Toi uu hoa NMLD
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Transcript of Toi uu hoa NMLD
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TI U HA NH MY LC DU(Dnh cho sinh vin ngnh Cng ngh ha hc Du v kh)
NGUYNNH LM
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Chng I: Cu trc hotng ca qu trnh lc du
Thit b v cc qu trnh
-Nghin cu: ch to v thay th.
-Cng trnh xy dng.
-Khai thc vn hnh thit b, bo dng thit b.
-Theo di hiu qu ca phn xng: Cht lng sn phm, tiu hao nguynliu, nng lng
-M hnh ho.
Nhn s
-c qun l bng h thng hnh chnh.
-Thit lp mi quan h tt trong cng vic.
-o to v thng tin
-Quan l don
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Chng I: Cu trc hotng ca qu trnh lc du
Nguyn liu v sn phm
Don thtrng K hoch cung cp
Xy dng chng trnh lc du
Mua sm: du th, bn sn phm, sn phm
Lu tr
Lc du
Vn chuyn
Xy dng ho n
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Chng I: Cu trc hotng ca qu trnh lc du
Hiu qu kinh t ca mt nh my lc du
u t
Chi ph du th, nguyn liu
Chi ph vn chuyn
Chi ph sn xut
Chi ph duy tu v sa cha thi t b
Chi ph chung
Chi ph ti chnh
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Chng I: Cu trc hotng ca qu trnh lc du
Cu trc hnh chnh ca mt cng ty lc du v phn phi sn phm
TNG GIMC
Chin lc v pht tri n
Nhn s, thng tin-Cht lng
Qun l v h thng tin hc
Hnh chnh chung v h thng ti chnh
Lc du
Phn phi sn phm vi s lng ln
Cc sn phm v nhu cu c bit
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Chng I: Cu trc hotng ca qu trnh lc du
Chin lc v pht tri n
-Lc du: Cc lnh vc cn pht trin hoc loi b
-Phn phi:
Mng li cho
Cc sn phm u tin
Cc phng ti n phc v: ngng, kho bi, phng ti n vn chuyn
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Chng I: Cu trc hotng ca qu trnh lc du
B phn nhn sv
thng tin
-Nhn s:
o to v pht trin nhn l c
Qun l don
Quyn li v quan h cng vic
Hu tr
Qun l hnh chnh v cc cng tc chung
-Thng tin:
Thit lp c mi quan h tt vi bn ngoi: bo ch, qung co
Thng tin ni b, ti liu
Thiua, th thao, gii tr
-Bo him cht lng:
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Chng I: Cu trc hotng ca qu trnh lc du
Qun l v h thng
tin hc
-Ngn sch
-Bo co
-n nh
-Tin hcng dng:
Qun l hnh chnh
Pht trin
Mng v h thng thng tin txa
Vn hnh
Tin hc cng nghip
C C
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Chng I: Cu trc hotng ca qu trnh lc du
Hnh chnh chung v b
phn ti chnh
-Ti chnh, thu quan
-K ton
Ngn sch
K ton chung
Kt qu ca cc chi nhnh
-Vn php l
Bo him
Tranh chp
Hp ng
Quyn li ca cng ty
-Hi quan
Ch I C t h t t h l d
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Chng I: Cu trc hotng ca qu trnh lc du
B phn lc du
-Cc nh my lc du
B phn khai thc: Ln chng trnh, Qun l cc phn xng, Nng
lngv h thng ph tr(hi, in, mi trng, n mn), Lu tr,
Trao i, Phi trn, Vn chuyn sn phm.
Cc phng ban chc nng: Duy tu, bo tr, cng trnh mi, Kthut (quy
trnh cng ngh, utilits, mi trng, PTN, Tin hc cng nghip, tin hc
qun l), An ton, Theo di vt liu, K ton-Qun l, Trao i thng tin.
-Raffinage opration - Pilotage
Tiu ho kt qu vn hnh: Cc nhu cu phn phi (s lng, cht
lng, khong thi gian cho php), Cng c lc du, Nhu cu v kho bi,
lu tr, Thtrng th gii (Chi mua bn /v nguyn liu, bases, snphm), Tnh li nhun trung v di hn.
Ch ng I C trc hot ng ca q trnh lc d
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Chng I: Cu trc hotng ca qu trnh lc du
B phn lc du
-Raffinage Exploitation
Tiu ho cc cng c, thit bca nh my.
Gim st cc hotng ca nh my (Kthut, cng ngh, An ton,Mi trng).
H trcho cc nh my m cng ty c hp tc lm vic.
Tham gia vo vic thi t lp ngn sch u t.-Kthut
Qun l cc dn, cng trnh (PXSX, offsites, nng lng, utilits,
cng trng...)Nhp, nh gi cht lng du th
Tng ho (iu khin, vn hnh, H thng gim st CL, AT, MT)
Cng ngh (ng c, luyn kim, vt liu, xy dng)
Quy trnh cng ngh (Rafinage, conversion)
Chng I: Cu trc hot ng ca qu trnh lc du
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Chng I: Cu trc hotng ca qu trnh lc du
B phn lc du
-Nghin cu pht trin, Qun l mi trng, An ton
Adaptation thng xuyn cc cng c sn c ca nh my vo spht
trin ca thtrng: Nghin cu trung v di hn.
Mi trng v cc nguy ccng ngh ch yu.
H trkthut xy dngv v n hnh cc cng vic ca cng ty c
lin quan n lc du.
-Kinh t, Qun l
Qun l kinh t: Kt qu SX v phn phi, Sn phm trn mng li
phn phi, Lp ho n.
Phn tch kinh t, Nghin cu thtrng.
Qun l ni b.
