Solution 1 Linear programming illustrating maximisation

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1 20 Solutions Lindo software at and similar proprietary products, which facilitate interactive use from the keyboard or customised subroutines linked directly to form an integrated program, have been adopted where appropriate in this text. Solution 1 Linear programming illustrating maximisation This type of problem can be readily solved using the linear programming mathematical technique developed to determine a maximum value such as profit or minimum cost, where availability of the associated resources is constrained. The descriptor linear implies proportionality while programming refers to the mathematical process of determining an optimum allocation of resources. Problem formulation The steps needed to be undertaken in solving a linear programming problem are as follows: (1) Define the decision variables, these being the controllable inputs in both the objective and constraint equations and which describe the number or amounts of each item to produce. (2) Describe the objective function in terms of the decision variables, this being the subject to be optimised, for example, profit, output, etc. expressed as a formula. (3) Establish the constraint value for each resource that has an imposed limit and express in terms of the decision variables. (4) Solve for the decision variables and substitute into the objective function and determine its quantity or amount. In this first problem the decision variables can be established as follows: Part (a) Decision variables: Let X = number of detached houses constructed Y = number of semi-detached houses constructed Modern Construction Management, Seventh Edition. Frank Harris and Ronald McCaffer with Francis Edum-Fotwe John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd. 108

2 Solutions 109 Row 1. Total profit objective function: MAX = 1000 X Y Rows 2 and 3. Constraints to prevent negative values of X and Y: Row 4. Constraint on the builder s capacity: Row 5. Constraint on detached: Row 6. Constraint on Type B design: X > 0 Y > 0 X + Y 400 X ( X + Y) 0. 4 X Y <= 200 The equations plotted graphically in Fig. S1.1 illustrate the area of feasible solutions and the optimum profit is the solution that maximises the objective function 1000X + 800Y. Clearly, various combinations of X and Y will lead to the same profit. For example, a profit of is obtained for all pairs of X and Y lying on the line 1000X + 800Y = Y Objective function Max = 1000X + 800Y 400 X + Y 400 2/5(X ) + 2/3(Y ) 200 TP = 300, 100 Feasible region ¼X ¾Y 0 Maximising 400 X Fig. S1.1

3 110 Self-learning exercises The profit will continue to grow with any line drawn parallel to the objective function, moving away from the origin until the line is just tangential to the feasible area, the objective function then being at the tangential point. The coordinates of X and Y at the tangent point (TP) are 300 and 100 respectively, which when substituted into the objective function yield a profit of In other words, to optimise profit, 300 detached and 100 semi-detached is the best mix. The amount of each design is shown below. Summary Objective value: Variable Value X 300 Y 100 Row Slack or surplus The above summary illustrates that the constraints for all rows were not violated in not being negative values. However, for Row 6, the constraint equation of 0.4X Y substituted with X, Y values of 300 and 100 respectively gives = 13, that is, produces a slack amount of 13 houses, meaning that this resource is underutilised. A negative, that is, surplus, value would have indicated insufficient availability of this resource. Part (b) Shadow prices or duality Clearly, there would be little point in increasing the limit of 200 houses on design Type B, as the constraints on Rows 4 and 5 are presently fully utilised, that is, no slack. However, it would be of interest to determine the effect on profit of increasing the limit on one or even both these resources, namely, the builder s total and type B design capacities. Called the shadow/dual price, this is the amount the objective function would increase by raising the constraint by one unit and is shown in tabular format below for each constraint manipulated independently. For example, Row 4: considering the constraint on total production capacity (Row 4), increase the capacity to produce to 401 houses. The new constraint becomes X + Y = 401( Fig. S1. 2) Again, the objective function 1000X + 800Y = profit and the optimum solution coordinates are X = and Y = New profit = = The extra profit generated by adding 1 house is 950, that is, 950/house.

4 Solutions 111 Y Objective function Max = 10X + 8Y 4000 X + Y 401 2/5(X ) + 2/3(Y ) 200 TP = , Feasible region ¼X + ¾Y 0 Maximising 4000 X Fig. S1.2 This policy is not justified as the extra resources required cost 1000/house. In a similar manner, the constraint of Row 5 produces a shadow price of 200 if the 75% limit on detached is raised as summarised below for each constraint respectively: Objective value: Variable Value Reduced cost X Y Row Slack or Dual price surplus It will be observed in the above summary that increasing production capacity improves the solution by the dual price values. Reduced cost The result also shows the reduced cost, which is the amount that would have to be added to, or subtracted from, the variable coefficient in the objective function in order for that variable to feature in the solution, meaning an increase in the coefficient when maximising, and a decrease when minimising. In the above example, X and Y are both used in the solution and

5 112 Self-learning exercises therefore have zero reduced cost values. Nothing further could be gained, since no other variable is involved in the analysis. Variable Value Reduced cost X Y Alternative analysis by the canonical method The primal arrangement of the problem is: Row 1 Max = 1000X + 800Y Row 2 X + Y < = 400 Row X 0.75Y < = 0 Row X Y 200 Row 5 X > 0 Row 6 X > 0 The canonical form takes the coefficients of the primal equations to form new equations whose variables (A, B and C) represent the shadow or dual values and minimises if the primal is a maximising and vice versa, thus: Canonical is: Solution Min = 400A + 200C A B C 1000 A 0. 75B C 800 Objective value: Variable Value A 950 B 0 C 200 These are the shadow prices and can be compared with the graphical solution shown above. Solution 2 Linear programming illustrating minimising Let XX = Number of parts Type A completed per day; Let YY = Number of parts Type B completed per day; Row 1. Production cost objective function: Row 2. Constraint on Type A proportion: MIN = 10XX + 8YY XX = ( XX + YY) 1 3

