D Networks MEI Conference 009: D Travelling Salesperson Problem 7 A B Route Inspection Problems (Chinese Postman) () A 7 B () 4 () C Identify the odd vertices in the network Consider all the routes joining pairs of odd vertices and select the shortest Find the sum of the weights on all the edges Shortest distance is the sum of all weights plus the extra that must be travelled Find a tour which repeats the edges found in step. Odd vertices A, B, C, D AB 7 AC AD 6 CD BD 4 BC 9 6 9 Choose to repeat AC and BD Length of all the edges is 4 Length of minimum tour 4 + 6 = 0 Possible tour ACDBCADBA 4 D () Upper Bounds: Nearest Neighbour Algorithm distance matrix complete matrix A B C D A - 7 - B 7-4 C - D - 4 - - Use nearest neighbour algorithm starting at A upper bound length 15 Lower Bounds 1 4 Delete a vertex and the edges incident on it Find a minimum spanning tree for the remaining network Reconnect the deleted vertex by the two shortest edges e.g. Delete B A C 7 B D A B C D A - 7 4 B 7-4 C - D 4 4 - A B C D A - 7 4 B 7-4 C - D 4 4-4 A complete matrix is required for NNA For a full solution, NNA should be repeated starting at each vertex in turn minimum connector + = 4 reconnect B (BC, BD) + 4 = 7 Lower bound 4 + 7 = 11 C D Repeat for all vertices. Greatest lower bound is the best lower bound Find a tour such that least upper bound > tour > greatest lower bound
Floyd s Algorithm There are circumstances in which we require all shortest paths in a network. Clearly we could achieve this by applying Dijkstra n times - one for each starting vertex. This would have complexity 0(n ). An alternative cubic algorithm is Floyd s algorithm which achieves the same ends. Example A B 7 1 6 C E The method successively modifies the following two matrices - 5 A distance (or weight) matrix. This gives the length of each arc in the network, and is symmetric unless we are dealing with a digraph. The distance from a vertex to itself is shown as 0. Where there is no direct connection (i.e. no edge) the distance is shown as α. A route matrix. The initial version of this has a column of 1 s followed by a column of s, etc. The Algorithm Step 1 Step Choose the first row and column, and highlight all elements in the chosen row and column. Note the iteration (via A) Compare every element in the chosen row or column with the sum of the corresponding entries in the chosen row and column. Initial Distance Matrix (D 0 ) Initial Route Matrix (R 0 ) destination destination A 7 5 A A B C D E start B 7 1 start B A B C D E C 6 C A B C D E D 5 D A B C D E E 1 6 E A B C D E These numbers represent distances by single stage routes. The highlighting is ready for the first iteration. D These letters represent the first vertex on corresponding single stage routes, so the first vertex encountered on the single arc route from E to D is D Step Step 4 If the calculated sum is smaller than the element then replace the element by the sum. If the element is replaced then we need to update the corresponding entry of the route matrix. The value to replace it with is found in the route matrix from the previous iteration - in the same row as the entry it is replacing, but in the column given by the iteration. D (A) R (A) Via A destination destination A 7 5 A A B C D E start B 7 14 1 1 start B A A C A E C 6 8 6 C A B A A E D 5 1 8 10 D A A A A E E 1 6 E A B C D E Step 5 Repeat steps 1 to 4 until all rows/columns have been chosen in step 1. Step 6 The distance matrix will now contain the shortest distances. The route matrix shows the next vertex en route from one to another along a shortest path. It can be used to construct complete shortest paths. D (B) Look at the highlighted row and column: 7 + 5 = 1 R (B) Via B destination destination A 14 7 5 8 A B B C D B start B 7 14 1 1 start B A A C A E C 6 8 4 C A B A A B D 5 1 8 10 D A A A A E E 8 1 4 E B B B D B 1 + = 4; 4 < 6 The route matrix shows that the first vertex on the route from E to A is vertex B
