
".2(0.55)+1.25 " " " "
" =" " " " "
"1.36 " " " " "
" " " " " "
" " " " " "
"Junio... " " " "
"=.2(69.95-69.75)+(1-.2) " " " "
"(1.36) " " " "
" = " " " "
".2(0.2)+1.088 " " " "
" =" " " " "
"1.13 " " " " "
" " " " " "
"Julio... = " " " "
".2(70.09-69.95)+(1-.2) (1.13)" " " "
" = " " " "
".2(0.14)+0.904 " " " "
" =" " " " "
"0.93 " " " " "
" " " " " "
" " " " " "
"Agosto... = .2(70.36-70.09)+(1-.2) (0.93) " " "
" = " " " "
".2(0.27)+0.744 " " " "
" =" " " " "
"0.80 " " " " "
" " " " " "
4.20 resuelva el problema 4.19 con a=.1 b=.8 empleando MSE cual constante
de suavizado proporciona mejor pronostico?
4.21 véase la ilustración suavisamiento exponencial con ajuste de tendencia
del ejemplo 7 en las Pág. 116 y117 con a=0.2 y B¿=04, pronostican las
ventas de 9 meses y mostramos l detalle en los cálculos para los meses 2 y
3. Con el problema 4.2 continuaremos el proceso para el mes 4.
En este problema muestre sus cálculos par los meses 5 y 6 para Fr Tr y Fiar
CALCULOS PARA LOS MESES 5 Y 6
Paso 1: F4= .2(20)+(1-.2) (15.18+2.10)
= 4 + (.8)(17-28)
= 4 + 13.824= 17.82
Paso 2: T4= .4(17.82-15.18) + (1-.4)(2.10
= .4(2.64)+.6(2-10)
= .1 056+1.26 = 2.32
Paso 3 FIT4= 17.82+2.32 =20.14
"Paso 1: F5= .4 (19.91-17.82) + (1-4) " " " "
"(2-32) " " " "
" 3.8 + (.8) " " " "
"(20.14) = 19.91 " " " "
" " " " "
". .4(2.09) " " " " "
"+ .6 (2-32) " " " " "
" " " " " "
".84+1.39 =2.23 " " " " "
" " " " " " "
" " " " " " "
"=22.14 " " " " " "
" " " " " " "
" "CALCULO PARA LOS MESES 7,8 Y 9 " " "
" " " " " " "
" " " " " " "
"Paso 1: F7= .2(21)+(1-2) "Mes 7 " " "
"(22.51+2.38) " " " "
" 4.2+.8(24.89) " " " "
" " " " " "
"= 24.11 " " " " "
" " " " " " "
"Paso 2: T7= .4(24.11-22.51) + (1-4) " " " "
"(2.38) " " " "
" " " " "
".4(1.6+)+.6(2.38) " " " "
" 0.64+1.43=2.07 " " " "
" " " " " " "
"Paso 3: FIT7= 24.11+2.07 " " " "
" " " " " "
"=26.18 " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Paso 1: F8= .2(31)+(1-.2) "Mes 8 " " "
"(24.11+2.07) " " " "
" 6.2+.8(26.18 " " " "
" " " " " "
"=27.14 " " " " "
" " " " " " "
"Paso 2: F8= " " " "
".4(27.14-24.11)+(1-.4)(2.07) " " " "
" .4 (3.03) + .6 " " " "
"(2.07) " " " "
" 1.21+1.24= 2.45 " " " "
" " " " " " "
"Paso 3:FIT= 27.14+2.45 " " " "
" " " " " "
"= 29.59 " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " "Mes 9 " " "
"Paso 1: F9= .2(28) + (1-.2) " " " "
"(27.14+2.45) " " " "
" 5.6+0.8(29.59) " " " "
" " " " " "
"=29.28 " " " " "
" " " " " " "
"Paso 2: T9= " " " "
".4(29.28-27.14)+0.6(2.45) " " " "
" .4 " " " "
"(2.14)+0.6(2.45) " " " "
" 0.85 + 1.47= 2.32" " " "
" " " " " " "
"Paso 3: FIT9= 29.28+2.32 " " " "
" " " " " "
"=31.60 " " " " "
4.23 la tabla muestra las ventas de deshidratadores de vegetales durante el
año pasado en la tienda departamental de descuentos Bud Banis de St. Louis.
