Linear programming. Debrecen, 2015/16, 1st semester. University of Debrecen, Faculty of Business Administration 1 / 46

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1 1 / 46 Linear programming László Losonczi University of Debrecen, Faculty of Business Administration Debrecen, 2015/16, 1st semester

2 LP (linear programming) example 2 / 46 CMP is a cherry furniture manufacturer they produce chairs and tables. Teir supply of wood is limited, their suppler provides them with 100 board feet of cherry wood weekly. Each chair requires 4 board feet of wood and a table needs 8 board feet. CMP can sell all chairs it can make but only 10 tables per week. CMP has at their disposal only 120 man-hours available per week for making the furniture. It takes 6 man-hours to make a chair and 4 man-hours to make a table. CMP makes $ 20 profit for every chair it sells and makes $ 30 profit for every table sold. Here CMP wants to decide how many chairs and tables it should make to maximize total profits.

3 LP example, continued 3 / 46 Solution: suppose that the number of chairs and tables produced are x 1 and x 2 respectively. The cherry wood constraint: 4x 1 + 8x 2 100, selling constraint: x 2 10, labor constraint: 6x 1 + 4x 2 120, nonnegativity constraint: x 1 0, x 2 0. The weekly profit: 20x x 2 dollars should be maximal. Summarizing: z = 20x x 2 maximum, subject to 4x 1 + 8x x 2 10 x 1 0, x x 1 + 4x 2 120

4 LP example, graphical solution 4 / 46 Graphical solution: sketch on the plane the (x 1, x 2 ) domain satisfying the constraints. The objective function with constant value of z are parallel lines which have to have common points with the above domain. Hence we shift these lines parallel until we get the maximum on z. In our case this is x 1 = 17.5, x 2 = 3.75, z max = The next figure shows the domain satisfying the constraints colored by green. The parallel lines z = 20x x 2 or x 2 = x z 30 are shown by red color for z = 0 and z = z max =

5 Graphical solution continued The solutions x 1 = 17.5, x 2 = 3.75 are not integers. Thus we have to find the point(s) in the green area which have integer coordinates and for which the value of z = 20x x 2 is the largest. This point is x 1 = 17, x 2 = 4, and z max = / 46

6 A little history of the simplex method 6 / 46 In 1939 L. V. Kantorovich noted the practical importance of a certain class of linear programming problems and in 1960 gave an algorithm for their solution. Unfortunately, for several years, his work was unknown in the West and unnoticed in the East. In 1947 when G.B. Dantzig invented the simplex method for solving the linear programming problems that arose in U.S. Air Force planning problems. The earliest published accounts of Dantzig s work appeared in 1951 his monograph published in 1963 remains an important reference. In the same year that Dantzig invented the simplex method, T.C. Koopmans showed that linear programming provided the appropriate model for the analysis of classical economic theories. In 1975, the Nobel Prize in economic science was awarded to L.V. Kantorovich and T.C. Koopmans for their contributions to the theory of optimum allocation of resources. Apparently the academy regarded Dantzig s work as too mathematical for the prize in economics (and there is no Nobel Prize in mathematics).

7 3 variable LP example The variables s 1, s 2, s 3 are called basic variables, x 1, x 2, x 3 are called nonbasic variables. 7 / 46 z = 5x 1 + 4x 2 + 3x 3 = maximum, subject to 2x 1 + 3x 2 + x 3 5 4x 1 + x 2 + 2x x 1 + 4x 2 + 2x 3 8 x 1, x 2, x 3 0 Introduce the slack variables s 1, s 2, s 3 as the difference of the right and left hand sides of the constraints. In this way we get another problem, equivalent to the previous one: z = 5x 1 + 4x 2 + 3x 3 = maximum, subject to s 1 = 5 2x 1 3x 2 x 3 s 2 = 11 4x 1 x 2 2x 3 s 3 = 8 3x 1 4x 2 2x 3 x 1, x 2, x 3, s 1, s 2, s 3 0

8 Start with the solution x 1 = x 2 = x 3 = 0 then s 1 = 5, s 2 = 11, s 3 = 8 and the value of the objective function z = 0. Try to find a better solution. As the coefficient of x 1 in the objective function is positive, thus increasing x 1 will increase z. However the value of x 1 cannot be increased arbitrarily since the slack variables must remain non-negative. If x 1 0, x 2 = x 3 = 0 then all three of the inequalities s 1 = 5 2x 1 0 x = 2.5 s 2 = 11 4x 1 0 x = 2.75 s 3 = 8 3x 1 0 x = must be satisfied, therefore 0 x i.e. x 1 can be increased at most to 2.5. Let therefore x 1 = 5 2, x 2 = x 3 = 0 then s 1 = 0, s 2 = 1, s 3 = 1 2 and the value of z increased to = / 46