6 Solutions 113 Row 3. Constraint on shutter Type B proportion: YY = ( XX + YY) 2 3 Row 4. Constraint on total number of parts per day: XX + YY 40 Row 5. Constraint on number of part Type A per day: Row 6. Constraint on daily production time: XX XX YY = 10 Rows 7 and 8. Constraints to prevent negative values of XX and YY: XX > 0 YY > 0 Solving graphically The equations plotted graphically below in Fig. S2.1 illustrate the area of feasible solutions, and the optimum cost is the solution that minimises the objective function 10 XX + 8 YY. 1/2(XX) + 1/5(YY) 10 YY 40 XX XX = 0.33YY TP = 11.11, XX + YY 40 Min = 10XX + 8YY Minimising 40 XX Fig. S2.1

7 114 Self-learning exercises The production cost will continue to reduce with any line drawn parallel to the objective function, moving towards the origin until the line is just tangential to the feasible area, the objective function then being at the tangential point. The coordinates of XX and YY at the tangent point (TP) are and respectively. In other words, the minimum cost is when approximately 11 Type A and 22 Type B are completed during the 8-hour working day. Objective value: Variable Value Reduced cost XX YY Row Slack/surplus Dual price Shadow/dual quantities Similarly to the maximising example, if XX and YY values are substituted into the constraint equations (Rows 1 to 8), it will be evident that Rows 2, 3 and 6 have no slack. However, if overtime is worked, the situation for Row 6 becomes: 0. 5 XX YY = 11 Once again, the equations can be redrawn graphically and the objective function MIN = 10 XX + 8 YY determined: Objective value: Variable Value Reduced cost XX YY Row Slack/surplus Dual price

8 Solutions 115 In this case, the optimum solution produces coordinates at the tangent point 12.22, and objective value The additional income generated from the overtime working: = This policy is not justified as the extra resources required cost 30. Solution 3 Simplex method constraints: equal to or less than Let x be hundreds of m 3 of concrete delivered each week in system A; Let y be hundreds of m 3 of concrete delivered each week in system B. Crew income objective function: Constraint on dumper production time: Constraint on crane production time: Constraint on crew production time: Max( M) = 50x + 60y 100/ 12. 5x + 100/ 25y / 25x + 100/ 20y / 2x y 40 5 Constraints to prevent negative values of x and y x 0 y 0 Solution by graphical or simultaneous equations is quite straightforward with two variables, and even three variables can be visualised in three dimensions, as shown in Fig. S3.1, where the maximum is given by x, y = 12 3, Max = However, more sophisticated methods of analysis are required for a many-variable problem, the most prominent being the Simplex method developed by G. B. Dantzig in the USA in 1947; although unknown elsewhere, L.V. Kantorovich was working on the problem in the USSR in The equations need to be written in a particular form to remove the less-than-or-equal-to constraints; this is achieved by introducing slack variables u, v, w.

9 116 Self-learning exercises y 50x + 13y TP = 1.67, 6.67 Feasible region Max = 50x + 60y 8x + 4y 40 4x + 5y x Fig. S3.1 x y u v w M 8x +4y +u = 40 E1 4x +5y +v = 40 E2 50x +13y +w = 200 E3 50x 60y +M = 0 E4 x 0, y 0, u 0, v 0, w 0 E5 All feasible solutions will lie in the region given by E1, E2, E3, E5. The optimal solution will be given by inclusion of the objective function, equation E4. The basic feasible solution is defined as that given by x = 0, y = 0, etc. Maximising The procedure to maximise M is to use E1, E2, E3 to modify E4 so as to convert the coefficients of x, y to zero or positive values. As x, y cannot be negative (E5), this will mean that any valid values of these variables would start to reduce the value of M. That is: the conditions just reached will represent the optimum, that is, maximum. The Simplex method is used as it leads to straightforward numerical calculation well suited to computers. First, write the equations as a matrix of the coefficients (called the Simplex Table in the method of solution to be followed later).

10 Solutions 117 The matrix represents a set of equations whereby: (a) (b) (c) An equation is valid if divided or multiplied through by a constant. Subtraction or addition of two equations leads to another valid equation. Right-hand-side values must not be negative numbers. x y u v w M E1 4 5* E E E4 Step 1. Operate on the objective function equation E4 and consider the most negative coefficient, 60 in column y, but it is equally possible to take any of the coefficients of E4. Take the positive coefficients of that column and divide into the column on the far right to produce the following quotients: 40 4 = 10, 40 = 8, 200 = Find the lowest quotient in this case 8 above and determine the position of the appropriate coefficient for this row, that is, the largest positive value namely, 5. This is called the pivot. Now divide E2 by 5 to reduce the pivot coefficient to 1, giving: 4 / 5 x 1y 0u 1 / 5 v 0w 0M 8 E6 The matrix is now rewritten as: x y u v w M E1 4 / / E l E l 0 E4 The coefficient of y is now removed from all equations by: (i) Multiplying E6 4 and subtracting from E1 to give E7: E / E / E7

11 118 Self-learning exercises (ii) Multiplying E6 13 and subtracting from E3: l E E E8 (iii) Multiplying E6 60 and adding to E4: l 0 E E E9 The new matrix becomes: 4 / 5 l 0 4 / E * / E E E9 Step 2. The new objective function equation is E9 and the most negative coefficient of E9 is 2. Dividing positive values of this column into the farthest right-hand column gives ,, The smallest value of this manipulation above produces coefficient 4 4 5, which becomes the new pivot. Dividing E7 by gives: x y u v w M E10 The matrix is now rewritten as: / E10 4 / 5 l 0 1 / E l 0 96 E E9 The coefficient of x is now removed from all equations by:

12 Solutions 119 (i) Multiplying E10 by 2 and adding to E9: ⅓ E E E13 (ii) Multiplying E10 3 / 5 and subtracting from E7: 4 / ¾ E10 4 / / E ⅓ E11 (iii) Multiplying E and subtracting from E8: E E E12 The new matrix is written as: ⅙ E ⅓ E E E13 Step 3. It can be seen that the objective function equation E13 has reached the optimum condition whereby the coefficients of x and y in E13 are now 0, thus values for M, x, y and w may be read from the table as follows: x = 1 2, y = 6 2, w = 30, M = x 0, y 0, u 0, v 0, w 0 i.e. x = 1.67, y = 6.67, profit = (compare with the graphical answer above). Solution 4 Simplex method constraints greater than or equal to The objective function to be maximised is 1.5x + y. i. e. M = 1. 5x + y Using a similar technique to that above, negative slack variables u and v are introduced as negative values together with artificial variables a and b in order to overcome the negativity elements. The equations may be written as:

13 120 Self-learning exercises 2x + 3y u + a = 14 8x + 5y v + b = 40 u 0, v 0, x 0, y 0, a 0, b 0 The solution proceeds in the normal manner using the Simplex method. However, since artificial variables have been introduced, these will have to be eliminated before an optimum solution can be obtained. This is achieved by either adding very large positive numbers as the coefficients of a and b in the objective function or alternatively designating the coefficients as p to represent such a number. The maximising objective function is thus written as: 1. 5x y + pa + pb + M = 0 p is the large positive number and the maximising table becomes: x y u v a b M E E p p 1 0 E3 u 0, v 0, x 0, y 0, a 0, b 0 It can be seen from the graph (Fig. S4.1) that a maximum solution to this problem is infinity as there is no limit on the upper boundaries of the constraints. A minimum value, however, is possible to calculate as shown in Solution 5. y 8 8x + 5y 40 Max = 1.5x + y 2x + 3y 14 Feasible region TP = 25/7, 16/7 5 Simplex method solution-maximising x Fig. S4.1

14 Solutions 121 Solution 5 Simplex method constraints: greater than or equal to (This process can also be similarly applied in minimising Question 3.) In minimising M, the procedure involves maximising M, that is, to minimise M merely maximise M or make M less negative. Step 1. Determine the table involving the main constraints written as: x y u v a b M E E p p 1 0 E3 As in the maximising problem, the artificial coefficients a and b are introduced to counter the negative u, v coefficients of the greater-than-or-equal-to constraints in E1 and E2. Again, p is introduced as the large positive number described previously and the theoretical basic feasible solution cannot reach a maximum value without a and b equal to zero. Thus, the first basic feasible table is provided with the a and b coefficients of E3 set to 0 to give E4 and E5, The initial maximising table thus becomes i. e. pa + pb = 0 x y u v a b M E E E4 p p 0 0 E5 u 0, v 0, x 0, y 0, a 0, b 0 From the above basic feasible solution, the Simplex method can now begin by first developing a new basis equation by subtraction or addition of two existing equations, which will lead to another valid equation. In this case, an initial basis equation is obtained by multiplying each of E1 and E2 by p and adding to E5 to give E6. This produces the following table: x y u v a b M E1 8* E E4 10p 8p 1p 1p p E6 Step 2. Select the most negative coefficient in the bottom line, 10p, is chosen, Row E2 produces the pivot to give E7 and the transforming procedure is executed as in the previous example:

15 122 Self-learning exercises x y u v a b M * 1 ¼ ¼ 0 4 E8 1 ⅝ 0 ⅛ 0 ⅛ 0 5 E E p ¼ p E10 Step 3. The most negative coefficient in the bottom line, 7/4p, is selected, Row E8 produces the pivot in Row E11 and the transforming procedure is continued to give the following table: x y u v a b M 0 l l ⅜ E14 The test criteria for the objective equation (E14) are satisfied, coefficients are 0 and the function is now maximised with x =, y =, M = that is, minimum M = + 107/14 = (compare Fig. S5.1 with graphical result in Fig. S4.1 above). y 8 8x + 5y 40 Min = 1.5x + y 2x + 3y 14 Feasible region TP = 25/7, 16/7 5 Simplex method solution-minimising Fig. S5.1

16 Solutions 123 Solution 6 Integer linear programming General integers The linear programming method previously described produces an optimal solution that may not be an integer value. However, in some practical situations, only whole number values may be acceptable, for instance, in the production of packaged quantities where an integer amount would be mandatory. While rounding to an integer solution by trial and error may be acceptable, the rounded-up or rounded-down values need to be examined, to determine whether an optimum integer solution has been found or is indeed feasible. The non-integer (relaxed version) optimum solution is illustrated in Solution 3 and is reworked as an integer solution as follows: Integer solution It can be observed in Fig. S6.1 that, within the feasible area for a relaxed solution, instead of moving the objective function to the tangential point, the line is restricted to touching the last (i.e. extreme) whole value (integer) coordinates, irrespective of whether it is tangential or actually within the boundary area. The dots shown imposed upon the diagram demonstrate that the optimum integer solution occurs at point X = 0, Y = 8 to produce a smaller objective value These X, Y values when substituted into the constraint equations produce the following summary: Y 50X + 13Y TP = 0, 8 Max = 50X + 60Y Feasible region 8X + 4Y 40 4X + 5Y 40 Integer solution 10 X Fig. S6.1