Los administradores prepararon un pronóstico empleando una combinación de
suavizamiento exponencial y su juicio colectivo para los siguientes 4 meses
(marzo, abril, mayo y junio del 2002)
"Mes "Ventas "Pronóstic"MAD. "MAPE " "
" "Unitarias "os " " " "
" "2001-2002 "Admón. " " " "
"Julio "100 " " " " "
"Agosto "93 " " " " "
"Septiembre"96 " " " " "
"Octubre "110 " " " " "
"Noviembre "124 " " " " "
"Diciembre "119 " " " " "
"Enero "92 " " " " "
"Febrero "83 " " " " "
"Marzo "101 "120 "19 "100(19/101)=18.8" "
" " " " "1 % " "
"Abril "96 "114 "18 "100(18/96)=18.75" "
" " " " "% " "
"Mayo "89 "110 "21 "100(21/89)=23.60" "
" " " " "% " "
"Junio "108 "108 "0 "100(0/108)=0% " "
" " " "58 "61.16% " "
" " " " " " "
" " " " " " "
" " " " " " "
"MAD "58/ 4 = " " " " "
" "14.5 " " " " "
" " " " " " "
" " " " " " "
"MAPE "61.16% = " " " " "
" "15.29% " " " " "
" "4 " " " " "
" " " " " " "
" " " " " " "
"Los resultados de la Administración superaron, tienen MAD de 14.5 "
"menor que MAPE "
"con 15.29 % " " " " "
" " " " " " "
"Recomendaría MAD por que sus error es mas pequeño 14.5 " "
" " " " " " "
" " " " " " "
" " " " " " "
4.24
"(Y) "(x) " " " " "
"Demanda de bombas "Aparicion"X2 "Xy " " "
" "es en TV." " " " "
" "de Green " " " " "
" "Shades " " " " "
" " " " " " "
"3 "3 "9 "9 " " "
"6 "4 "16 "24 " " "
"7 "7 "49 "49 " " "
"5 "6 "36 "30 " " "
"10 "8 "64 "80 " " "
"8 "5 "25 "40 " " "
" EX=39 "EY=33 "EX2=1"EXY=2" " "
" " "99 "32 " " "
" " " " " " "
" " " " " " "
"Regresión de Mínimos cuadrados para obtener una ecuacion de " "
"pronostico " "
" " " " " " "
" " " " " " "
"X= 33/6 = 5.5 " " " " " "
" " " " " " "
"Y= 39/6= 6.5 " " " " " "
" " " " " " "
" " " " " " "
"b= 232- (6) (5.5) (6.5) " "17.5 "1 " " "
" 199-(6) (5.5"2) " "17.5 " " " "
" " " " " " "
"a= 6.5 - 1 (5.5) " "1 " " " "
" " " " " " "
" " " " " " "
"La Ecuación de mínimos cuadrados para la tendencia es Y= 1 + 1 " " "
"x " " "
" " " " " " "
" " " " " " "
"La estimación de las ventas de bombos si Green Shades hubiese " "
"aparecido " "
"9 veces en TV el mes anterior " " " " "
" " " " " " "
"Y= 1 + 1 (9) " " " " " "
" " " " " " "
"Y= 1 + 9 " " " " " "
" " " " " " "
"Y= 10 " " " " " "
" " " " " " "
"Seria de 10 Bombos " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"4.25 " " " " " "
" " " " " " "
"Mes " " " " " "
"Mes " " " " " "
"(X) " " " " " "
"X2 " " " " " "
"XY " " " " " "
" " " " " " "
" " " " " " "
"Y " " " " " "
" " " " " " "
"No. De Accidentes " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"1 " " " " " "
"Enero " " " " " "
"30 " " " " " "
"900 " " " " " "
"30 " " " " " "
" " " " " " "
" " " " " " "
"2 " " " " " "
"Febrero " " " " " "
"40 " " " " " "
"1600 " " " " " "
"40 " " " " " "
" " " " " " "
" " " " " " "
"3 " " " " " "
"Marzo " " " " " "
"60 " " " " " "
"3600 " " " " " "
"60 " " " " " "
" " " " " " "
" " " " " " "
"4 " " " " " "
"Abril " " " " " "
"90 " " " " " "
"8100 " " " " " "
"90 " " " " " "
" " " " " " "
" " " " " " "
"EY=10 " " " " " "
" " " " " " "
"EX=220 " " " " " "
"EX2=14,200 " " " " " "
"EXY=220 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Regresion de Minimos cuadrados para " " " " " "