Chng I: Cu trc hot ng ca qu trnh lc du
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Chng I: Cu trc hotng ca qu trnh lc du
B phn lc du
-Cc trung tm nghin cu
Cc qu trnh lc du: xc tc, cng ngh, thit b.
Cc loi nhin liu
Cc nghin cu c bit, toxixologie.
In n, ph bin ti liu.
Bng pht minh, Hp ng nghin cu.
Chng I: Cu trc hot ng ca qu trnh lc du
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Chng I: Cu trc hotng ca qu trnh lc du
B phn phn phi sn
phm s lng ln
-Maketing.
-B phn qun l khu vc.
-B phn qun l mng li phn phi.
-Nhin liu v chtt.
-Sn phm en.-Logistique.
Chng I: Cu trc hot ng ca qu trnh lc du
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Chng I: Cu trc hotng ca qu trnh lc du
B phn phn phi sn
phm c bit
-Gaz.
-Dung mi.
-Ho du.
-Hng khng.
-Du nhn.-Paraffine.
-Soude
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
Gii thiu
QHTT Nghin c u vn hnh Gi i php tiu
Tin hnh thng xuyn trong cng nghip: Hp kim, phi liu CN thc
phm, SX -t, Tiu ho qu trnh nhp liu, sx v phn phi trong LD
S pht trin ca vic p dng QHTT
QHTT: thc hin nhiu tinh ton tr gip ca my tnh: T/g hp l
Lc du: M hnh cbn (Modle de base) i din cho 1 nh my
(khong 10 rng buc).
M hnh a nh my (Multiraffineries).
M hnh a nh my c tnh n sthayi cc H cung cp.theo thi gian (Multiraffineries-Multipriodes): hng ngn rng
buc.
Thut ton n hnh (Simplexe) Bell LaboratoriesG.B. Dantzig et Von Neumann 1947 M. Karmarkar, 1980
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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g Q y y p g q
S pht trin ca vic p dng QHTT
Knh thc bi ton thng gp hin nay: vi ngn rng buc v n.Nabisco (M): 30.000 rng buc, 300.000n.
Cng ty hng khng: 850 rng buc, 5.500.000n.
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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g y y p g q
Bi ton u tin ca qu trnh lc du, lp phng trnh v gii
-Bi ton: (n gin: 3 sp, khng hao ht, khng tiu th ni nh my)Xl 2 loi du th A v B sx xng, GO v FO vi hiu sut:
Du th A B
Xng 0,2 0,4
GO 0,4 0,2
FO 0,4 0,4
Rng buc lu tr:
Xng: 1.200 t nGO: 1.200 t n
FO: 1.400 t n
Hiu qu kinh t (li nhun):
140 USD/1 tn du th A150 USD/1 tn du th B
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Bi ton u tin ca qu trnh lc du, lp phng trnh v gii
-Lp phng trnh:PA xl ring l tng loi du th:
Du th A: xl tia 3000 tn (Rng buc lu trGO): 420.000USD
Du th B: xl tia 3000 tn (Rng buc lu trXng): 450.000USD
Xl Kt hp hai loi du th: Hiu qu?
X1 lng du th A xl, X2: lng du th B cn xl, Mcch: Tia li nhun.
Phng trnh: Max(Z) 140X1 + 150X2
0,2X1 + 0,4X2 1200
0,4X1 + 0,2X2 1200
0,4X1 + 0,4X2 1400
X1 0, X2 0, X1, X2: Bin cu trc (bin chnh)
B sung cc bin sai khc (variables dcart) X1, X2 v X3 h trn trthnh:
Max(Z) 140X1 + 150X20,2X1 + 0,4X2 + X1 = 12000,4X1 + 0,2X2 + X2 = 12000,4X1 + 0,4X2 + X3 = 1400
X1 0, X2 0, X1 0, X2 0, X3 0
X1, X2, X3: Chnh lch gia s x ti a Xng, GO v FO.
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Bi ton u tin ca qu trnh lc du, lp phng trnh v gii
i vi bi ton c m rng buc v n n ta c dng:
Max(Z)
miBXA
XC
ij
n
jij
j
n
j
j
=
=
=
1
1
1
Hoc: Min(Z)
miBXA
XC
ij
n
j
ij
j
n
j
j
=
=
=
1
1
1
Bi ton vn cn c th chac c dng rng buc tuyn tnh khc: , , =
Tt c cc rng buc bt phng trnh u c th chuyn sang dng PT bng cch thm cc
bin ph:Max(Z)
miBXXA
XC
iij
n
j
ij
j
n
j
j
==+
=
=
1'
1
1
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phng php Simplexe: phng n xut pht, Bin c s
Bi ton xem xt c 3 PT v 5 n (X1, X2, X1, X2, X3)
H ny s giai c khi cnh 2 n, h PT c th vit lai:
X1=1200-0,2X1-0,4X2 (a)
X2=1200-0,4X1-0,2X2 (b)
X3= 1400-0,4X1-0,4X2 (c)
v d: X1= 1000, X2= 1000 ta c X1=600, X2=600, X3=600
y l 1 phng n (PA) v tho mn tt c cc rng buc vi gi tr cahmmc tiuZ=290.000USD Khng phi PA tiu!