17 124 Self-learning exercises Branch count: 2 Variable Value Reduced cost X 0 50 Y 8 60 Row Slack or Surplus Dual price Branch and bound calculation of integer variable values The branch count above is the number of times the integer values needed to be forced to an integer value in arriving at the optimum solution via the branch and bound process of calculation, this being a specifically devised algorithm that enumerates all combinations of the integer variables (e.g. the whole numbers seen on the grid in the graph). The best feasible combination of variables is selected within the constraints permitted as defined in the linear programming problem. To illustrate the kind of difficulty that may arise, consider: Subject to constraints: Maximum = Y Y + Z = The relaxed solution is Y Z Y = Z = 10. 5/ 2 = In determining an integer solution by smoothing the graph manually and rounding up or down according to a guess might result in Z = 6 and Y = 4.5 which breaks the constraint Y Z and so the solution is infeasible. The correct answer to comply with the constraints is clearly Y = 5. 5 and Z = 5 and the Maximum = Obviously, when the number of variables exceeds three in total, the problem lies outside a three-dimensional configuration and a graph cannot be drawn. The branch and bound algorithm carries out this calculation, first by using the relaxed solution values of the variables (determined by, for example, the Simplex method described earlier) to compute the best feasible combination of integers. Because of the tedium involved in such calculations, enumeration by computer is normally adopted. However, when a computer is not available, the most convenient approach is to eye in the integer variables and select the combination that almost produces the relaxed value of the solution.

18 Solutions 125 Reduced costs The table also shows reduced costs and that variable X is not used in the solution, thereby indicating that its coefficient in the objective function could be reduced by 50 without causing a change in income penalty. This is illustrated by superimposing MAX = 0 X + 60 Y on the graph, which becomes tangential at point X, Y (0, 8) when moving away from the origin. Solution 7 Binary integer linear programming (i) The inclusion of binary integer constraints may be specified as: A = 1 if system A is acceptable, A = 0 if not B = 1 if system B is acceptable, B = 0 if not and as before: Let X = hundreds of m 3 /week of concrete delivered with system A; Let Y = hundreds of m 3 /week of concrete delivered with system B. Row 1. The objective function now includes the binary variables in the set-up costs which will have 0, 1 values applied, or not, depending on whether the particular system A or B is operational: Subject to the following constraints: MAX = 50 X + 60 Y 50 A 125 B Row 2. Constraint on dumper production time: Row 3. Constraint on crane production time: Row 4. Constraint on crew production time: 8 X + 4 Y 40 4 X + 5 Y X + 13 Y 200 Rows 5 and 6. Constraints to prevent negative values of X and Y: X 0 Y 0

19 126 Self-learning exercises Plus the following extra constraints: Rows 7 and 8 are constraints that prevent concrete being produced by system A when production exceeds the 175 m 3 /week permitted, that is, A = 0, or by system B when production exceeds 700 m 3 /week, that is, B = 0: X Y A BIN( BIN( B); Note: In a solution using Lindo, function denotes a shortcut, setting the selected variables to be binary integers. A 0/1 binary integer variable is a special case of an integer variable that is required to be either zero or one, and is used as a switch to model Yes/No decisions. Computer solution Objective value: Branch count: 1 Variable Value Reduced cost X Y A 1 50 B As can be seen from the above table, both A and B systems are needed and must operate within their respective capacities. Clearly the new optimum is the same as for Question 3 minus the set-up costs, that is, = Alternative equipment options Other situations may be possible, for example, changed set-up costs, which then become critical in determining the viability of a particular system for inclusion in the solution. Carrying out a number of trials for a range of set-up cost estimates enables evaluation of the alternative candidates. (ii) As an example, system B becomes critical (B = 0) at 396, but results in a correspondingly much-reduced concrete output produced by system A (A = 1) alone, as follows: With the set-up cost of system B = 396, the solution is:

20 Solutions 127 The routine has tried Objective value: Branch count: 1 Variable Value Reduced cost X Y A B A = 1 with B = 0 A = 0 with B = 1 A = 1 with B = 1 and found the optimum at A = 1 combined with B = 0. (iii) Similarly, A becomes critical at a set-up cost of A = 64 where A = 0 with B = 1: Objective value: 295 Branch count: 1 Variable Value Reduced cost X 0 0 Y 7 0 A B (iv) Alternatively, the capacity limits on A or B may be evaluated where B becomes critical at Y limited to a capacity of 2.08 m 3 /week, that is, A = 1 with B = 0: Objective value: Branch count: 0 Variable Value Reduced cost X Y 0 0 A B (v) Similarly, A becomes critical at X = 1 where A = 0 with B = 1 as follows:

21 128 Self-learning exercises Objective value: 295 Branch count: 0 Variable Value Reduced cost X 0 0 Y 7 0 A 0 0 B These examples illustrate that, by carrying out simulation routines, it is readily possible to determine which resources to adopt simply by attaching 0, 1 variables to the appropriate elements in the objective function. Notably, the original constraint equations were unaffected other than that the model was expanded by the inclusion of additional constraints to control the on/off parameter of the production systems A and B within their specified capacities. Solution 8 Transportation problem Expressed as a typical linear programming problem of maximising or minimising a supply/ demand situation, the problem would be formulated as follows: 3 Suppliers 3 Consumers Capacity (m ) Requirements (m ) 60 S 1 C S 2 C S 3 C 3 40 Transport costs S 1 C 1 = 70p per item S 1 C 2 = 40p per item, etc. Decision variables: Xi, j = amount sent from Supplier ( row) i to Consumer ( column) j The objective function is to minimise the total cost of transport cost as follows (i.e. the sum of the amount multiplied by its cost for every cell in the matrix), thus: Minimise 70X + 40X + etc 1, 1 1, 2. Subject to constraints on supplier capacity and customers requirements: Supplier constraints (i.e. can supply less or equal to its capacity): X1, 1 + X1, 2 + X1, X2, 1 + X2, 2 + X2, X3, 1 + X3, 2 + X3,

