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prelim1_sol

# prelim1_sol - l ENGRI 110 Engineering Applications of OR...

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Unformatted text preview: l ENGRI 110] Engineering Applications of OR Fall 2008 Prelim 1 I | I Name: $0 (/l/l Tl 0N5 Problem Points Possible Points Received H l 40 2 30 * L 3 30 t | Total 100 l 1' You have ] hour and 30 minutes to answer all the questions in the exam. The total number oi points is 100. For questions with multiple parts. the point. value oi each part is given at, the start of that part. Be sure to give full explanations for all of your answers. In most cases, the correct reasoning alone is worth more credit than the correct answer by itself. Your explanation need not be word): but it should be sufﬁcient to justify your answer. You receive much more credit ior an incorrect, answer if you recognize that something that you have computed is not correct than it you merely pretend that it, is correct. Not all the parts are in the same level of difficulty? Please make sure that you do not spend too much time on one part. Good Luck! ® 1. For both of parts (a) and (b). you must, use one of the algorithms given in this course. 9 (a) (12) In the graph above. ﬁnd a minimum spanning tree; Show your work Clearly. am: brleﬁy gxplan‘lxeéach step. ( O Krggkal s 3 qreétizg Qéjw‘ipﬁm > 9 ”km 3 Mazw’mm: “‘ ‘J J , x, Q\3€‘ﬂ WOAE (D %9 he *‘n-a maﬁa “ma, (b) (15) For the same graph, compute the shortest path tree starting from node 1 (which contains the shortest part from node 1 to each other node in the graph); Show your work, and brieﬂy explain how you computed your aHSWer. (Note that this is an undirected graph; having an undirected edge of length [(i, 3') between two nodes i and j is the same thing as having a directed edge from 'i to j and a directed edge from j to 1', each of length €(i,j) .) awww ﬁr \ X h 5'6 ’ on ,., . , ¥ £3 ' ' 3 )\\ met E l {m5} “ﬁg :1 3"? L; 5- b ' {g} n‘awefaél m‘x a 3 :3»? we x» AQ- l l a 3 3 it e 1 3 . E , wage“ a (c) (10) Suppose that 17 is added to the ieiiet h (it (with edge in this graph. Does. the, answer to either part (a) or (1)) remain the mint“? Does the answer to either part (at) or 03) change? (Notit'e that the answer to Path of these is a tree. not the total \ "5% :3?“ >3 i W L . . . (d) (3) C01151de1" your answers to part} (c‘).‘°}o(-iis on the question(s) ior Wthh you ﬁgured out that the optimal solution iei'imiiied optimal Would this still be true ﬁlo f a out if we added a iitm'iher diﬂei‘ent, from 17 09.11.74. 10000001. or 5% )7 Justify youro answers. t \\ x _ M/ﬁt m; ‘\ \J‘EV 21:? \ “YMQHS 13 {\me‘tm sixpfgt w‘t o @Q ' \ 11953.3 “5:358; f “twee v V 1‘“ ”GR, \ o. \ mm m 3%$ng {fig-EN 3% game: «if» m: 39% ﬁve \m’z e; r-‘tefeém 954‘ iength of the tree.) Explaiiii 9 ® 2. Consider the following graph with source node 1. Sink node (i. such that each are (7.51”) is labelled with two numbere a capacity 1113' and a flow value f(7',j). where the flow value is boxed and the capacity is not boxed: SOUTCC (a) (5) ls the current, ﬂow valid? Explain. In case it is valid. what is the value of the flow? In case it. is not valid correct, it to a valid flow and give the value of the corrected flow. vahé {%%a§9e]i “a , ram COOSWM :2 w: w 1/: 2 ® Non-neml‘v’fﬁ Cf; {8%}! J < pl l: U(\/W\ i g will mm 0mm“ I / em ‘i 9“” a. a -h soda : {Eig‘ {mfg .zzs'iiﬁ‘s Elﬁn? %{ {295; h Cwiiﬂm 5‘ «a; ”” “‘ M @ “l loud i» M \ m Z“ 