quantrisanxuat

GII THIU V QUN TR SN XUT

1. MT S KHI NIM

1.1 Khi nim v sn xut

Theo quan nim ph bin trn th gii th sn xut c hiu l qu trnh to ra sn phm hoc dch v.

nc ta lu nay c mt s ngi thng cho rng ch c nhng doanh nghip ch to, sn xut cc sn phm vt cht c hnh thi c th nh xi mng, tivi, my git,... mi gi l cc n v sn xut. Nhng n v khc khng sn xut cc sn phm vt cht u xp vo loi cc n v phi sn xut. Ngy nay trong nn kinh t th trng, quan nim nh vy khng cn ph hp na.

Mt h thng sn xut s dng cc yu t u vo l nguyn vt liu th, con ngi, my mc, nh xng, k thut cng ngh, tin mt v cc ngun ti nguyn khc chuyn i n thnh sn phm hoc dch v. S chuyn i ny l hot ng trng tm v ph bin ca h thng sn xut. Mi quan tm hng u ca cc nh qun tr h thng sn xut, l cc hot ng chuyn ha ca sn xut.

S 1.1:

Qu trnh sn xut.

Nh vy, v thc cht sn xut chnh l qu trnh chuyn ha cc yu t u vo bin chng thnh cc sn phm hoc dch v u ra. Ta c th hnh dung qu trnh ny nh trong s 1.1. S khc bit gia qu trnh sn xut sn phm hu hnh v sn phm v hnh (dch v) c th hin nh sau:

u vo - Nguyn vt liu - My mc thit b - Cng ngh - Ngun nhn lc - Vn - Thng tin

Chuyn ha - Lm bin i - Tng thm gi tr

u ra - Sn phm hu hnh- Sn phm v hnh

San xuat san pham hu hnh Sn xut sn phm v hnh

(dch v) 1. To ra sn phm vt cht 2. C th d tr (tn kho) 3. t tip xc vi khch hng

trong qu trnh sn xut 4. Cn nhiu my mc thit b 5. Thng thng cn s vn ln 6. Cht lng sn phm d nh

gi 7. Sn phm c phn phi

khng gii hn a l.

1. Khng to ra sn phm vt cht 2. Khng d tr c 3. Thng xuyn tip xc vi

khch hng. 4. Cn nhiu nhn vin 5. Thng thng cn s vn t hn6. Cht lng sn phm dch v

kh nh gi 7. Vic phn phi dch v c gii

hn v a l. Theo ngha rng, sn xut bao hm bt k hot ng no nhm tha mn nhu

cu ca con ngi. N c th phn thnh: sn xut bc 1; sn xut bc 2 v sn xut bc 3.

Sn xut bc 1 (sn xut s ch): l hnh thc sn xut da vo khai

thc ti nguyn thin nhin hoc l nhng hot ng s dng cc ngun ti nguyn c sn, cn dng t nhin nh khai thc qung m, khai thc lm sn, nh bt hi sn, trng trt,...

Sn xut bc 2 (cng nghip ch bin): l hnh thc sn xut, ch to, ch bin cc loi nguyn liu th hay ti nguyn thin nhin bin thnh hng ha nh g ch bin thnh bn gh, t gingqung m bin thnh st thp. Sn xut bc 2 cn gm c vic ch to cc b phn cu thnh c dng lp rp thnh sn phm tiu dng v sn phm cng nghip.

Sn xut bc 3 (cng nghip dch v): Cung cp h thng cc dch v nhm tha mn nhu cu a dng ca con ngi. Trong nn sn xut bc 3, dch v c sn xut ra nhiu hn cc hng ha hu hnh. Cc nh sn xut cng nghip c cung cp nhng iu kin thun li v dch v trong phm vi rng ln. Cc cng ty vn ti chuyn ch sn phm ca cc nh sn xut t nh my n cc nh bn l. Cc nh bn bun v nh bn l cung cp cc dch v n ngi tiu dng cui cng. Ngoi ra cn nhiu loi dch v khc nh: bc d hng ha, bu in, vin thng, ngn hng, ti chnh, bo him, y t, gio dc, nh hng, khch sn,...

1.2 c im ca sn xut hin i Qun tr sn xut ngy cng c cc nh qun tr cp cao quan tm, coi

nh l mt v kh cnh tranh sc bn. S thnh cng chin lc ca doanh nghip ph thuc rt nhiu vo s nh gi, to dng, pht trin cc ngun lc

t chc nng sn xut. Sn xut hin i c nhng c im: Sn xut hin i yu cu phi c k hoch hp l khoa hc, c i ng

k s gii, cng nhn c o to tt v thit b hin i. Quan tm ngy cng nhiu n thng hiu v cht lng sn phm. y

l mt tt yu khch quan khi m tin b k thut ngy cng pht trin vi mc cao v yu cu ca cuc sng ngy cng nng cao. Cng nhn thc r con ngi l ti sn qu nht ca cng ty. Yu cu ngy

cng cao ca qu trnh sn xut, cng vi s pht trin ca my mc thit b, vai tr nng ng ca con ngi tr nn chim v tr quyt nh cho s thnh cng trong cc h thng sn xut. Sn xut hin i ngy cng quan tm n vn kim sot chi ph. Vic

kim sot chi ph c quan tm thng xuyn hn trong tng chc nng, trong mi giai on qun l. Sn xut hin i da trn nn tng tp trung v chuyn mn ha cao. S

pht trin mnh m ca khoa hc k thut lm cho cc cng ty thy rng khng th tham gia vo mi lnh vc, m cn phi tp trung vo lnh vc no mnh c th mnh ginh v th cnh tranh. Sn xut hin i cng tha nhn yu cu v tnh mm do ca h thng

sn xut. Sn xut hng lot, qui m ln tng chim u th lm gim chi ph sn xut. Nhng khi nhu cu ngy cng a dng, thay i cng nhanh th nhng n v va v nh, c lp mm do c u th nht nh. S pht trin ca c kh ho trong sn xut t ch thay th cho lao ng

nng nhc, n nay ng dng nhiu h thng sn xut t ng iu khin bng chng trnh. Ngy cng ng dng nhiu thnh tu ca cng ngh tin hc, my tnh tr

gip c lc cho cc cng vic qun l h thng sn xut. M phng cc m hnh ton hc c s dng rng ri h tr cho vic

ra quyt nh sn xut kinh doanh.

1.3 Khi nim v qun tr sn xut Qun tr sn xut v tc nghip bao gm tt c cc hot ng lin quan n

vic qun tr cc yu t u vo, t chc, phi hp cc yu t nhm chuyn ha chng thnh cc sn phm vt cht hoc dch v vi hiu qu cao nht.

to ra sn phm v dch v cc doanh nghip u phi thc hin 3 chc nng c bn: Marketing, sn xut v ti chnh. Cc nh qun tr Marketing chu trch nhim to ra nhu cu cho sn phm v dch v ca t chc. Cc nh qun tr ti chnh chu trch nhim v vic t c mc tiu ti chnh ca doanh nghip. Cc doanh nghip khng th thnh cng khi khng thc hin ng b cc chc nng ti chnh, Marketing v sn xut. Khng qun tr sn xut tt th khng c sn phm hoc dch v tt; khng c Marketing th sn phm hoc dch v cung ng khng nhiu; khng c qun tr ti chnh th cc tht bi v ti chnh s din ra. Mi chc nng hot ng mt cch c lp t c mc tiu ring ca mnh ng thi cng phi lm vic cng nhau t c mc tiu

chung cho t chc v li ch, s tn ti v tng trng trong mt iu kin kinh doanh nng ng.

Do c th ni rng qun tr sn xut v tc nghip c tm quan trng c bit trong hot ng ca doanh nghip. Nu qun tr tt, ng dng cc phng php qun tr khoa hc th s to kh nng sinh li ln cho doanh nghip. Ngc li nu qun tr xu s lm cho doanh nghip thua l, thm ch c th b ph sn.

2. CC BC PHT TRIN CA QUN TR SN XUT

Khoa hc v qun tr sn xut v dch v pht trin lin tc nhanh chng cng

vi vic pht trin khoa hc v cng ngh. Xt v mt lch s, chng ta c th chia thnh 3 giai on chnh sau:

2.1 Cch mng cng nghip Anh vo nhng nm u th k XVIII, khoa hc k thut pht trin mnh

ko theo s bng n cch mng cng nghip. Vic pht minh ra ng c hi nc ca Jame Watt vo nm1764, to iu kin cho ra i hng lot nhng my mc khc trong k ngh. H qu tt yu l s thay th rng ri lc lng lao ng th cng bng my mc c nng sut cao hn, cng vi s thit lp h thng nh xng v cc pht minh khc ca thi i. Tnh sn c ca my hi nc v my mc sn xut to iu kin cho vic tp hp cc cng nhn vo nh my. S tp trung ny to ra mt nhu cu v vic sp xp h li mt cch hp l sn xut ra sn phm .

Tc phm ca Adam Smith S giu c ca quc gia vit nm 1776, chng minh cho s cn thit ca phn cng lao ng, hay cn gi l chuyn mn ha ca lao ng. Vic sn xut sn phm c phn chia ra thnh tng b phn nh, nhng nhim v chuyn bit c phn cng cho cng nhn theo qui trnh sn xut. V th, cc nh my vo cui thi k ny khng nhng ch ch n vic trang b my mc thit b cho sn xut, m cn cch thc hoch nh v qun l cng vic sn xut ca cng nhn.

Cch mng cng nghip lan truyn t Anh sang Hoa K. Nm 1790 Eli Whitney, nh pht minh Hoa K, thit k mu sng trng sn xut theo dy chuyn em li hiu qu rt ln v thi gian v chi ph.

Nm 1800 nhng ngnh cng nghip khc pht trin ln cng vi s pht trin ca ng c xng du v in, nhu cu v sn phm phc v cho chin tranh thc y s thnh lp nhiu nh my hn na. H thng sn xut th cng c thay th bi h thng nh xng vi nhng my mc hin i vo thi k to nn nhng thay i ln i vi nh my ni ring v c ngnh cng nghip ni chung.

K nguyn cng nghip mi Hoa k xut hin ngay khi bt u th k 20, to ra mt giai on m rng ln lao v nng lc sn xut. S chm dt vic s dng lao ng n l, s di chuyn ca lc lng lao ng trong nng thn vo

cc thnh th v s nhp c cung cp mt lc lng lao ng ln cho s pht trin nhanh chng ca trung tm cng nghip thnh th. S pht trin ny dn n hnh thc mi ca ngnh cng nghip l gii quyt vn vn thng qua vic thit lp cc cng ty c phn. T , c th nh qun l tr thnh ngi lm thu cho x nghip v c tr lng t nh ti chnh, hay ngi lm ch u t.

2.2 Qun tr khoa hc Frederick W.Taylor c xem nh l cha ca phng php qun tr khoa hc. ng nghin cu cc vn thuc v nh my vo thi i ca ng mt cch khoa hc, ch trng n tnh hiu qu vi mong mun t c kt qu v vic tit kim thi gian, nng lc v nguyn vt liu.

H thng hot ng ca Taylor nh sau: - K nng, sc lc v kh nng hc tp c xc nh cho tng cng nhn

h c th c n nh vo cc cng vic m h thch hp nht. - Cc nghin cu v theo di ngng lm vic c tin hnh nhm a ra kt

qu chun cho tng cng nhn tng nhim v. Kt qu mong mun i vi tng cng nhn s c s dng cho vic hoch nh v lp thi gian biu, so snh vi phng php khc thc thi nhim v.

- Cc phiu hng dn, cc kt qu thc hin v c im ring bit ca tng nguyn vt liu s c s dng phi hp v t chc cng vic, phng php lm vic v tin trnh cng vic cng nh kt qu lao ng c th c chun ha.

- Cng vic gim st c ci tin thng qua vic la chn v hun luyn cn thn.

Taylor thng xuyn ch ra rng qun tr khng quan tm n vic i mi chc nng ca n. ng tin rng qun tr phi chp nhn vic hoch nh, t chc, qun l v nhng phng php xc nh trch nhim hn l nhng chc nng quan trng ny cho chnh cng nhn.