22 Solutions 129 Customer constraints (i.e. requires the particular amount): and X1, 1 + X2, 1 + X3, 1 = X1, 2 + X2, 2 + X3, 2 = X1, 3 + X2, 3 + X3, 3 = X i, j 0 Alternative solution by the tabular method Clearly, to solve the problem by hand calculation using, for example, the Simplex method would be very time-consuming, and a tabular minimisation method is available to quickly derive a manual solution even for very large tables. (The table can also be maximised by transforming each cell into subtracted values from a base number, e.g. 100, and then minimising. The optimum results in the cells would thereafter be retransformed to enable calculation of the maximum.) Solution Total available for supply = = Total required to be delivered = = m 3 It is first necessary to get some idea of the most likely routes. This can be guessed simply from looking at the costs and routes involved sometimes. Alternatively, the following method has been shown to be very quick and in many instances gives the optimum solution also. The analysis is recommended for complex problems. Preliminary stage Step 1 The data are arranged in a suitable table, such as that shown in Fig. S8.1. The route costs are indicated in the bottom right-hand corner of each square. In this instance the material available for supply exceeds that required, so a dummy site requiring Fig. S8.1 Step 1.

23 130 Self-learning exercises the excess is invented in which the transport costs from supplier to site are zero. In other words, the material is never moved at all. Now look along each row and deduct the lowest cost from the second lowest cost, repeat the procedure for each column. Allocate materials to the cheapest route along the row or column with the greatest difference in costs. The dummy site is fully supplied and so this column can now be eliminated. (All material quantities are in m 3.) Step 2 Repeat Step 1 procedure with the routes remaining. Site B is fully supplied and eliminated. See Fig. S8.2. Fig. S8.2 Step 2. Step 3 Supplier 2 is empty and may be eliminated, but Site A still requires m 3 and Site B, m 3. See Fig. S8.3. Fig. S8.3 Step 3. Step 4 No clear choice is possible, so the material may be distributed along any combination of the available routes. See Fig. S8.4.

24 Solutions 131 Fig. S8.4 Step 4. Final guess Note from Fig. S8.5 that the sum of the quantities in the routes used along a row or a column must be the amount available from a supplier or to the supplier respectively. It is now necessary to test if the guess is in fact the optimum solution. Fig. S8.5 Final guess. Final stage Test for optimality Consider the costs of the routes used in two components: (1) cost of despatch from the source of supply, (2) cost of reception at the destination. Start with one of the components arbitrarily chosen (usually zero). The remaining despatch and reception costs are determined by considering each used route in turn and testing the equation. Despatch cost + reception cost = actual cost along the route. Note that in the example (Fig. S8.6), the top left-hand corner is chosen. Unfortunately, in this example the five routes used are less than the number of rows (m) plus columns (n) minus l. m + n 1 = = 6 It is therefore impossible to generate all the despatch and reception costs. In order to overcome this difficulty, a small quantity q is introduced into an unused route so that further row and column costs may be generated (Fig. S8.7).

25 132 Self-learning exercises Fig. S8.6 Despatch cost + reception cost. Fig. S8.7 Introduction of q. It is now possible to test for optimality. Compare for each unused route the actual costs with the sum of costs for its corresponding despatch and reception costs. If the actual cost is greater than, or equal to, the sum for every unused route, then the optimum has been found. If not, the solution can be improved by using that route (marked x) for which the actual cost is less than the sum (Fig. S8.8). Clearly, we have a route which is cheaper and should be used. But because an amount x is introduced, the balance of supply and requirement is upset and thus it is necessary to deduct the quantity x from some of the routes (Fig. S8.9). The route containing q also contains a quantity x and so q can be dropped from further calculations as it is relatively small. Fig. S8.8 Route x.

26 Solutions 133 Fig. S8.9 Balance of supply and requirement. To reduce the solution to m + n 1 routes, it is necessary to remove a route containing x. Obviously by choosing x = minimum (1 or 8), that is, x = 1, the new route pattern becomes as shown in Fig. S8.10. Note: m + n 1 = 6. This equals routes used and so the new despatch and reception costs may be generated. The optimising test procedure is now repeated (Fig. S8.11). The actual cost is greater than, or equal to, the sum of the despatch and reception costs for each unused route, therefore the optimum is found. Fig. S8.10 New route pattern. Fig. S8.11 Repeat of optimising test procedure.