9 How can we continue? Since now s 1 = x 2 = x 3 = 0 the role of x 1 was taken over by s 1, we have to modify the objective function and the constraints accordingly. From the definition of s 1 we get 9 / 46 x 1 = s 1 1.5x 2 0.5x 3 substituting this into the objective function and into s 2, s 3 we get z = 5 ( s 1 1.5x 2 0.5x 3 ) + 4x 2 + 3x 3 = s 1 3.5x x 3 s 2 = 11 4 ( s 1 1.5x 2 0.5x 3 ) x 2 2x 3 = 1 + 2s 1 + 5x 2 s 3 = 8 3 ( s 1 1.5x 2 0.5x 3 ) 4x 2 2x 3 = s x 2 0.5x 3 Our problem with the new variables: z = s 1 3.5x x 3 = maximum, subject to x 1 = s 1 1.5x 2 0.5x 3 s 2 = 1 + 2s 1 + 5x 2 s 3 = s x 2 0.5x 3 s 1, x 2, x 3, x 1, s 2, s 3 0.

10 10 / 46 We see again that with s 1 = x 2 = x 3 = 0 we have x 1 = 2.5, s 2 = 1, s 3 = 0.5 and z = Here in the objective function only the coefficient of x 3 is positive, increasing x 3 will again increase the objective function. By how much can we increase x 3? With x 3 0, s 1 = x 2 = 0 from the conditions x 1, s 2, s 3 0 x 1 = x 3 0 x 3 5 s 2 = 1 0 is satisfied for all x 3 s 3 = x 3 0 x 3 1 therefore x 3 = 1 s 3 = 0 and s 1 = x 2 = 0, the objective functions increases by = 0.5 to 13. The new nonbasic or independent variables s 1, x 2, s 3, the role of x 3 is taken over by s 3.

11 11 / 46 From the equation s 3 = s 1 + 0, 5x 2 0.5x 3 we get x 3 = 1 + 3s 1 + x 2 2s 3, substituting this into z, x 1, s 2 (please do the calculations) we get our problem written by the help of the new variables: z = 13 s 1 3x 2 s 3 = maximum, subject to x 1 = 2 2s 1 2x 2 + s 3 s 2 = 1 + 2s 1 + 5x 2 s 3 = 1 + 3s 1 + x 2 2s 3 s 1, x 2, s 3, x 1, s 2, x 3 0 Now we do not have positive coefficients in z meaning that we cannot increase z anymore. As s 1, x 2, s 3 0 it follows that z = 13 s 1 3x 2 s 3 13, but with s 1 = x 2 = s 3 = 0 we get z = 13 thus the optimal solution is z = 13 (the values of x 1, s 2, x 3 can be calculated from the previous formulae).

12 Simplex tableau The problem we have dealt with (after introducing the slack variables s 1, s 2, s 2 ) is 2x 1 + 3x 2 + x 3 + s 1 = 5 4x 1 + x 2 + 2x 3 + s 2 = 11. 3x 1 + 4x 2 + 2x 3 + s 3 = 8 5x 1 4x 2 3x 3 + z = 0 where we seek the maximum of z with x 1, x 2, x 3, s 1, s 2, s 3 0. The simplex tableau of this consists of the matrix x 1 x 2 x 3 s 1 s 2 s 3 s s s z The notations of rows and columns, the objective function and the right hand side column was separated here by line segments. 12 / 46

13 13 / 46 FIRST STEP: FINDING THE PIVOT ELEMENT. First we find the most negative element of the last line in our example this is 5. (if there are more such element we can choose any of them). The column of this element is the pivot column (in our case this is the first column). Then we divide all elements of the last (denoted as RHS= right had side) column by the corresponding elements of the pivot column (we divide only by positive elements, we neglect the other rations, the ratios form the last column. basis x 1 x 2 x 3 s 1 s 2 s 3 RHS ratio 5 s = 2.5 pivot line s = 2.75 s =