"obtener una ecuacion de pronostico " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"X= 220/4 = 55 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Y= 10/4= 2.5 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"b= 220- (4) (55) (2.5) " " " " " "
"330 " " " " " "
"0.16 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"14,200-(4) (55"2) " " " " " "
"2100 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"a= 2.5 - 0.16 (55) " " " " " "
"6.3 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"La Ecuacion de minimos cuadrados para la " " " " " "
"tendencia es Y= 2.5 + 6.3 x " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"4.26 " " " " " "
" " " " " " "
" " " " " " "
"Años " " " " " "
"Demanda Promedio " " " " " "
"Demanda Promedio " " " " " "
"Índice Estacional " " " " " "
" " " " " " "
"Estaciones " " " " " "
"1 " " " " " "
"2 " " " " " "
"2 Años " " " " " "
"Mensual " " " " " "
" " " " " " "
" " " " " " "
"Otoño " " " " " "
"20 " " " " " "
"250 " " " " " "
"225 " " " " " "
"83 " " " " " "
"2.71 " " " " " "
" " " " " " "
"Verano " " " " " "
"300 " " " " " "
"285 " " " " " "
"292.5 " " " " " "
"83 " " " " " "
"3.524 " " " " " "
" " " " " " "
"Invierno " " " " " "
"350 " " " " " "
"300 " " " " " "
"325 " " " " " "
"83 " " " " " "
"3.916 " " " " " "
" " " " " " "
"Primavera " " " " " "
"150 " " " " " "
"165 " " " " " "
"157.5 " " " " " "
"83 " " " " " "
"1.898 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Promedio total de Demanda Anual " " " " " "
"1,000.00 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Demanda Promedio Mensual " " " " " "
" " " " " " "
"1,000.00/12 meses " " " " " "
"83 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Demanda 1200 Próximo Año " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Estaciones " " " " " "
"Demanda " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Otoño " " " " " "
"1200/12x2.710= 271 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Verano " " " " " "
"1200/12x3.524= 352 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Invierno " " " " " "
"1200/12x3.916= 392 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Primavera " " " " " "
"1200/12x1.898= 190 " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
4.27
"Días/Semana "Demanda Promedio "Demanda Promedio "Índice "
" " "Diaria "estacional "
"Lunes "85 "94 "0.904 "
"Martes "74 "94 "0.787 "
"Miércoles "87 "94 "0.926 "
"Jueves "97 "94 "1.032 "
"Viernes "134 "94 "1.423 "
"Sábado "139 "94 "1.479 "
"Domingo "42 "94 "-447 "
" " " " "
" " " " "
" " " " "
" " " " "
"Promedio Total de Demanda " " "
"Semanal…………………………………………..658 " "
" " " " "
" " " " "
"Demanda "658/7 días de la "94 Diarios " "
"Promedio "semana " " "
"Diario " " " "
" " " " "
4.28
"Estaciones"Trimestres "Demanda "Demanda Promedio "Indice "
" " "Promedio "Trimestral " "
" "2001 "2002 "2003 " " " "
"Invierno "73 "65 "89 "76 "110 "0.691 "
"Primavera "104 "82 "146 "111 "110 "1.009 "
"Verano "168 "124 "205 "166 "110 "1.509 "
"Otoño "74 "52 "222 "87 "110 "0.791 "
" " " " " " " "