27 134 Self-learning exercises Summary Total cost = = = = = Minimum transport cost is using routes A2, A3, B1 and C3. Solution 9 Assignment problem This is a problem characterised by the supplier only being able to assign a single task at each despatch location to a single receiver/consumer at its respective destination, which can also proceed as a linear program solved by the Simplex method as follows. Decision variables: Xi, j = amount sent from Supplier ( column) i to Consumer ( row) j i. e. X amount sent from Manager ( column ) i to Project ( row) j i, j = Manager performance rating R i,j out of 100 points is: S 1 C 1 = 75 points S 1 C 2 = 78 points, etc. The objective function is to maximise the total amount or, where the problem requires, alternatively minimise the total amount as follows (i.e. the sum of the amount multiplied by its cell value for every cell in the matrix), thus the objective function is to maximise the total points for the six managers, written as: Maximise = R X + R X + etc. + R X + R X + R X + etc. + R 1, 1 1, 1 1, 2 1, 2 1, 5 1, 5 2, 1 2, 1 2, 2 X + R X + etc 2, 2 3, 1 3, 1 Subject to constraints: Variable constraints (rows) Project constraints (columns) X + X + X + X + X 1, 1 1, 2 1, 3 1, 4 1, 5 1 X + X + X + X + X 2, 1 2, 2 2, 3 2, 4 2, 5 1 X + X + X + X + X 3, 1 3, 2 3, 3 3, 4 3, 5 1 X + X + X + X + X 4, 1 4, 2 4, 3 4, 4 4, 5 1 X + X + X + X + X 5, 1 5, 2 5, 3 5, 4 5, 5 1 X + X + X + X + X 6, 1 6, 2 6, 3 6, 4 6, 5 1 X + X + X + X + X = 1, 1 1, 2 1, 3 1, 4 1, 5 1 X + X + X + X + X = 2, 1 2, 2 2, 3 2, 4 2, 5 1 2, 2 2, 5

28 Solutions 135 and X + X + X + X + X = 3, 1 3, 2 3, 3 3, 4 3, 5 1 X + X + X + X + X = 4, 1 4, 2 4, 3 4, 4 4, 5 1 X + X + X + X + X = 5, 1 5, 2 5, 3 5, 4 5, 5 1 X + X + X + X + X = 6, 1 6, 2 6, 3 6, 4 6, 5 1 X i, j 0 Alternative solution by the tabular assignment method In a similar fashion to the transportation problem, a quick manual procedure involving matrix reduction has been devised to give solutions for this even more special-purpose type of problem where supply and demand equals 1. The transportation method may be used, but as only five routes are possible in the solution, and there are m + n 1 = = 9 routes* needed to generate the despatch and reception costs, the procedure would be lengthy. Thus, the assignment method was developed to test for optimality. In this example, it is first necessary to subtract all the figures from 100 since it is required to maximise the points scores for the managers (i.e. minimise 100 minus points score). * m = number of rows and n = number of columns in the matrix. Initial stage Subtract points scores from 100 (Fig. S9.1). Fig. S9.1 Subtract point scores from 100. Step 1 Subtract the lowest number in each row from every number in its row (Fig. S9.2). Fig. S9.2 Step 1.

29 136 Self-learning exercises Step 2 Subtract the lowest number in each column from every number in its column (Fig. S9.3). Fig. S9.3 Step 2. Step 3 Test for an assignment. An assignment is possible if the minimum number of horizontal and vertical lines drawn through the rows and columns needed to cover all the zeros equals the number of rows or columns in the table (Fig. S9.4). In this case, a minimum of four lines covers all zeros, thus an assignment is not possible. Fig. S9.4 Step 3. Step 4 Find the smallest uncovered number remaining after Step 3 is undertaken, then (1) subtract this number from every number in the table (Fig. S9.5); (2) add the same number back to Fig. S9.5 Step 4.

30 Solutions 137 each number covered by Step 3 lines (i.e. a number covered by both a horizontal and vertical line will be added twice) (Fig. S9.6); (3) try the assignment test again. (1) Smallest number is 3. (2) Fig. S9.6 Illustration of (2). (3) The zeros cannot be covered by fewer than five lines now, therefore a zero assignment is possible. If not, repeat procedure from Step 4. Step 5 In order to reach the optimal solution, it is only possible to use one route for each row and column (Fig. S9.7). Fig. S9.7 Step 5. Columns 2, 4 and 5 each have a single zero, so these routes must be used. The extra zeros in Rows B and D can therefore be discarded. Hence, the zero in Column 3 must be used and the zero in Column 1 now discarded, leaving the only possible route in Column 1 as the zero in Row C. Final solution This is shown in Fig. S9.8. The final solution converted back to points is shown in Fig. S9.9.

31 138 Self-learning exercises Therefore, the best combination of manager to project is: Manager Project Points 1 C 73 2 B 71 3 A 61 4 E 74 5 E Fig. S9.8 Final solution. Fig. S9.9 Final solution converted back to points. Solution 10 Assignment when number of agents and tasks differ The client needs to allocate the four managers best suited to the five projects available. The assignment method overcomes this difficulty by including a DUMMY Manager but who has ZERO points for each project as follows: Five rows by f ive columns as in Q10.1 In Fig. S10.1 a DUMMY column (Manager) with zero cell rating values is introduced. Then each cell rating value is deducted from 100 (Fig. S10.2) and minimised (Fig. S10.3). Using Lindo, the cell allocations are:

32 Solutions 139 Managers Project Amount A B l 35 C D E Fig. S10.1 Assignments uneven. Managers Dummy Project Amount A B l 35 0 C D E Fig. S10.2 Managers Project A B C D E S1 S2 S3 S4 Dummy Amount Fig. S10.3 Minimise. X = 1 X = 1 X = 1 X = 1 X = 1 ( 1, 3) ( 2, 2) ( 3, 1) ( 4, 5) ( 5, 4) R = 39 R = 29 R = 27 R = 100 R = 26 ( 1, 3) ( 2, 2) ( 3, 1) ( 4, 5) ( 5, 4) Total = 221 or, when deducted from 500, maximum = 279 points. Solution 11 Goal programming Problems of the LP type can have multiple criteria with different levels of priority in making a decision, and can be solved by the goal programming technique. The procedure requires the following steps:

33 140 Self-learning exercises (1) Identify each goal and its target. (2) Prioritise each goal in the order of importance, high to low, for example, P1, P2, etc. (3) Define the decision variables, these being the controllable inputs that describe the number or amounts of each item to produce, and formulate in the conventional LP manner. (4) Develop an equation for each goal, involving the decision variables and goal deviation (amount α above or below the target value) on one side, and the goal target on the other. (5) Write the objective function in terms of minimising the deviation value. Thus, in the above problem, the goals and priority levels are as follows: Goal 1 and priority level 1: Build 200 houses with Type B designs. Goal 2 and priority level 2: Complete no more than 400 houses. The decision variables are: X = Number of detached houses to build Y = Number of semi-detached houses to build Hence for Goal 1 the LP problem can be stated as: Minimise α 1 Constraint on proportion of detached houses: Goal 1 equation for Type B design limit: X 0. 75( X + Y) 0. 4X Y + α = 200 Constraints to prevent negative values of X, Y and α 1 : X 0 Y 0 α 1 0 Drawn graphically (Fig. S11.1), the approach is similar to the LP approach except that each priority level is considered separately, thus beginning with priority level 1 as follows. Substituting the G1 point = 321.4, into the Goal 1 equation: 0. 4X Y + α = = 200 α 1 = 0, i. e. the goal can be achieved. Goal 2 is a secondary goal and must not degrade the above solution obtained with the priority level 1 conditions, hence the LP for the priority level 2 situation can be written as: Minimise α + α

34 Solutions 141 Y 400 2/5*(X) + 2/3*Y + a1 = 200 G1 point = 321.4, Feasible region ¼X ¾Y X Fig. S11.1 Step 1. Constraint on detached houses proportion: Goal 1 equation for Type B design limit: Goal 2 equation for total housing limit: X 0. 75( X + Y) 0. 4X Y + α = 200 X + Y + α 2 = 400 Constraints to prevent negative values of X, Y, α 1 and α 2: X 0 Y 0 α 1 0 α 2 0 The graphical interpretation is shown in Fig. S11.2 as follows: In order for priority level 1 to be achieved, the G1 point cannot be violated, hence, substituting the G1 point value into the Goal 2 equation: X + Y + α 1 = α = α 2 =

35 142 Self-learning exercises Y 400 X + Y + a2 = 400 2/5*(X) + 2/3*Y + a1 = 200 G1 point = 321.4, Feasible region ¼X ¾Y X Fig. S11.2 Step 2. Priority level 2 goal cannot be achieved, that is, 428 houses will have to be built. Notes: (i) If the above priorities were switched, the solution would be different. (ii) Several goals may have the same priority level as appropriate. (iii) In more complex problems involving multiple decision variables, goals and priorities solutions, a computer program becomes essential. Solution 12 Non-linear programming Equations of this form arise in a construction-industry situation, for example, where statistical data indicate that, say, the profit (in s) achieved in groundworks per m 3 construction is governed by the above-stated objective function variable relationship between excavation (x) and associated temporary works (y). A cubic metre of excavation takes 2 hours and fixing 1 m 2 of temporary works requires 1 hour, and an 8-hour day is available. The problem becomes one of determining the minimum and maximum profit expectations in a given period (Lootsma, 1972). Graphical Solution (Fig. S12.1) (a) Maximising Let x be the amount of excavation in m 3. Let y be the amount of temporary works in m 2. The object is to maximise profit, thus, objective function is written as: 2 2 Row 1 MAX ( ) = x 2x + y 6y

36 Solutions 143 y 10 Contour direction Feasible region 2x + y 8 x* x-2* x + y* y-6* y = Max 4, 0 10 x Non-Linear maximising Fig. S12.1 Subject to constraint Row 2 2x + y 8; Row 3 x 0; Row 4 y 0; The objective function can be rearranged to represent a circle with the centre at (1, 3) and radius (r) as shown in Fig. S12.1, that is: where: Hence 2 2 ( x 1) + ( y 3) 10 = MAX r = ( x 1) + ( y 3) r 2 10 = MAX In order to find the maximum, the contours of the circle need to be moved away from the centre as shown in Fig S12.1. It can be observed that the last position of a contour of centre (1, 3) falling inside the constraint region coincides with it at point x = 4, y = 0. Substituting x, y values in the objective function gives: MAX = 8 Hence, r 2 = 18, that is, r = radius = 4.24 with the circle centre at x = 1, y = 3.

37 144 Self-learning exercises Note that the alternative option of point x = 0, y = 8 would result in the larger contour falling outside the feasible region at point (4, 0) and would therefore be invalid. Computer solution Objective value: 8.00 Variable Value Reduced cost x 4 0 y 0 9 The result shows that the best strategy is to work only on projects not requiring a temporaryworks element. Also: Row Slack or Surplus Dual price (b) Minimising This time, the objective is to find the minimum profit, hence the objective function is to minimise: x 2 2x + y 2 6y Again, this is the equation of a circle, which can be rearranged as: with the centre at (1, 3) and radius (r) where: Thus: 2 2 ( x 1) + ( y 3) 10 = MIN r = ( x 10) + ( y 3) r 2 10 = MIN This time, as clearly shown in Fig. S12.2, for the minimum contour of the possible circles to fall wholly within the constraint region, the condition must be that x = 1, y = 3, that is, it coincides with the centre of the circle. Substituting x, y values in the objective function gives: MIN = 10 Hence, r 2 = 0, that is, r = radius = 0 with the circle centre at x = 1, y = 3.