H thng tr thng khuyn khch c s dng gia tng hiu qu v lm gim i trch nhim truyn thng ca nhng ngi qun l l n c cng nhn.

Henry L.Gantt lm vic cng vi Taylor nh my Midvale, ni chung ng c cng quan im vi Taylor, ngoi tr vic ch n ngi thc hin cng vic hn l bn thn cng vic. ng t ra hiu bit tm l cng nhn hn Taylor v tha nhn tm quan trng ca tinh thn v li ch ca phn thng tinh thn i vi vic ng vin cng nhn.

Frank v Lillian Gilbreth, l nh thu thnh t, ngi quan tm n phng php lm vic khi mi bt u lm th ph. Sau ny ng c nhiu ci tin trong phng php xy v cc ngh khc ca ngnh xy dng. ng quan nim vic lp k hoch cng tc v hun luyn cho cng nhn nhng phng php lm vic ng n khng ch nng cao nng sut, m cn m bo sc khe v an ton cho cng nhn.

Sau chin tranh th gii II, cc gio trnh v qun tr tc nghip c gii thiu trong cc trng i hc, cc t chc t vn v nghin cu tc nghip...m

ngy nay chng ta c bit nh l k thut nh lng, qui hoch tuyn tnh, PERT/CPM v cc m hnh d bo.

Nghin cu tc nghip tm kim vic thay th cc quyt nh phc tp bng mt phng php ch r nhng kh nng ti u thng qua vic phn tch.

2.3 Cch mng dch v Mt trong nhng s pht trin khi u trong thi i ca chng ta l s n r

ca dch v trong nn kinh t Hoa K. Vic thit lp cc t chc dch v pht trin nhanh chng sau th chin th II v vn cn tip tc m rng cho n nay.

Cc nhn t nh hng n qun tr sn xut v dch v ngy nay:

- Cht lng, dch v khch hng v cc thch thc v chi ph. - S pht trin nhanh chng ca cc k thut sn xut tin tin. - S tng trng lin tc ca khu vc dch v. - S khan him cc ti nguyn trong sn xut. - Cc trch nhim x hi trong hot ng sn xut. Cc nh qun tr cn nhn thy tm quan trng ca dch v khng ch trong

phm vi quc gia m cn tm quc t. C nh vy, kh nng cnh tranh s tt hn hiu qu phc v khch hng s cao hn v khi t chc s c c nhng li th quan trng so vi cc i th cnh tranh.

3. HNG NGHIN CU QUN TR SN XUT

3.1 Sn xut nh l mt h thng Russel Ackoff nh tin phong trong l thuyt h thng, m t h thng nh

sau: H thng l mt tng th khng th chia nh c m khng lm cho n mt i nhng nt c trng, v v th n phi c nghin cu nh l mt tng th.

H thng sn xut tip nhn u vo cc hnh thi nh nguyn vt liu, nhn s, tin vn, cc thit b, cc thng tin... Nhng yu t u vo ny c chuyn i hnh thi trong h thng to thnh cc sn phm hoc dch v theo mong mun, m chng ta gi l kt qu sn xut. Mt phn ca kt qu qun l bi h thng qun l nhm xc nh xem n c th c chp nhn hay khng v mt s lng, chi ph v cht lng. Nu kt qu l chp nhn c, th khng c s thay i no c yu cu trong h thng; nu nh kt qu khng chp nhn c, cc hot ng iu chnh v mt qun l cn phi thc hin. M hnh h thng sn xut: (S 1-2)

u vo

Chuyn ha

u ra

Yu t ngoi vi

- Php lut - Chnh tr - Kinh t - VH-XH - K thut

Yu t th trng - S cnh tranh - Thng tin sn

phm - Nghin cu

khch hng

Ngun lc s cp - Vn - Vt t - Nhn s - Tin ch khc

Yu t l tnh - Ch to - Khai m

Dch v nh v - Vn chuyn

Dch v trao i- Bn s - Bn l

Dch v khc - Bo him - Ti chnh - Y t - Gio dc - ..

u ra trc tip- Sn phm - Dch v

u ra gin tip- Thu - Tin lng - S pht trin

cng ngh - S tc ng ca:

+ mi trng + cng nhn + x hi +

B phn kim sot

Quan tri san xuat

8

D BO NHU CU SN PHM

A. Khi nim : 1. Khi nim : D bo nhng vn xy ra trong tng lai da vo nhng s liu hin ti, xu hng .

c im chung ca d bo - Khi tin hnh d bo cn gi thit: h thng cc yu t nh hng n gi tr ca i

lng d bo trong qu kh s tip tc cho nh hng trong tng lai. - Khng c mt d bo no hon ho 100% - D bo da trn din i tng kho st cng rng, cng a dng th cng c nhiu kh

nng cho kt qu chnh xc hn V d: D bo v gi xng du trong thi gian ti - chnh xc ca d bo t l nghch vi khong thi gian d bo. - D bo ngn hn thng chnh xc hn d bo trung v di hn.

2. Cc loi d bo :

a. Cn c vo thi gian : D bo di hn : C thi gian ln hn 3 nm D bo trung hn: C thi gian t 3 thng n 3 nm D bo ngn hn: C thi gian nh hn 3 thng

b. Cn c vo ni dung: D bo kinh t :Thng l d bo chung v tnh hnh pht trin kinh t ca mt ch th

(DN, vng, quc gia, khu vc hay kinh t th gii), Do cc c quan nghin cu, vin, trng H c uy tn thc hin

D bo k thut cng ngh: D bo cp n mc pht trin ca khoa hc cng ngh trong tng lai. Loi d bo ny c bit quan trng vi cc ngnh c hm lng k thut cao nh: nng lng nguyn t, v tr, in t, nhin liu Cu hi: theo bn cng ngh no l cng ngh ca tng lai?

D bo nhu cu : D kin, nh gi nhu cu trong tng lai ca cc sn phm, gip Dn xc nh c chng loi, s lng sn phm cn sn xut v hoch nh ngun lc cn thit p ng

D bo dn s , thi tit ..

B. Cc phng php d bo : - Phng php nh tnh : D bo da trn kin ca ch quan ca cc ch th c kho sat

nh: gii qun l, b phn bn hng, khch hng hoc ca cc chuyn gia - Phng php nh lng : D bo da trn s liu thng k trong qu kh vi s h tr ca

cc m hnh ton hc. 1. CC PHNG PHP D BO NH TNH

a. Ly kin ban Lnh o, ngi i trc : - Ni dung: D bo v nhu cu SP c xy dng da trn kin d bo ca cn b

qun l cc phng, ban chc nng ca DN. - u im: S dng ti a tr tu v kinh nghim ca cn b trc tip hot ng trn

thng trng. - Nhc im: nh hng quan im ca ngi c th lc. Vic gii hn trch nhim

d bo trong mt nhm ngi d lm ny sinh t tng li, tr tr. b. Ly kin nh phn phi, b phn bn hng

- Ni dung: Nhn vin bn hng s a ra d tnh v s lng hng bn trong tng lai lnh vc mnh ph trch. Nh qun l c nhim v thm nh, phn tch, tng hp a ra mt d bo chung chnh thc ca DN.

- u im: Pht huy c u th ca nhn vin bn hng. - Nhc im: Nhn vin bn hng thng hay nhm ln trong xc nh: nhu cu t

nhin (need) nhu cu (requirement) nhu cu c kh nng thanh ton (demand) . Kt qu ph thuc vo nh gi ch quan ca ngi bn hng.

c. Ly kin ngi tiu dng, khch hng - Ni dung: iu tra kin khch hng a ra d bo v nhu cu sn phm. Cch

lm: phiu iu tra, phng vn - u im:Hiu r thm yu cu ca khch hng hon thin sn phm.

Quan tri san xuat

9- Nhc im:Cht lng d bo ph thuc nhiu vo trnh chuyn nghip ca ngi

iu tra;Hiu ng m ng. d. Da vo kin cc chuyn gia trong ngnh (Phng php Delphi) Ni dung D bo c xy dng trn kin ca cc chuyn gia trong hoc ngoi

doanh nghip. Thnh phn tham gia thc hin:

* Nhng ngi ra quyt nh; * Cc chuyn gia xin kin; * Cc nhn vin iu phi.

Cc bc thc hin: 1. Thnh lp ban ra quyt nh v nhm iu phi vin 2. Xc nh mc ch, nhim v, phm vi v thi gian d bo 3. Chn chuyn gia xin kin 4. Xy dng bn cu hi iu tra, gi chuyn gia (ln 1) 5. Nhn, phn tch, tng hp cu tr li 6. Vit li bn cu hi cho ph hp hn, gi chuyn gia (ln 2) 7. Tip tc nhn - tng hp phn tch lm mi -gi 8. Thc hin cc bc 6-7 v ch dng li khi kt qu d bo tho mn yu cu v

mc ch ra. u im:

- Khch quan hn, trnh c mi quan h trc tip gia cc c nhn - c bit hiu qu trong lnh vc d bo cng ngh. (V sao?)

Nhc im: - i hi trnh tng hp rt cao - Ni dung cc cu hi c th c hiu theo nhiu ngha khc nhau -> ni dung tr

li khng tp trung - Thnh phn cc chuyn gia d thay i v thi gian tin hnh thng khng di 1

nm - Vic n danh ngi tr li c th lm gim tin cy v trch nhim ca ngi a

ra kin. Phng php Delphil ln u tin c tp on Rand (M) ng dng nm 1948 khi h mun d bo v kh nng M b tn cng bng v kh ht nhn.

2. CC PHNG PHP D BO NH LNG Da trn cc s liu thng k trong qu kh vi s h tr ca cc m hnh ton hc tin

hnh d bo. Hai m hnh ton thng dng nht thng dng trong d bo l: d bo theo chui thi gian

v hm nhn qu. I. Phng php bnh qun di ng:

1. Bnh qun di ng gin n : D bo nhu cu ca k tip theo da trn kt qu trung bnh ca cc k trc .

n

AF

n

ii

n

=

+ = 11 Trong :

- Fn+1 l nhu cu d bo cho giai on n+1; - Ai l nhu cu thc t ca giai on i; - n S giai on c nhu cu thc t dng quan st

u im: - Chnh xc hn phng php gin n - Ph hp vi nhng dng yu cu u c xu hng n nh.

Nhc im: - Phi lu tr mt s lng d liu kh ln.

Quan tri san xuat

10 2. Bnh qun di ng c trng s : Ni dung: L phng php trung bnh ng c tnh n nh hng ca tng giai on khc

nhau n nhu cu thng qua s dng trng s.

Nhu cu d bo =

=

n

i 1[Trng s thi k n X nhu cu thi k n]

Cc trng s u im: Cho kt qu st vi thc t hn so vi pp tbd gin n v c s dng h s Nhc im

D bo khng bt kp xu hng thay i ca nhu cu; i hi ghi chp s liu chnh xc v ln.

II. Phng php san bng s m:

1. Ni dung: Nhm khc phc nhc im ca phng php trc, pp san bng m cho rng d bo mi bng d bo ca giai on trc cng vi t l chnh lch gia nhu cu thc v d bo ca giai on qua, c iu chnh cho ph hp.

2. Cng thc: Ft = Ft-1 + ( A(t-1) - F(t-1) ) vi F l d bo

A l thc hin l h s san bng

3. La chn h s Ch s th hin nhy cm ca sai s d bo, nn ph thuc nhiu vo loi hnh sn

phm v kinh nghim ca ngi kho st; 0 1 MAD : lch tuyt i bnh qun ( cng nh cng tt)

n

FAMAD

n

iii

=

= 1 n : s giai an kho st Vi mi : MAD min c tnh chnh xc nht

III. Phng php d bo theo ng khuynh hng:

Phng trnh ng khuynh hng : y = ax +b vi x : s th t thi gian ; s giai on kho st

y : s thc t trong qu kh v s d bo trong tng lai H s a,b tnh theo cng thc:

= 22 )( xnx

yxnxya xayb =

IV. D bo theo ng khuynh hng c ch s thi v :

C 2 trng hp : 1. Khng n nh ch tiu :

Bc 1 : D bo theo ng khuynh hng (yi) Bc 2 : Xc nh ch s thi v theo tng thi k (Is) Bc 3 : iii Isyy = ( iy : d bo theo ng khuynh hng c ch s thi v)

2. C n nh ch tiu : Bc 1 : Xc nh d bo bnh qun tng thi k( y )

Quan tri san xuat

11

Bc 2 : Xc nh ch s thi v theo tng thi k (Is) Bc 3 : ii Isyy =

QUN TR TN KHO I. Khi nim , chc nng , chi ph tn kho:

1. Khi nim : Qun tr tn kho l qun tr qu trnh bo m mc tn kho ti u v ngun lc p ng kp thi nhu cu sn xut, tha mn yu cu ca khch hng v gim ti a chi ph tn kho cho DN.