Bt u bng PA xut pht sau ci thin dn kt qu ca phng n chn t n PA ti u
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phng php Simplexe: phng n xut pht, Bin c s
PA cc bin xut pht [PA(0)]:
X1=0, X2=0, X1=1200, X2=1200, X3=1400, Khng lm g c v Z=0
Trong PA ny X1, X2 v X3 0 gi l b in c s
X1=X2=0 l cc bin ngoi c s (hors base)
Ci thin hm mc tiu ca PA cc bin xut pht [PA(1)]:Tng gi trca X2c h s cao nht trong hm mc tiu (150 so vi 140)
Gi tr ti a ca X2 theo phng n ny phi tho mn:
(a): X10: 0,4X21200: X2 3000
(b): X20: 0,2X21200: X2 6000(c): X30: 0,4X21400: X2 3500
X2=3000, X1=0, X1=0, X2=600 v X3=200
PA(0) PA(1)
X1=0 X1=0 (Bin ngoi cs)
X2=0 X2=3000
X1=1200 X1=0 (Bin ngoi cs)
X2=1200 X2=600
X3=1400 X3=200
Z=0 Z=3000*150=450.000
PA(1) tt hn PA(0)
PA(1) tiu?
M t hm mc tiu theo hai bin
mi X1 v X1 (bin ngoi csca
PA(1)) v xt du ca cc h s ca
chng
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phng php Simplexe: Qu trnh lp (Itration), Phng nti u
Z: 140X1 + 150X2X1=1200-0,2X1-0,4X2 (a)
X2=1200-0,4X1-0,2X2 (b)
X3= 1400-0,4X1-0,4X2 (c)
M t hm mc tiu theo hai bin mi X1 v X1: s dng Phng trnh (a):
X2=3000-0,5X1-2,5X1, thay X2 trong hm mc tiu Z v PT (b) v (c):
Z: 450000 + 65X1 -375X1
X2=3000-0,5X1-2,5X1 (a)
X2=600-0,3X1+0,5X1 (b)X3=200-0,2X1+X1 (c)
Tng X1 cho php ci thin c
hm mc tiu Zv X1 b gii hn bi:
(a): X2 0: X1 6000(b): X2 0: X1 2000
(c): X3 0: X1 1000
X1=1000, X2=2500 (a), X1=0 (Khng thayi gi tr ), X2=300 (b), X3=0 (c)
Z = 450000+65*1000 = 1000*140+2500*150 = 515000USD
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phng php Simplexe: Qu trnh lp (Itration), Phng nti u
PA(1) PA(2)
X1=0 X1=1000
X2=3000 X2=2500
X1=0 X1=0 (Bin ngoi cs)
X2=600 X2=300
X3=200 X3=0 (Bin ngoi cs)
Z=450000 Z=515000
Z: 450000 + 65X1 -375X1
X2=3000-0,5X1-2,5X1 (a)
X2=600-0,3X1+0,5X1 (b)
X3=200-0,2X1+X1 (c)
M t hm mc tiu theo hai bin ngoi cs
X1 v X3 tPT (c):
X1=1000+5X1-5X3 thay vo Z, (a) v (b):
Z: 515000 50X1 325X3X2=2500-5X1+2,5X3 (a)
X2=300-X1+1,5X3 (b)
X1=1000+5X1-5X3 (c)
Phng n ti u
Bng n hnh ti u (Tableau simplexe loptimum)
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phng php Simplexe: Phn tch phng nti u
ngha vt l ca PA ti u: PA ti u l phi x l 1000 tn du th A v 2500 tn du th B
Du th A: 1000 Du th B: 2500
Hiu sut S lng Hiu sut S lng Tng
Xng 0,2 200 0,4 1000 1200
GO 0,4 400 0,2 500 900
FO 0,4 400 0,4 1000 1400
1,0 1000 1,0 2500 3500
SX thc t SX ti a Chnh lch
Xng 1200 1200 0
GO 900 1200 300
FO 1400 1400 0
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phng php Simplexe: Gnralisation
Sau khi a cc bin ph vo, ta s c mt h PT c m PT v n n (m
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Phn tch cc gi tr marginales (cots marginaux)
-Xc nh cc gi tr marginales:PA ti u ca bi ton kho st: 1000 tn du th A v 2500 tn du th B
Pbl: ci g s xy ra khi ta thay i nh s SX ca xng, GO hoc FO?
PA ti u trong bng simplexe:
Z: 515000 50X1 325X3
X2=2500-5X1-2,5X3 (a)
X2=300-X1+1,5X3 (b)
X1=1000+5X1-5X3 (c)
Gi s rng chng ta c th sx 1 tn xng nhiu hn, lc rng buclu tr xng c vit li nh sau:
0,2X1+0,4X2+X1=1201 0,2X1+0,4X2+(X1-1)=1200
PT ban u trong X1 c thay th bng X1-1Gia tng sx xng 1 /v Gim 1 /v bin ph X1: Hm ti u tng50. Tng t:
Gia tng sx FO 1 /v Gim 1 /v bin ph X3: Hm ti u tng 325.
Cot marginal Rng buc: Thay i gi tr ca h m mc tiu
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phn tch cc gi tr marginales (cots marginaux)
PA ti u trong bng simplexe:
Z: 515000 50X1 325X3
X2=2500-5X1-2,5X3 (a)
X2=300-X1+1,5X3 (b)X1=1000+5X1-5X3 (c)
Gi s rng chng ta c th sx 1 tn xng nhiu hn, lc rng buclu tr xng c vit li nh sau:
0,2X1+0,4X2+X1=1201 0,2X1+0,4X2+(X1-1)=1200
PT ban u trong X1 c thay th bng X1-1
Gia tng sx xng 1 /v Gim 1 /v bin ph X1: Hm ti u tng50. Tng t:
Gia tng sx FO 1 /v Gim 1 /v bin ph X3: Hm ti u tng 325.
Cot marginal Rng buc: Thay i gi tr ca hm mc tiu
-Xc nh cc gi tr marginales:PA ti u ca bi ton kho st: 1000 tn du th A v 2500 tn du th B
Pbl: ci g s xy ra khi ta thay i nh s SX ca xng, GO hoc FO?
S phn tch ny ch c gi tr khi s thay i ca sx Xng v FO nh
Phm vi ng dng?