" " " " " " " "
" " " " " " " "
"Promedio total " " " " " "
"de " " " " " "
"Demanda "440/4 = " " " "
"Trimestral "110 " " " "
4.29
"RECTA DE DEMANDA D= 77 + 0.43Q " " " " " " "
" " " " " " " " "
" " " " " " " " "
"Trimestre "Factor Índice " " " " " " "
"Invierno "0.8 " " " " " " "
"Primavera "1.1 " " " " " " "
"Verano "1.4 " " " " " " "
"Otoño "0.7 " " " " " " "
" " " " " " " " "
" " " " " " " " "
" " " " " " " " "
"I TRIMESTRE " "II TRIMESTRE "III TRIMESTRE "IV TRIMESTRE "
"D= 77 + 0.43 (1) " "D= 77 + 0.43 "D= 77 + 0.43 "D= 77 + 0.43 "
" " "(2) "(3) "(4) "
" " " " " " " " "
"D= 77 + 0.43 " "D= 77 + 0.86 "D= 77 + 1.29 "D= 77 + 1.72 "
" " " " " " " " "
"D= 77.43 " "D= " "D= " "D= " "
" " "77.86 " "78.29 " "78.72 " "
" " " " " " " " "
" " " " " " " " "
"Pronostico para los cuatro Trimestres" " " " " " "
"del 2005 " " " " " " "
" " " " " " " " "
"Trimestre "Demanda Kwts. " " " " " "
"I TRIMESTRE "77 + 0.43(29)= 89.47 " " " " " "
"II TRIMESTRE "77 + 0.43(30)= 89.9 " " " " " "
"III TRIMESTRE "77 + 0.43(31)= 90.33 " " " " " "
"IV TRIMESTRE "77 + 0.43(32)= 90.76 " " " " " "
" " " " " " " " "
" " " " " " " " "
" " " " " " " " "
"Pronostico Ajustado " " " " " " " "
"Trimestre "Demanda Kwts. " " " " " "
"I TRIMESTRE "(.8) (89.47) = 71.58 " " " " " "
"II TRIMESTRE "(.8) (89.9) = 98.89 " " " " " "
"III TRIMESTRE "(.8) (90.33) = 126.46 " " " " " "
"IV TRIMESTRE "(.8) (90.76) = 63.53 " " " " " "
" " " " " " " " "
" " " " " " " " "
" " " " " " " " "
" " " " " " " " "
4.30
" " " " " " " " "
" " " " " " " " "
" " " " " " " "
" " " " " " " "
" " " " " " " " "
" " " " " " " "
" " " " " " " " "
" " " " " " "
"Año "Estaciones "Ventas "X2 "Xy " "
"1 "Primavera/Vera"26,825 "1 "26,825 " "
" "no " " " " "
"2 "Otoño/Invierno"5,722 "4 "11,444 " "
"3 "Primavera/Vera"28,630 "9 "85,890 " "
" "no " " " " "
"4 "Otoño/Invierno"7,633 "16 "30,532 " "
"5 "Primavera/Vera"30,255 "25 "151,272 " "
" "no " " " " "
"6 "Otoño/Invierno"8,745 "36 "52,470 " "
" EX=21 " "EY=107,810 "EX2=91 "EXY=358,433" "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Regresión de Mínimos cuadrados para obtener una ecuación de " "
"pronostico " "
" " " " " " "
" " " " " " "
"X= 21/6 = 3.5 " " " " " "
" " " " " " "
"Y= 107,810/6= 17,968.33 " " " " "
" " " " " " "
" " " " " " "
"b=358,433- (6) (3.5) (17,968.33)" "18,901.9"10,801.11 " "
" " "3 " " "
" 91-(6) " " "17.5 " " "
"(3.5"2) " " " " " "
" " " " " " "
"a= 17,968.33 - 10801.11 (3.5) " "19,835.5" " "
" " "5 " " "
" " " " " " "
" " " " " " "
"La Ecuación de mínimos cuadrados para la tendencia es Y= 10,801.11 + "
"19,835.55 x "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Error " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Sy,x = " " "
"2,622,101,548-19,835.55(107,810)-10,801.11(358,433) " " "
"6 menos 2 " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Sy,x = 2,622,101,548- " " " "
"2,138,470,646-3,871,474.26 " " " "
"4 " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Sy,x = 479,759,427.7 " " " " "
"4 " " " " " "
" " " " " " "
"Sy,x = 119,939,856.9 " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"Sy,x = $ 10,951.71 Error Estándar de la Estimación de las " "
