CRITICAL PATH ANALYSIS

Transcription

1 UMMRY H IION MH RIIL PH NLYI he main ideas are covered in efore the exam you should know Q dexcel MI OR When you need to use dummy activities. he main ideas in this topic are rawing ctivity or Precedence Networks Performing orward and ackward Passes and Identifying ritical ctivities rawing ascade harts and Resource Levelling How to draw precedence networks. as you possibly can. How to perform forward and backward passes on a precedence network to calculate early and late start times. How to find the critical activities. How to calculate the various types of float. How to draw a cascade chart and construct a resource histogram. Where resource levelling is required and how to make effective use of float to improve efficiency. What is meant by crashing a network. erminology n activity is a task which needs to be done and takes an amount of time/resources to complete. Precedence tables show the activities that need to be done together with their duration and their immediate predecessors. Precedence networks show the sequence of the activities. he network must have one start node and one end node. n event is the start/finish of one or more activities. ummy activities are used to keep the correct logic and to ensure each activity is uniquely defined by (i, j) where i is its starting event and j is the finishing event. his is incorrect his is correct It can be a good idea to do an initial sketch as it s often possible to make your diagram clearer by repositioning activities to avoid them crossing over one another. orward pass establishes the earliest times that events can happen. ackward pass establishes the latest time that an event can happen. ritical activites are those whose timing is critical if the project is to be completed in the minimum time. he critical activities will form a path through the network loat is the amount of time by which an activity can be delayed or extended. Independent float does not affect other activities. Interfering float is shared between two or more activities. xample: he table shows the activities involved in creating a small patio in a garden. ctivity Name ask ime (hrs) Preceding ctivities lear Garden Measure area esign Patio Purchase fencing uy pots and plants, Plant all pots G Purchase paving H onstruct Garden,,G he network for this precedence table () () () () G() () H() () isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

2 he forward and backward pass his is the earliest start time for the next activity his is the latest start time for the next activity () 7 9 () () G() () () H() he duration of the project is 0 hours he critical activities are,,, G and H () loat activity float type hours independent hours Interfering (with ) hour Interfering (with ) In this example there are two hours of float shared between activities and ascade hart and Resources levelling G H ascade hart shows each activity set against a time line. loat time is sometimes shown by using shading. ependencies are shown by vertical lines. he cascade chart can be adjusted by using the float times to make use of resources more efficient. If activity needs two people and all the rest can be done by one person, then the resource histogram looks like this (note that people are needed in the second hour). G H If only three people are available for the first three hours, but a fourth friend can then come and help for an hour, we could move activity within its float time to make this possible. his would make the cascade chart look like this he resource histogram would now look like this isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

3 RVIION H IION MH YNMI PROGRMMING he main ideas are covered in Q dexcel MI OR he main idea in this topic is: inding the shortest (or longest) route through a network by working backwards from to efore the exam you should know: hat dynamic programming is a technique for solving multi-stage decision making problems by working backwards. What is meant by stage and state variables, actions and values (or costs). he meaning of sub-optimisation. How to set up a dynamic programming tableau. How to calculate the sub-optimal strategy at each stage. hat a minimax route is one on which the maximum route is as small as possible. hat a maximin route is one on which the minimum route is as large as possible. ynamic programming is used to solve some optimisation problems modelled y networks, though the solution is usually presented in a table. In a dynamic programming network, the nodes are referred to as states, directed arcs are called actions and the transition from one state to the next is a stage. he method starts at (state 0) and considers all the nodes joined directly to (, and in the diagram), which are called stage nodes. We find the best route from each of these to then move on to the stage nodes (in this case, and ). he optimal route from each of these to is found using the best routes from stage and so on until we reach. t each stage you work out the best strategy from that point. his is called the sub-optimal strategy. xample : he network below shows the weight limit, in tonnes, on vehicles that use the road. transport company making regular deliveries from to wants to use the largest possible vehicle in order to minimise their outgoings. What is the heaviest vehicle they can use? Optimal route Maximum weight of lorry is tonnes tage tage tage he problem is about finding the route with the greatest minimum weight from to so it is an example of a maximin problem. or each node we are choosing the minimum arc on the current route to that node. olution stage tate (node) ction (into node) Value (Route min) urrent max 0 () 0 * () 0 * () 0 * 0 () 0 min (,)= min (,)=* () 0 min (,)= min (,)= min (,)=* () 0 min (,)= min (,)=* 0 () 0 min (,)= min (,)= min (,)=* isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

