VISHVVEER 44766
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Transcript of VISHVVEER 44766
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GaGanttntt PERPER
TT
&
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TT WHAT ISSCHEDULING ?
Scheduling refers to-
1.Determination of rate at which the work shallbe performed in order to its priority.
2.Determination of proper times when work willbe released to the plant
3.Determination of proper sequence
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TECHNIQUES FORSCHEDULING
GRA!"#A$ %&'!(D
A$G&)RA"# %&'!(D
%(D" %&'!(D
$"*&AR R(GRA%%"*G
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WHAT ARE THE?
'hese are tools to+
mana,e the tasks inoled in bi, and
comple pro/ects let pro/ect mana,ers or,anise time0
people0 equipment and money
ensure the ri,ht people and equipment arein the ri,ht place and the ri,ht time
allow mana,ers to monitor the pro,ress ofa pro/ect
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Henry Laurence Gantt (1861-1919)
Gantt
Charts
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!"'(RDesi,ned by !"# HENR LGANTT in 990 to meet therequirements of army for
an improed method of
lannin, 4 #ontrollin, theproduction of ordnance
factories.6
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"*'R(D#'"(*Displays the schedule as wellas makes a comparision
between actual performance4 ori,inal schedule.
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%A5"*G
AGA*''#!AR'
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tep 1 6 $ist the tasks in the pro/ect
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tep 2 6 Add task durations
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tep 3 6 Add dependencies 7which taskscannot start before another task 8nishes9
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"%(R'A*' ("*'
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•The arrows indicate dependencies.
•Tas 1 is a predecessor o! tas 2 " i.e. tas 2 cannot start #e!ore tas 1 ends.
•Tas 3 is dependent on tas 2. Tas 7 is de$endent on two other tass
•%&ectrics' $&#in* and &andsca$in* are concurrent tass and can ha$$en at
the sae tie' so the+ o,er&a$ on the chart. && 3 can start a!ter tas 4 ends.
•aintin* st wait !or #oth e&ectrics and $&#in* to #e !inished.
•Tas 9 has /ero dration' and is a milestone
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tep : 6 ;ind the c"itical $ath
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The critica& $ath is the seence o! tass !ro #e*innin* to end that taes
the longest time to co$&ete.
t is a&so the shortest possible time that the $roect can #e !inished in.
n+ tas on the critica& $ath is ca&&ed a critical task.
o critica& tas can ha,e its dration chan*ed withot a!!ectin* the end
date o! the $roect.
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'he len,th of the critical path is the sum of
the len,ths of all critical tasks 7the red tasks102030:03>1>1.2>1 ?1@.< days.
"n other words0 the minimum amount oftime required to ,et all tasks completed is1@.< days
'he other tasks 70B9 can each run oer-time
before aCectin, the end date of the pro/ect 14
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'he amount of time a task can be etendedbefore it aCects other tasks is called slack 7orfoat 9.
'ask can take an etra day and a half
before it aCects the pro/ects end date0 soeach has %#& da'() )lac* .
#ritical tasks0 by de8nition0 can hae *(slack.
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Can task X’s duration
be changed withoutaecting the end date
o the project ?If it is a critical task the answeris always
NOE
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%TCharts
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"*'R(D#'"(*tands for P"+g"a, E-aluati+nRe-ie. Techni/ue
how order of actiities ordependencies between actiities
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&R' charts
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This %T chart !o&&ows the cti,it+ on rrow st+&e.
•The tass are shown #+ arrows. Tas nae are shown #+ &etters'
in this case.
•The circ&es are ca&&ed nodes. The nodes indicate the start or end o!
tass.
•Tas drations are the shown #+ the n#ers.
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%T %:
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1+ Fhich tasks are on the critical path2+ Fhat is the slack time for tasks #0 D and G
3+ 'ask # is delayed by one day. Fhat impact wouldthis hae on the completion date of the pro/ectFhy
:+ 'ask A will be delayed by 2 days because someequipment has arried late. "f the pro/ect mana,erwants to 8nish the pro/ect on time he will need toshorten the duration of one or more of the tasks.!ow can he achiee this
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1+ Fhich tasks are on the critical path
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ANSWER A!"!#!$!%
ossi#&e $aths; ''C'%' < 2=3=1=4=3 < 13 da+s
''>'?' < 2=3=3=3=3 < 14 da+s
'G'@' < 2=2=5=3 < 12 da+s
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2+ Fhat is the slack time for tasks # and G
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ath C'% < da+s' ath >'? < ' da+s
ath 'C'% < ( da+s. ath ' >' ? < ) da+s
ath G' @ < * da+s.
#i++erence ,slack- . / day +or tasks 0 or E
compared to #!$
So G or H ha1e 2 days3 slack between them4
"!0 or E ha1e / day3s slack4
"!#!$ ha1e no slack4
5AS6S 07
5AS6 G7
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3+ 'ask # starts one day late. Fhat impactwould this hae on the completion date of thepro/ect Fhy
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No impact! because task 0 has
one day3s slack
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:+ 'ask A will be delayed by 2 days because someequipment has arried late. "f the pro/ect mana,er stillwants to 8nish the pro/ect within the ori,inal time frame0he will need to shorten the time for one or more of thetasks. Fhat steps can he take to reduce the number ofdays allocated to a task
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5he answer has N85H%NG to do with the
chart9 :ust say how ;obs can be +inishedmore
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ESTI!ATION OFACTI0IT TI!E
"n &R'0 these three estimate times are deried foreach actiity.
OPTI!ISTIC TI!E 1to9
"t is the shortest possible time in which an actiitycan be completed.
!OST LI2EL TI!E 1tm9
"t is the time in which the actiity is normally
epected to complete under normal contin,encies.PESSI!ISTIC TI!E 1tp)
"t is the maimum time required to complete anactiity.
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ESTIMATED TIME
te = to + (4 x tm) + tp
6
STANDARD DEVIATION=S
t
= (t
p
- t
o
)/6
VARIANCE=
V
t
= [(t
p
- t
o
)/6
!
= (S
t
)
!
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I" t
= #$ t
= 6 % t
p
= &$t'e
*+- ) Expe,te+ T*me
!) St+.+ De*t*o
) V.*,e
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ANS1ER2
te = 6
S
t
= /6
V
t
= /6
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'!A*5 (
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