14 We find the smallest ratio (if the are more than one such ration we can take any of them) its line will be the pivot line in our case the smallest ratio is 2.5 in the first line, thus the pivot line in the first one. The pivot element is the element in the pivot line and column in our case the element 2. The incoming variables is the variable corresponding to the pivot column (x 1 ), the outgoing variables is the variable corresponding to the pivot line (s 1 ). 14 / 46

15 15 / 46 SECOND STEP: PIVOTING. Divide all elements of the pivot line by the pivot element: x 1 x 2 x 3 s 1 s 2 s 3 RHS ratio 5 s = 2.5 pivot line s = 2.75 s = then adding (or subtracting) suitable multiples of this line we make the other elements of the pivot column to be zero. In our case 2nd line - 4 times first line, 3rd line - 3 times first line, finally 4th (last) line + 5 times first line gives the required result. The outgoing variable should be replaced by the incoming variable.

16 16 / 46 The new tableau basis x 1 x 2 x 3 s 1 s 2 s 3 RHS x s s We repeat the first and second step with the new tableau until all elements will be nonnegative or zero. Then the optimal solution can be read from the column on the right. In the new tableau the column of -0.5 is the pivot column, the pivot line is the line where we have the smallest ratio, now it is the third line. The incoming variable is x 3 ), the outgoing variable is s 3.

17 17 / 46 basis x 1 x 2 x 3 s 1 s 2 s 3 RHS ratio x = 5 s s = 1 pivot line

18 18 / 46 Divide the third line by 0.5 then subtracting its multiple by 0.5 from the first line, and adding its multiple by 0.5 to the last line we get the tableau) from which we omitted the ratios) basis x 1 x 2 x 3 s 1 s 2 s 3 RHS x s x Since there is no negative coefficient in the last line, the solution process ended. The maximum value of z is 13, the optimal values of the basic variables are in the RHS column i.e. x 1 = 2, s 2 = 1, x 3 = 1 the optimal values of the nonbasic variables are zero, i.e. x 2 = s 1 = s 3 = 0.

19 19 / 46 REMARKS. The above method called primal simplex method can be applied to the solution of standard normal maximum problems. This can be formulated as c 1 x 1 + c 2 x c n x n = z max subject to a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b 2.. a k1 x 1 + a k2 x a kn x n b k x 1, x 2,..., x n 0 where the c i, a ij, b j are known numbers (and we may assume that among the c i s there are positive values, otherwise x 1 = x 2 = = x n = 0 would give the optimal solution).

20 LP example:eload1b.lpp, with fives variables 20 / 46 x 1, x 2, x 3, x 4, x 5 0, x 1 + 2x 3 2x 4 + 3x 5 60 x 1 + 3x 2 + x 3 + x 5 12 x 2 + x 3 + x x 1 + 2x x 1 + 4x 2 + 5x 3 + 3x 4 2x 5 = z max or min

21 LP example: data entry 21 / 46 WinQSB= Windows based (Quantitative System for Business Data entry into WinQSB software in matrix form:

22 LP example:solution The table of solution: Combined Report for eload1b 13:07:30 Sunday February Decision Variable Solution Value Unit Cost or Profit c(j) Total Contribution Reduced Cost Basis Status Allowable Min. c(j) Allowable Max. c(j) 1 X1 10,00 3,00 30,00 0 basic 2,00 M 2 X2 0,67 4,00 2,67 0 basic 3,00 12,00 3 X3 0 5,00 0-1,00 at bound -M 6,00 4 X4 9,33 3,00 28,00 0 basic 2,00 4,00 5 X5 0-2,00 0-2,33 at bound -M 0,33 Objective Function (Max.) = 60,67 Constraint Left Hand Side Direction Right Hand Side Slack or Surplus Shadow Price Allowable Allowable Min. RHS Max. RHS 1 C1-8,67 <= 60,00 68,67 0-8,67 M 2 C2 12,00 <= 12,00 0 0,33 10,00 40,00 3 C3 10,00 <= 10,00 0 3,00 0,67 M 4 C4 20,00 <= 20,00 0 1, ,00 22 / 46

23 LP example:explanation of the solution 23 / 46 The reduced cost appears at decision variables which have (the optimal value) zero, and it shows the change of the objective function when we require positive value for that decision variable (instead requiring it to be nonnegative only). For example at x 3 = 0 the reduced cost is 1, which means that requiring x 3 a 3 (> 0) instead of x 3 0 the objective function will (approximately) change by ( 1) a 3. The shadow price appearing at a constraint shows that the change of the constant at the right hand side of the constraint how influences the value of the objective function. For example at the constraint C 3 the shadow price is 3, this means that increasing the right hand side of C 3 by b 3 (in our case to 10 + b 3 ) the value of the objective function will (approximately) change by 3b 3 (positive sign means increase, negative decrease).