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phn tch cc gi tr marginales (cots marginaux)
-Gi tr marginales v h s thay th:
X2=2500-5X1-2,5X3 (a)
X2=300-X1+1,5X3 (b)X1=1000+5X1-5X3 (c)
X2=2500+5 = 2505 (a)
X2=300+1 = 301 (b)X1=1000-5 = 995 (c)
SX tng 1 tn xng
X1 X1-1
Du th A Du th B Chnh lch
-5 +5Xng -1 (-5*0,2) +2 (5*0,4) +1
GO -2 (-5*0,4) +1 (5*0,2) -1
FO -2 (-5*0,4) +2 (5*0,4) 0
Li nhun -140*5 +150*5 +50
Bi tp: Sn xut FO tng thm 1 tn
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phn tch cc gi tr marginales (cots marginaux)
-Phm vi ng dng ca gi tr marginales v bin thin h s thay th:
PA ti u
Gi tr marginales
nh gi s thay i cc gi tr ti u khi thay i nh cc rng buc
Phm vi thay i
ca cc rng buc?
Nu nhu cu xng=0, X1=-1200
X2=2500-5*1200=-3500, X1=7000: PA
khng chp nhn c (X20 hoc X1
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Phn tch cc gi tr marginales (cots marginaux)
-Phm vi ng dng ca gi tr marginales v bin thin h s thay th:
900 SX Xng 1400 Bin cs: X1, X2 v X2, ngoi cs: X1 v X3,
Ga trmarginaux khngi
Cc gi trmi ca bin cs(X1, X2, X2) vn c
th tnh ton c tbng simplexe tiu.
Hm mc tiu bthayi
Bi tp: Kho st khong lm vic ca FO
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Phn tch cc gi tr marginales (cots marginaux)
-Phm vi ng dng ca phng n: bin thin h s ca hm mc tiu
t vn :
Khi gi tr ca du th thay i Hm kinh t thay i
Vt qua mt gii hn no : Bin c s Bin ngoi c s
V d: Li nhun t du th A = 0, v du th B = 150USD ch x l du th B, X1 ngoi c s
Phm vi thayi ca gi tr du th A m khng lm thayi cu trc bng
Simplexe tiu Cots marginaux ca cc bi n ngoi csvn cn 0
V1: gi tr ca du th A
X1=-1 Hm mc tiu gim 5V1, tng 5*150: cot marginal:750-5V10, V1150.
Tng t: X3=-1 Hm mc tiu tng 5V1, gim 2.5*150:cot marginal: 5V1-2.5*1500, V175.
75 V1 150:PA ti u khng thay i (X1=1000, X2=2500), gi tr cacc bin khc cng gi nguyn:X1=X3=0, X2=300.
Gi tr ca hm ti u v cots marginaux thay i theogi tr ca A
BT: Xc nh khong gi trca du th B V2.
PA ti u trong bng simplexe:
Z: 515000 50X1 325X3
X2=2500-5X1-2,5X3 (a)
X2=300-X1+1,5X3 (b)
X1=1000+5X1-5X3 (c)
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Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Bi ton th hai ca nh my lc du: Ti thiu chi ph
Mt nh my cn SX ti thiu 3 sn phm:
Xng: 1600 tn
GO: 2000 tn
FO: 2800 tn
t 3 loi du th A, B v C vi cc hiu sut (%m)
nh sau:
A B C
Xng 0,2 0,25 0,4
GO 0,4 0,25 0,2
FO 0,4 0,5 0,4
Gi du 150 140 160
t phng trnh:
Min: 150X1 + 140X2 + 160X3
0,2X1 + 0,25X2 + 0,4X3 1600
0,4X1 + 0,25X2 + 0,2X3 2000
0,4X1 + 0,5X2 + 0,4X3 2800
Min: 150X1 + 140X2 + 160X3
0,2X1 + 0,25X2 + 0,4X3 X1 = 1600 (1)
0,4X1 + 0,25X2 + 0,2X3 X2 = 2000 (2)
0,4X1 + 0,5X2 + 0,4X3 X3 = 2800 (3)
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
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Bi ton th hai ca nh my lc du: Ti thiu chi ph
Min: 150X1 + 140X2 + 160X3
0,2X1 + 0,25X2 + 0,4X3 X1 = 1600 (1)
0,4X1 + 0,25X2 + 0,2X3 X2 = 2000 (2)
0,4X1 + 0,5X2 + 0,4X3 X3 = 2800 (3)PA xut pht (PA(0)):
X2=X3=0
Xc nh X1 ti thiu ccbin ph 0 X1=8000, ta c:
X2=X3=0
X1=8000
X1=0
X2=1200
X3=400
Min: 1200000 47,5X2 140X3 + 750X1
X1 = 8000 1,25X2 2X3 + 5X1 (1)
X2 = 1200 0,25X2 0,6X3 + 2X1 (2)
X3 = 400 - 0,4X3 + 2X1 (3)
Phng n tip theo (PA(1)): tng gi tr ca X3X3 4000 (1), X3 2000 (2), X3 1000 (3) X3=1000
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
Bi ton th hai ca nh my lc du: Ti thiu chi ph
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Bi ton th hai ca nh my lc du: Ti thiu chi ph
PA(1): tng gi tr ca X3
X3 4000 (1), X3 2000 (2), X3 1000 (3) X3=1000
(3) X3 = 1000 + 5X1 2,5X3
Min: 1060000 47,5X2 + 50X1 + 350X3
X1 = 6000 1,25X2 5X1 + 5X3 (1)
X2 = 600 0,25X2 X1 + 1,5X3 (2)
X3 = 1000 + 5X1 2,5X3 (3)
Min: 94600 + 240X1 + 190X2 + 65X3
X1 = 3000 + 5X2 2,5X3 (1)
X2 = 2400 4X1 - 4X2 + 6X3 (2)
X3 = 1000 + 5X1 2,5X3 (3)
PA(1) PA(2): tng gi tr ca X2X2 4800 (1), X2 2400 (2), X2 (3)
X2=2400
(2) X2 = 2400 4X1 - 4X2 + 6X3
Phng n ti u
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
i ngu (Dualit)
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i ngu (Dualit)
y l mt nguyn l cbn ca QHTT:
Kt hp bi ton tiu tuyn tnh cc hotng vt l (Problme primal) vi mt bi
ton tuyn tnh khc, i xng, tiu v chi ph hoc gi c, li nhun (Dual).