2. Chc nng ca qun tr tn kho p ng y , chnh xc cc yu cu sn xut v nguyn vt liu. Bo m ngun tn kho qu trnh sn xut din ra lin tc, hiu qu thng qua vic to ngun

tn kho ti u (bufer). Ngn nga kh nng cn kit ngun lc SX v cc l do bt kh khng. Ngn nga nhng bin ng bt thng ln gi thnh sn phm (tch tr, phng trt gi). Gim ti a chi ph sn xut thng qua vic ti u ha chi ph tn kho.

3. Chi ph tn kho : 4 nhm chi ph c bn a. Chi ph mua hng ( Cmh ) :L chi ph mua mt lng hng mi.Tuy nhin chi ph ny

khng lin quan nhiu n cc m hnh tn kho. Cmh = S lng x n gi

b. Chi ph t hng ( Cdh )(ordering cost) :L chi ph thc hin n hng, l (s tin thanh ton cho t hng trong 1 nm)

QDSCdh =

Trong : S : Chi ph cho 1 ln t hng D : Nhu cu vt t trong 1 nm Q : S lng cho 1 ln t hng

Chi ph lp, gi, nhn n t hng; Chi ph nhn hng: vn chuyn, bc d; Chi ph giao nhn, kim tra cht lng hng ha; Chi ph thanh quyt ton l hng; Nhng chi ph ny thng c tnh chung theo tng l hng. T l thun vi s ln t v nhn hng, t l nghch vi s lng SP trong mt n hng.

c. Chi ph tn tr ( Ctt ) : L chi ph lin quan n vic gi v bo qun hng ha trong kho trong mt khong thi gian xc nh.

2QHCtt = H : Chi ph tn tr chi 1 n v sn phm trong 1 nm

Chi ph thu kho, bi; Chi ph dch v lu kho, CP bo qun hng ha; Chi ph pht sinh trong qu trnh bo qun; Chi ph lin quan n hng ha: bo him, thu, khu hao; Chi ph c hi do vn ng trong hng tn kho. Chi ph ny t l thun vi s lng hng ha tn kho.

Tng chi ph tn kho TC = Cdh + Ctt 2 TC QH

QDS +=

Quan tri san xuat

12 Vn : gim chi ph tn tr th nn t hng nhiu ln vi s lng t, nhng lm nh th li lm tng chi ph t hng. d. Chi ph tn kho

Chi ph pht sinh do khng ngun hng tn kho L chi ph xut hin trong trng hp cu vt cung (mt khch hng v khng p ng

kp, nhu cu). Chi ph loi ny kh nh gi v mang tnh ch quan.

4. H thng qun tr tn kho : a. Phi tr li hai cu hi chnh t hng khi no? S lng bao nhiu?

b. C hai h thng qun tr tn kho c bn: Ti to tn kho nh k theo thi gian, vi s lng khc nhau m hnh P; Ti to tn kho theo s lng khng ph thuc vo thi gian m hnh Q.

5. Hiu qu hot ng ca h thng qun tr tn kho : qun tr tn kho hiu qu DN cn

quan tm hn: D bo nhu cu; Kim sot thi gian thc hin n hng; Kim sot, ti u ha chi ph tn kho, ch trng chi ph t hng v chi ph lu kho. i vi DNNVV p dng hnh thc kim tra nh k, ti to tn kho theo thi gian; p dng hnh thc qun tr tn kho n gian: thng hai ngn. S dng m s, m vch qun tr tn kho. Tm hiu thc t qun tr tn kho DN.

II. Cc m hnh tn kho : 1. M hnh lng t hng kinh t ti u ( Economic Order Quality model EOQ )L m

hnh ti to tn kho theo s lng cho php xc nh s lng tn kho ti u vi chi ph thp nht c th m vn m bo DN hot ng hiu qu.

a. Gi thit ca m hnh: Nhu cu bit trc v khng i; Nhu cu phn b u trong; Thi gian thc hin n hng bit trc v khng i; n hng ca cc ln t hng u nh nhau; Ch tnh hai loi chi ph c bn: CPt hng v chi ph tn tr; Tnh ton ch vi 1 loi hng ha.

b. M hnh tng qut :

t1 t2 t3

Q1 Q2 Q3

t1=t2=t3; Q1Q2Q3

M hnh P

t1 t2 t3

Q

Q1 Q2 Q3

Q1=Q2=Q3; t1t2t3

M hnh Q

Q0

Quan tri san xuat

13

Tm gi tr Q* ti u cho 1 ln t hng chi ph tn kho l b nht?

Tc tm Q* CDT = Cdh +Ctt -> min Ta c: TC = Cdh + Ctt

TC = Sx D/Q + HxQ/2 TC min khi Cdh = Ctt hay HxQ/2 = SxD/Q TC

HSDQ 2* =

Khi : *

*

*

*

*min 222

2HQQH

QDSQH

QDSTC ===+=

S ln t hng trong nm :

Q Qmax

im t hng

Qmin t0 t1 t2

Tc xut hng

Thi im nhn hng

Q

TC

Q*

Ctt

Cdh

Quan tri san xuat

14

*QDN = (ln)

N lun lun lm trn s ln

Khong cch gia 2 ln t hng:

T= S ngy lm vic trong nm N Thi im t hng (ROP -reorder point):

ROP =d x L

Trong : d nhu cu trong mt n v thi gian ( vd: 1 ngy ..)

d= D S ngy lm vic trong nm L thi gian thc hin n hng d tr (t lc t hng n lc nhn c hng)

V d : Doanh nghip A trong nm ti s bn c khong 9600 sp. Chi ph lu kho cho 1vsp loi ny/1nm l $16, chi ph mt ln t hng d tnh l $75. DN lm vic 288ngy /nm.Thi gian t hng l 5 ngy

1. Tnh s lng t hng ti u Q*. 2. DN cn t hng bao nhiu ln trong 1 nm? 3. Khong thi gian gia 2 ln t hng l bao nhiu? 4. Thi im t hng ?

Gii

D = 9600sp 1) S lng t hng ti u:

Mc nhu cu cao nht c th

Mc nhu cu d t h

ROP

Thi gian

Mc d tr d phng (an ton)

t hng Nhn hng

Quan tri san xuat

15H = $16 S = $75 L = 10 ngy S ngy lm vic trong nm : 288

Q* HDS2

=16

7596002 = 300 sp

2) S ln t hng :

*QDN = =

3009600

= 32 ln

3) Chu k t hng :

T =N

288 =

32288

= 9 ngy

4) Thi im t hng :

ROP = d x L = LD 288

= 10288

9600 = 334 sp

2. M hnh cung ng theo nhu cu sn xut (Production Order Quality model POQ ) M hnh lng t hng theo sn xut c p dng trong trng hp lng hng c a

n mt cch lin tc, hng c tch lu dn cho n khi lng t hng c tp kt ht. M hnh ny cng c p dng trong trng hp doanh nghip va sn xut va bn hoc doanh nghip t sn xut ly vt t dng. Trong nhng trng hp ny cn phi quan tm n mc sn xut hng ngy ca nh sn xut hoc mc cung ng ca nh cung ng.

Trong m hnh POQ, cc tc gi thit k v c bn ging nh m hnh EOQ, im khc bit duy nht l hng c a n nhiu chuyn.. Bng phng php ging nh EOQ c th tnh c lng t hng ti u Q*. Nu ta gi:

p Mc sn xut (Mc cung ng hng ngy) d Nhu cu s dng hng ngy

=pdH

SDQ1

2*

Khi :

+=

pdQH

QDSTC 1

2

*

*min

V d : Bi tp 11 :

Mt doanh nghip sX hng may mc c nhu cu c nm 2000 tn vi. Chi ph t hng cho mi n hng l 100.000 . Chi ph tn tr hng l 10.000 /tn/nm .

Hy xc nh : 1. Theo m hnh POQ, tnh sn lng t hng ti u 2. Tng chi ph tn kho ti thiu 3. S ln t hng ti u trong nm 4. S ngy cch qung gia 2 ln cung ng.

Bit rng mc sn xut bnh qun 1ngy m l 10 tn v DN hot ng 250 ngy/nm Gii

D = 2000 tn 1) S lng t hng ti u:

Quan tri san xuat

16H = 10.000 /tn/nm S = 100.000 / n hng P = 10 tn/ ngy S ngy lm vic trong nm : 250

d = 82502000

250==D tn/

ngy

Q* = )1(

2

pdH

DS

=

)1081(000.10

000.10020002

= 447,21 tn

2) Tng chi ph tn kho ti thiu :

+=

pdQH

QDSTC 1

2

*

*min=

+

1081

221,447000.10

21,4472000000.100

= 894,428 3) S ln t hng :

*QDN = =

21,4472000

= 5 ln

4) S ngy cch qung gia 2 ln cung ng :

T =5

250 = 50 ngy

3. M hnh lng t hng li (Back Order Quality model BOQ ) Trong hai m hnh d tr trn, chng ta khng chp nhn c d tr thiu ht trong ton b

qu trnh d tr. Trong thc t c nhiu trng hp, trong doanh nghip c nh trc v s thiu ht v nu duy tr thm mt n v d tr th chi ph thit hi cn ln hn gi tr thu c. Cch tt nht trong trng hp ny l doanh nghip khng nn d tr thm hng theo quan im hiu qu.

M hnh BOQ c xy dng trn c s gi nh rng doanh nghip ch nh d tr thiu ht v xc nh c chi ph thiu ht do vic li mt n v d tr ti ni cung ng hng nm. Ngoi ra, chng ta cn gi nh rng doanh thu khng b suy gim v s d tr thiu ht ny. Nh vy, m hnh ny ging vi cc m hnh trc y, duy ch thm mt yu t b sung l chi ph cho mt n v hng li ni cung ng hng nm. Nu gi:

B Chi ph tn tr cho mt n v sn phm i vi hng d tr ( li ni cung ng) hng nm; *1Q _ Lng t hng s dng *2Q _ Lng t hng d tr

Ta c :

*2

*1

* QQQ +=

BBH

HSDQ += 2*

BHBQQ +=

**1

Thng thng : *1Q > *2Q v B > H

V d : Bi tp 15 Mt DN kinh doanh go nhu cu c nm l 1000 tn . Chi ph cho mi n t hng l 100.000 . Chi ph tn tr cho mi tn hng trong nm l 5000 . Chi ph cho 1 tn hng li ni cung ng l 50.000 . Theo m hnh BOQ , lng t hng kinh t l bao nhiu ? Sn lng li ni cung ng l bao nhiu ?

Gii

Quan tri san xuat

17D = 1000 tn H = 5.000 /tn/nm B = 50.000 /tn/nm S = 100.000 / n hng

1) Sn lng n hng ti u:

Q* = B

BHHDS +2 =

000.50000.55

000.5000.10010002

= 209,76 tn

2) Sn lng li ni cung ng *2Q

*1

**2

*2

*1

* QQQQQQ =+= M BH

BQQ +=**

1

69,190000.55000.5076,209*1 ==Q

*2Q = 209,76 190,69 = 19,07 tn li sau mi chu k cung ng

4. M hnh khu tr theo s lng

khuyn khch tiu dng nhiu DN p dng chnh sch gim gi theo s lng mua hng.Nhim v ca ngi mua l phi xc nh c s lng t hng ti u va tha hng li ch do gim gi m khng lm tng tng gi tr chi ph d tr.