24 LP example:explanation of the solution 24 / 46 The shadow price appearing at a constraint shows that the change of the constant at the right hand side of the constraint how influences the value of the objective function. For example at the constraint C 3 the shadow price is 3, this means that increasing the right hand side of C 3 by b 3 (in our case to 10 + b 3 ) the value of the objective function will (approximately) change by 3b 3 (positive sign means increase, negative decrease).

25 LP example:explanation of the solution 25 / 46 The slack or surplus appearing at a constraint is the difference between the two sides of that constraint (at the optimal values of the decision variables). The upper 1-5 lines of the last two columns show the lower and upper bounds of the coefficients in that decision variable in the objective function by which there is still optimal solution of the problem. The last 4 lines of the last two columns show the maximum and minimum values of the right hand sides of the constraints by which there is still optimal solution.

26 Network Modeling 26 / 46 This programming module, Network Modeling, solves network problems including transportation, assignment, shortest path, maximal flow, minimal spanning tree, traveling salesperson and capacitated network (transshipment) problems. A network consists of nodes and connections (arcs/links). Each node may have a capacity (in case of the network flow and transportation problems). If there is a connection between two nodes, there may be a cost, profit, distance, or flow capacity associated with the connection. Based on the nature of the problem, NET solves the link or shipment to optimize the specific objective function.

27 Transportation problem 27 / 46 The objective is to ship goods from supply sources to demand locations at the lowest total cost. The transportation problem could be balanced (the supplies and demands are equal) or could be unbalanced. CMP has production facilities in Nashville and Atlanta and markets its products in New York, Miami, and Dallas. Production capacities and demands and transport costs are given at the table below (supply demand in some units, cost in $). From/to New York Miami Dallas Supply Nashville Atlanta Demand How many unit need to be shipped from Nashville and Atlanta to New York, Miami, and Dallas at the lowest total cost.

28 Transportation problem, LP model Total supply is 800 units, total demand is 800 units hence the supply covers the demand (balanced problem). Denote the number of units shipped from Nashville to the 3 destinations by x 11, x 12, x 13 respectively, and the number of units shipped from Atlanta to the 3 destinations by x 21, x 22, x 23 respectively. Then we have to minimize the total cost being z = 10x x x x x x 23 subject to the constraints: x 11 + x 12 + x supply constraint in Nashville x 21 + x 22 + x supply constraint in Atlanta x 11 + x 21 = 150 demand satisfaction constraint in NY x 12 + x 22 = 300 demand satisfaction constraint in Miami x 13 + x 23 = 350 demand satisfaction constraint in Dallas Solution by WinQSB, either LP problem as above or by Network Modeling.(TRANSPORT.NET) 28 / 46

29 Optimal working order A supermarket is open from Monday to Saturday closed on Sunday and has 46 workers. Each worker has five days of duty and has one free per week. The daily wage of the workers is the same, 6000 HUF(=Hungarian forint), except for Saturday when it is 50 percent more. It is known that on each day of the week how many workers are needed. The present working order and daily work requirement is given by the next table (e.g. the work order I. means that the worker gets a free Monday and works on Saturday): W.o. Free day Workers M Tu W Th F Sa I Monday II Tuesday III Wednesday IV Thursday V Friday VI Saturday Alltogether: Workers required: / 46

30 30 / 46 Wage per day 6000 HUF, total daily wage of workers = HUF/workday, weekly wage according to the present work order, taking into consideration of the increased Saturday wage: = HUF. We have to find the optimal work assignment which means that the number of workers and their work order should be such that every day the required number of workers do work and the total weekly wage is minimal. Solution. Let x 1,..., x 6 be the number of workers working according to work orders I.,...,VI. x i are nonnegative integers, for which x 2 + x 3 + x 4 + x 5 + x 6 30 x 1 + x 3 + x 4 + x 5 + x 6 32 x 1 + x 2 + x 4 + x 5 + x x 1 + x 2 + x 3 + x 5 + x 6 34 x 1 + x 2 + x 3 + x 4 + x 6 44 x 1 + x 2 + x 3 + x 4 + x 5 40 The objective function is the weekly wage z = (x 1 + x 2 + x 3 + x 4 + x 5 + x 6 )+3000(x 1 + x 2 + x 3 + x 4 + x 5 ) = 33000(x 1 + x 2 + x 3 + x 4 + x 5 ) x 6.

