Topic 10 Directed graphs and networks

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1 Topic 1 irected graphs and networks TOPI 1 irected graphs and networks XIS 1. 1 xercise 1. ritical path analysis 1 arliest completion time is minutes., 1 1,,,,,,, 11 L, 1,, 1 The earliest completion time = 1 days.,, 1, 1,,, 1,, Q,, L, P,, 11, 1 1, 1, 1 1, 1, With an earliest completion time of : L = + = 1 minutes, compared to = + = minutes So a difference of 1 = Thus can increase by ; + = 1 can increase to 1. With an earliest completion time of 1: L P L P = = 1 days Q = + + = days So a difference of 1 = If path and stay the same, Q can increase by Q :+ = days,,,, 1 1,, L, 1,, 1 11, , 1 a ritical path = L b loat() = 1 1 = 1 loat() = 1 = 1 loat() = 11 1 = loat() = 1 =, 1, 11 loat() = 1 11 = loat() = 11 = loat() = 1 = loat() = 1 = loat() = 11 = days loat() = 1 1 = days loat() = 1 = days loat() = 1 1 = days loat() = 1 = days loat() = 11 = days loat() = 11 = days loat() = 11 = 1 day loat(q) = 1 11 = 1 day ctivity Immediate predecessor, ctivity has no predecessor and is the first edge. ctivity and have as an immediate predecessor. ctivity has and as immediate predecessors. ctivity Immediate predecessor O P O, T Q P S T S, Y U O, T V O, T W V X Y Y Z X irst edges and have no immediate predecessors. is a predecessor for activity O and S. O S aths Quest 1 urther athematics V Units and ifth dition Solutions anual

2 1 TOPI 1 irected graphs and networks XIS 1. is a predecessor for activity Y. O and,, S Y Y is a precedessor for activity X, S and Y are precedessors for activity T. O S T X Y O and T are precedessors for activities P, U and V. O P V S T U X Y P is a precedessor for activity Q, V is a precedessor for activity W. O P S T W Q U V Y X Z a,, 1,, 11 = + = mins = mins and can be delayed. = + = 1 mins = mins = 11 mins,, and can be delayed. 11 The critical path for the network is. 1 a, b,, Select maximum value. + = 1 The answer is.,, 1 + = 1 The answer is. c The earliest completion time for all tasks is:,,,, Which of the following statements is true? ctivity is an immediate predecessor of. alse, is a predecessor for. ctivity is an immediate predecessor of. True. ctivity must be done before activity. alse, after. ctivity must be done before activity. alse. ctivity is an immediate predecessor of. o, can occur simultaneously. The answer is. b inimum time to complete all activities follows path = = minutes The answer is. 1 a arliest completion time fill in triangles with maximum time to each vertex/node 1,,,,,, 1, 11 The earliest completion time is minutes. b Tasks which can be delayed. Identify sections of the network where there was a choice.,, 1 a = = minutes The answer is = = = = 1 = + = = + 1= = 1+ = *Take maximum value 1, aths Quest 1 urther athematics V Units and ifth dition Solutions anual

3 TOPI 1 irected graphs and networks XIS = + = 1 = = 1 L L= + + = *Takemaximumvalue = + 1 = 1 = + *Take maximum value = 1 = + = IL OUTO:, 1,,, 11 L,,, 1, 1, 1, 1, 1, b The earliest completion time for the project = = 1 minutes 1 Working backwards to find predecessors. and are predecessors of. is a predecessor of and. is a predecessor of. and have no predecessors. 1 1 ctivity letter Immediate predecessor Time 1, 1 Working backwards to find predecessors. is a predecessor of and. is a predecessor of. and are predecessors of. is a predecessor of. is a predecessor of., and have no predecessors. ctivity letter Immediate predecessor Time, 1 1 Working backwards to find predecessors. and L are predecessors of. L is a predecessor of. is a predecessor of and L. L aths Quest 1 urther athematics V Units and ifth dition Solutions anual

4 1 TOPI 1 irected graphs and networks XIS 1., and are predecessors of and. is a predecessor of. ctivity has as an immediate predecessor. is a predecessor of. is a predecessor of, and. has no predecessor. ctivity letter Immediate predecessor Time ,,,, 1 11 L, L 1 a ritical path follows activities that cannot be delayed. The path that takes the largest time is = = minutes b and c loat time is the maximum time that an activity can be delayed without delaying a subsequent activity on the critical path. ctivity can be delayed 1 minutes, ctivity can be delayed 1 minute, ctivity can be delayed 1 minute, ctivity can be delayed 1 minute, ctivity can be delayed 1 minute and ctivity can be delayed minutes. 1 a ritical path = longest path = = 1 minutes b ctivities which have float times are not on the critical path. These are therefore,,,,, L and. 1 a ctivity Immediate predecessor ctivity and have no predecessor, so they become the first edge. b ctivity Immediate predecessor,,,, I, irst edge has no predecessor. is a predecessor for activity and. is a predecessor for and. is a predecessor for activity. is a predecessor for activity., and are predecessors for activities and. and are predecessors for activity I. I aths Quest 1 urther athematics V Units and ifth dition Solutions anual