Tn trng cc rng but ca bi ton u
VD: Bi ton nh my lc duu tin:
Primal
Max(Z) 140X1 + 150X2
0,2X1 + 0,4X2 1200
0,4X1 + 0,2X2 1200
0,4X1 + 0,4X2 1400
Dual
Min (Z) 1200U1 + 1200U2 + 1400U30,2U1 + 0,4U2 + 0,4U3 140
0.4U2 + 0.2U2 + 0.4U3 150
Kinh tthtrng
Ti thiu gi bn
Li nhun Gi trdu th
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
i ngu (Dualit)
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i ngu (Dualit)
Primal
Max (Z) 140X1 + 150X20,2X1 + 0,4X2 + X1 = 1200
0,4X1 + 0,2X2 + X2 = 1200
0,4X1 + 0,4X2 + X3 = 1400
Dual
Min (Z) 1200U1 + 1200U2 + 1400U3
0,2U1 + 0,4U2 + 0,4U3 U1 = 140
0.4U2 + 0.2U2 + 0.4U3 - U2 = 150
Primal
Z: 515000 50X1 325X3
X2 = 2500 - 5X1 + 2,5X3X2 = 300 - X1 + 1,5X3
X1 = 1000 + 5X1 - 5X3
Dual
Z: 515000 + 300U2 + 1000U1 + 2500U2
U3 = 325 1.5U2 +5U1 2.5U2
U1 = 50 + U2 5U1 + 5U2
Cng hai gi tr ti u
Gi tr ti u ca Dual = gi tr marginal ca primal
S lng ti u ca primal = gi tr marginal ca Dual
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
i ngu (Dualit)
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i ngu (Dualit)
Primal
Max (Z) 140X1 + 150X20,2X1 + 0,4X2 + X1 = 1200
0,4X1 + 0,2X2 + X2 = 1200
0,4X1 + 0,4X2 + X3 = 1400
Dual
Min (Z) 1200U1 + 1200U2 + 1400U3
0,2U1 + 0,4U2 + 0,4U3 U1 = 140
0.4U2 + 0.2U2 + 0.4U3 - U2 = 150
Primal
Z: 515000 50X1 325X3
X2 = 2500 - 5X1 + 2,5X3X2 = 300 - X1 + 1,5X3
X1 = 1000 + 5X1 - 5X3
Dual
Z: 515000 + 300U2 + 1000U1 + 2500U2
U3 = 325 1.5U2 +5U1 2.5U2
U1 = 50 + U2 5U1 + 5U2
Phn tch cc gi tr marginales ca Dual: U1, U2 v U2
Chng II: Quy hoch tuyn tnh p dng vo qu trnh lc du
p dng vo nh my lc du
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p dng vo nh my lc du
Mc ch ca nh my lc du: (Mi trng cnh tranh)
Ti a li nhun ca nh my: Maxi s chnh lch gia gi tr bn SP v chi ph:
Nguyn liu: du th v cc nguyn liu khc
Vn hnh: Chi ph c nh, chi ph thay i
c im ca NMLD:
SX nhiu loi sn phm t nhiu loi nguyn liu khc nhau.
T l tng i gia cc SP ph thuc vo:
S a dng ca nguyn liu
Hot ng ca cc PX bo m cho s chuyn ho
Cc iu chnh c th c ca PX
Khng th xc nh chi ph SX trn mt sn phm (mt cch ton hc)
Nhiu kh nng la chn PA SX trong mt bi cnh cho trc: PA ti u?
Biu din cc hot ng ca nh my lc du thnh cc PT tuyn tnh. S dng m hnh tuyn tnh thu c xc nh PA ti u.
Chng III: M hnh ha shotng ca nh my l c du
C s qu trnh m hnh ho s hot ng ca nh my lc du
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q g y
3 HOT NG CHNH:
Phn tch cc phn on Chng ct Chuyn ho Cht lng SP (reforming, x l bng Hydro...), Hiu sut SP (FCC...) Phi trn Sn phm cui
I. Phn tch v chuyn ho:
c thc hin bi cc PX sn c trong NMLD
Xem xt hot ng ca PX
- Ch lm vic n nh
- Lu lng v c tnh ca nguyn liu- Cc thng s lm vic: T, P, T s hi lu, VVH...