Tng chi ph d tr trong trng hp ny c tnh nh sau: TC = Cmh + Ch + Ctt Cn xc nh Q0 CDT = min? ng dng m hnh EOQ gii :

Bc 1 : Tnh i

i IPSDQ 2* =

Vi I l t l chi ph tn tr 1 vsp/ n gi 1vsp Pi l n gi chit khu th I tc Hi = Pi x I

Bc 2: iu chnh *iQ Nu *iQ nm trong mc khu tr Gi nguyn Nu *iQ nm cao hn mc khu tr Loi b Nu *iQ nm di mc khu tr iu chnh ln bng mc thp nht

ca mc khu tr tng ng.

Bc 3 : Tnh Tci iiii

i DPPIQ

QDSTC ++=

2

*

*

Chn minTC Kt lun : lng t hng ti u mc chi ph tng ng minTC V d : bi tp 20 (n v : ngn ng )

D = 1.000 Tn S = 100/n I = 10 %/ nm

Mc khu tr T l khu tr % n gi (/tn)001 150 0 50151 200 10 45201 250 15 42251 300 20 40

> = 301 30 35Bc 1 :

*1Q = 501,0000.110022

1 = PI

SD = 200 Tn *2Q = 451,0

000.110022

2 = PI

SD = 211 Tn

*3Q = 421,0

000.110022

3 = PI

SD = 217 Tn *4Q = 401,0

000.110022

4 = PI

SD = 224 Tn

Quan tri san xuat

18

*5Q = 351,0

000.110022

5 = PI

SD = 239 Tn

Bc 2: iu chnh Q* *1Q = 200 tn ln hn mc khu tr(150)

*1Q => b

*2Q = 211 tn ln hn mc khu tr (200)

*2Q => b

*3Q = 217 tn nm trong mc khu tr ( 201 250) *3Q = 217 tn *4Q = 224 tn nm di mc khu tr (251 300) iu chnh *4Q = 251 tn *5Q = 239 tn nm di mc khu tr (301 tr ln) iu chnh *5Q = 301 tn

Bc 3 :Tng chi ph vi mi *iQ c chn iiii

i DPPIQ

QDSTC ++=

2

*

*

Mc khu tr

Gi v P Q*

CP mua hng Cmh

Chi ph t hng Cdh

Chi ph tn tr Ctt

Tng chi ph TC

201 - 250 42.5 217 42500 461 461 43422251 - 300 40 251 40000 398 502 40900

> = 301 35 301 35000 332 527 35859Vy chn Q* = 301 tn

HOCH NH CNG SUT Khi qut chung

1. Cng sut l g? Kh nng sn xut ti a ca mt i tng sn xut. i vi DN l khi lng sn phm m DN c th sn xut c trong mt n v thi gian.

2. Phn loi cng sut Cng sut thit k : Cng sut ti a theo thit k Cng sut hiu qa Cng sut ti a trong iu kin lm vic c th. Cng sut thc t : Cng sut thc t t c.

3. nh gi cng sut: Mc hiu qu = Cng sut thc t/Cng sut hiu qu Mc s dng = Cng sut thc t/cng sut thit k

4. Cc yu t nh hng n cng sut Yu t bn ngoi : Th trng, chnh sch, php lut, tiu chun, mi trng Yu t bn trong : Con ngi - Cng ngh - Sn phm - Nng lc sn xut v trnh qun l. Cc quyt nh v hoch nh nng lc sn xut c th c phn tch bng nhiu phng php

khc nhau nh: phn tch im ha vn, thng c s dng so snh hm s chi ph ca 2 hay nhiu phng tin sn xut khc nhau, s hnh cy (cy quyt nh) cng rt hu hiu trong phn tch cc quyt nh v hoch nh nng lc sn xut. I. L thuyt cy qun tr :

Bc 1: V cy qun tr Bc 2: Tnh gi tr mong i EMV

EMVi = (Tin t thanh ton)i X (Xc sut)i Bc 3: Chn phng n c EMV Max V d: Bi 1 ( n v : ngn USD)

Phng n cng sut E1 : Th trng tt E2 : Th trng xuS1: XDNM ln 25.000 T/nm 100 - 90S2: XDNM va 10.000 T/nm 60 - 10S3: XDNM nh 5.000 T/nm 40 - 5

Quan tri san xuat

19S4: Khng lm g 0 0Xc xut 0,4 0,6

V cy quyt nh : E1 (0,4) 100

E2 (0,6) - 90 E1 (0,4) 60 E2 (0,6) -- - 10 E1 (0,4) 40 E2 (0,6) - 5

Tnh gi tr tin t mong i EMV : EMV1 = (100 x 0,4) + (-90 x 0,6) = - 14 EMV2 = (60 x 0,4) + (-10 x 0,6) = 18 EMV3 = (40 x 0,4) + (- 5 x 0,6) = 13 Max EMV = EMV2 = 18

Cng ty nn chn phng n S2 Xy dng nh my c quy m cng sut va (10.000 T/nm) . Nh vy th li nhun mong i trong 1 nm ca cng ty l 18.000 USD

II. Phn tch im ho vn : Cng sut ti thiu l cng sut tng ng vi im ho vn .( Khng th chn cng sut thp hn cng sut ho vn v nh vy s l. Gi P : Gi bn 1 n v sn phm TR Tng doanh thu x Lng sn phm sn xut F Tng bin ph VC Tng nh ph V Bin ph1 tnh cho 1 n v SP Ti im ho vn th tng doanh thu bng tng chi ph (TR = FC + VC)

Do ta c : Px = F + Vx hay Sn lng ho vn BEP(x) = VP

F

im ho vn bng vtt BEP() =

PV

F

1

(Px) Chi ph (F + Vx) Vng li BEP

(V) (F) Vng l

Quan tri san xuat

20

0 Cng sut V d : Bi 6 : (n v : ngn ng)

F = 100.000 V = 15 + 7,5 =22,5 P = 40

1) im ho vn bng ng :

BEP() =

PV

F

1 =

405,221

000.100

= 228571,4

2) im ho vn bng sn lng

BEP(x) = VP

F = 5714 chic

HOCH NH LCH TRNH SN XUT Trong qu trnh sn xut hoc thc hin cc dch v, chng ta cn tin hnh nhiu cng vic khc

nhau. iu ny i hi s iu hnh, sp xp sao cho khoa hc, hp l, cht ch vo nhng lc cao im v ngay c nhng lc rnh ri.

iu sn xut nhm mc tiu m bo cc cng vic c thc hin vi hiu qu cao nht, c th l thi gian thc hin nhanh nht, t tn km nht, mang li li nhun cao nht ng thi gi c mc phc v khch hng tt nht.

Chng ta s kho st di y cc ni dung lin quan n vic sp xp th t ti u trong sn xut, dch v v phng php phn cng cng vic i vi h thng sn xut theo qu trnh.H thng sn xut theo qu trnh l mt hnh thc t chc theo chc nng vi cc b phn sn xut hoc trung tm sn xut trn c s cc loi thit b hoc tc nghip chuyn bit.

V d: Khoan, rn, tin hay lp rp. Dng sn phm qua cc b phn theo l ph thuc vo cc n hng ring l (c th cc n hng lu kho hay cc n hng do khch hng t). Vic hoch nh v kim sot hot ng ca h thng ny bao gm cc cng vic sau:

Xc nh th t u tin cho tng n hng v o lng tm quan trng ca n nhm sp xp th t cc n hng cn sn xut tng my, tng b phn sn xut.

Lp danh sch cc cng vic cn gii quyt tng my, tng b phn sn xut, gip cho cc b phn gim st bit c n hng c thc hin u, khi no, u tin ra sao v lc no cn hon thnh.

Kim sot u vo, ra tt c cc b phn sn xut, iu ny c ngha l pht trin thng tin v cch thc cng vic lu chuyn gia cc b phn sn xut.

o lng hiu qu, mc s dng my mc tng b phn sn xut v sc sn xut ca cc cng nhn.

1. Trng hp ch c mt my hoc mt dy chuyn sn xut:

Ngay sau khi my mc hoc dy chuyn sn xut c chun b xong sn sng vn hnh th vn t ra l nn lm cng vic no trc, cng vic no sau?

Quan tri san xuat

21C nhiu nguyn tc sp xp th t cng vic:

- Cng vic t hng trc lm trc FCFS : Bo m tnh cng bng khch hng , tuy nhin cha u tin khch hng ln, khch hng chin lc , thn thch

- Cng vic c thi hn giao hng trc lm trc EDD : C u im l mc tr trung bnh tnh cho mi cng vic thp nht, khch hng tng i chp nhn. Thng c s dng.

- Cng vic c thi gian thc hin ngn nht lm trc SPT: C u im thi gian ch i t hn, khch hng t phin h tuy nhin nhc im l cha cng bng v khng tp trung vo khch hng ln

- Cng vic c thi gian di nht lm trc LPT: t c hiu qu v thi gian hon tt trung bnh thng ln, thi gian tr trung bnh cho mi cng vic rt ln . C u im gi chn khch hng ln.

Trong kinh doanh nn chn phng php LPT v khch hng ln s l mi lm n lu di mang li nhun cho doanh nghip

Trong sn xut nn chn phng php EDD hay SPT - Mt s nguyn tc khc: Khch hng quan trng nht; cng vic c li nhun cao nht.

i n quyt nh l nguyn tc no thch hp cho mt nhm cc cng vic ch thc hin, ngi ta thng s dng cc ch tiu sau y:

Bi 1: Cng ty X c cc cng vic c cc thng s sn xut, kinh doanh nh sau

Cng vic Thi gian SX ( ngy ) Thi im phi hon

thnh yu cu ( ngy th ) A B C D E

62839

8 6

18 15 23

Cng 28 Hy tnh :

Thi gian hon tt trung bnh, s cng vic trung bnh v s ngy chm tr trung bnh theo nguyn tc FCFS,EDD, SPT, LPT

Bi gii Theo nguyn tc 1 FCFS :

Cng vic

Thi gian SX ( ngy )

Thi im phi hon thnh yu cu ( ngy th )

Thi gian hon thnh k c ngy ch i (ngy )

Thi gian chm tr so vi yu cu ( ngy )

A B C D E

6 2 8 3 9

86

181523

68

161928

02045

Cng 28 77 11 Thi gian hon tt trung bnh mt cng vic ( ttb )

= Tng s dng thi gian

=77

= 15,4 ngy S cng vic 5 S cng vic trung bnh nm trong h thng (Ntb )


=77

= 2,75 Tng thi gian SX 28

S ngy tr hn trung bnh (THtb )

= Tng s ngy tr hn

=11

= 2,2 ngy S cng vic 5

Theo nguyn tc 2 EDD : Cng vic

Thi gian SX ( ngy )




B A D C

2 6 3 8

68

1518

2 8

11 19

0001

Quan tri san xuat

22E 9 23 28 5



=68



=68



= Tng s ngy tr hn

=6


Theo nguyn tc 3 SPT : Cng vic

Thi gian SX ( ngy )




B D A C E

2 3 6 8 9

6158

1823

2 5

11 19 28

00315



=65

= 13 ngy S cng vic 5 S cng vic trung bnh nm trong h thng (Ntb )


=65



= Tng s ngy tr hn

=9


Theo nguyn tc 4 LPT : Cng vic

Thi gian SX ( ngy )




E C D A B

9 8 6 3 2

23188

156

9 17 23 26 28

00

151122

Cng 28 103 48

Thi gian hon tt trung bnh mt cng vic ( ttb )


=103


S cng vic trung bnh nm trong h thng (Ntb )


=103



= Tng s ngy tr hn

=48


Bng so snh : Cc

nguyn tc

Thi gian hon tt trung bnh mt cng

vic ( ttb )

S cng vic trung bnh nm trong h

thng (Ntb ) S ngy tr hn

trung bnh (THtb )

1. FCFS 15,4 2,2 2,752. EDD 13,6 1,2 2,423. SPT 13,0 1,8 2,32

Quan tri san xuat

234. LPT 20,6 9,6 3,68

Nguyn tc th 2 c s cng vic bnh qun trn dy chuyn l nh nht trong khi nguyn tc th 3 c s ngy trung bnh tr hn nh nht v thi gian hon thnh cng vic nh nht. Tu theo thc t tng t chc kinh doanh hay sn xut , quan h vi khch hng . M nh qun l chn nguyn tc thch hp.