5 TOPI 1 irected graphs and networks XIS 1. 1 a irst edge has no predecessor is a predecessor for and. is a predecessor for. is a predecessor for. and are predecessors for. and are predecessors for. is a predecessor for.,,,,,,, 1, 1 b inimum time in which all tasks could be completed follow the path = = minutes,, 1,,,, xercise 1. ritical path analysis with backward scanning and crashing 1,,,,,,,,, orward Scanning,,, ( + ), 1, = + = 1 = + = = + = 1* take maximum value,, ackward Scanning (1 ),, (1 ) = =,, 11 = 11 =, * = = take maximum value ( ) (1 ) ritical path = loat time for the non-critical activities : hours : hours : hours : hours a nter 1 for the last node. nter for the next node before activity. nter for the node before activity. nter either (path, ) or 1 1. (path, ) for the node before activities,. eject the path,. nter.. = for the node before activity. nter. 1 =. for the node before activity. nter. =. for the node before activity. omplete by entering a at the start node., 1 1.,,, 1,., 1., , 1.,., The critical path is where triangle number = box number:. b loat times for non-critical activities: loat() =. =. hours loat() =. 1 = 1. hours loat() =. 1 = 1. hours loat() =. 1 = 1. hours loat() =.. = hours a ctivity letter Immediate predecessor Time 1 1 1, 1 b irst edge (no predecessor)., aths Quest 1 urther athematics V Units and ifth dition Solutions anual

6 1 TOPI 1 irected graphs and networks XIS 1. is a predecessor for and., 1,, 1 is a predecessor for. is a predecessor for and., 1,,, 1, 1, and are predecessors for.,,, 1, 1, 1,, 1 c ( + 1) = = 1 *take max value 1 = 1 = 1, ( ),, ( 1) = + = *take max value = + 1 = (1 1), 1, 1, 1 (1 + 1), = + 1 = *take max value = + = = 1 = *take max value = = The earliest completion time is minutes. d ritical path = ctivities with float time are not on the critical path. These activities would be, and. The answer is. reate dummy activities,, since has both and as immediate predecessors and has both and as immediate predecessors. lternatively, could have a dummy activity instead of, and could have a dummy activity instead of.,,, 1,,,,, aths Quest 1 urther athematics V Units and ifth dition Solutions anual

7 TOPI 1 irected graphs and networks XIS 1. 1 ctivity letter Immediate predecessor Time (h), 1, and are first edges and have no predecessors. is a predecessor of. is a predecessor of and. also a predecessor of. (to skip a parallel edge use dummy edge ) and are predecessors of.,, ʹ,, 1 1 The critical path is L P.,, 1 1 1, ,, 1,, 1,,,,, 1, L,, 1, 1, , orward Scanning 1 1, 1 1 (1 + ), 11,, 11 Q, 11, P,, = 1 + = 1 = + 11 = *take maximum value,,,, 11 ackward Scanning, 1,, ',, 1, 1 (1 ) ( ),, 1 1 = 1 take maximum value* =, 11 y examining outf low at (1 + + ) and inf low at ( ), the maximum possible f low could be. a orward scan shows that the earliest completion time is minutes., 1,, 1, 1,,,, ( 11) ritical path is loat times for the non-critical activities :1 minute :1 minute :1 minute 1 a b ackward scan shows that the critical path is. c loat times for non-critical activities: loat() = 1 = minutes loat() = = 1 minute, 11,,,, 1,, arliest completion time = 1 days aths Quest 1 urther athematics V Units and ifth dition Solutions anual