T l tuyn tnh gia lu lng SP v NL:
X1 = 1X
X2 = 2X
X3 = 3X
Sn phm
X1 m3/h
X2 m3/h
Nguyn liu
X m3/hX3 m
3/h
Ch lm vic ca PX 1,
2,
3
Chng III: M hnh ha shotng ca nh my l c du
Nh my lc du n gin
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y g
V d: Xt mt NMLD n ginc homt giai on no :-1 PX chng ct x l 2 loi du th A (X1) v B (X2) thu 3 SP: Xng (a1, a2), GO (b1, b2) v FO (c1, c2)
-1 PX chuyn ho GO (X3) thnh Xng (a3) v FO (b3)
-Khng xem xt cht lng SP
-Tiu th nng lng ni b: s dng FO (Y) vi d1, d2, d3 (% nguyn liu)
-Kh nng SX ti a: Chng ct = Q, Chuyn ho = Q3
-Yu cu cung cp Xng, GO v FO: E, G v F
-Gi du th + chi ph SX bin i: du th A: 1 USD/m3, du th B: 2 USD/m3
-S SX c th > yu cu cung cp:
- Lu lng sn phm sn xut vt k hoch v gi bn: Xng: m3 ( USD/m3), GO: m3 ( USD/m3),
FO: m3 ( USD/m3)
-Chi ph SX bin i ca phn xng chuyn ho 3 USD/m3
-Gi bn ca cc SP trong k hoch: Xng e USD/m3, GO g USD/m3, FO f USD/m3,
Chng III: M hnh ha shotng ca nh my l c du
S n gin ca qu trnh lc du
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Bn SX tha
a1
b1
c1
E
Xng
a2
b2
c2
Du th B
X2
d2
Du th A
X1 a3
c3
X3
Gd1
GOd3
FFO
Du th Chng ct Chuyn ho NL ni b Sn phm cui
Chng III: M hnh ha shotng ca nh my l c du
Xy dng ma trn bi ton
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Xy dng ma trn bi ton:Sn xut - S dng ni b - SX d = Nhu cu
Danh sch cc n s: X1, X2, X3, , ,
CN BNG VT CHT:
Cn bng xng: a1X1 + a2X2 +a3X3 = E +
Cn bng GO: b1X1 + b2X2 X3 = G +
Cn bng FO: c1X1 + c2X2 +c3X3 -Y = F + CN BNG NHIT LNG:
Nng lng PX: d1X1 + d2X2 +d3X3 = Y
RNG BUC KH NNG SX:
Kh nng chng ct: X1 + X2 Q
Kh nng chuyn ho: X3 Q3
HM KINH T (HM MC TIU)
Min: 1 X1 + 2 X2 + 3X3 - - - - e E g G - f F
Chng III: M hnh ha shotng ca nh my l c du
Kt qu
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Ma trn bi ton :
X1 X2 X3 Y RHS
Cn bng Xng a1 a2 a3 -1 = E
Cn bng GO b1 b2 -1 -1 = GCn bng FO c1 c2 c3 -1 -1 = F
Cn bng NL NL ni b d1 d2 d3 -1 = 0
Kh nng chng ct 1 1 QKh nng chuyn ho 1 Q3
Hm kinh t (ti u) 1 2 3 Min
Cn bng sn
phm
Rng but KNSX
Chng III: M hnh ha shotng ca nh my l c du
Phi trn cc bn sn phm (bases)
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-Sn phm cui Tiu chun SP: Cc rng but ti a hoc ti thiu ca cc tnh cht vt l.-Cc tiu chun SP tho mn yu cu khch hng hoc cc tiu chun nh nc.
-Sn phm cui ca nh my lc du: Phi trn cc bases
-Phng php d on (mthodes de prdiction) cc tnh cht ca SP phi trn
Sn phm cui
(Q1)Base 1 Z1
(Q2)Z (Q)Base 2 Z2
(Q3)Base 3 Z3
=
i
i
i
ii
Z
ZQ
Q
Z: s lng ca cc bases v SP cui
(th tch hoc khi lng)Q: Tnh cht cn kho st hoc cc gitnh cht (pseudo-qualit)
Chng III: M hnh ha shotng ca nh my l c du
Phi trn cc bn sn phm (bases)
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Mt sn phm p c tnh cht Q thu c t s phi trn ca i bases c th tch vi v c tnh cht qi
S phi trn ny tun theo Quy tc trn ln th tch:
maxS
i
ii
vvq
=
i
i
i
ii
v
vq
Q Nu Q Smax: imaxii vSvq
( ) 0vS-q imaxi
S phi trn ny tun theo Quy tc trn ln khi lng:
( ) 0vS-q imaxi id
=Vvi
maxSVvq ii
Cng thc m rng: = 0Vvi
0max VSvq ii
Chng III: M hnh ha shotng ca nh my l c du
Phi trn cc bn sn phm (bases)
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V d 1: Phi liu SX hai loi xng supercarburant (SU) v Xng thng (CO) t cc bases sau y:
Li nhun 8USD/m3 SU, 2USD/m3 CO
Base K hiu Sn c (m3) p sut hi Ch s OctaneButane C4 A1 t1 r1
Xng FCC EC A2 t2 r2
Rformat ER A3 t3 r3
C4
(A1)
EC
(A2)
ER
(A3)
SU
CO
C4SU
C4CO
ECCO
ECSU
ERCO
ERSU
VSU
VCO
Chng III: M hnh ha shotng ca nh my l c du
Phi trn cc bn sn phm (bases)
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V d 1: Phi liu SX hai loi xng supercarburant (SU) v Xng thng (CO)
Bng Ma trn phi trn
C4SU C4CO ECSU ECCO ERSU ERCO RHSLu tr C4 1 1 A1Lu tr EC 1 1 A2Lu tr EC 1 1 A3TVMaxSU t1-TVMax1 t2-TVMax1 t3-TVmax1 0
TVminSU t1-Tvmin1 t2-Tvmin1 t3-Tvmin1
0OCminSU r1-OCmin1 r2-OCmin1 r2-OCmin1 0TVMaxCO t1-TVMax2 t2-TVMax2 t3-TVMax2 0TVminCO t1-Tvmin2 t2-Tvmin2 t3-Tvmin2 0OCminCO r1-OCmin2 r2-OCmin2 r2-OCmin2 0
Hm kinh t 8 2 8 2 8 2 Max
Chng III: M hnh ha shotng ca nh my l c du
Phi trn cc bn sn phm (bases)
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V d 1: Phi liu SX hai loi xng supercarburant (SU) v Xng thng (CO)
