115_Lab_3_Solutions - ENGR‘I 115 Engineering Applications...

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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] 1-1 - Point house (in and out) 15 - Install lighting and outlets lfi - install trim l? - Landscape MLOMM—IH‘hfl-mnmhfifiwwfioh 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 house-building 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 house-building 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 zrl-lrirlw r. ell-m 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! l-m"Fl’fi'fl/lE-m'r‘bflch‘l J'.'r‘v""-’."3 .l-{u [permit-"hrNils-Myriad H of)?” {Ilsa} all mtg-:5 lanai/r "-fl-mr; o r . . .- “in i-"rt li‘l} ( C’l" WVllMllfl {Lari-lrmif- -.l--r/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 Flo-C; 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 lfii-ij‘illloé lmMrS‘IL faifi‘l 11(5) Fifi/65 ‘H‘W +07%} \il‘ll'lififlC-lzfl/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 before-it's too late to start the activiti that follow it. 'I'herefore, latetimeU-B) - m { 2-1-31. 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? satin-V molt S h I gall/ll“; "i'Jijjfi’J. “Thus! V'fo Mal-.3 “7} A0 w r :eL-Egirt 132;,”th I g Egg-Jiwgiiaifiumgnlg without delaying the completion of the project? Canwewait more than this? This quantityiaoften called the moi flmgofthe‘ ivity. A J 4 IL 2} Yip/9 {nggu,,m3_ 4H- PWAEngprs 063 i ‘5-5145, MW tea-Wm stew “Ln—mas} IUD Si-Wifi’ ‘i'h91h+\4/5' willr Sim/’4 ‘i’bé (ff-"J'VN-L? i712: ALP/5' 5 w 13- .4- ~.a,t. ~l r'fi ‘l‘ Whatcenyofisey about the [:35 whooe+wtilqflofiismhddbéja§fjo (’9 a: 0;: nine, awf‘i-r'cfidi eggiivjvisliegi Hwy!” {if-tied Vh U 5. 3+5?de (1+ hiked» {Tarr'iffo’57Lj moyL Wimfi (3p ‘fr’i 5 :9 )0 Vii) W l'L/H/ b9 J'C-f'rzy'fl 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 howtostortAMPLandi-eod inapmblem Emmafile. rrlTF-Sl'fl-‘jfi Hun-J llii’l‘ llflllffl; I . 'Jfi‘i‘c Fl'it’lla' 'r'll l l’lJ’lf'. ) 5 (lo. ‘13) C. - ‘ I _ r I log-{Carl} rim-air -- rim-1'1 rm W GOMU'J-tw'; all 1.9.4 'lhfi'fjid/I t” I T'. 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 Elihu-H 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- afar-i! 4W:- ”"--,"i .-"rv;-"ei--imm 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 re-solve 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:{ excl-m'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!" (rill-UIJI U1] I INK; Ll. _.I Eff“; \ - .r: r - .I J. . r .r'l." 1:3. «J0 FlWJ-J I 7' M:- i 1m? ‘:' .r': "'r—J- _: r I ’.. ' '-' r“ -x-.. l.-‘- , ‘E “mil” '- ' '-"’ '- . '31P “lo-r3 ._l L ...
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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).

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115_Lab_3_Solutions - ENGR‘I 115 Engineering Applications...

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