8 1 TOPI 1 irected graphs and networks XIS 1. b ritical path, from the network, where the triangle numbers are equal to the box numbers. c If ctivity is reduced to days, the path is reduced to days, which is still greater than the other possible path ( ). 1 loat time for activity = 1 = days The answer is. 1 a, and L are predecessors of. L is a predecessor of. ctivity letter Immediate predecessor Time 1, 11 1 L,,, L b and c, and are predecessors of.,,,,, 1, 11,,, 1 L, and are predecessors of L. L is a predecessor of. is a predecessor of and. is a predecessor of and. is a predecessor of., and have no predecessors. ctivity letter Immediate predecessor Time (continued ),, The earliest completion time is minutes. The critical path is. (Where the triangle numbers are equal to the box numbers.) d on critical activities are:,,,,,,, L and. loat times are : = : = :1 = :1 = : 1 = :11 = : 1 = L: 1 = : = 1 a and are first edges and have no predecessors. is a predecessor for and. is a predecessor for. is a predecessor for. aths Quest 1 urther athematics V Units and ifth dition Solutions anual

9 TOPI 1 irected graphs and networks XIS 1. 1 and are predecessors for. is a predecessor for and. and are predecessors for., 11,,,,,,,,, b 11 = 1 = * ( + ) ( + ) 11 ( ),, + = = * ( ) =,,,,,,,, = * = ( + ) 1 1 (1 ) (1 ) (1 + ) 1 1 = 1* = = 1* = 1 + = * 1 + = arliest completion time = hours. c ritical path d ctivity Time ST T loat Time a ctivity letter Immediate predecessor Time (h) 11 (continued ) aths Quest 1 urther athematics V Units and ifth dition Solutions anual

10 1 TOPI 1 irected graphs and networks XIS 1. ctivity letter and are first edges and have no predecessors. Immediate predecessor Time (h) 1,, is a predecessor of and. is a predecessor of. is a predecessor of. is a predecessor of. is a predecessor of and. and are predecessors of as well (as,,, but also needs, use a dummy edge ). (11 + ) 1 = 1* 1 = 1 ( + 11) 11 1, (1 ) (11 + ), 1, 1 (1 ),, ʹ, 1 1, 1 + = = = 1* 1 = = *,, 1, ( + ) (1 1) ( + 1) 1 1 orward/ackward Scanning b arliest completion time = hours c ritical path = d ctivity Time ST T loat Time = 1* = 1 + = 1 + = aths Quest 1 urther athematics V Units and ifth dition Solutions anual

11 TOPI 1 irected graphs and networks XIS a and b 11 = 1 = * 1 = 1 ( + ) = 1 + X = 1* (1 ), ( ) 1 + = ( + ), 1 + = *, 1 X, 1, 1 ( P) 1 = 1* 1 X = 1 P, 1,,, 1,, ( + ), (11 + ) 1, 1 1 Q, 1, (1 ) (1 + ) L, ( Q) 1 + = 1* 1 + = 1 ( L) = = 1* = arliest completion time = days ritical path = Q c loat time for activity X= T ST Time = 1 1 = days = * = + = * 1 + = + P = + = + Q = * d When is reduced by days to days, the earliest completion time is reduced to days. The new critical path becomes P. is no longer a critical activity. 1 a LST for = = 1 So, LST for I = 1 So, activity time I = 1 = LST for = = LST for = 1 = 1 So, LT for and = 1 LST for = 1 1 = 1 So, LST for = 1 So, activity time = 1 = LST for and = 1 1, 1 1, 1,, 1,, 1,, 1, I,, b ritical path = I c loat time for = = a ctivity time for = = LST for = = 1 LST for = 1 LST for = 1 = So, LST for = 1 So, activity time = 1 = 11 LST for = 1 = 1 LST for = = 1 So, LT for = 1 aths Quest 1 urther athematics V Units and ifth dition Solutions anual

12 1 TOPI 1 irected graphs and networks XIS , 1, 1 ʹ,, 1, 1 1 1,, 1, 11, 1 1 I,, 1, b ritical path = I c loat time for = 1 1 = 1 1 a Immediate predecessors of and are, and respectively. ST for = hours ST for = + + = 1 hours b ST for = 1 and ST for = So, activity time X = 1 = hours,,,,, X, I,, S 1 1 I,,,, c ritical path = X d arliest completion time = LST for = = 1 LST for = 1 = hours after start 1 a weeks b inimum time is 1 weeks. c ritical path is I d Slack time is 1 = weeks e Stages along the critical path can be shortened: I (from the critical path) f fter stages and being reduced to weeks the new critical path will be I with a minimum time for completion 1 weeks xercise 1. etwork flow 1 rom To low capacity S S T T U 1 T V 1 U V 1 S T U 1 1 S T U low apacity S minimum = 1 S T 1 minimum = 1 T U 1 1 V V rom To low capacity S I 1 1 minimum = 1 a low capacity = 1 b This does meet the demand as V requires 1. aths Quest 1 urther athematics V Units and ifth dition Solutions anual

