Toi uu hoa NMLD

8/2/2019 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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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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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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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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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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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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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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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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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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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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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.



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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.



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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.



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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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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



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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.



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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



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-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.



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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



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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



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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


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



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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



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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


X2=600 X2=300


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)



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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



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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



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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?



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-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



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-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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-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



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-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.


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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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)



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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




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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


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


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


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


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


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,


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


Xy dng ma trn bi ton


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44

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


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


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)




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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




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48

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




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49

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




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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


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


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


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)


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


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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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


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




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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




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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


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


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


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


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


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


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

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