Bi 2: C 5 hp ng c lm trn 1 my c cc thng s sn xut, kinh doanh nh sau

Cng vic Thi gian thc hin ( ngy ) Thi gian giao hng

( ngy th ) A B C D E

924142218

26 43 20 34 30

Cng 87 Hy tnh :

Thi gian hon tt trung bnh, s cng vic trung bnh v s ngy chm tr trung bnh theo nguyn tc EDD, SPT

Bi gii

Theo nguyn tc EDD : Cng vic

Thi gian SX ( ngy )




C A E D B

14 9

18 22 24

2026303443

14 23 41 63 87

00

112944



=228



=228



= Tng s ngy tr hn

=84


Theo nguyn tc SPT : Cng vic

Thi gian SX ( ngy )




A C E D B

9 14 18 22 24

2620303443

9 23 41 63 87

03

113144



=223



=223



= Tng s ngy tr hn

=89


Quan tri san xuat

24

Bng so snh : Cc

nguyn tc

Thi gian hon tt trung bnh mt cng

vic ( ttb )

S cng vic trung bnh nm trong h

thng (Ntb ) S ngy tr hn

trung bnh (THtb )

1. EDD 45,6 2,62 16,82. SPT 44,6 2,56 17,8

Bi 3: C cng vic sau y c tun t c cc s liu v yu cu sau : Cng vic Ngy cn hon thnh Thi gian gia cng ( ngy )

A B C D E

313312325314314

8 16 40 5 3

Cng Xp th t gia cng cc cng vic ny nh th no tun t theo cc nguyn tc a. FCFS b. EDD c. SPT d. LPT S th t c nh s theo lch cng tc t u nm. Bit rng 5 cng vic n tun t trong ngy th 275.

Bi gii

Theo phng php FCFS Cng vic

Ngy cn hon thnh

S ngy phi hon thnh (ngy)

Thi gian gia cng ( ngy )

A B C D E

313312325314314

3837503939

8164053

Theo phng php EDD

Cng vic

Ngy cn hon thnh



B A E D C

312313314314325

3738393950

16835

40 Theo phng php SPT

Cng vic

Ngy cn hon thnh



E D A B C

314314313312325

3939383750

358

1640

Theo phng php LPT Cng vic

Ngy cn hon thnh



C B A D E

325312313314314

5037383939

4016853

2. iu n cng vic trn 2 my:

Quan tri san xuat

25Mc tiu b tr cc cng vic sao cho tng thi gian thc hin cc cng vic trn cc my l nh

nht. Song trong thc t, thi gian thc hin trn mi my l c nh, do c thi gian thc hin nh nht ta phi sp xp cc cng vic sao cho tng thi gian ngng vic trn cc my l nh nht.

p dng nguyn tc Johnson gm cc bc sau: Bc 1 : Lit k tt c cc cng vic theo thi gian tng dn. Bc 2,3 : B tr cc cng vic theo nguyn tc Jonhson:

- Cng vic no c thi gian nh nht nm trn my 1 th xp bn tri ( u) - Cng vic no c thi gian nh nht nm trn my 2 th xp bn phi ( cui)

Bc 4 : Cui cng ta v biu thy tng thi gian hon thnh cc cng vic. Bi tp 4: C 5 cng vic phi thc hin ln lt trn my khoan v my tin c thi gian sau :

Cng vic Thi gian thc hin cc cng vic1. My khoan 2. My tinA B C D E

538

107

2647

12Hy xp th t cc cng vic c tng thi gian thc hin l nh nht.

Bi gii Bc 1 : Lit k tt c cc cng vic theo thi gian tng dn ( bi sp xp ) Bc 2,3 : B tr cc cng vic theo nguyn tc Jonhson

B E D C A My 1 3 7 10 8 5 My 2 6 12 7 4 2

Bc 4 : V biu 0 3 10 20 28 33

B=3

E=7 D=10 C=8 A=5

B=6 E=12 D=7 C=4 A=2

0 3 9 10 22 29 33 35 Thi gian lm vic t nht l 35 gi

Bi tp 5: Mi ngy bnh vin An Bnh cn gic 5 loi khn khc nhau . Bnh vin ch c 1 my git v 1 my sy .Thi gian git v sy trn 2 my theo bng sau :

Loi khn Thi gian thc hin cc cng vic1. git ( pht ) 2. Sy ( pht )A B C D E

3050901020

4020702030

1. Hy xp th t sao cho cc cng vic xong sm nht. 2. Nu hng ngy bt u git lc 9 gi sng th khi no git,sy xong ? 3. Thi gian bnh qun cho mi loi khn l bao nhiu ?

Bi gii Bc 1 : Lit k tt c cc cng vic theo thi gian tng dn

Loi khn Thi gian thc hin cc cng vic1. git ( pht ) 2. Sy ( pht )D B E A C

1050203090

2020304070

Bc 2,3 : B tr cc cng vic theo nguyn tc Jonhson

D E A C B

Quan tri san xuat

26Git 10 20 30 90 50 Sy 20 30 40 70 20

Bc 4 : V biu

0 10 30 60 150 200

D = 10

E = 20 A = 30 C = 90 B = 50

D = 20 E = 30 A = 40 C = 70 B = 20

0 10 30 60 100 150 220 240 1. Thi gian lm vic t nht l 240 pht = 04 gi 2. Nu bt u t 9 gi sng th n 13 gi th git sy xong 3. Dng thi gian trung bnh cho mi loi khn l : 240 pht / 5 = 48 pht

Bi tp 6: C 6 cng vic phi lm tun t trn 2 thit b . Thit b th nht l phun ct , thit b th hai l sn. Hy lp th t gia cng v v s iu cc s liu cho nh sau

Cng vic Thi gian thc hin cc cng vic1. Phun ct (gi) 2. Sn ( gi )A B C D E F

1075324

547863

Bi gii Bc 1 : Lit k tt c cc cng vic theo thi gian tng dn

Cng vic Thi gian thc hin cc cng vic1. Phun ct (gi) 2. Sn ( gi )F B A E C D

47

10253

345678

Bc 2,3 : B tr cc cng vic theo nguyn tc Jonhson

E D C A B FPhun 2 3 5 10 7 4 Sn 6 8 7 5 4 3

Bc 4 : V biu

0 2 5 10 20 27 31

E = 2

D = 3 C = 5 A = 10 B = 7 F = 4

E = 6 D = 8 C = 7 A = 5 B = 4 F=3

0 2 8 16 23 28 32 35

Thi gian lm vic t nht l 35 gi

Bi tp 7: C 6 cng vic phi thc hin ln lt trn 2 thit b I v II vi s gi gia cng nh sau :

Cng vic Thi gian thc hin cc cng vicThit b 1 Thit b 2A 10 6

Quan tri san xuat

27B C D E F

67836

127498

Hy xp sao cho thi gian gia cng thc hin l ngn nht.

Bi gii

Bc 1 : Lit k tt c cc cng vic theo thi gian tng dn

Cng vic Thi gian thc hin cc cng vicThit b 1 Thit b 2D A C F E B

8107636

46789

12 Bc 2,3 : B tr cc cng vic theo nguyn tc Jonhson

E F B C A D TB 1 3 6 6 7 10 8 TB 2 9 8 12 7 6 4

Bc 4 : V biu

0 3 9 15 22 32 40 49

E= 3

F = 6 B = 6 C = 7 A= 10 D = 8

E = 9 F = 8 B = 12 C = 7 A = 6 D = 4

0 3 12 20 32 39 45 49 Thi gian lm vic t nht l 49 gi

Bi tp 8: C 5 cng vic phi thc hin tun t trn 2 my iu khin s gi gia cng nh sau :

Cng vic S gi gia cngMy 1 My 2A B C D E

2,53,82,25,84,5

4,21,53,04,02,0

Hy xp sao cho thi gian gia cng thc hin l ngn nht. Bi gii

Bc 1 : Lit k tt c cc cng vic theo thi gian tng dn v i n v tnh bng pht

Cng vic Thi gian thc hin cc cng vicThit b 1 Thit b 2B E C D A

228270132348150

90120180240252

Bc 2,3 : B tr cc cng vic theo nguyn tc Jonhson C A D E B

My 1 132 150 348 270 228 My 2 180 252 240 120 90

Bc 4 : V biu

0 132 282 630 900 1128

Quan tri san xuat

28

C = 132

A = 150 D = 348 E = 270 B = 228

C = 180 A = 252 D = 240 E = 120

B = 90

0 132 312 564 630 870 900 1020 1128 1218 Thi gian lm vic t nht l 1218 pht = 20 gi 18 pht gi

Bi tp 9: Cc cng vic phi thc hin tun t trn 2 my vi thi gian hao ph c cho trong bng di y :

Cng vic Thi gian hao phMy 1 My 2A B C D E

77258

86194

Hy xp sao cho thi gian gia cng t ngn nht v l bao nhiu. Bi gii

Bc 1 : Lit k tt c cc cng vic theo thi gian tng dn Bc 2,3 : B tr cc cng vic theo nguyn tc Jonhson

D A B E C My 1 5 7 7 8 2 My 2 9 8 6 4 1

Bc 4 : V biu 0 5 12 19 27 29

D = 5

A = 7 B = 7 E = 8 C = 2

D = 9 A = 8 B = 6 E = 4 C=1

0 5 14 22 28 32 33 Thi gian gia cng t ngn nht 33 gi

3. iu n cng vic trn 3 my: Bi 10 : C 4 cng vic c thc hin trn 3 my vi thi gian thc hin cho di y. Hy xp th t cng vic sao cho tng thi gian thc hin l nh nht.

My t1 My t2 My t3 A 13 5 9B 5 3 7C 6 4 5D 7 2 6

Bc 1: Xt bi ton tha nguyn tc Jonhson khng ?

Thi gian ngn nht trn my 1 phi ln hn hay bng thi gian di nht trn my 2 Thi gian ngn nht trn my 3 phi ln hn hay bng thi gian di nht trn my 2 Nu tho mn mt trong hai iu kin trn, ta tin hnh gii bi ton theo trnh t : Ta c : t1 min = 5 >= t2 max = 5;

t3 min = 5 >= t2 max = 5

Bc 2: Lp ma trn mi vi d liu l t1+t2 v t2+t3 My t1+t2 My t2+t3

A 18 14B 8 10C 10 9D 9 8

Quan tri san xuat

29 Bc 3: Sp xp cng vic theo th t thi gian min tng dn

My t1+t2 My t2+t3 D 9 8B 8 10C 10 9A 18 14

Bc 4: Sp xp th t thc hin theo nguyn tc Johnson : BACD

B A C D

t1 5 13 6 7 t2 3 5 4 2 t3 7 9 5 6

Bc 5: V biu v tnh tng thi gian thc hin

5 18 24 31 t1 B = 5 A = 13 C = 6 D = 7 t2 B=3 A = 5 C = 4 D = 7 t3 B = 7 A = 9 C = 5 D = 6

8 15 23 32 37 43 Tng thi gian hon tt cng vic l 43 gi. Kt qu ny ch l gn ng, nhng c dng tt trong thc t.

Bi 11 : C 5 cng vic c thc hin trn 3 my vi thi gian thc hin cho di y. Hy xp th t cng vic sao cho tng thi gian thc hin l nh nht.

My t1 My t2 My t3 A 7 5 8B 7 4 8C 8 2 14D 12 6 11E 11 5 10

Bc 1: Xt bi ton tha nguyn tc Jonhson khng ?

Thi gian ngn nht trn my 1 phi ln hn hay bng thi gian di nht trn my 2 Thi gian ngn nht trn my 3 phi ln hn hay bng thi gian di nht trn my 2 Nu tho mn mt trong hai iu kin trn, ta tin hnh gii bi ton theo trnh t : Ta c : t1 min = 7 >= t2 max = 6;

t3 min = 8 >= t2 max = 6 Bc 2: Lp ma trn mi vi d liu l t1+t2 v t2+t3

My t1+t2 My t2+t3 A 12 13B 11 12C 10 16D 18 17E 16 15

Bc 3: Sp xp cng vic theo th t thi gian min tng dn

My t1+t2 My t2+t3 B 11 12A 12 13C 10 16E 16 15D 18 17

Bc 4: Sp xp th t thc hin theo nguyn tc Johnson : CBADE

Quan tri san xuat

30

C B A D E t1 8 7 7 12 11 t2 2 4 5 6 5 t3 14 8 8 11 10

Bc 5: V biu v tnh tng thi gian thc hin 8 15 22 34 45

My 1 C = 8 B = 7 A = 7 D = 12 E = 11 My 2 C=2 B = 4 A = 5 D = 6 E = 5 My 3 C = 14 B = 8 A = 8 D = 11 E = 10 24 32 40 51 61 Tng thi gian hon tt cng vic l 61 gi. Kt qu ny ch l gn ng, nhng c dng tt trong thc t. Bi 12 : C 5 cng vic c thc hin trn 3 my vi thi gian thc hin cho di y. Hy xp th t cng vic sao cho tng thi gian thc hin l nh nht.