D (C) R (C) Via C destination destination A 6 6 5 7 A C C C D C start B 6 6 11 1 start B C C C C E C 6 8 4 C A B A A B D 5 11 8 10 D A A A A E E 7 1 4 E B B B D B D (D) R (D) Via D destination destination A 6 6 5 7 A C C C D C start B 6 6 11 1 start B C C C C E C 6 8 4 C A B A A B D 5 11 8 10 D A A A A E E 7 1 4 E B B B D B D (E) This shows that there is a path of length 7 from vertex E to vertex A R (E) This entry came from row 5, column in the previous route matrix Via E destination destination A 6 6 5 7 A C C C D C start B 6 4 1 start B C E C E E C 6 7 4 C A B A B B D 5 4 7 6 D A E E E E E 7 1 4 E B B B D B The final matrix D(E) shows all of the shortest distances, so the shortest distance from vertex C to vertex D is 7. The R (E) matrix shows that the first vertex on the route is vertex B. Now use the R (E) matrix again to find the first vertex on the shortest route from vertex B to vertex D. This is vertex E, so the route is now seen to be C, B, E,... Finally the first vertex on the shortest route from vertex E to vertex D is vertex D. So the final route is C, B, E, D. D: Linear Programming Introduction Many problems in Decision Maths are to do with finding the best solution to a problem, given that the solution must satisfy a number of constraints. Linear programming is, perhaps, the most obvious example of this type of problem. Linear programming was developed during the Second World War to solve logistical problems. The term LINEAR refers to the functions defining the problem which must be linear i.e. the output is proportional to the input. The simplex algorithm was developed by Dantzig in the 1940s in order to solve particular types of Linear Programming problems. The Syllabus The Simplex Algorithm: Be able to solve simple maximisation problems with < constraints and with two or more variables. Use the tabular form of the algorithm Geometric interpretation: Be able to identify tableaux (initial, intermediate and final) with points, particularly in the case of problems involving two or three variables. Show alternative feasible points and their costs. Greater than ( > ) inequalities: Understand the use of two stage simplex, and of the big M methods to construct an initial feasible solution to problems involving > constraints, including the possibility that there is no feasible solution Equality constraints: Understand how to model an equality constraint using a pair of inequality constraints. Getting started: Formulation Translating a real life problem into a linear programming problem is called formulating the problem and is an example of mathematical modelling. Each problem must have clearly defined variables, an objective function and is subject to certain constraints Slack Variables In order to enable problems to be converted into a format that can be dealt with by computer, slack (or surplus) variables are introduced to change the constraint inequalities into equalities. Each vertex of the feasible region would then be defined by the intersection of two lines where the variables equal zero. The Simplex Method The Simplex Method starts at one vertex and systematically moves round all the vertices, increasing the objective function as it goes, until it reaches the one with the optimal solution. This is easy to visualise on a dimensional problem, but can be generalised to include more variables. Once there are more than two variables, a graphical approach is no
longer appropriate, so we use the simplex tableau, a tabular form of the algorithm which uses row reduction (remember Gaussian elimination?) to solve the problem. Adaptations of the algorithm Integer programming: In many practical situations, the solution to a problem must take an integer value. This situation is much more difficult to guarantee a correct solution, but since the IP solution is almost always close to the LP solution a heuristic approach can be used. Greater than Inequalities: these lead to problems which involve minimisation. In dimensions these problems like this can be solved graphically in the same way as maximisation problems. There are also adaptations of the Simplex algorithm to deal with this type of problem. Negative values: The simplex algorithm requires all variables to be nonnegative. If a variable, x say, has a negative value, it will be replaced in the formulation by x 1 - x where x 1 < x. Other types of problems: Simplex can be adapted to solve a range of problems, such as shortest path (minimisation), maximum flow (maximisation) and game theory (maximin). Formulation Define Variables (x, y, z or x 1, x, x,...) which are usually restricted to positive values Define Objective Function (Maximise/Minimise