31 Solution by WinQSB LP ILP modul. 31 / 46

32 32 / 46 The solution table is The optimal weekly wage is HUF, 7200 HUF less than before, 44 workers are sufficient, the optimal working order can be read from the solution table.

33 An assignment problem 33 / 46 Assignment problem is a special type of network or linear programming problem where objects or assignees are being allocated to assignments on a one-to-one basis. The object or assignee can be a resource, employee, machine, parking space, time slot, plant, or player, and the assignment can be an activity, task, site, event,asset, demand, or team. Our terminology: we assign agents to do tasks. The problem is linear because the cost function to be optimized as well as all the constraints contain only linear terms. The assignment problem is also considered as a special type of transportation problem with unity supplies and demands and is solved by the network simplex method.

34 Example: assignment problem 34 / 46 Example. A factory unit has 4 agents (workers) A 1 (Tim), A 2 (Peter), A 3 (John), A 4 (Rudy) and they have to make 4 different type of tasks (jobs): J 1, J 2, J 3, J 4. The next table shows how many hours each of them needs to do the same jobs. From/to J 1 J 2 J 3 J 4 A 1 (Tim) A 2 (Peter) A 3 (John) A 4 (Rudy) Which worker should be given the task (job) J 1, J 2, J 3, J 4 if all the 4 jobs should be done in minimal time and each worker can be given only one type of job.

35 35 / 46 Let the variable x ij represents the assignment of agent A i to the job J j taking value1 if the assignment is done and 0 otherwise. This formulation allows also fractional variable values, but there is always an optimal solution where the variables take integer values. The objective function (cost) z = 3x 11 +6x 12 +7x x 14 +5x 21 +6x 22 +3x 23 +8x x 31 +8x 32 +4x x 34 +8x 41 +6x 42 +5x 43 +9x 44 = min. subject to the constraints: x 11 +x 12 +x 13 +x 14 = 1 x 21 +x 22 +x 23 +x 24 = 1 x 31 +x 32 +x 33 +x 34 = 1 x 41 +x 42 +x 43 +x 44 = 1 x 11 +x 21 +x 31 +x 41 = 1 x 12 +x 22 +x 32 +x 42 = 1 x 13 +x 23 +x 33 +x 43 = 1 x 14 +x 24 +x 34 +x 44 = 1 agent A 1 performs one of the 4 jobs agent A 2 performs one of the 4 jobs agent A 3 performs one of the 4 jobs agent A 4 performs one of the 4 jobs job J 1 is done by one of the 4 agents job J 2 is done by one of the 4 agents job J 3 is done by one of the 4 agents job J 4 is done by one of the 4 agents Solution by WinQSB, select the Network Modeling, (ASSIGN.NET)

36 Shortest path problem 36 / 46 The shortest path problem includes a set of connected nodes where only one node is considered as origin node and only one node is considered as destination node. The objective is to determine a path of connections that minimizes the total distance from the origin to the destination. The shortest path problem is solved by a Labeling Algorithm. Let say CMP s trucks can only travel between CMP headquarters and its manufacturing plants in Nashville and Atlanta and its markets in Dallas, Miami and New York as shown in the next figure.

37 37 / 46

38 The owner wants to know the shortest distance from headquarters to Miami. One would use the shortest path model to find the answer. Select Network Modeling from the WinQSB menu, and then select Shortest Path Problem from the Problem Type option. Objective Criterion is set to Minimization. Numbering the nodes are Dallas (1), Nashville (2), CMP (3), NYC (4), Atlanta (5), Miami (6). (SHORTEST.NET) 38 / 46

39 Capacities of roads in terms of the number of vehicles per minute. In WinQSB, select Network Modeling. In Problem Type, select Maximal Flow Problem. The Objective Criterion is by default Maximization. (EVACUATE.NET) Graphic solution. 39 / 46 Maximal flow problem You are in charge of civilian defense of a city. You need to evacuate the city as quickly as possible. The road map for removing the citizens is shown in the next figure.

40 40 / 46 Minimal spanning tree problem CMP needs to install a sprinkler system in their office. The layout of the office shown in the next figure.