4 xample company is planning to build three new houses, and at the rate of one every three months. hey can build the houses in any order but the profit will be affected by the workers available and the supply costs, which are subject to variation. he expected profits in thousands of pounds are shown in the table ime period lready completed Profits ( 000) and - - and and (a) Represent the information on a network such that the optimal strategy will correspond to the longest path through the network. (b) Use dynamic programming to determine the order in which they should build the houses in order to maximise profit olution he letters in the nodes represent the houses already built at that stage of the project 0 7, 7 none 0, 9,, , (b) tage (time period) tate (houses built) ction (house to be built) Value (profit) max 0 (, ) 0 () 7* (, ) 0 () 9* 9 (, ) 0 () 7* 7 0 () 0 () 0+7=7* 7 () +9 () 0 () 7+7=0* 0 () +7=8 () 0 () 70+9=9* 9 () +7= 0 0 () +7= () 0+0=90* 0 () 8+9=87 It can be very helpful to write what is happening in the state and action columns. In this case the state is the houses that have already been built and the action is the house that is about to be built. he houses should be built in the order then then he maximum profit is isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

5 RVIION H IION MH GM HORY he main ideas are covered in Q dexcel MI OR he main ideas in this topic are: inding out whether a two person zero sum game has a stable solution. If a stable solution does not exist, using a graphical method to find an optimal mixed strategy. as the other player s loss. In a two person, zero sum game, one player s gain is the same efore the exam you should know: What is meant by a two person, zero sum game. How to interpret a payoff matrix. How to find a playsafe strategy. What is meant by a stable solution. How to find the value of a game. How to simplify games using dominance. How to find an optimal mixed strategy for a game with no stable solution. How to convert a two person zero sum game into a linear programming problem. pay-off matrix represents the gain for one of the players for each combination of strategies for the two players in a two-person game. inding a play-safe strategy for a zero-sum game In the pay-off matrix for, find the minimum entry in each row; will use the strategy that involves the greatnesses of these values (maximin).hen find the maximum entry in each column, will use the strategy that involves the lowest of these values (minimax). stable solution occurs if and only if the maximum(row minimum) = the minimum(column maximum). assumes uses his play-safe strategy and cannot do better by using an alternative strategy and assumes uses his play-safe strategy and cannot do better by using an alternative strategy. stable solution is sometimes called a saddle point. xample: nna and arry play a zero sum game. he game can be represented by the following payoff matrix for nna - - (a) how that there is not stable solution olution Row max - - ol min - - maximin minimax so solution is not stable ominance: row/column can be eliminated if all the entries in that row/column are less than or equal to the corresponding entries of another row/column because the player would never choose that row/column. inding a mixed strategy using a graph: If player has two possible strategies, assume he adopts the first strategy with probability p and the second with probability -p. he expected payoff for will depend on which strategy chooses so plot s expected payoff for each of s strategies as lines on a graph. s best strategy is to play with the value of p which gives the highest point at the intersection of the lines. his will maximise s minimum return (maximin). he value of the game can be calculated using this value of p. xample (cont) (b) xplain why nna should never play strategy (c) ind the best mixed strategy for nna and give the value of the game for her. - - isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

6 the urther Mathematics network V 07 olution (b) nna should not play since row is dominated by row since >, > and >. (c) Let nna play with probability p - - nna s expected payoff if arry plays strategy: with probability ( p) nna s optimal strategy p p = p : ( ) : p+ ( p) = p : p p = 7p ( ) olving for p : p = p p = p = rom the graph you can see that the optimal expected return for nna occurs at the intersection of the lines for arry playing strategies and. o nna should play with probability and with probability Value of the game: 0 = = = inding a mixed strategy using the implex algorithm he graphical method cannot be used if player has more than two strategies so we must use the simplex method. p p Let the probabilities of each of s strategies be,, p etc. If there are any negative elements in the payoff matrix, add k to make them all positive. Model the situation as a linear programming problem: Maximise V-k (this is the expected payoff from the original game) ubject to V expected payoff from playing strategy V expected payoff from playing strategy etc p + p + p +... (because the sum of the probabilities cannot exceed ) p,, p 0 (no negative probabilities) p his information can now be put into an initial simplex tableau and solved in the usual way. isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