My t1 My t2 My t3 A 22 8 10B 18 6 5C 16 3 3D 20 12 17E 15 14 12

Ta c : t1 min = 15 >= t2 max = 14;

My t1+t2 My t2+t3 A 30 18B 24 11C 19 6D 32 29E 29 26

D E A B C t1 20 15 22 18 16 t2 12 14 8 6 3 t3 17 12 10 5 3

0 20 35 57 75 91 My 1 D=10 E = 15 A = 22 B = 18 C = 16 20 32 35 49 57 65 75 81 91 94 My 2 D=12 E = 14 A = 8 B = 6 C= 3 32 49 61 65 75 81 86 94 97 My3 D =17 E = 12 A = 10 B =5 C = 3 Tng thi gian hon tt cng vic l 97 gi. Bi 13 : C 5 cng vic c thc hin trn 3 my vi thi gian thc hin cho di y. Hy xp th t cng vic sao cho tng thi gian thc hin l nh nht.

My t1 My t2 My t3 A 4 1 6B 5 3 8C 5 1 8D 6 4 7E 7 1 6

Ta c : t1 min = 4 >= t2 max = 4;

Quan tri san xuat

31

My t1+t2 My t2+t3 A 5 7B 8 11C 6 9D 10 11E 8 7

A C B D E

t1 4 5 5 6 7 t2 1 1 3 4 1 t3 6 8 8 7 6

0 4 9 14 20 27

My 1 A=4 C = 5 B = 5 D = 6 E = 7 4 5 9 10 14 17 20 24 27 28

My 2 A = 1 C= 1 B = 3 D = 4 E= 1 5 11 19 27 34 40

My3 A = 6 C =8 B = 8 D = 7 E = 6Tng thi gian hon tt cng vic l 40 pht. Bi 14 : C 5 cng vic c thc hin tun t trn 3 my chuyn dng vi thi gian thc hin cho di y. Hy xp th t cng vic sao cho tng thi gian thc hin l nh nht.

My t1 My t2 My t3 A 23 9 27B 19 15 24C 25 10 22D 21 13 18E 26 17 29

Ta c : t1 min = 19 >= t2 max = 17;

My t1+t2 My t2+t3 A 32 36B 34 39C 35 32D 34 31E 43 46

A B E C D t1 23 19 26 25 21 t2 9 15 17 10 13 t3 27 24 29 22 18

0 23 42 68 93 114

My 1 A=23 B = 19 E = 26 C = 25 D = 21 23 32 42 57 68 85 93 103 127

My 2 A = 9 B= 15 E = 17 C = 10 D= 13 32 59 83 114 136 154

My3 A = 27 B = 24 E = 29 C = 22 D=18Tng thi gian hon tt cng vic l 154 n v.

Bi 15 : C 5 cng vic c thc hin tun t trn 3 my chuyn dng vi thi gian thc hin cho di y. Hy xp th t cng vic sao cho tng thi gian thc hin l nh nht.

My t1 My t2 My t3 A 3 1 4

Quan tri san xuat

32

T 5 2 3S 4 1 2Y 6 2 8N 2 1 5

Ta c : t1 min = 2 >= t2 max = 2;

t3min = 2 >= t2 max = 2

My t1+t2 My t2+t3 A 4 5T 7 5S 5 3Y 8 10N 3 6

N A Y T S t1 2 3 6 5 4 t2 1 1 2 2 1 t3 5 4 8 3 2

0 2 5 11 16 20

My 1 N = 2 A = 3 Y = 6 T = 5 S = 4 2 3 5 6 11 13 16 18 20 21

My 2 N = 1 A = 1 Y = 2 T = 2 S = 1 3 8 12 13 21 24 26

My3 N = 5 A = 4 Y = 8 T = 3 S=2 Tng thi gian hon tt cng vic l 26 pht.

Bi 16 : C 6 cng vic c thc hin tun t trn 3 my chuyn dng vi thi gian thc hin cho di y. Hy xp th t cng vic sao cho tng thi gian thc hin l nh nht.

My t1 My t2 My t3 A 15 6 10B 13 9 12C 21 2 15D 14 8 11E 17 10 19F 23 5 13

Ta c : t1 min = 14 >= t2 max = 10;

t3min = 10 >= t2 max = 2

My t1+t2 My t2+t3 A 21 16B 22 21C 23 17D 22 19E 27 29F 28 18

E B D F C At1 17 13 14 23 21 15 t2 10 9 8 5 2 6

Quan tri san xuat

33t3 19 12 11 13 15 10

0 17 30 44 67 88 103

M 1 E = 17 B = 13 D = 14 F = 23 C = 21 A = 15 17 27 30 39 44 52 67 72 88 90 103 109

M 2 E = 10 B = 9 D = 8 F = 5 C=2 A=6 27 46 58 69 72 90 105 109 119

M 3 E = 19 B = 12 D=11 F = 13 C=15 A= 10 Tng thi gian hon tt cng vic l 119 pht.

4. Sp xp lch trnh N cng vic N my ( Phn cng cng vic cho cc my )

4.1 : Bi ton cc tiu Cc my u c kh nng thay th ln nhau Mi cng vic ch b tr trn mt my v mi my ch ph trch mt cng vic Chi ph (hoc thi gian) thc hin mi cng vic ca mi my u khc nhau. Ngi ta c

th b tr mi cng vic trn mi my sao cho tng chi ph thc hin (hoc thi gian) hon thnh l nh nht.

Bc 1 : Chn trong mi hng 1 s min, ly cc s trong hng tr i s min Bc 2 : Chn trong mi ct 1 s min , ly cc s trong ct tr i s min Bc 3 :

o Chn hng no c mt s 0, khoanh trn s 0 v k ng thng xuyn sut ct. o Chn ct no c mt s 0 , khoanh trn s 0 v k ng thng xuyn sut hng. Nu tng s s 0 khoanh trn bng s p n th bi ton gii xong. Nu tng s s 0 khoanh trn cha bng s p n th phi thc hin tip bc 4

Bc 4 : Ta to thm s 0 bng cch : o Chn trong cc s khng nm trn cc ng thng 1 s min o Ly cc s khng nm trn ng thng tr i s min . o Ly s min cng vo cc s nm trn giao im ca cc ng thng o Sau li b tr cng vic nh trnh by bc 3 o Khi cc s 0 khoanh trn bng s p n cn tm th bi ton gii xong

Bi tp 17: c 3 cng vic R-34, S-66; T-50 c thc hin trn 3 my A,B,C vi cc chi ph khc nhau cho trong bng s liu sau. Hy phn cng cng vic sao cho tng chi ph thc hin l nh nht VT : USD

My A B C Cng vicR-34 11 14 6S-66 8 10 11 T-50 9 12 7

Bc 1 Bc 2

My A B C My A B C Cng vic Cng vicR-34 5 8 0 R-34 5 6 0 S-66 0 2 3 S-66 0 0 3 T-50 2 5 0 T-50 2 3 0

Bc 3 Bc 4

My A B C My A B C Cng vic Cng vicR-34 5 6 0 R-34 3 4 0S-66 0 0 3 S-66 0 0 3+2=5

Quan tri san xuat

34T-50 2 3 0 T-50 0 1 0

Kt qu : Cng vic R-34 b tr vo my C = 6 USD Cng vic S-66 b tr vo my B = 10 USD Cng vic T-50 b tr vo my A = 9 USD

Tng chi ph thc hin cc cng vic : 25 USD l chi ph ti thiu Bi tp 18: c 4 cng vic A, B, C, D; c thc hin trn 4 my W, X, Y, Z vi cc chi ph khc nhau cho trong bng s liu sau. Hy phn cng cng vic sao cho tng chi ph thc hin l nh nht VT : pht

My W X Y Z Cng vicA 47 97 26 74 B 45 87 26 74 C 38 82 13 62 D 59 96 37 66

Bc 1 Bc 2

My W X Y Z My W X Y Z Cng vic Cng vicA 21 71 0 48 A 2 12 0 19 B 19 61 0 48 B 0 2 0 19 C 25 69 0 49 C 6 10 0 20 D 22 59 0 29 D 3 0 0 0

Bc 3 Bc 4

My W X Y Z My W X Y Z Cng vic Cng vicA 2 12 0 19 A 2 10 0 17 B 0 2 0 19 B 0 0 0 17 C 6 10 0 20 C 6 8 0 18 D 3 0 0 0 D 3+2 0 0+2 0

Bc 5

My W X Y Z Kt qu: Cng vic A b tr vo my W = 47 pht Cng vic B b tr vo my X = 87 pht Cng vic C b tr vo my Y = 13 pht Cng vic D b tr vo my Z = 66 pht

Tng thi gian thc hin l : 123 Pht

Cng vic A 0 8 0 15B 0 0 0+2 17 C 4 6 0 16 D 5 0 2+2 0

Bi tp 19: c 5 cng vic I; II; III; IV; V c thc hin trn 4 my A, B, C, D, E vi cc chi ph bng USD khc nhau cho trong bng s liu sau. Hy phn cng cng vic sao cho tng chi ph thc hin l nh nht VT : pht

My A B C D E Cng vicI 5 6 4 8 3 II 6 4 9 8 5 III 4 3 2 5 4 IV 7 2 4 5 3 V 3 6 4 5 5

Bc 1 Bc 2

My A B C D E My A B C D E Cng vic Cng vicI 2 3 1 5 0 I 2 3 1 3 0 II 2 0 5 4 1 II 2 0 5 2 1 III 2 1 0 3 2 III 2 1 0 1 2

Quan tri san xuat

35IV 5 0 2 3 1 IV 5 0 2 1 1 V 0 3 1 2 2 V 0 3 1 0 2

Bc 3 Bc 4

My A B C D E My A B C D E Cng vic Cng vicI 2 3 1 3 0 I 1 3 1 2 0II 2 0 5 2 1 II 1 0 5 1 1 III 2 1 0 1 2 III 1 1 0 0 2 IV 5 0 2 1 1 IV 4 0 2 0 1 V 0 3 1 0 2 V 0 3+1 1+1 0 2+1

Kt qu: Cng vic I b tr vo my E = 3 USD Cng vic II b tr vo my B = 4 USD Cng vic III b tr vo my C = 2 USD Cng vic IV b tr vo my D = 5 USD Cng vic V b tr vo my A = 3 USD

Tng chi ph thc hin l 17 USD Bi tp 20: c 4 cng viccn phn cho 4 cng nhn c tay ngh cao l i, Bnh, Chinh , Duyt vi chi ph nh sau. Hy phn cng cng vic cho tng ngi sao cho tng chi ph thc hin l nh nht VT : 10.000 ng

Thit b i Bnh Chinh Duyt Cng vic1 40 60 50 45 2 50 90 60 70 3 30 80 40 40 4 45 85 50 65

Bc 1 Bc 2 Thit b i Bnh Chinh Duyt Thit b i Bnh Chinh DuytCng vic Cng vic

1 0 20 10 5 1 0 0 5 0 2 0 40 10 20 2 0 20 5 15 3 0 50 10 10 3 0 30 5 5 4 0 40 5 20 4 0 20 0 15

Bc 3 Bc 4 Thit b i Bnh Chinh Duyt Thit b i Bnh Chinh DuytCng vic Cng vic

1 0 0 5 0 1 0+5 0 5+5 0 2 0 20 5 15 2 0 15 5 10 3 0 30 5 5 3 0 25 5 04 0 20 0 15 4 0 15 0 10