13 TOPI 1 irected graphs and networks XIS low apacity minimum = 1 minimum = 1 a low capacity = 1 b o this does not meet the demand as requires 1. a 1 S T U 1 1 V b low apacity S 1 T U T 1 V (inimum = ) 1 T U 1 (inimum = 1) Total low apacity = 1 a 1 V b low apacity Total low apacity = 1 (as shown in question ) + 1 ( ) = onsidering outflow at and inflow at, the largest possible flow is 1. ut 1 ut ut 1 ut 1 ote that cut is in fact not a cut as it fails to stop all flow. Some possible cuts are: UT1= + = 11 UT= + + = 1 UT= + = 1 ny other cut is more than 11. inimum cut = maximum flow = 11. a ut is invalid as it does not stop all flow from to. b UT1= = UT= = UT= = c 1 1 a inimum cut is = Therefore the maximum flow is. Q 1 11 P O b aximum flow = 1 ( ) 1 a Q P 1 O b aximum flow = ( ) 11 a rom To low capacity b 1 1 rom To low capacity Q O 1 Q 1 O P 1 P 1 aths Quest 1 urther athematics V Units and ifth dition Solutions anual

14 1 TOPI 1 irected graphs and networks XIS 1. 1 Q 1 1 O 1 1 a The inflow of is. 1 b dge capacity flowing out of is 1. 1 c The outflow from is the minimum of a and b, so 1. 1 a The inflow of is + =. b dge capacity flowing out of is. c The outflow from is the minimum of a and b, so. 1 a 1 low capacity outf low } P meets demand of so can only have i low capacity = ii This doesn t meet the demand as requires. b 1 1 O P 1 1 Q low apacity 1 Q 1 minimum = P 1 Q 1 minimum = 1 1 O P i low capacity = ii This does meet the demand as P requires. 1 a b c rom To low capacity rom To low capacity aths Quest 1 urther athematics V Units and ifth dition Solutions anual

15 TOPI 1 irected graphs and networks XIS 1. 1 d 1 a b rom To low capacity 1 rom To low capacity 1 (, but ) (, but 1) () but doesn t use all, so extra from not required and extra from not required low capacity 1 (minimum flow = ) apacity met on this flow, so flow from and not required. c d 1 a i (inimum flow = ) (inimum flow = ) Total low apacity = + + = 1 (inimum = ) (inimum = ) (inimum = ) not required (inimum = ) Total low apacity = 1 (inimum = ) (inimum = ) 1 (inimum = to and ) (inimum = ) Total low apacity = 1 1 aths Quest 1 urther athematics V Units and ifth dition Solutions anual

16 1 TOPI 1 irected graphs and networks XIS 1. ii low apacity 1 1 (inimum = ) (inimum = ) Total low apacity = b i 1 1 O P 1 1 Q ii low apacity 1 1 O P (inimum = 1, excess ) P 1 Q 1 (inimum = 1, excess ) P (inimum = ) Total low apacity = 1 In question 1b, the outflow from is. The answer is. 1 1 a ut is invalid since it does not stop all flow from to. b ut 1 = = ut = = ut = = ut = = c inimum cut = maximum flow = = a i Q P 1 1 O ii aximum flow = 1 ( ) 1 b i 1 1 Q O ii aximum flow = 1 ( + + ) 1 reeway 1 reeway P I a There would be a traffic jam because inflow > outflow. ode can only handle ( + ) = 1. b t node, the traffic should flow smoothly as the inflow (1) is less than the capacity of flows leading from. ( + + = 1) c I 1 1 reeway reeway s node has the potential to carry another cars, then join a road between and. xercise 1. ssignment problems and bipartite graphs 1 lectricity produced = supply kwh, kwh, and kwh Total = 1, kwh Towns supplied = demand Town = % of 1 = kwh Town = % of 1 = kwh Town = 1% of 1 = kwh Town = 1 ( + + ) = kwh S p1 p p aths Quest 1 urther athematics V Units and ifth dition Solutions anual

17 TOPI 1 irected graphs and networks XIS 1. 1 a s1 s S 1 1 b Send from S1 to, from S to, 1 from S to, from to, from to and from to. (This may not be the cheapest method.) S S ased on information in question, rian and hris between them have more different dishes than avid and arl. The answer is. rian and hris have ish, Soup, eef and essert while avid and arl have essert, eef and ish. X Y Z Subtract Smallest in ow= in ow= in ow= X Y Z 1 umber of lines = number of columns X Y Z There is only 1 possible allocation: Y, X, Z Total time = + + = hours W X Y Z Subtract Smallest in ow= in ow= in ow= in ow= W X Y Z umber of lines = number of columns W X Y Z llocation: Y, W, X, so Z Total time = = 1 hours 1 ow reduction Subtract smallest in ow1= = = = 1 1 olumn reduction Subtract smallest in olumn 1= = = = Smallest uncovered number = Subtract overall smallest number = lines required inimum total allocation = Solved by ungarian lgorithm ow reduction subtract Smallest numberinow 1= = = 1 = olumn reduction subtract Smallest numberinolumn1= = = = aths Quest 1 urther athematics V Units and ifth dition Solutions anual