5%{\N,\/\ (”{5} ll v 6 N; 2‘ €(v3w wlwxdih / a; ‘ I l w (\AW)€A A g“: , i if; W3!” ‘ W 6 ¥ .. l3 (b) (8) “The down the residual graph with respect, to the turrem flow (it yun <~u1‘1‘e('1ed the ﬂow in (a) above refer 10 the corrected flow). (C) (8) ls there any augmenting path (a path from source (node 1) to sink (node 6)) on the residual graph G“, ‘7 In case there is. augment the flow accordingly, what is The value of the new flow? jmcreme 4 ' ~ 1 , l emfﬁ? x , '\ ‘ A»; W“ W Q (d) (5) What. is the capacity of the L6 cut deﬁned by S 2: {1, 2, 3, 4} ? ( H .5) a , Q: “3' @0033} ~93? {337% 03—33% 6 3' 3 3" 7 (e) (4) W hat can you conclude on the 1-6 cut 5’ from (d) and the ﬂow from (c) 17 Be sure to explain your answer. I 3 «Q {3m {7; Q“ E) . 2 3 ‘mg \N‘a‘i \nowe, may} <2 ‘K be; V V ’ 1 3 “‘3 ‘0‘ :4 f‘ﬁxﬁa‘ g ‘jé3‘ 133$” S in “a- Q a»)? {‘32 K > 395* *hﬂz f} i - ~ \ _ j éhgi ‘ E A ‘11 “y 6:} M>R: 3 i ”xii ‘3 ‘1 " E) i); Ow 4% i3 @QKQQE kg Eh‘é L 3 \a Q \ K? ‘ ‘ ‘E \i 5' \ 2$ “*“é 2‘ 2; 3% 3( igbnu 33mm "C 3 one, w C/Ofxuogﬁe ‘ 0‘ ‘ 5% { E; M é\ ‘ i; «We ~ @W (agamﬂ we we mgyix mom \g 39%; mwm \Y 3. (a) (10) Solve the following input for the assignment problem. Use an algorithm tal'lght in this class, and explain your work. 3 5 6 5 6 6 5 9 7 Wm ”Hi \emﬂ t} 2% ESWWH Rim“) 6"“ E 3i M (if \\ ii E 4’: \ “‘4; WE \i 2‘ Re§> mo 0 i E 0’? it i o O i a i 21 O % a “M 1;: . ’J :2 33; My "a“ Es\§?\~§\$ (Ava I); 3&9353431- ‘3 0223 i ,‘ é/ (b) (5) Give a simple explanation, other than you applied the right algorithm, of why the answer you have just found for the input above is correct. \ E . 2 i % 20' 59mg, gas}? mcf‘ai ‘X villa (c) (15) Consider the semi-assygmnem problem. In this problem. the input consists of an 7? by n array of numbers c7]. 7 : l ..... w. J = 1 ..... n (just as in the assignment problem). Here. a leasihle solution consists of 7') entries in the array such that. for each column. exactly one entry is selected: the cost, of a, solution is the sum of the entries selected: the objective is to ﬁnd the minimum~cost feasible solution. Explain why, for any input. the optimal value, for the semioassignment . problem is a lower bound on the optimal value for the assignment problem. (Be as careful as 'Nou can not to make (my unjustiﬁed statements.) it‘SE'z’ljV « ll a; Ntaf‘ﬂ ‘2’} . i ﬁg: “"2 ago .3 Q i A g”: 5?»ng SQKW’RG"? , \ mm “fl-ﬁg a l: I u .4; stem» asst) t I . E l ’2 i r w”) B 2 g M ,ng‘xlﬂ § g é‘dﬂxn ’ \i} \ ﬁg?“ ‘4 Elgar“ "3‘5 SAW Q‘wm “ ‘Jr Wows O J A? \ g\OC€ ‘ A” ”\Dt‘QXCNQM gm\ waiﬁ‘é‘lww (3 Moe. )s {\ i a“ \uf: <3 )4 Br V a: 3%,; KW\ \Jo xi AEVQV Q Rene 6, so . e VCR [O L \e s3 New at 3}? gﬂg \t}\\€3\z’\ k5 ‘ R as ”GEM (a < K m g Q90 C Q ’ on? \ (I 233ng \ mb\@ {e . i \OWW .y \ “a \ \ ark ‘ 3“ :vz’ u: any an m \J \ U : \bkgm Kw ‘ «8AA WBVJV‘ / . f. ﬁttest QPQ I Q Stan {A 391% 2,3}: 5:; K 2 M 39 *»6 x l ‘ xiv b, ‘s . lam oz; 3 a We at; (\ «saw \ \ 7‘ :' 3Q»; ewe. >9 ms?- ._ «tee am. ( Vﬂﬂmgl V)»; ill) s, 9 CO\ \A CC} \w {plume o \ @024 «9%? Q a . \l ’prQ \o\e,m w a E W §€M\ . CU; 3‘ all mg l ‘ m»; em to we} \ Kw; \A (333\ a n me. x \(\ A “6 \j ,. R 1 i) ax}; \, {3’23 9 X Of} \3 on. t’; €; («3 kpdk} 7R \ E :4 k ‘3 («I A f, {n 4:“, Al; (320;: XQ\\¢Q {\ . . ‘ , CU 3x vi. a . ﬂea Q2, be cw v Elk} («A 3v Am 6 t . A . \e 43; stab w. e}. s < "3N5; at s k; ...
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prelim1_sol - l ENGRI 110 Engineering Applications of OR...

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