Kt qu: Cng vic 1 Cho Bnh = 600.000 ng Cng vic 2 Cho i = 500.000 ng Cng vic 3 Cho Duyt = 400.000 ng Cng vic 4 Cho Chinh = 500.000 ng Tng kinh ph : 2.000.000 ng

Bi tp 21: Phn xng c kh c 4 cng nhn u c th ng c c 4 loi my phay : Ging, ng, Ngang, Rng. Nhng do mc lng v trnh thnh tho ca h khc nhau nn chi ph ng my c phn b nh sau:

Thit b Ging ng Ngang Rng C nhn An 25 30 15 20

Bnh 25 10 5 15 Cng 30 10 25 10 Dn 20 15 10 5

Bc 1 Bc 2

Thit b Ging ng Ngang Rng Thit b Ging ng Ngang Rng C nhn C nhn

Quan tri san xuat

36An 10 15 0 5 An 0 15 0 5

Bnh 20 5 0 10 Bnh 10 5 0 10 Cng 20 0 15 0 Cng 10 0 15 0 Dn 15 10 5 0 Dn 5 10 5 0

Bc 3 Bc 4

Thit b Ging ng Ngang Rng Kt qu: An ng my phay Ging Bnh ng my phay Ngang Cng ng my phay ng Dn ng my phay Rng

Tng chi ph l : 25 + 5 + 10 + 5 = 45 ngn ng

C nhn An 0 15 0 5

Bnh 10 5 0 10 Cng 10 0 15 0Dn 5 10 5 0

Bi tp 22: Cng ty t vn v qun tr cht lng SMETEC c 4 cng vic(A, B, C, D) cho 4 cng tc vin An, Gia, K, Cnh. Ty theo kinh nghim i vi tng cng vic m 4 chuyn gia ny c kh nng gii quyt trong s gi c cho trong ma trn sau

CTV An Gia K Cnh Cng vicA 5 12 12 14 B 7 15 20 15 C 5 10 14 5 D 20 12 10 7

Nn phn cng sao cho tng s gi gii quyt t nht.

Bc 1 Bc 2 CTV An Gia K Cnh CTV An Gia K Cnh Cng vic Cng vic

A 0 7 7 9 A 0 2 4 9 B 0 8 13 8 B 0 3 10 8 C 0 5 9 0 C 0 0 6 0 D 13 5 3 0 D 13 0 0 0

Bc 3 Bc 4

CTV An Gia K Cnh CTV An Gia K CnhCng vic Cng vic A 0 2 4 9 A 0 0 2 7 B 0 3 10 8 B 0 1 8 6 C 0 0 6 0 C 0+2 0 6 0 D 13 0 0 0 D 13+2 0 0 0

Kt qu: An lm cng vic B Gia lm cng vic A K lm cng vic D Cnh lm cng vic C

Tng chi ph l : 7+ 12 + 10 + 5 = 34 gi Bi tp 29: C 3 cng nhn c th lm 3 vic vi thi gianhao ph nh sau : ( ngy )

Cng vic X Y Z Cng nhnA 17 21 5 B 15 7 23 C 19 29 9


Bc 1 Bc 2

Quan tri san xuat

37Cng vic X Y Z Cng vic X Y Z Cng nhn Cng nhn

A 12 16 0 A 4 16 0 B 8 0 16 B 0 0 16 C 10 20 0 C 2 20 0

Bc 3 Bc 4

Cng vic X Y Z Cng vic X Y Z Cng nhn Cng nhnA 4 16 0 A 2 14 0 B 0 0 16 B 0 0 16+2 C 2 20 0 C 0 18 0

Kt qu: A lm cng vic Z B lm cng vic Y C lm cng vic X Chi ph thi gian = 5 + 7 + 19 = 31 ngy

Bi tp 30: C 4 cng nhn c th lm 3 vic vi thi gian hao ph nh sau : ( gi )

Cng vic X Y Z T Cng nhnA 5 9 6 7 B 4 5 1 2 C 3 2 5 9 D 5 5 1 7


Bc 1 Bc 2 Cng vic X Y Z T Cng vic X Y Z T Cng nhn Cng nhn

A 0 4 1 2 A 0 4 1 1 B 3 4 0 1 B 3 4 0 0 C 1 0 3 7 C 1 0 3 6 D 4 4 0 6 D 4 4 0 5

Bc 3

Cng vic X Y Z T Kt qu: A lm cng vic X B lm cng vic T C lm cng vic Y D lm cng vic Z Chi ph thi gian 5 +2+ 2+1 = 10 gi

Cng nhn A 0 4 1 1 B 3 4 0 0C 1 0 3 6 D 4 4 0 5

Bi tp 31: C 4 cng nhn c th lm 4 vic vi thi gian hao ph nh sau : ( ngy )

Cng vic X Y Z T Cng nhnA 5 23 9 8 B 11 7 29 39 C 17 15 19 34 D 21 19 14 49

Hy dng thut ton Hugary b tr cng vic sao cho tng thi gian gii quyt t nht.

Bc 1 Bc 2 Cng vic X Y Z T Cng vic X Y Z T Cng nhn Cng nhn

A 0 18 4 3 A 0 18 4 0 B 4 0 22 32 B 4 0 22 29 C 2 0 4 19 C 2 0 4 16 D 7 5 0 35 D 0 18 4 0

Quan tri san xuat

38Bc 3 Bc 4

Cng vic X Y Z T Cng vic X Y Z T Cng nhn Cng nhnA 0 18 4 0 A 0 20 4 0 B 4 0 22 29 B 2 0 20 27 C 2 0 4 16 C 0 0 2 14 D 0 18 4 0 D 0 20 4 0

Bc 5

Cng vic X Y Z T Kt qu: A lm cng vic T B lm cng vic Y C lm cng vic Z D lm cng vic X Chi ph thi gian 8 +7+ 19+21 = 55 ngy

Cng nhn A 2 22 4 0B 2 0 18 25 C 0 0 0 12 D 0 20 2 0

Bi tp 33: C 4 sinh vin c th lm 4 vic vi thi gian hao ph nh sau : ( gi) Hy dng thut ton Hugary b tr cng vic sao cho tng thi gian gii quyt t nht

Cng vic X Y Z T Sinh vinHng 18 16 10 46 Xun 58 78 22 14 Trng 38 68 34 30 Minh 28 98 42 38

Bc 1 Bc 2 Cng vic X Y Z T Cng vic X Y Z T Sinh vin Sinh vinHng 8 6 0 36 Hng 8 0 0 36 Xun 44 64 8 0 Xun 44 58 8 0 Trng 8 38 4 0 Trng 8 32 4 0 Minh 0 70 14 10 Minh 0 64 14 10

Bc 3 Bc 4 Cng vic X Y Z T Cng vic X Y Z T Sinh vin Sinh vinHng 8 0 0 36 Hng 12 0 0 38 Xun 44 58 8 0 Xun 44 54 4 0 Trng 8 32 4 0 Trng 8 28 0 0 Minh 0 64 14 10 Minh 0 60 10 10

Kt qu: Hng lm cng vic Y Xun lm cng vic T Trng lm cng vic Z Minh lm cng vic X Chi ph thi gian16 +14+ 34+28 = 92 gi

Bi tp 37: Hy phn 5 Xe ti (1,2,3,4,5) i theo 5 con ng khc nhau (A,B,C,D,E)sao cho c chi ph thp nht . Tnh tng chi ph khi chi ph n v l 10.000 c cho trong ma trn sau y

Con ng A B C D E Con ng A B C D EXe Xe 1 4 5 9 8 7 1 0 1 5 4 3 2 6 4 8 3 5 2 3 1 5 0 2 3 7 3 10 4 6 3 4 0 7 1 3 4 5 2 5 5 8 4 3 0 3 3 6 5 6 5 3 4 9 5 3 2 0 1 6

Con ng A B C D E Con ng A B C D EXe Xe

Quan tri san xuat

391 0 1 5 4 1 1 0 1 5 4 1 2 3 1 5 0 0 2 3 1 5 0 0 3 4 0 7 1 1 3 4 0 7 1 1 4 3 0 3 3 4 4 3 0 3 3 4 5 3 2 0 1 4 5 3 2 0 1 4

Con ng A B C D E Con ng A B C D EXe Xe

1 0 1 5 3 0 1 0 1 5 3 0 2 4 2 6 0 0 2 4 2 6 0 0 3 4 0 7 0 0 3 4 0 7 0 04 3 0 3 2 3 4 3 0 3 2 3 5 3 2 0 0 3 5 3 2 0 0 3

C 2 kt qu: Xe 1 A; Xe 2 E; Xe 3 D; Xe 4 B; Xe 5 C = 4+5+4+2+3 = 180.000 Xe 1 A; Xe 2 D; Xe 3 E; Xe 4 B; Xe 5 C = 4+3+6+2+3 = 180.000

Bi tp 38: Hy phn 5 cng vic (1,2,3,4,5) cho 5 my (A,B,C,D,E)sao cho t thi gian gia cng ngn nht c tnh bng pht c cho trong ma trn sau y

my A B C D E my A B C D E Cng vic Cng vic1 14 18 20 17 18 1 0 4 6 3 4 2 14 15 19 16 17 2 0 1 5 2 3 3 12 16 15 14 17 3 0 4 3 2 5 4 11 13 14 12 14 4 0 2 3 1 3 5 10 16 15 14 13 5 0 6 5 4 3

My A B C D E my A B C D E Cng vic Cng vic1 0 3 3 2 1 1 0 3 3 2 1 2 0 0 2 1 0 2 0 0 2 1 03 0 3 0 1 2 3 0 3 0 1 2 4 0 1 0 0 0 4 0 1 0 0 0 5 0 5 2 3 0 5 0 5 2 3 0

Kt qu: 1 A; 2 B; 3 C; 4 D; 5 E = 14+15+15+12+13 = 69 pht Bi tp 26: C 3 cng nhn c th lm 3 cng vic vi chi ph nh sau .Nn phn cng sao cho tng chi ph nh nht(10.000 )

Cng vic X Y Z Cng vic X Y Z Cng nhn Cng nhnA 3 9 6 A 0 6 3 B 12 8 4 B 8 4 0 C 5 10 15 C 0 5 10

Cng vic X Y Z Cng vic X Y Z Cng nhn Cng nhnA 0 2 3 A 0 1 2B 8 0 0 B 9 0 0C 0 1 10 C 0 0 9

Kt qu: A: Cng vic X; B: Cng vic Z; C: Cng vic Y = 3+4+10 = 170.000 Bi tp 27 : nh trn nhng A khng lm c cng vic X

Cng vic X Y Z Cng vic X Y Z Cng nhn Cng nhnA X 9 6 A X 3 0 B 12 8 4 B 8 4 0 C 5 10 15 C 0 5 10

Cng vic X Y Z Kt qu: A: Cng vic Y; B: Cng vic Z; C: Cng vic X = 9+4+5 = 180.000

Cng nhn A X 0 0 B 8 1 0

Quan tri san xuat

40C 0 2 10

Bi tp 39: Hy phn 3 cng vic (1,2,3) cho 4 my (A,B,C,D) sao cho chi ph nh nht c tnh bng 10.000 c cho trong ma trn sau y vi iu kin mi cng vic ch thc hin trn 1 my

my A B C D my A B C D Cng vic Cng vic1 12 16 14 10 1 2 6 4 0 2 9 8 13 7 2 2 1 6 0 3 15 12 9 11 3 6 3 0 2 0 0 0 0 0 0 0 0

my A B C D my A B C D Cng vic Cng vic1 2 6 4 0 1 1 5 4 0 2 2 1 6 0 2 1 0 6 0 3 6 3 0 2 3 5 2 0 2 0 0 0 0 0 0 1 1

Kt qu : 1-D; 2-B; 3-C = 10+8+9 = 270.000 Bi ton cc i : Bi tp 23: Cng ty mi gii tm bn (CMT) c nhn yu cu ca 4 nam l Nhn, Tm, Phong, Gip vi yu cu ca 4 n nh , Xun, Lan, Mai. Sau khi phn tch cc iu kin trong yu cu, cng ty CMT c lp ma trn sau

N yu cunh Xun Lan Mai Nam yu cu

Nam 30 20 10 40 Tm 70 10 60 70

Phong 40 20 50 40 Gip 60 70 30 90

Sp xp kh nng kt hp cc cp vi nhau sao cho tng kh nng xng hp ny cao nht.