18 1 TOPI 1 irected graphs and networks XIS Solved by olumn reduction. inimumtotal allocation= = Produces supply produces 1 copies per month actory 1 actory istributed emand Queensland = Victoria = 1 = f1 Q S f V 1 rom the bipartite graph it can be said that: al and rances, in total, have more hobbies than Sam and Will. The answer is. Sam nitting rances al atherine ogging ooking hess Will Photography 11 W X Y Z Subtract Smallest in ow= 1 in ow= 1 in ow= 1 in ow= W X Y Z W X Y Z Z, W X (as no other to X) Y Total time = 1 hours ame oll Truck 1 Subtract Smallest in ow1= ow = ow = ame oll Truck inimum cost 1 = $ etween Store and it is cheaper to buy the oll from Store T Store Item 1 ame oll Truck + + = $11 1 Subtract Smallest in ow= in ow= in ow= 1 llocation.,, + + = 11 The answer is. 1 a 1 ar ow eduction Subtract Smallest in ow1= 1 = 1 = 1 = 1 = 1 aths Quest 1 urther athematics V Units and ifth dition Solutions anual

19 TOPI 1 irected graphs and networks XIS a ar olumn eduction Subtract Smallest in olumn 1= = = = = 1 1 ar1 b ar 1 ar ar ar ar c i 1, 1,,, or 1, 1,,, ii Total = $ ow reduction Subtract smallest in ow1= 1 = 1 = 1 = 1 olumn eduction Subtract smallest in olumn 1= = = 1 = 1 inimum number of lines required = Solved at olumn reduction. inimumtotal allocation= = b ow reduction subtract Smallest numberinow 1= 1 = 11 = 1 = = Solved by row reduction inimumtotal allocation= = 1 a 1 ow eduction Subtract Smallest numberinow = = = = olumn eduction Subtract Smallest numberinolumn 1= = = 1 = b ungarian lgorithm Smallest uncovered number = Overall smallest number = (One possible result) aths Quest 1 urther athematics V Units and ifth dition Solutions anual

20 1 TOPI 1 irected graphs and networks XIS 1. c 1 d i,,, 1 or,,, 1 ii Total= = 1 minutes 1 1 T 1 U V W ow eduction Subtract Smallest in owt= U = V = W = 1 T 1 U 1 1 V 1 W olumn eduction Subtract smallest number fromolumn1= 1 = = = 1 T U V W ungarian lgorithm Smallest uncovered number = 1 T U 1 V W 1 Smallest number = 1 T U V W 1 a T, U 1, V, W b Time = = 1 minutes 1 a L irst modify the minimisation problem, by subtracting each number by the overall largest value,. 1 L 1 1 ow eduction subtract Smallest fromrow = 1 L = = = 1 L olumn eduction subtract Smallest fromcolumn = 1 = = 11 = b 1 L c To solve, continue with ungarian lgorithm Smallest uncovered number = L Subtract smallest number from all = L 1 epeat Smallest uncovered number = L Subtract smallest number from all = 1 L en Louise ark lgebra alculus unctions ancy eometry en lgebra, Louise eometry, aths Quest 1 urther athematics V Units and ifth dition Solutions anual

21 TOPI 1 irected graphs and networks XIS 1. 1 ark unctions, ancy alculus d verage score = = 1.% P1 P P P V1 V V V ow eduction subtract Smallest number from ow 1= 1 = = = 1 P1 P P P 1 1 V1 V 1 V V 1 olumn eduction subtract Smallest number fromolumn1= = = = 1 V1 V V V P1 P P P Smallest uncovered number = 1 P1 P P P V V V 1 V Subtract Smallest number = 1 P1 P P P V1 V V V 1 1 a V1 P, V P, V P1, V P or V1 P, V P, V P1, V P b Total= = km aths Quest 1 urther athematics V Units and ifth dition Solutions anual

Time Planning and Control

Time Planning and Control Time Planning and ontrol Precedence iagram G44 NGINRING MNGMNT 1 Precedence iagramming S ctivity I LS T L n important extension to the original activity-on-node concept appeared around 14. The sole relationship