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## Quantitative Reasoning and Problem Solving

by: Zechariah Hilpert

18

0

7

# Quantitative Reasoning and Problem Solving MATH 141

Marketplace > University of Wisconsin - Madison > Mathematics (M) > MATH 141 > Quantitative Reasoning and Problem Solving
Zechariah Hilpert
UW
GPA 3.8

Gabriele Meyer

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COURSE
PROF.
Gabriele Meyer
TYPE
Study Guide
PAGES
7
WORDS
KARMA
50 ?

## Popular in Mathematics (M)

This 7 page Study Guide was uploaded by Zechariah Hilpert on Thursday September 17, 2015. The Study Guide belongs to MATH 141 at University of Wisconsin - Madison taught by Gabriele Meyer in Fall. Since its upload, it has received 18 views. For similar materials see /class/205279/math-141-university-of-wisconsin-madison in Mathematics (M) at University of Wisconsin - Madison.

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Date Created: 09/17/15
8 Hospital emergency room activities and police officer actiVities may have to be interrupted to start new tasks For manufacturing machines in standard production runs the assumption is reasonable 18 a T6 T2 T4 T3 T1 T5 b Processor 1 T6 idle 2022 Processor 2 T2 T4 T5 Processor 3 T3 T1 idle 1822 This schedule is optimal for 3 processors c It is no better than the best schedule found there 28 Tasks could be rearranged in order of increasing time on each machine39 tasks could be reordered so those that when completed resulted in a cash payment were competed as early as possible 56 If one has to pack trucks of the same capacity using a single loading dock it might be natural to send a a truck 0 its way when the next item would not fit into the space left in the partiallyfilled truck Any situation where keeping many bins up simultaneously is costly or storing an item temporarily is costly will encourage using an approach where when the next item will not fit into the currently open bin that bin is closed 4 9 a The critical path has length 32 b 1 Processor 7quot T3 T5 idle30to44 ProcessorZ T2 T4 T6 T7 idle36to38 2 Processorl T2 T5 T6 739 Processor217 7 T4 idle 33 Hull 0 No d Ere sum of the task time divided by 2 is 37 Hence no schedule can nish earlier than time a Processor 1 7 T6 idle 15 t021 T7 idle27 to31 Processor 2 T2 T5 7 Processor 3 7 73 idle from 13 to 31 b Processorl I 7 idle 15 t02 T7 idleZ7 to31 PrOCessor 2 7quot 7 idlefrorn 13 to 21 TB Processor 3 T2 T5 idle om 21 to 31 c Processor 1 7 idle lOto ll 7 idle 18 to 21 T3 Processor2 2 T5 T7 idle27 to 31 Processor 3 7 idle 11 to 31 1 n 14 a The critical path which has length 17 is 7 T1 7 b 7 7 T5 T2 T6 T7 T3 is the list to be used The one processor would have the tasks scheduledonit 7 7 T5 T2 T6 T7 T3 c T6 1 T7 T2 T4 7 7 would be the list The resulting schedule on one processor would be T1 T7 T T2 T5 T6 7 d No idle time Their completion times are the same e Earlier completion of tasks giving rise to cash payments f The required schedule is Processor 1 TI T6 7Processor 2 T T5 T1 T g The completion time does halve 4020 As the number of hroeessors goes up the completion time may decrease but at some point the length of the critical path will gOVem the completion time rather than the number of processors h i Completion time goes down by 7 ii Completion time is 19 for two processors using the decreasing time list 19 a b 5 120 No Whatever list is used 7 only task ready at time 0 No First while Processor 1 works on 7 Processor 2 must be idle Second the task times are integers with sum 31 If there are two processors one 39 I I of the processors must have time since when 2 d1v1des 31 there is a remainder of l Idle must be assigned to the rst machine at time 0 because it is the No i Processorl 7 T3 T5 T7 idle from 16m 20Processor2 T2 T4 T6 T3 ii Processor 1 7 T5 7 7 ProcessorZ T7 T6 7 T2 The schedule in ii is optimal With this list the order of the tasks on the processors is Machine 1 8 17 16 5 3 7 2 l idle 59 to 61 Machine 2 ll 14 9 2 l 18 6 The completion time is 61 The list is 18 17 16 14 ll 9 8 7 6 5 3 2 2 1 1 On two processors we get the schedule Machine 1 18 14 ll 7 6 2 2 Machine 2 17 16 9 8 5 3 l l The completion time is 60 on both machines The answer in b is one optimal schedule Using the original list we get the schedule Machine 1 19 20 1 2 3 5 11 17 18 2 16 Machine 2 19 20 1 2 3 5 ll 18 17 16 2 Both machines nish at time 114 The decreasing list is 20 20 l9191818l7171616111155332 2 2211 Machine 120 1918 1716 11 5 3 2 2 1 Machine 2201918171611 5 3 2 2 1 Both machines nish at time 114 Both lists lead to optimal schedules 44 The bins have capacity 135 First t uses 5 bins Bin 1 80gt 50 Bin 2 90 20 Bin 3 130 Bin 460 30 30 Bin 5 90 40 The decreasing list is 130 90 90 80 60 50 40 30 30 20 First t decreasing uses 5 bins Bin 1 130 Bin 2 90 40 Bin 3 90 30 Bin 4 80 50 Bin 5 60 30 20 Either of these packings is optimal Using rstfit on the list 60 50 40 40 60 90 90 50 20 30 30 50 yields bin capacity 135 39 0 50 20 Bin 2 40 40 50 Bin 3 60 30 30 Bin 4 90 Bin 5 90 Bin 6 50 The decreasing list is 90 90 60 60 50 50 50 40 40 30 30 20 Using rst t on this list yields Bin 1 90 40 Bin 2 90 40 Bin 3 60 60 Bin 4 50 50 30 Bin 5 50 30 20 The decreasing time solution is optimal and uses one fewer bin than rst t applied to the original list f C039nno k 01m M 92000 CwM l WOW 4 791074 Camqmo amp 00144 wit ZcWo ccauoz i maWhoa mmAU WM 0n Mr oe romm fc Ivamg a 5 lgrm c an t 0 dwmc i I7th 2 Canm f tdmu M QcamZMWJ eam 2 DrWC CWMIJMW 9 z 1 Wm C vfamaft crw m r f CanMM 1m 05 040 072 WM V W W7 7 rmmm m 2 397 quotCWfWO fwvc aw mawrnI WM 0 2 3 mm 4 L thl W 2U L Q COWM MM 6 39 roman mm W HMO ad a 4 V WJWQ W L We f Mid Aramaic fluMb4f 5 6 9 791 W daaoyn lv he conform paw Anhog M 0M 0sz L amx onol pm aw fa waM m 0010 01m fmIA ie C07me tr07404 W 00 39 4Q am K 0 nut J wa gtnud Woeam L fa JapanM 8 V wa 71903 Nam 153 39 mQ Jfro 77m 7 a cor6394 Mm Xmm Ila xjJM MAWWA 5145 A WWWWW A mmmmmmm g M L h z 3 b 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