an expression in terms of variables) Define Constraints (inequalities) restrictions on variable values Simplex Algorithm (Tableau) Our example: Let x be... Let y be... Maximise x+y Subject to x + y 16 x + y 4 y 6 x, y 0 1. Represent the problem in a tableau. Use the objective row to find the pivot column. Use the ratio test to find the pivot element 4. Divide through the pivot row by the pivot element 5. Add/subtract multiples of the transformed pivot row to/from the other rows to create zeros in the pivot column 6. Repeat until no negatives in objective row 7. Set basic variables (with columns containing a 1 and 0s) to corresponding RS, and non-basic variables to zero. Note on finding pivot column (step ) Can choose any variable in objective row with negative coefficient, but usual to pick largest magnitude, and priority is to increase original rather than slack variables (NB for two stage simplex will sometimes have minimise objective - if so then pick positive and decrease). Note on ratio test (step ) Divide each R.H.S. value by the corresponding element in the pivot column - ignore negative ratios and division by zero - choose row giving smallest and mark pivot column element as pivot. Worked example P x y s 1 s s RHS 1-1 -1 0 0 0 0 R1 0 1 1 0 0 16 R 0 0 1 0 4 R 0 0 1 0 0 1 6 R4 1-1 0 0 0 1 6 +R4 0 0 1 0-1 10 -R4 0 0 0 1-6 -R4 0 0 1 0 0 1 6 R4/1 1 0 0 0 ½ -½ 9 +R 0 0 0 1-1 4 -R 0 1 0 0 ½ -/ R7/ 0 0 1 0 0 1 6 same 1 0 0 ¼ ¼ 0 10 +½R 0 0 0 ½ -½ 1 R10/ 0 1 0 ¾ -¼ 0 6 +1½R 0 0 1 -½ ½ 0 4 -R Simplex: Two Stage and Big M methods When? For constraints Minimisation problems Two Stage Simplex Simplex relies on (0,0) being a feasible solution. If it is not, then we introduce a set of artificial variables a 1, a, etc. to all constraints (to ensure that none of the variables have a negative value) and a new objective function Q = a 1 + a + The artificial variables effectively move us from (0,0) into the feasible region.
We now need surplus (as opposed to slack) variables which are subtracted from the constraints. In a similar problem to our earlier one, the constraints could be Maximise x+y Subject to x + y 16 x + y 4 x, y 1 Firstly we need to add the slack variables to the constraints and subtract the surplus variables from the constraints. Since clearly x=y=0 is not a feasible solution we add artificial variables and form a supplementary problem giving new constraints: x - s + a 1 = 1 and y - s 4 + a = 1 The additional objective Q = a 1 + a is written in terms of the other variables and rearranged Q + x + y - s - s 4 =. We then aim to minimise Q = a 1 + a to ensure that we are in the feasible region before completing the Simplex in the normal way Q P x y s 1 s s s 4 a 1 a v 1 0 1 1 0 0-1 -1 0 0 0 1-1 -1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 16 0 0 0 1 0 0 0 0 4 0 0 1 0 0 0-1 0 1 0 1 0 0 0 1 0 0 0-1 0 1 1 1 0 0 1 0 0 0-1 -1 0 1 0 1 0-1 0 0-1 0 1 0 1 0 0 0 1 1 0 0-0 14 0 0 0 0 1 0-0 0 0 1 0 0 0-1 0 1 0 1 0 0 0 1 0 0 0-1 0 1 1 Q P x y s 1 s s s 4 a 1 a v 1 0 0 0 0 0 0 0-1 -1 0 0 1 0 0 0 0-1 - 0 0 0 0 1 0 1 - -1 1 0 0 0 0 0 1 - - 19 0 0 1 0 0 0-1 0 1 0 1 0 0 0 1 0 0 0-1 0 1 1 Since the artificial variables are now 0, these, and the supplementary objective can be removed and the remaining tableau solved ( iterations) to give the solution as before. The big M method Using the same problem as for Two Stage Simplex, we have Maximise x+y Subject to x + y + s 1 = 16 x + y + s = 4 x - s + a 1 = 1 y - s 4 + a = 1 Our aim is to maximise P=x + y - M(a 1 + a ) where M is an arbitrarily large value and hence to achieve our maximisation we require a 1 and a to be zero. Re-writing without the artificial variables: P - x - y + M(1 - x + s + 1 - y + s 4 )=0 P - (1+M)x + (1+M)y + Ms + Ms 4 = -M P x y s 1 s s s 4 a 1 a RHS 1 -(1+M) -(1+M) 0 0 M M 0 0 -M 0 1 1 0 0 0 0 0 16 0 0 1 0 0 0 0 4 0 1 0 0 0-1 0 1 0 1 0 0 1 0 0 0-1 0 1 1 P x y s 1 s s s 4 a 1 a RHS 1 0 -(1+M) 0 0-1 M 1+M 0 1-M 0 0 1 1 0 0-0 14 0 0 0 1 0-0 0 1 0 0 0-1 0 1 0 1 0 0 1 0 0 0-1 0 1 1 P x y s 1 s s s 4 a 1 a RHS 1 0 0 0 0-1 -1 1+M 1+M 0 0 0 1 0 1 - -1 1 0 0 0 0 1 - - 19 0 1 0 0 0-1 0 1 0 1 0 0 1 0 0 0-1 0 1 1 Our artificial variables are no-basic (zero) and the M only appears in their columns. They can therefore be deleted and the remaining tableau is as before for further iteration.