Bng Ma trn phi trn m rng
C4SU C4CO ECSU ECCO ERSU ERCO VSU VCO RHSLu tr C4 1 1 A1Lu tr EC 1 1 A2Lu tr EC 1 1 A3
VolSU 1 1 1 -1 = 0
TVMaxSU t1 t2 t3 -TVMax1 0TVminSU t1 t2 t3 -Tvmin1 0OCminSU r1 r2 r2 -OCmin1 0
VolCO 1 1 1 -1 = 0TVMaxCO t1 t2 t3 -TVMax2 0
TVminCO t1 t2 t3 -Tvmin2 0OCminCO r1 r2 r2 -OCmin2 0Hm kinh t 8 2 Max
Chng III: M hnh ha shotng ca nh my l c du
Phi trn sn phm theo cc cng thc
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Nu cht lng SP theo cng thc trn ln tuyn tnh sai khc nhiu vi thc t
Cng thc c kim tra ti phng th nghim
La chn gia cc cng thc c thit lp
V d: SX mt loi xng c bit vi nhucu Q bng hai cng thc 1 v 2, tho mn tt c cc tiuchun cn kim tra
% th tch iC5 (Chng ct) Reformat Xng FCCFormule 1 3 60 37Formule 2 4 50 46
Chng III: M hnh ha shotng ca nh my l c du
Phi trn sn phm theo cc cng thc
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V d: SX mt loi xng c bit vi nhucu Q bng hai cng thc 1 v 2, tho mn tt c cc tiuchun cn kim tra
Chct Reforming FCC CASF1 CASF2 RHSCn bng iC5 a -0.03 -0.04 = 0Cn bng Reformat b -0.6 -0.5 = 0Cn bng Xng FCC c -0.37 -0.46 = 0
Nhu cu CA 1 1 = QChi ph C1 C2 C3 Min
Chng III: M hnh ha shotng ca nh my l c du
S m phng qu trnh lc du v phi trn sn phm
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Chng IV: Sdng phn mm Lingo trong tinh ton tiu
Cu trc ma trn ca bi ton thng gp:
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g gp
Cu trc bin
Gi tr cc bin
RHSDU
Cn bng SP v bn SP
Rng buc khng ch cht lng
Cc rng buc c bit
Rng buc v kh nng SX, lu tr
HM KINH T (MC TIU)
Chng IV: Sdng phn mm Lingo trong tinh ton tiu
Cc bc cn tin hnh khi gii bi ton ti u bng phn mm Lingo
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M hnh ha qu trnh sn xut
Xc nh cu trc cc bin v rng buc
Xy dng ma trn ca bi ton trn bng tnh Excel
Khai bo cc mng cha bin, cc kiu rng buc, RHSv cc h s ca bi ton
Lin kt d liu gia Excel v Lingo
Gii ti u bng Lingo v trao i kt qu
Chng IV: Sdng phn mm Lingo trong tinh ton tiu
V d v phng php khai bo v lin kt d liu!Bai Tap so 1;MODEL:
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MODEL:SETS:
Ban_SP : Zero;SP_NB : Nhu_cau;RB_min : GT_min;RB_Max : GT_Max;Ten_bien : GT_bien, chi_phi;LINKS1(Ban_SP,Ten_bien) :VOLUME1;LINKS2(SP_NB,Ten_bien) :VOLUME2;LINKS3(RB_min,Ten_bien) :VOLUME3;LINKS4(RB_Max,Ten_bien) :VOLUME4;
ENDSETS
MIN=@SUM(Ten_bien(J): GT_Bien(J)*Chi_phi(J));@FOR(Ban_SP(I):
@SUM(Ten_bien(J):VOLUME1(I,J)*GT_Bien(J))=Zero(I));
@FOR(SP_NB(K):@SUM(Ten_bien(J):
VOLUME2(K,J)*GT_Bien(J))=Nhu_cau(K));@FOR(RB_min(L):
@SUM(Ten_bien(J):VOLUME3(L,J)*GT_Bien(J))>=GT_Min(L));
@FOR(RB_Max(M):
@SUM(Ten_bien(J):VOLUME4(M,J)*GT_Bien(J))
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Chng IV: Sdng phn mm Lingo trong tinh ton tiu
V d v bo co kt qu ca bi ton ti u
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Chng V: Cc chtiu kinh tca snng cao gi trsn phm
Mc ch:
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- nh gi v so snh nhanh vo mi thi im hiu qu kinh t cacc phng n hot ng khc nhau ca mt nh my lc du
- Ti a li nhun ca thi im kho st
Nguyn tc:
- Da trn mc hot ng c s
- Chi ph cnh c trang tri mc hot ng c s-Gi tr ca cc phng n hot ng khc nhau c tnh t mc hot ng c s v ch xt cc chi ph bin i i km
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Cc phng php tip cn: 4 giai on
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Cc phng php tip cn: 4 giai on
1. Xc nh trng hp c s
2. Lit k cc phng n hot c th c bng cch s dng kh nng uyn
chuyn ca nh my v ccng buc3. nh gi hiu qu kinh t ca cc phng n khc nhau:
- Cn bng vt cht v cc s thay i kim km vi sn phm xem xt
- Chi ph bin i i km- Gi tr ca sn phm theo c ch ti cn bng ca nh my lc du
4. Phn loi cc hot dng theo th t hiu qu kinh t v quyt nh
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Tnh ton hiu qu ca nh my lc du
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Bi ton: Nh my Lc du Basse Seine, s dng nguyn liu du thArabe light 34API vi FCC. nh gi gi tr ca du th i ra khi nh mynh sau:
162.48100
00.40Hao ht
4.725.980Tiu th ni b
19.22480FO
57.5730.3190GOM