41 CMP wants to minimize the total length of pipes to be installed. The minimal spanning tree procedure is to be used here. From the WinQSB menu se Network Modeling. In NET Problem Specification select Minimal Spanning Tree. The Objective Criterion by default is Minimization. Select Spreadsheet Matrix Form. Problem name is CMP Safety. There are a total of nine nodes: Entrance, Lobby, Office, Control Room, Computer Room, Shop 1, Shop 2, Restroom, and Storage. Click OK. Now you will see the spreadsheet matrix input form. The node names are edited from default to actual names. Enter the appropriate distances in the from/to cells. After entering all data, click on the Solve and Analyze button on the toolbar. The next screen will display the solution in terms of sprinkler piping connection between the locations. The total length of pipe is 507 feet. The graphical solution option gives you the graphical layout of different nodes and the piping connections between them.(sprinkler.net) (symmetric arcs) 41 / 46

42 42 / 46 Traveling salesman problem CMP has sales representatives visiting between headquarters, the manufacturing facilities, and their customers. A sales representative starts the sales call from headquarters and must visit all the locations without revisiting them and then return to headquarters. This is the classical traveling salesman problem. The next figure displays the location of all the facilities and their distances.

43 In WinQSB select Network Modeling, then click on Traveling Salesman Problem. The Objective Criterion is to minimize the total distance a sales representative has to travel in visiting all of the places. We will use Spreadsheet Matrix form for data input. Enter the name of the problem in the Problem Title space. There are all together six places, hence enter Number of Nodes equal to six. Click OK. In input form edit the node variables from the Edit menu in the Toolbar and enter the names of nodes. Enter the distance data in appropriate cells. Click on Solve and Analyze. From popup menu, select appropriate solution method (here we have selected Branch and Bound Method) and click Solve. The computer will display the solution on the screen. The sales person should travel from CMP headquarters to New York and from New York to Atlanta, from Atlanta to to Miami, from Miami to Dallas, from Dallas to Nashville, from Nashville to CMP, with a total of 5330 miles traveled. There is also a graphical solution option output. 43 / 46

44 44 / 46 A network flow (transshipment) problem In the next transshipment problem there are two supply points S 1, S 2, three transshipment points T 1, T 2, T 3 and two demand points D 1, D 2. Supply capacities and demand requirements are shown in the circle and respective shipping costs are shown along the arrows in squares. We want to ship the goods through transshipment points to demand points at the least possible cost. The total supply = demand is 1200 units.

45 Denote by x 11, x 12, x 13 the number of units shipped from S1 to T1, T2, T3 respectively, x 21, x 22, x 23 be the number of units shipped from S2 to T1, T2, T3 respectively. Further let y 11, y 12 be the number of units shipped from T1 to D1, D2 respectively, y 21, y 22 the units shipped from T2 to D1, D2 respectively, and y 31, y 32 the number of units shipped from T3 to D1, D2 respectively. The cost function 45 / 46 z = 4x 11 +5x 12 +3x 22 +6x 23 +3y 11 +4y 21 +2y 22 +3y 32 should be minimized subject to the constraints x 11 + x 12 + x supply constraint at S1 x 21 + x 22 + x supply constraint at S2 y 11 + y 21 + y 31 = 800 demand at D1 must be satisfied y 12 + y 22 + y 32 = 400 demand at D2 must be satisfied x 11 = y 11 what goes in T1 also goes out x 12 + x 22 = y 21 + y 22 what goes in T2 also goes out x 23 = y 32 what goes in T3 also goes out and all variables x ij, y ki should be nonnegative.

46 Click on WinQSB, and in the popup menu select Network Modeling. From there select Problem Type: Network Flow. The Objective Criterion is Minimization. Select Spreadsheet Matrix Form for data entries. Enter Problem Title: Transshipment Problem. In our problem we have seven nodes, hence enter Number of Nodes equals seven. Click OK. The screen will display a spreadsheet form for inputs with default node names. Click on EDIT on toolbar, select Node names and input appropriate node names, one to be replaced by S1 and so on and click OK. Input the data, note that blank cells represent no connections.save it as TRSSHIPM.NET Now click on the Solve and Analyze button on the toolbar and select Solve Problem. Minimal cost is 7600 and in the solution the amount of units shipped (in various ways) will be displayed. If you want to see another form of the solution, click on result and then select graphical solution. The computer will display the graphical representation of the solution. 46 / 46