7 RVIION H IION MH LLOION PROLM N H HUNGRIN LGORIHM he main ideas are covered in Q dexcel MI OR he main ideas in this topic are: xtending the ideas introduced in matchings to find a maximum matching and its associated cost. Using the Hungarian algorithm to find an optimal, minimum cost solution. efore the exam you should know: hat you will be working on a matrix or array of numbers called the payoff matrix. How to reduce a matrix to one containing zeros and interpret this to give an optimal allocation. How to apply the Hungarian algorithm to minimise costs. hat the matrix of numbers must be square before you can apply the Hungarian algorithm. How to deal with maximisation problems by subtracting all the entries in the matrix from a constant. xample olution: irstly reduce the rows P Q R ubtract 0 from each entry XYZ taxis have four drivers, lan, etty, hris and 0 ubtract 0 from each entry ave. One morning they have four bookings to collect people (P, Q, R, ) and take them to the ubtract 7 from each entry station for the 0800 train. he table shows the time 0 ubtract from each entry in minutes it will take for each driver to drive form their house to the customer s house and then his is not a maximum P Q R transport each customer to the station. matching since all the P Q R zeros can be covered with lines Reduce columns llocation problems are solved by reducing the payoff matrix P by Q subtracting R the least his value is in now each a maximum row (or column) form all the entries in that row (or column). You now have a matrix 0 0 showing 8 relative costs matching. and the least entry in each he row manager (column) of will XYZ now taxis be wants zero. hese to minimise zeros will the represent a minimum 0 cost solution. lan good P, way etty to see Q whether you total have time a maximum taken, who matching should is he to allocate find the to minimum each number of lines needed to cover all the hris zeros R, in ave the matrix journey? otal time = = 8 minutes he Hungarian algorithm. If the payoff matrix is not square, add in dummy row(s) or column(s) of equal numbers to make it square.. ubtract the minimum entry from each row from all the entries in the row.. If necessary repeat step for the columns.. raw the minimum number of horizontal and/or vertical lines to cover all the zeroes.. If the number of lines is equal to the number of columns in the matrix, the positions of the zeroes indicate an optimal matching; if not, go to step.. ugment the matrix: find the smallest element not covered; subtract it from the non-covered elements and add it to any elements covered by lines. Go to step. isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

8 Note: If you need to maximise, subtract every number in the original payoff matrix from the largest number in the matrix before applying the algorithm. he final payoff is then found by adding the original payoffs in the cells used in the optimal matching. xample: he average scores for five members of a quiz team are shown in the table: music sport geography history General knowledge lan renda ally 7 7 avid 8 dwin different person must be chosen to answer questions in each of the five rounds in the final. Using their past performance, who should do each round in order to maximise score? (a) xplain why you should replace each entry x by 0 x before using the Hungarian lgorithm (b) orm a new table by subtracting each number from 0. Use the Hungarian algorithm to allocate the sports team members. (c) tate the expected score for the team based on their practice scores. olution: (a) ecause the problem is a maximisation problem and 0 is the largest number in the matrix. (b) music sport geography history General knowledge lan renda 0 ally 7 avid 8 dwin Reduce rows music sport geography history General knowledge lan 0 renda 0 ally 0 0 avid 0 dwin 0 0 Reduce columns music sport geography history General knowledge lan 0 renda 0 ally 0 0 avid 0 dwin olution is not optimal as all zeros can be covered with lines ugment by music sport geography history General knowledge lan 0 renda 0 0 ally avid 0 dwin 0 0 It now takes lines to cover the zeros so matching is optimal lan sport, renda music, ally History, avid Geography, dwin General Knowledge (c) xpected scores: = 88 isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

