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Samenvatting Opgeloste Examenvragen: Operationeel Management (KUL antwerpen)
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- Institution
Opgelost examen vragen van voorbije jaren incl voorbeeld exam van dit jaar, 2021. Prof Ocampo y Vilas Carlos
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By: maud18072000 • 2 year ago
By: student12321 • 2 year ago
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KU Leuven, campus Antwerpen,Handelswetenschappen
Operationeel Management 2021
Prof. Ocampo y Vilas Carlos
, Examenvragen
D= 40 stuks 52=2080
gegeven
:
per week ✗ per jaar
( = 20€
☐
(
n
=
0,25 per C-
aankoop kost
EOQ
=d×☐ →(
.
Q
=
=
ets d-
Bestel hvlheid < 300 2×2080×20
voor = =
182,43
0,25×10
voor 300 f Bestandheid = 2×2080×20 = 192,30 → 3005 192,30 ✗
0125×9 Dus EOQ aanpassen
naar 300 .
'
Aantallen CP EOQ EOQ TC
1) 0 -
299 10 182,43 182143 21266,07
9
2) vanaf 300 192,30 300 19 196,17 → optimaal
TCIQ/ D. (p D- .co
f. in
= + +
Q
1) Telt ) = 2080×10 + -2€ ✗ 20 + 182-141 ✗ 2,5 = 21
266,07
2
182,43
2) TCCZ ) = 2080 ✗ 9 +
20-80 ✗ 20 + 30J ✗ 2,5 = 19 196,17
2
300
conclusie :
vanaf 300 is de optimale bestandheid ,
daar z n de TC de
laagste .
ij ij
, TK =
aankoop kost + bestel kost + voorraad kosten
+ te
µ
Cp D- (n
✗ D + + ✗
§
↳ ✗ Q
EOQ
d×D →(
'
=
Q =
D= 1000 bekers CSL = 88.1
gegeven : .
1000 normaal verdeeld
µ
:
0 = 25
L 4
=
dagen
L 2- Waarde : P ( DDLT ( op ) = 88%
µ DDLT
= ×
µ
.
D= 4 ✗ 1000=4000
= 0
,
88
0
DDLT
= [ ✗
% =
V4 ✗ 25 = 50
↳ in tabel
gegeven ,
zoeken
omgekeerd
OP = DDLT + VV 4000 + 11,171×100=4058,75 ligt tussen :
1,17 en 1,18
↳ 1,17
2- waarde ✗ 100
↳
Dichterb dus z =
-
komt in op exam want je moet
veel combinaties maken
ij ij
, gegeven
ta = 24 min → 6=24
to = 16 min → 6=16
Ns = 10 eenh .
( om de 10min
defect)
ts = 12 min → 0=12
1000 ( machine stilstand)
mf
=
mr
=
200 ( machine repair)
a)
/ herstellen 2de
1st in preemptief om stellen
preemptief machine stilstonden
-
= =
te to 16 17,2 ) A te 83%
MI
+ E 4
1) ±
= =
= = =
+ =
'^ '
'°
Ns 1000+200
mf + mr
of -16s
' '
( Ns -11 ts
+
283,36
[ µ,
✗
2) = -
=
'
Ns Ns 5) te = te µ, =
17-12 = 20,64 min
16147 A 0183
3) (
en )
=
¥ = 28-3,36 =
te
'
17,2 6) { '
Is te
taal
= + 2 × ✗ mr ✗
,
'
(A)
Daarna delen F. (W ) 2831L
H gen
voor +2 ✗
17,2 ✗ 200 ✗
'
( 0183 / in
7) vergeten
Cel
→
= =
1820,48 = 4,27
j
'
te ! =
1820,48min
0186
20216µL
8) te
Te = = =
ta
a) Flw ) ( cez )
' '
: te
È
en + ✗ ✗ te +
,
/
2
,
P
4*-4127 20164 354,73min
91¥.gg
= ✗ ✗ 20,64 + =
2
b)
geg :
effectieve gem bewerkingst d =
te = 8min
(
[ = 3
↳ ie
Ë 3×82=192
'
=
c- ) de =
de ✗ te =
1) ta
td of
te § 0,333
= = = =
Ii =p ? e! ) / @ werken )
' '
( Ii -111 11 '
ca ce ca
2) met getallen
✗
× +
van
= -
; :
= c
} (z)
=
0,8602 × 4,27 + ( 1 -
0,8602 ) ✗ 1
= 3 , 42
ENK 13,42 3) ÏO , 82
12
3) = ✗ + ✗ 0,333-+8+8 = min
1- 0,333
Totaal F. (W ) = 354,73 + 20,82 min = 375,55 min
ij ijij
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