D: Logic and Boolean Algebra Boolean variables: p, q Truth tables NOT AND OR p ~p 1 0 1 0 0 1 0 1 p q p q 1 0 0 0 1 0 0 0 0 IMPLIES IMPLIED BY EQUIVALENCE p q p q 1 0 0 0 1 1 0 0 1 e.g. Prove ~ (a b) = ~a ~b e.g Prove a (b c) (a b) (a c) Laws of Boolean algebra p q p q 1 0 1 0 1 0 0 0 1 p q p q 1 0 1 0 1 1 0 0 0 p q p q 1 0 0 0 1 0 0 0 1 LHS RHS a b a b ~(a b) a b ~a ~b ~a ~b 0 1 1 0 0 0 1 0 1 0 1 0 0 1 0 0 1 1 0 0 1 1 0 0 0 0 0 1 0 0 LHS RHS a b c b c a (b c) a b c a b a c (a b) (a c) 1 1 1 1 0 1 1 1 1 0 1 0 1 1 0 1 0 1 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The most obvious way to simplify Boolean expressions is to manipulate them in the same way as you would manipulate algebraic expressions. A set of rules for is needed for this and these are shown in the table p ~p = 1 p ~p = 0 ~(~p) = p p p = p p p = p p 0 = p p 1 = p p 1 = 1 p 0 = 0 p q = q p p q = q p p (q r) = (p q) (p r) p (q r) = (p q) (q r) p (p q) = p ~(p q) = ~p ~q ~(p q) = ~p ~q e.g. Simplify p (~p q) Circuits = (p ~p) (p q) = 1 (p q) = p q a Complement laws Double complement Indempotent laws Identity laws Dominance laws Commutative laws Distributive laws Absorption law DeMorgan's laws this circuit can be written as b c a (b c) Logic Gates Students need to be able to apply their knowledge of truth tables and circuits to simple circuits involving logic gates: NOT, AND, OR, NAND, NOR. D: Decision Analysis Terminology Decision trees can be drawn in situations where the outcome of events is uncertain. EMV (expected monetary value) can be calculated for various courses of action to assist decision-making in situations where probabilities can be assigned to the outcomes. The decision tree has three types of nodes
Decision node Chance node Payoff node Example The organiser of a tennis tournament has to decide whether to take out insurance against bad weather. For a premium of 150, she will receive a payout of 10000 if it rains on the day of the tournament. The organiser calculates that if it rains, the losses will total 1000 but if it is fine they will make 9500 profit. The weather forecast estimates that there is a 5% chance of rain on the day of the tournament. What would you advise the organiser to do? ( 450 X 0.5) + (800 X 0.75) rain (0.5) 10000 150 1000 insurance premium losses - 450 516.50 + 6500 insure (- 150) 516.50 fine (0.75) 800 7500 1000 profit - insurance select the larger figure don t rain (0.5) insure 875-1000 ( 1000 X 0.5) + (9500 X 0.75) fine (0.75) 9500 Solution Advise the organiser to take out insurance EMV 516.50
Teaching D using the online resources Chapter 1: Linear programming, the simplex method This chapter is quite challenging for students. It can be difficult to teach and off-putting for students. It is worth taking some time to make sure that they can interpret tableaux at every stage of the working including giving values of the slack/surplus variables. Don t be tempted to take too long on this chapter it can easily become death by simplex. Students are expected to have a good understanding of what the method is doing but are not going to be asked to perform long and complicated simplex tableaux in the exam. When starting out on basic simplex and again when doing stage simplex and big M method, it is worth tackling a reasonably complex variable problem and working the simple alongside a graph. This gives students a much clearer picture of how the simplex algorithm works and what it is actually doing. This can make tacking or 4 variable problems less intimidating. Don t forget to subtract the surplus variables when tackling stage simplex. Some find it difficult to subtract multiples of M and hence make algebraic errors in the big M method Students may be asked to do an ordinary simplex but it s unlikely that they will be asked to do a full -stage or big M Two stage simplex has come up on the exam more often that big M Chapter : Networks Route inspection is the topic which students seem to find easiest so this could be tackled first. TSP uses the MST algorithms learned in D1so make sure the students are