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Unformatted text preview: ENGR‘I 115 Engineering Applications of OR Fall 2006 The Critical Path Method Lab 3 Name: Objectives:
0 Introduce the Critical Path Method for project sdioduling as used in industry.
0 Demonstrate the concept of semltiviw analysis. 0 Practice formulation of network problems Key Ideas:
0 (directed) networks. paths in networks. longest and sham path
0 project scheduling. activities. critical activities
I sensitivity analysis Reading: DO NOT read section 8.4 in Winston, we will discuss the critical path method diﬁerently. Prelah Exercise:
Please write you: answer on the book of this sheet or an attached paper. You are the owner of Bill D. House Construction Company. You are interested in submitting a bid for
the construction of a house in a nearby town, but ﬁrst you want to figure out how long it will take you to
build the house. After looking at the blueprints, you come up with the following estimates 501 the length of
time necessary to complete ml) task involved in building the home: his Complete Task (days) 1  Clear lot 2  Lay foundation 3  Drill well 4  lnstall septic system 5 — Put up frame 5  Install windows and doors ‘1’  Put on roof 8  Install electric wirng 9  Install plumbing lD — Put on aiding (outside walls} 11 v Install ﬂooring 1‘2  Install insulation 13  Put up drywall [inside walls] 11  Point house (in and out) 15  Install lighting and outlets lﬁ  install trim l?  Landscape MLOMM—IH‘hﬂmnmhﬁﬁwwﬁoh Some or these tasks can be done at the same time (like installing insulation and landscaping), while
some tasks require other tasks to complete before they can start (e.g.. cannot start laying the foundation
before the lot is cleared). The following is a list of some (but not all) such prudence relationships for the
housebuilding problem: 0 you cannot install plumbing until after the well and septic system are ﬁnished. I you cannot put up the siding until after the windows and doors. a you cannot do ﬂooring until after the electric and plumbing are installed, I you cannot insulate until after the ﬂooring and siding are on. I the insulation must be installed before the sidewall. 0 you cannot start landscaping until all work on the exterior of the house has been completed. o you cannot install trim until the interior of the house has been completed. Your job is to draw a network model of the problem. Assign nodes to the tasks. and connect two nodes
with a directed arc if the task at the tail of the arrow is a direct predecessor of the task at the head of the
arrow. (Hint: you might want to include dummy nodes that represent the beginning and the end of the
housebuilding process.) Th ﬁgure out how to incorporate the task lengths into the problem. In the next lab we are going to use this network model to ﬁnd the minimum time required for building
the house so that all the precedence relations are observed. We have already come up with a network that
we think represents the precedence relations well. It is alright if you draw a network slightly diﬁerent than m (are zrllrirlw r. ellm fir) Part #1 — Solving the problem with rmursive equations Take a look at the attached network. Usually we have to include two dummy nodes in a netwrk like this,
one to represent a dummy task that precedes all other tasks. and one that comes after all tasks. Here, we
didn't need a dummy starting node since activity 1 (clear the lot} has to be done before anything else can
start. The numbers on the arcs represent the time it takm to complete the task that corresponds to the node at
the tail of the are. So if a node has many out—going arcs (arcs that point away from the node} then we write
the length of the task on each of those areas. We will denote the numbers on the area with durationti.j). Our network is a directed graph. is it acyclic? How can you check this quickly (without looking ﬁor cycles)? A (NT: trim/“f (ill! lm"Fl’ﬁ'fl/lEm'r‘bﬂch‘l J'.'r‘v""’."3
.l{u [permit"hrNilsMyriad H of)?” {Ilsa} all mtg:5 lanai/r "ﬂmr; o r . . . “in i"rt li‘l} ( C’l" WVllMllﬂ {Larilrmif .lr/o (fr/:3 0"“0 “75””
Our goal is to ﬁnd the minimum time required to complete the project. For each activity we will determine 3 ; p“ I '1 "(" the earliest possible time that this activity can start. This will answer our question since the earliest possible "  —' i
time to start the activity "end" is exactly the earliest time to complete the project. So let us denote the earliest possible start time for activity 1 by earlytineti) (this is also called the early
event time). The ﬁrst task can start at once. so aarlytimﬂn 0. What is the earliest time to start activities 2. 3. 4, 5. B. 1', a ‘? Write these numbers down next to
the nodes on the ﬁrst copy of the network. Why was this so easy to do for these tasks? Edi \J V.” 1/3? '".’T K JG had Only FloC; l} in?» . My :1 2 ~._._.v New to compute earlytime{9) we have to look at which tasks need to be ﬁnished before we can start 9.
What are these tasks? When do these tasks complete? numbers computed above. That is.
earlytimetsl w{ 7 . _? . IL‘T} How can we write this down formally? For each node we consider the immediate predecessors of the node,
and compute when the corresponding activities complete. That is. we have the following recursive equations aarlytimau)  0
earlytimeﬂ) = mum“; {aariytm(ji+dumt1on<1.1)} 1  2. 5. and. continue the computation for the rest efthe nodes. At each step highlight the m that points from the task
that you need to wait for. What. kind of subgraph did you get? l
A Spanﬂlf’g +W§ Poo? “rs lljiﬁiyiffur (I
is lla j we. How long does it take to build the house? Which tasks cannot be delayed if you want to ﬁnish the project
on time? These are called the critical activities. Observe that they lie on a path from 1 to and. this path is
called the critical path. 93 ‘4”ng Ac’i'iviliw W W; H; is}, ism/J5me e
newer be Mat/M The recursive equations above look much like the ones you have seen for the shortest path distances on
lecture. with the small exception that we take the maximum of some numbers instead of the minimum.