"Ventas " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
" " " " " " "
"4.32 " " " " " " "
" " " " " " " "
" "(X) "(Y) " " " " "
"Semanas "Huespedes "Ventas del "X2 "Xy " " "
" " "Bar " " " " "
"1 "16 "$ 330 "256 "5,280 " " "
"2 "12 "$270 "144 "3,240 " " "
"3 "18 "$380 "324 "6,840 " " "
"4 "14 "$300 "196 "4,200 " " "
"n=4 " EX=60 "EY=1,280 "EX2=920 "EXY=19,560" " "
" " " " " " " "
" " " " " " " "
" " " " " " " "
" " " " " " " "
"Regresión de Mínimos cuadrados para obtener una " " "
"ecuación de pronostico " " "
" " " " " " " "
" " " " " " " "
"X= 60/4 = 15 " " " " " "
" " " " " " " "
"Y= 1,280/4= 320 " " " " " "
" " " " " " " "
" " " " " " " "
"b=19,560- (4) (15) " "360.00 "18 " " "
"(320) " " " " " "
" 920-(4) " "20 " " " "
"(15"2) " " " " " "
" " " " " " " "
"a= 320 - 18 (15) " "50.00 " " " "
" " " " " " " "
" " " " " " " "
"La Ecuación de mínimos cuadrados para la tendencia es " " "
"Y= 50 + 18 x " " "
" " " " " " " "
" " " " " " " "
"B " " " " " " "
" " " " " " " "
"Pronostico es 20 " " " " " "
" " " " " " " "
"Y= 20 + 18 (20) "Y= 410 " " " " "
" " " " " " " "
" " "Se espera "$410 " " " "
" " "que las " " " " "
" " "ventas sean " " " " "
4.33
(x) (y)
errores
Año Transistores X2 XY Pronóstico
cuadráticos
1 140 1 1 80.36 59.64=3556.93
2 160 4 8 130.18 29.82=889.23
3 190 9 27 180 10=100
4 200 16 64 229.82 (29.82)=889.23
5 210 25 125 279.64 (-69.64)=4879.73
EX= 15 EY =900 EX2=55 EXY= 225 10,285.12
MAPE
100(59.64/140)=42.6
100(29.82/160)=18.64
100(10/190)=5.26
100(29.82/200)=14.91
100(69.64/210)=33.16
114.57%
"X= 15/5 = 3 " " " " " " " "
" " " " " " " " "
" " " " "
"Transitares para el Próximo " " " " " " "
"año " " " " " " "
" " " " " " "
"Y= 80.36 " "Y= 130.18" "Y=180 " "Y= 229.82 " "
"Error Medio Cuadrático MSE usando la Regresión " " " " "
"Lineal " " " " "
" " " " " " " "
" " " " " " "
" " " " " "
" " " " " "
" "No, de "Ventas de "Cantidad de " "
" "Automóviles "Bebidas "lluvia " "
" "Registrados en"Alcohólicas "pulgadas " "
" "Miles "unidades " " "
" "X1 "X2 "X3 " "
"(a) "2 "3 "0 " "
"(b) "3 "5 "1 " "
"( c ) "4 "7 "2 " "
" " " " " "
" " " " " "
"FORMULA " " " " "
" " " " " "
"Y = a + b1 x1 + b2 x2 +" " " "
"b3 x3 " " " "
" " " " " "
"Donde Y= Numero de Accidentes " " "
"Automovilísticos " " "
" " " " " "
"a= " " " " "
"7.5 " " " " "
" " " " " "
"b1= " " " " "
"3.5 " " " " "
" " " " " "
"b2= " " " " "
"4.5 " " " " "
" " " " " "
"b3= " " " " "
"2.5 " " " " "
" " " " " "
" " " " " "
"CALCULAR NO. DE ACCIDENTES ESPERADOS EN LAS " "
"CONDICIONES A, B, C " "
" " " " " "
"A) " " " " "
"Y= 7.5 + 3.5 (2) + 4.5" " " "
"(3) + 2.5 (0) " " " "
" " " " " "
"Y= 7.5 + 7 + 13.5 + 0" " " "
" " " " " "
"Y= 28 " " " " "
" " " " " "
" " " " " "
"B) " " " " "
"Y= 7.5 + 3.5 (3) + 4.5" " " "
"(5) + 2.5 (1) " " " "
" " " " " "
"Y= 7.5 + 10.5 + 22.5 " " " "
"+ 2.5 " " " "
" " " " " "
"Y= 43 " " " " "
" " " " " "
" " " " " "
" " " " " "
"C) " " " " "
"Y= 7.5 + 3.5 (4) + 4.5" " " "
"(7) + 2.5 (2) " " " "
" " " " " "
"Y= 7.5 + 14 + 31.5 + " " " "
"5 " " " "
" " " " " "
"Y= 58 " " " " "
" " " " " "
" " " " " "