18.69.3200JetA1
45.7620.8220Xng super
10.085.6180Naphtha
3.852.2175Butane
2.71.5180PropaneGi tr
c
a phn
o
n (USD)Hi
u su
t (%)Gi tr
s
n ph
m (USD/t
n)S
n ph
m
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Tnh ton hiu qu ca nh my lc du
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Gi tr ca cc sn phm theo c ch ti cn bng:
120 CAF120 CAFNhpFO (1%S)
80 FOB85 CAFXut sang AnhFO (3.5%S)
190 CAF190 CAFNhpGasoil (0.2%S)
200 FOB210 CAFXut sang RotterdamKerosene220 FOB230 CAFXut sang RotterdamXng super
180 FOB190 CAFXut sang RotterdamXng nng
175 CAF130 FOBXut sang RotterdamButane
180 CAF135 FOBNhpPropane
Sn phm
Ch s th trng du m (Rotterdam)C ch ti cn bng
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Tnh ton hiu qu ca nh my lc du
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Chi ph du th v chi ph sn xut (USD/tn)
143.2Chi ph tng
5.0Chi ph bin i (nh my vi FCC)
4.7Chi ph nhin liu
2.0Chi ph vn chuyn t terminal n NMLD131.5Gi tr du th ti cng (CAF)
0.5Hao ht khi vn chuyn(0.35%)
13.4Chi ph vn chuyn + bo him
117.6Gi mua du th (FOB)
Hiu qu ca NMLD khi x l du th (USD/tn)
162.5 - 143.2 = 19.3 USD/tn
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Chn im ct ca phn on Xng v Kerosene
Nguyn l:
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-y l tham s iu chnh v to nn mt s uyn chuyn quan tr ng i vi nhmy lc du
-Do cc rng buc v cht lng sn phm Ti thiu l 145-150oC v ti a l185oC
C ch v cn bng vt cht:
S thay i im ct gia phn onxng nng v kerosene c thc
hin trong gii hn trn v phi luntn trng tiu chun v im vn cca gasoil
C ch ny cho php gia tng 2 tnkerosene v 0.4 tn VGO cracking
trong khi gim 1.4 tn gasoil v 1 tnxng nng.
Xng nng
Kerosene
Gasoil
VGO
Xng nng
Kerosene
Gasoil
VGO
=1
=1
=0.4
1 tn kerosene = 0.7 tn gasoil + 0.5 tn xng 0.2 tn VGO
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Chn im ct ca phn on Xng v Kerosene
C s nh gi (USD/tn) :
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C s nh gi (USD/tn) :
150VGO193.4Gasoil
Nhp khu c d=0.845, xt chnh lch t trng ta c:190Gasoil
(gi xut khu)200Kerosene
(gi xut khu + 6USD)186Xng nng
1 tn kerosene = 0.7 tn gasoil + 0.5 tn xng nng 0.2 tn VGO
Chi ph sn xut thm 1 tn kerosene (marginal) (USD/tn)0.7*193.4 + 0.5*186 0.2*150 = 198.4
Hiu qu:
Tng li nhun: 200-198.4=1.6 USD
Sn xut ti a Kerosene
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Xt hiu qu ca phn xng reforming xc tc
Xng nng chng ct
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Reformingxc tc
Mua trc tip
0.11 C2-
0.07 C3
0.09 C4
0.73 Reformate0.08 Nhin liu
213Reformate
155.50.73 tn Reformate
38.9
15.7Butane 0.09*175
12.6Propane 0.7*180
10.6C2-
=0.11*80*1.2
Cc sn phm ph
194.4
6.4Nhin liu (0.08*80)
2.0Chi ph vn hnh186Xng nng nguyn liu
USD/tn
Xc nh gi tr ca Reformate
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Xt hiu qu ca phn xng reforming xc tc
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Xc nh gi tr ca Xng Super c ch
210.9Xng Super c ch
4.0Ch (0.15g/l)7.0Butane 0.04*175
25.2Xng nh 0.14*180
174.7Reformate 0.82*213
USD/tn
Gi FOB ca xng Super: 220 USD/tn (xut khu)
Vn hnh ch ti a ca phn xng Reforming
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Xt hiu qu ca phn xng FCC
Hiu sut v gi tr cc sn phmca phn xng FCC, nguyn liu 2%S
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156.2100Tng
000.5Mt mt
006.5Cole
19.368024.2Du nng (HTS)
66.519035Gasoil
52.822024Xng Super
10.81806Xng nh
3.151751.8Butane
3.61802Propane
Gi tr ca phn on(USD)
Gi tr(USD/tn)
Hiu sut(%m)
Sn phm
Nh my thiu VGO cracking v phi nhp t Rotterdam
Chng III: Cc chtiu kinh tca snng cao gi trsn phm
Xt hiu qu ca phn xng FCC
Cht pha long gim nht ~ 30%GO+70%FO 2 8%S
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Gi tr ca VGO cracking
Kh nng cracking: 6 USD/tn
Gi tr tng ng fluxant:
116.5 - 4.2=112.3USD/tn
Vn chuyn: 10 USD/tn
VGO cracking:112.3 + 6 + 10 = 128.3 (USD/tn)
Cht pha long, gim nht 30%GO+70%FO, 2.8%S
(Fluxant)
Gi tr nht: 0.3*190 + 0.7*85 = 116 USD/tn
Khu tr hm lng S: (2.5 so vi 2.8):
(2.8-2.5)*(120-85)/(3.5-1) = 14 USD/%S
Hiu qu ca FCC: 156.2 128.3 = 23.9 (USD/tn)
Tng cng sut ca phn xng FCC