9 RVIION H IION MH MHING he main ideas are covered in Q dexcel MI OR he main ideas in this topic are: Modelling real situations using bipartite graphs. Using the maximum matching algorithm to solve problems efore the exam you should know: What is meant by a bipartite graph. hat a matching maps vertices in one set to vertices in a second set. No vertex may be used more than once. or a complete matching the two sets must have the same number of vertices. complete matching pairs every vertex in the first set to one in the second set. maximal matching is one where there is no solution that uses a greater number of edges. bipartite graph has two sets of vertices, X and Y such that the edges only connect vertices in set X to those in set Y and never to vertices in the same set.,,, are the vertices in set X xample P Q R P, Q, R, are the vertices in set Y college has to fit rench, geography, History, Maths and cience into a single timetable slot. here are five teachers available all of whom can teach two or more of these subjects. nn can teach rench and Geography ob can teach rench, Maths and cience arol can teach Geography and History avid can teach Geography, Maths and cience laine can teach History Maths and cience How should the college allocate the staff so that all subjects are covered? complete matching is always maximal, but a maximal matching is not necessarily complete. he algorithm for finding a maximum matching. lways start with an initial matching. If the matching is not maximal it can be improved by finding an alternating path. n alternating path: tarts on an unmatched vertex on the right hand side onsists of edges alternately not in and in the matching inishes on an unmatched vertex in the second set. If every vertex is now matched so we have a complete matching. If it is not, then repeat step.. he solution consists of: edges in the alternating path but not in the initial matching; edges in the initial matching but not in the alternating path. isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

10 olution tart by drawing a bipartite graph to model the situation G H hese vertices represent the teachers M hese vertices represent the subjects tart with an initial matching: G, M, H, this is not a maximum matching since laine has not been allocated a subject and there is no-one to teach rench. G H M We must try to find an alternating path tart on an unmatched vertex on the right hand side () hoose an edge which is not in the initial matching () hoose an edge which is in the initial matching (G) G * hoose an edge which is not in the initial matching (G) hoose an edge which is in the initial matching (H) H hoose an edge which is not in the initial matching (H) M We have now reached which was not in the initial matching so we have breakthrough. * he laternating path is G H he solution consists of: edges in the alternating path but not in the initial matching:, G, H edges in the initial matching but not in the alternating path: M, so the solution is: nn teaches rench ob teaches Maths arol teaches Geography avid teaches cience laine teaches History isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

11 he main ideas are covered in Q dexcel MI OR he main ideas in this topic are: Modelling flows using bipartite graphs. RVIION H IION MH inding the maximum flow through a network. he maximum flow-minimum cut theorem NWORK LOW efore the exam you should know: What is meant by source, sink and capacity. What a cut is. How to find an initial flow. he meaning of a flow augmenting path and how to find them. How to use the labeling procedure. What is meant by excess capacity and back capacity. What is meant by a saturated arc. he maximum flow minimum cut theorem. How to insert a super-source and super-sink into a network. he algorithm for finding a maximum flow. lways start with an initial feasible flow, found by inspection.. Label each arc with the flow along it, shown by an arrow pointing back towards the source the excess capacity, which is the amount by which the flow could be increased, shown by an arrow pointing forward towards the sink. ystematically look for flow augmenting paths and mark these on your network using the labelling procedure. When all paths are blocked by saturated arcs you have found the maximum flow. xample In this the directed network a) What is the maximum flow along the path? b) ind an initial flow of value 7. c) ind the maximum flow in the network 8 d) What are the capacities of these cuts olution a) he maximum flow along is (this is determined by the arc of least capacity on the path). b) flow of 7 (shown on the diagram) is with capacity with capacity isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.

12 c) low augmenting paths 0 with capacity 0 with capacity with capacity iagrams showing the flow his gives a maximum total flow 0 of 0. he flow is shown on this diagram, along with the saturated arcs. augmenting paths can be very messy. ry to keep yours as tidy as possible and always list the flow augmenting paths you have used uts cut partitions the vertices into two sets, one containing the source and one containing the sink. he capacity of a cut is the total of all the cut edges with direction going from source to sink ind the capacity of the cuts shown on the directed network: Note that only three cuts have been shown here, but there are many more cuts in this network. is the cut {}, {,,,, } It has capacity + = is the cut {, }, {,,, } It has capacity = is the cut {,, }, {,, } It has capacity + + = 0 Note that we do not add the capacity of arc as it is directed from the sink side of the cut to the source side 8 Maximum flow- minimum cut theorem. he theorem states that the maximum flow in a directed network is equal to the capacity of the minimum cut. In the example above the cut is the minimum cut and it has a value 0. his confirms that the flow of 0 found in (c) above is the maximum flow. Networks with many sources and sinks If there is more than one source ( and on the diagram) or sink ( and on the diagram) you must introduce supersource () and/or supersink (). must have a capacity + = 9 must have capacity + = 0 must have capacity + = 8 must have capacity 8 + = isclaimer: very effort has gone into ensuring the accuracy of this document. However, the M Network can accept no responsibility for its content matching each specification exactly.