38 Solutions 145 y 10 2x + y 8 Contour direction Feasible region is point (1, 3) Non-Linear minimising x* x-2* x + y* y-6* y = Min 10 x Fig. S12.2 Computer solution In the same way as previously, the Lindo analysis produces corresponding results: 2 2 MIN = x 2x + y 6y 2x + y 8 x 0 y 0 Objective value: Variable Value Reduced cost x y E-07 This case shows that the project is unprofitable. Also: Row Slack or Surplus Dual price E Lagrangian multiplier and Kuhn Tucker method When the equations cannot be graphically interpreted, the analysis needs more sophisticated processes using, for example, the Lagrange-multiplier method combined with the Kuhn Tucker

39 146 Self-learning exercises theorem, these being reasonably efficient developments to help find the saddle points, that is, the turning points of the objective function curve. The approach broadly builds on the principles described earlier in the basic LP, where it follows that the linear/non-linear problem for maximising f(z) subject to constraints g i (z) = 0 (i = 1,, m) can be formulated as a Lagrangian function of the type: L( z, u) = f ( z) u g ( z) z being the variables, and u i unknown vectors called the Lagrangian multipliers. Solutions for z can be found among the extreme points of L(z, u) by means of partial differential calculus, that is, differentiating the Lagrangian function separately with respect to z and u to determine the gradients as follows: and L( z, u) z = f ( z) z + u g ( z) z = gradient L( z, u) u = g ( z) = gradient The optimum value (Kuhn Tucker point) can be found at the stationary/turning point of the Lagrangian function where the gradient is zero, that is, L(z, u)/ z = 0, and L(z, u)/ u = 0. There are a variety of methods available, developed by different authors, for determining the Lagrangian-multiplier values (u) see Lootsma. However, the equations can be quite intricate to manipulate in resolving the values for u in complex problems, and are more readily calculated with modern computer software packages. By way of illustrating the Lagrangian principles for maximising the simple-circle example, maximise: x 2 2x + y 2 6y subject to the constraints 2x + y 8 and x 0, y 0. The Lagrangian function for the maximising situation gives the following formulation: L( z, u) = x 2 2x + y 2 6y + u ( 2x y + 8) + u x + u y Partially differentiating with respect to x and y gives: and adding them together gives: i i i i i L/ x = 2x 2 2u + u = L/ y = 2y 6 u + u = x 2 + 2y 6 3u + u + u = 0 Partially differentiating with respect to u gives: L/ u1 = 2x y + 8 = 0 L/ u2 = x = 0 L/ u3 = y = 0

40 Solutions 147 The Kuhn Tucker point is characterised by determining the appropriate positive Lagrangian multipliers to solve the partial differential equations and finding the stationary point (z, u) where the gradient is zero. In this example, it can be found that for multipliers set at u 1, u 2, u 3 = 0, the maximum point x, y = 4, 0 satisfies all the above partial equations and therefore yields the L(z, u) solution, Max = 8. A similar calculation would demonstrate the minimising case: L( z, u) = x 2 2x + y 2 6y u ( 2x y + 8) u x + u y Local optimum solutions While in the above simple case the global maximum or minimum values can be readily visualised, solutions to non-linear problems commonly produce local optima. Indeed, the available mathematical optimisation models generally operate by converging to a local optimum point. Hence, other and better solutions may thus exist. For example, solutions of the objective function = xsin(πx) (i.e. non-linearly sinusoidal shaped) have different turning points or saddle points (i.e. local maximum and minimum) depending on the constraint limits chosen, as shown in Fig. S12.3. With constraint, say, x < 6, a plot of this function produces positive turning points of 0.58, 2.52 and 4.51 at x = 0.645, x = 2.54 and x = 4.52 respectively, and a minimum value of 1.53 and 3.51 at x = 1.56 and x = Other turning points exist for different limits on the constraint, thus, clearly, considerable trial and error would be necessary to establish the required result in analysing more complex problems. Indeed, the behaviours of the equations need to be understood and carefully considered in order to interpret the results meaningfully, as slight changes in the parameters can have marked effects. OF x x x x X x 1.53 Fig. S12.3 Local optimum solutions.

41 148 Self-learning exercises Solution 13 Quadratic assignment problem It will be observed that the problem consists of two variables, amount of items transferred from each lorry (variable L) and transfer time from one bay to another bay (variable B), thus, the objective function is quadratic, that is, multiplication of the two variables is involved and therefore not soluble by linear methods. However, since all items in the table are already assigned, the problem becomes merely calculation of the totals from which the minimum value is determined. The problem is of the 0/1 binary integer variable type, where the variable can be either on or off (i.e. zero or one), which in this example would be according to the truck-to-bay designation and solved as follows: Number of items transferred (variable L): to truck 1 2 from truck Transfer time in minutes from one bay to another bay (variable B): to Bay A B from A 0 5 Bay B 15 0 By inspection in this simple case, the trucks directed as above would yield the minimum total transfer time calculated from the amount transferred from each truck to the other truck multiplied by the corresponding bay time as follows: and substituting values from the tables: Min = L B + L B + L B + L B 1, 1 1, 1 1, 2 1, 2 2, 1 2, 1 2, 2 2, 2 Min = = 450 mins Obviously, however, if the trucks arrivals were switched so that truck 1 arrived in Bay B and truck 2 arrived in Bay A and the amounts transferred were unchanged, the situation would now become: and substituting values from the tables: Min = L B + L B + L B + L B 1, 1 1, 1 1, 2 1, 2 2, 1 2, 1 2, 2 2, 2 Min = = 550 mins