Bi gii

Cch sp xp cho trong bng trn c nh gi t 0 n 100% kh nng hai ngi xng hp nhau c th tin ti hn nhn.

Cch tnh lm sao c tng s im t ti a. V chi ph v tin li ch khc nhau con du ( chi ph = - tin li ). Nn ta chuyn bi ton nh l tnh chi ph nh nht ri gii nh bi ton giao vic.

Bc 1 Bc 2 N nh Xun Lan Mai N nh Xun Lan Mai Nam Nam

Nam -30 -20 -10 -40 Nam 10 20 30 0Tm -70 -10 -60 -70 Tm 0 60 10 0

Phong -40 -20 -50 -40 Phong 10 30 0 10 Gip -60 -70 -30 -90 Gip 30 20 60 0

Bc 3 A Bc 3 B

N nh Xun Lan Mai N nh Xun Lan MaiNam Nam Nam 10 0 30 0 Nam 10 0 30 0 Tm 0 40 10 0 Tm 0 40 10 0

Phong 10 10 0 10 Phong 10 10 0 10 Gip 30 0 60 0 Gip 30 0 60 0

Kt qu:

Theo phng n A : nh Tm; Xun Gip; Lan Phong; Mai Nam 70 + 70 + 50 + 40 = 230

Theo phng n B : nh Tm; Xun Gip; Lan Phong; Mai Nam 70 + 20 + 50 + 90 = 230

Quan tri san xuat

41 Bi tp 28: Mt cng ty xy dng c 3 i thi cng . Cc i k hp ng thc hin 3 cng trnh vi s tin cho bng sau ; Nn phn cng mi i thc hin 1 hp ng sao tng s tin thu c cao nht Bc 1

Hp ng I II III Hp ng I II III i iA 3 9 7 A -3 -9 -7 B 6 11 16 B -6 -11 -16 C 14 10 6 C -14 -10 -6

Bc 2+3

Hp ng I II III Kt qu : i A : Hp ng II i B : Hp ng III i C : Hp ng I Tng s tin = 9+16+14 = 33 triu

i A 6 0 2 B 10 5 0C 0 4 8

Bi tp 32: C 4 cng nhn c th lm 4 cng vic vi nng sut nh sau : Nn phn cng tng nng sut ln nht.

Bi gii

Cng vic X Y Z T Cng vic X Y Z T Cng nhn Cng nhn

A 18 0 14 15 A 1 0 4 15 B 28 32 10 0 B 11 32 0 0 C 17 19 15 0 C 0 19 5 0 D 28 30 35 0 D 11 30 25 0

Kt qu :

Cng nhn A : Cng vic Y Cng nhn B : Cng vic Z Cng nhn C : Cng vic X Cng nhn D : Cng vic T Tng sn phm = 23+29+17+49 = 118 SP/ ngy

Bi tp 35: C 3 cng nhn c th lm 4 cng vic vi nng sut nh sau : Nn phn cng tng nng sut ln nht. Vi iu kin mi cng nhn ch lm 1cng vic m thi

Bi gii

Cng vic X Y Z T Cng vic X Y Z T Cng nhn Cng nhn

A 18 0 14 15 A 8 0 4 15 B 28 32 10 0 B 18 32 0 0

Cng vic X Y Z T Cng vic X Y Z T Cng nhn Cng nhnA 5 23 9 8 A -5 -23 -9 -8 B 11 7 29 39 B -11 -7 -29 -39 C 17 15 19 34 C -17 -15 -19 -34 D 21 19 14 49 D -21 -19 -14 -49

Cng vic X Y Z T Cng vic X Y Z T Cng nhn Cng nhnA 5 23 9 8 A -5 -23 -9 -8 B 11 7 29 39 B -11 -7 -29 -39 C 17 15 19 34 C -17 -15 -19 -34 0 0 0 0 0 0 0 0

Quan tri san xuat

42C 17 19 15 0 C 7 19 5 0 0 0 0 0 0 10 0 10

Kt qu :

Cng nhn A : Cng vic Y Cng nhn B : Cng vic Z Cng nhn C : Cng vic T Tng sn phm = 23+29+34 = 86 SP/ ngy

Bi 25 : Cng ty c 4 tho chng vin l Bnh, Chnh, Thn , i . C 4 u c th vit c bt k 1 trong 4 chng 1,2,3,4. Kh nng tng ngi i vi vic thc hin chng trnh cho trong bng sau : V : pht Sp xp cng vic mi ngi cho mi chng trnh sao cho :

1. Tng thi gian thc hin nh nht 2. Thi gian thc hin cng vic mi ngi di 120 gi

Bi gii

1. Tng thi gian thc hin nh nht

2. Thi gian thc hin cng vic mi ngi di 120 gi

Chng 1 2 3 4 Tho c vinBnh 80 120 125 140

Chnh 20 115 145 60 Thn 40 100 85 45

i 65 35 25 75

Chng 1 2 3 4 Chng 1 2 3 4 Tho c vin Tho c vinBnh 0 40 45 60 Bnh 0 30 45 55

Chnh 0 95 125 40 Chnh 0 85 125 35 Thn 0 60 45 5 Thn 0 50 45 0

i 40 10 0 50 i 40 0 0 45

Chng 1 2 3 4 kt qu: Bnh Chng 2; Chnh Chng 1; Thn Chng 4; i Chng 3; Thi gian = 120 + 20 + 45 + 25 = 210 pht

Tho c vin Bnh 0 0 15 55

Chnh 0 55 95 35 Thn 0 20 15 0

i 70 0 0 75

Chng 1 2 3 4 Chng 1 2 3 4 Tho c vin Tho c vinBnh 80 X X X Bnh 0 X X X

Chnh 20 115 X 60 Chnh 0 95 X 40 Thn 40 100 85 45 Thn 0 60 45 5

i 65 35 25 75 i 40 10 0 50

Chng 1 2 3 4 Chng 1 2 3 4

Quan tri san xuat

43

Bi tp 40: C 5 k s c phn vit 5 chng ca 1 n. Mi k s phi ph trch 1 chng . S ngy m mi k s hon thnh chng c cho bng sau : Vy nn phn cho ai vit chng no thi gian hon thnh sm nht ( ngy)

Chng 1 2 3 4 5 Chng 1 2 3 4 5 K s K sA 46 59 24 62 67 A 22 35 0 38 43 B 47 56 32 55 70 B 15 24 0 23 38 C 44 52 19 61 60 C 25 33 0 42 41D 47 59 17 64 73 D 30 42 0 47 56 E 43 65 20 60 75 E 23 45 0 40 55

Chng 1 2 3 4 5 Chng 1 2 3 4 5 K s K s

A 7 11 0 15 5 A 4 8 0 12 2 B 0 0 0 0 0 B 0 0 3 0 0 C 10 9 0 19 3 C 7 6 0 16 0D 15 18 0 24 18 D 12 15 0 21 15 E 8 21 0 17 17 E 5 18 0 14 14

Chng 1 2 3 4 5 Chng 1 2 3 4 5 K s K s

A 0 4 0 8 2 A 0 2 0 6 2 B 0 0 7 0 0 B 2 0 9 0 2 C 3 2 0 12 0 C 3 0 0 10 0D 8 11 0 17 15 D 8 9 0 15 15 E 1 14 0 10 14 E 1 12 0 8 14

Chng 1 2 3 4 5 Chng 1 2 3 4 5 K s K s

A 0 2 1 6 2 A 0 0 1 4 0B 2 0 10 0 2 B 4 0 12 0 2 C 3 0 1 10 0 C 5 0 3 10 0 D 7 8 0 14 14 D 7 6 0 12 12 E 0 11 0 7 13 E 0 9 0 5 11

Chng 1 2 3 4 5 K s

A 0 0 1 4 0 B 4 0 12 0 2

Tho c vin Tho c vinBnh 0 X X X Bnh 0 X X X

Chnh 0 85 X 35 Chnh 0 40 X 35 Thn 0 50 45 0 Thn 0 5 0 0

i 40 0 0 45 i 85 0 0 90

Chng 1 2 3 4 kt qu: Bnh Chng 1; Chnh Chng 4; Thn Chng 3; i Chng 2; Thi gian = 80 + 60 + 85 + 35 = 260 pht

Tho c vin Bnh 0 X X X

Chnh 0 5 X 0Thn 35 5 0 0

i 120 0 0 90

Quan tri san xuat

44C 5 0 3 10 0 D 7 6 0 12 12 E 0 9 0 5 11

kt qu: C 2 phng n : Phng n 1 Phng n 2

Chng 1 2 3 4 5 Chng 1 2 3 4 5 K s K sA 46 59 24 62 67 A 46 59 24 62 67 B 47 56 32 55 70 B 47 56 32 55 70 C 44 52 19 61 60 C 44 52 19 61 60D 47 59 17 64 73 D 47 59 17 64 73E 43 65 20 60 75 E 43 65 20 60 75

Thi gian = 67+55+52+17+43 =234 ngy Thi gian = 59+55+60+17+43 =234 ngy

HOCH NH NHU CU VT T Bi 1: sn xut 1 n v sn phm U cn 3 n v hng D v 2 n v hng Q. Mi n v hng D cn 2 n v hng M v 2 n v hng N . Mi n v hng Q cn 1 n v hng N v 4 n v hng Z. Mi n v hng N cn 1 n v hng M v 2 n v hng T. Hy v s cu trc sn phm , tnh ton nhu cu cc loi hng lp rp. Thi gian lp rp cc loi hng nh sau : Hng U D Q M N T Z Thi gian 1 3 2 2 1 2 4 Xy dng s cu trc sn phm theo thi gian

Bi gii 1. S cu trc sn phm : Cp 0 U(1) 10 D(3) 30 Cp 1 Q(2) 20

M(2) 60 N(2) 60 Cp 2 N(1) 20 Z(4) 80 Cp 3 M(1) 60 T(2) 120 M(1) 20 T(2) 40

a. S trn c 4 cp t cp 0 n cp 3 Hng gc ( cu trc 2 chi tit tr ln) (U,D,Q,N) Hng pht sinh ( cu trc nn hng gc ) (D,Q,M,N,Z,T)

3. Nhu cu cc loi hng nh sau :

Hng Tnh ton S lng U 1x10 10 D 3x10 30 Q 2x10 20 M 30x2 + 60x1 + 20x1 140 N 30x2 + 20 x1 80 Z 20x4 80 T 60x2 + 20x2 160

Quan tri san xuat

45

4. S cu trc SP theo thi gian:

M 60 D 30

M 60 N T 60 U

120 10 M

20 N T 20 40 Q Z 20 80

0 1 2 3 4 5 6 7

Bi 2: sn xut 1 n v sn phm Y cn 3 H;4I v 2J. Mi H cn 1K v 1L. . Mi I cn 2M v 4N. Mi J cn 1O v 1P Mi M cn 2Q v 1K Mi O cn 1R v 1S Mi Q cn 2L v 4T Mi R cn 1U v 2V

Thi gian phn phi cc loi hng nh sau (tun) Hng Y H I J K L M N O P Q R S T U V Tgian 1 2 2 2 1 3 2 3 3 1 2 2 2 3 3 1

1. Hy v s cu trc sn phm Y. 2. S c bao nhiu cp 3. Xy dng s cu trc sn phm theo thi gian

Bi gii

1. s cu trc sn phm Y.

Cp

0 Y(1) 7

Cp 1 H(3) 21 I(4) 28 J(2) 14

Cp 2 K(1) 21 L(1) 21 M(2) 56 N(4) 112 O(1) 14 P(1) 14

Cp 3 Q(2) 112 K(1) 56 R(1) 14 S(1) 14

Cp 4 L(2) 224 T(4) 448 U(1) 14 V(2) 28

Quan tri san xuat

462. S c 5 cp (0,1,2,3,4) 3. s cu trc sn phm theo thi gian

K H L L Q T M K I Y N U R V O S J P

0 1 2 3 4 5 6 7 8 9 10 11

Documents

quantrisanxuat