Convince yourself that eerlytmeti) is really the length of n longest path from 1 to 1. Explain in terms of
the pmble why does it make sense to compute the longest paths here. The lﬁiij‘illloé lmMrS‘IL faiﬁ‘l 11(5) Fifi/65 ‘H‘W +07%}
\il‘ll'liﬁﬂClzfl/Ill'iotze 1L0 (453m lﬁﬂz’} SL553 not} 5571/: 4):??? ?
{lﬂicrmlmi 44% [mg 5} ideas/is? lul/fgfﬂ' iq/ (29/7
clc’ir'v‘rl’l' W19 tor/43m {h Now we would like to ﬁnd out by how much activities can be delayed so that the project is still ﬁnished at
the time originally planned. First we will tryr to determine the latest possible starting time For the activities
{this is also called the late mt time). We will use the notation lamina“) for the latest possible start
time for activity 1. We don’t. want to £19133 the project. so leteuueund)  as, the length of the project. We will work our
way back until we reach the ﬁrst node. What is the latest time activities 1?. 16. 1.5, 14 can be started? (Merl: these on a new copy of the network.) (5'55" CO/J‘U ﬂtc ﬂﬁ‘iwfm To compute the latest time activity 13 can be started is not quite so obvious. We have to take into account.
that it has me than one immediate successor. So this activity has to ﬁnish beforeit's too late to start the activiti that follow it. 'I'herefore,
latetimeUB)  m { 2131. 193}  We can write this down formaliy with recursive equations. laminatend)  23
latetiueU) = mjnmweg {laminath wuton‘LjJ} 1  17. 16. .. 1 Why do these recursive equations give the latest possible starting times? Explain this with a few #0
M6? ~ T W lair/5+ 7W) )6 “if in l/li‘i y COW/0M V
0! Moire/ma; S‘in V‘i’l's Hag MW 9W9 Wmfwr i7 ail/57w
a i l m.“ Hag ’1 m ; WI im’ (3 2—3 Lia? satinV molt S h I gall/ll“; "i'Jijjﬁ’J. “Thus! V'fo Mal.3 “7} A0
w r :eLEgirt 132;,”th I g EggJiwgiiaiﬁumgnlg without delaying the completion of the
project? Canwewait more than this? This quantityiaoften called the moi ﬂmgofthe‘ ivity. A J 4 IL 2}
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Part #2  Solving the problem with AMPL AMPL is a software package that can be used to solve various optimization pmblm You: TA will explain
howtostortAMPLandieod inapmblem Emmaﬁle. rrlTFSl'ﬂ‘jﬁ
HunJ llii’l‘ llﬂllfﬂ; I .
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log{Carl} rimair  rim1'1 rm W
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Read. in the ﬁle 115_cpm_1ab.mod the co ' the problem description with the recursive equations For
earlytime. Display the problem with oak , and then solve it with NFL and minpere your compu— tation with the computer output. Did you get the same earlytine values? (Just disregard the zeros at the
end of the output.) Note that the node and has been substituted with 18. I
"l i; a: r? r N I}? £44124 AHMEst ElihuH be ‘5 SW10 0'3 'l 1‘ 6% _ i As the next step, read in 3 more oophiatieated formulation of the problem that contains the latetinae and
totai ﬂoats as well. 115_cpo_1ah2.nod. Point out both sets of recursive equations and the deﬁnition of the
total ﬂoats in the i’orroulation. Now solve it with AMPL. Do your results match AMPL‘a output? :4 mi; .71 5r afari! 4W: ”","i ."rv;"eiimm if?""’I’J’i“ﬂ'fl
{3’39"} "’3?an l llf Shrab’nd a}: M?” "7/"? (“E/1+7! (2.4“ Assume you could shorten the length of time it takes to install the plumbing from 3 days to 1 day. Modify
the problem to incorporate this change, and resolve it with AMPL. How did the solution change? Which
activities are critical now? How much time does it take to build the house? 3 7 Cl 3 33 "I—Ca'll (‘1 sh} (1.1.641 n i". 7}an l3 rlﬂlxl/
r ,.:4.rr:{ exclm'i‘fm' Iii}{Tibialoll‘ll3113/”ljl€;jlr§ You are thinking about purchasing a diﬁ'erent kind of siding which is less expensive than the one you have
originally planned to buy, but is more diﬂ'nmlt to put up {takes an additional day). Can you go ahead with
the new plan without delaying the completion of the project? I "/ ' I '  w ﬂ ] \I r I I .' " ' .'
. Th!" (rillUIJI U1] I INK; Ll. _.I Eff“;
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This note was uploaded on 11/20/2011 for the course OPERATION 1101 taught by Professor Trotter during the Spring '11 term at Cornell University (Engineering School).
 Spring '11
 TROTTER

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