secure in this first. It is important to distinguish between TSP and RI and which to use for what type of problem. When teaching TSP, make sure students understand the distinction between the classical problem, which uses a complete network, and the practical problem, which almost certainly doesn t. When finding a lower bound for TSP, we use the method where a vertex is deleted. The resulting lower bound DOES NOT give a tour, except in exceptional circumstances. Students don t understand why it is the GREATEST lower bound and LEAST upper bound. Floyd s algorithm is straightforward but takes a bit of time to do thoroughly. When applying Floyd s algorithm students can easily get lost using the distance and the route matrices side by side. Draw attention to the way to change the route matrix Use colours highlighter pens are indispensable when doing Floyd s algorithm. Chapter : Decision analysis Decision analysis is a dynamic tool. The power of decision analysis is not in the numbers at the decision node, it is the ability to change the utilities and probabilities and see how this affects the decision node. When drawing a decision tree you can use increase in wealth (profit) or total wealth but it is important to be consistent. EMV can be calculated in the same way as expectation in statistics. You do not need to have drawn a decision tree in order to calculate the EMV for a situation. When considering utility always use total wealth, since utility is a function of wealth. The utility function is specific to a person or situation and is given for each problem. It can be loosely described as a measure of happiness Chapter 4: Logic and Boolean algebra Students usually prefer to use truth tables to prove logical propositions. They need to know, or be able to work out, the truth tables. They do need to know how to use the laws of Boolean algebra as well it can be specified on the exam paper. Students must be able to distinguish between a combinatorial circuit and a switching circuit. They can be asked to produce tables of outputs for circuits They are not required to memorise the half adder. D Online resources
Teaching D using the online resources Every chapter contains A glossary of terms An end of chapter assessment the answers to this are in the teachers resources Every section within the chapter contains A study plan Crucial points that students need to be aware of when learning the material Notes that explain the topic and more examples than are in the textbook. Worked solutions to some of the questions in the textbook. A multiple choice test which can printed off and used in class or be done online by registered students. They will receive instant feedback. The additional resources are being added to all the time. Below is a list of some of the things currently available Topic section about Additional resources Linear The simplex method The initial simplex tableau, Interactive excel Programming Manipulating systems of equations, The simplex method Problems involving Terminology, Post-optimal analysis, Interactive excel constraints, Two-stage simplex, The big M method Problem solving Equalities and practicalities Networks Route inspection Euler Cycles. Solving route inspection problems Power point Travelling salesperson Hamiltonian cycles. The travelling Power point salesperson problem, The practical problem and the classical problem, Upper and lower bounds, Tour building Floyd s algorithm Drawing an activity network, identifying a critical path, earliest and Power point Decision analysis Logic Decision analysis Propositional calculus and truth tables Boolean algebra Switching circuits latest event times, float Construct and interpret Decision trees. Calculate EMV. Understand Utility functions and use utility to compare alternatives Be able to understand and form compound propositions using the correct symbols. Use truth tables to analyse propositions Manipulate Boolean expressions and use the laws of Boolean algebra Be able to model compound propositions with simple switching and combinatorial circuits.