Demo 1: Solving LP problem graphically and with Excel Solver

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1 MS-C2105 Introduction to Optimization Solutions 2 Ehtamo Demo 1: Solving LP problem graphically and with Excel Solver Solve the linear optimization problem graphically and with Excel Solver. a) max 8 + 3x 2 s.e x x x 2 1 0, x 2 0 b) min + x 2 s.e x x 2 1 0, x 2 0 c) Write problems of sections a) and b) in matrix form: min c T x s.e. Ax b x 0 Solution a) Draw on a, x 2 plane the lines according to the inequality constraints. An easy way to do this is to calculate values of and x 2 in two different points and draw a line through them. The feasible area is between the four lines. Optimal solution can be found by drawing level curves of the objective function (lines in which the objective function has the same value). Move with the level curves in the direction of the objective function (the perpendicular of the line) and the last feasible point is the optimum. The optimum is in the intersection of the constraints 4 + 3x 2 21 and 2x 2 1 which is = 4, 09 and x 2 = 1, 55. The optimal objective function value is 37,36. Remember that the optimum of a linear optimization problem can be found in one of the feasible area corners. 1

9 8 7 6 5 3 + 8x 2 48 x2 4 3 0 4 + 3x 2 21 2 1 2x 2 1 0 x 2 0 1 1 0 1 2 3 4 5 The use of Excel Solver is similar to Exercise Session 1.

2 x 2 48 x x x x The use of Excel Solver is similar to Exercise Session 1. The Excel cells can also be arranged using the matrix form of the LP problem (see. section c). The decision variables are placed in the cells B2 and C2. The coefficients of the decision variables in objective function are inserted in the cells B4 and C4. The objective function values is calculated in the cell E4 with formula =SUMPRODUCT($B$2:$C$2;B4:C4) which sums the products of the corresponding variables. Sign $ is placed befor the character and the number of the cell reference so that the Excel doesn t change the reference cell if you copy the cell formula downward. Write the coefficients of the decision variables in the first constraint, 3 and 8, in the cells B7 and C7 and calculate the constraint value with the formula =SUMPRODUCT($B$2:$C$2;B7:C7) in the cell E7. Place the constant value of the constraint in the G7. Repeat the procedure to all of the constraint by using cells G8 and H8. Use these cells in the Solver to set the constraints and the objective function in the cell E7 in the Aseta tavoite (Set target) selection. b) Draw the lines of the inequality constraints in the graph. Draw the level curves of the objective function to find the optimum. The optimum solutions are on the edge of the area, in which + x 2 = 2, 0 and x

3 2.5 2 x2 1.5 1 x 2 1 0.5 0 2 + 2x 2 4 3 0 0.5 x 2 0 1 1 0 1 2 3 4 Insert the information in the Excel as in the section a.

3 x x x x Insert the information in the Excel as in the section a. By changing the initial values you realize that the Solver gives different solutions in which +x 2 = 2, 0 and x 2 1. c) a) Decision variables are in a column vector x [ ] x1 x = and the objective function coefficients of the objective function are in a column vector c [ ] 8 c =. 3 Now the objective function can be presented as max c T x = [ 8 3 ] [ ] = 8x + 3x 2. 2 The constraints coefficients of the decision variables are in a matrix A 3 8 A = and the constant terms of constraints in a column vector b 48 b = x 2

4 Now the constraints can be presented as 3 8 [ ] 3 + 8x 2 48 Ax b 4 3 x1 = 4x + 3x x 2 1 Non-negativity constraints for decision variables are [ ] [ ] x1 0 x 0 0 c) b) Now the objective function is max c T x = [ 1 1 ] [ ] x 2 x 2 = + x 2. There are constraints among the constraints so those have to be changed to constraints by multiplying the both sides of the inequality with -1 to turn the sign of the inequality: 2 + 2x x 2 4 and x 2 1 x 2 1. Now the constraints are 2 2 [ ] 2 2x 2 4 Ax b 1 0 x1 = x x 2 1 and non-negativity constraints are x 0 [ x1 x 2 ] [ ] 0 0 Demo 2: Formulation of linear optimization problem The Finnish Steel Co. has decided to optimize its production which consists of carbon steel, stainless steel and tool steel. In one week the company can use 100 tons of iron and 10 tons of chrome in its production. Thanks to double shift work week there are 80 h of work time. One ton of carbon steel requires 1,1 tons of iron and 1 h of work time. One stainless steel ton requires one ton of iron, 0,1 tons of chrome and 1 h 15 min of work time. One tool steel ton needs one ton of iron, 0,15 tons of chrome and 1 h 45 min of work time. The company has also made a deal with United Components Ltd. to deliver 25 tons of carbon steel to United Components each week. Current prices for the steels are 400 eper ton for carbon steel, 500 eper ton for stainless steel and 700 eper ton for tool steel. How much of each of the steels should the company produce to maximize its profits? Formulate as LP problem and solve with Excel Solver. Solution The decision variables are the quantities of the steels: := x 2 := x 3 := quantity of carbon steel (ton) quantity of stainless steel (ton) quantity of tool steel (ton) 4

The objective is to maximize the profits: max 400 + 500x 2 + 700x 3.

5 The objective is to maximize the profits: max x x 3. There is 100 tons of iron available in a week and carbon steel requires 1,1 tons, stainless steel 1 ton and tool steel also 1 ton: 1, 1 + x 2 + x The stainless and tool steels require also chrome of which there are 10 tons available in a week. The stainless steel requires 0,1 tons and tool steel 0,15 tons: 0, 1x 2 + 0, 15x There are 80 h of work time and carbon steel requires 1 h, stainless steel 1,25 h and tool steel 1,75 h: + 1, 25x 2 + 1, 75x The deal made forces the company to produce at least 25 tons of carbon steel: 25. The quantities of the steels cannot be negative: 0, x 2 0 and x 3 0. Solve the problem with Excel Solver. Optimally the company produces only the demanded 25 tons of carbon steel and uses the rest of its capacity to produce 31,42 tons of tool steel. This way the company gains profits of e. Problem 1: Solving LP problem graphically and with Excel Solver Solve the linear problem graphically and with Excel Solver. a) max 6 + 5x 2 s.e x x x x 2 1 0, x 2 0 5

b) min 6 + 5x 2 s.e. 3 + 5x 2 15 7 + 2x 2 14 + 3x 2 9 0, x 2 0 Solution a) Draw the lines of constraints to perceive the feasible area and draw the objective function level curves to find the optimum.

6 b) min 6 + 5x 2 s.e x x x 2 9 0, x 2 0 Solution a) Draw the lines of constraints to perceive the feasible area and draw the objective function level curves to find the optimum. The optimum is the intersection of the constraint lines 5 + 3x 2 12 ja 3 + 8x 2 24 which is = 0, 774 and x 2 = 2, 710. The objective function value is 20,13. The same solution can be found with Excel Solver x x x x x 2 1 x b) Again, draw the constraint lines and the level curves of the objective function. The optimum is = 0 and x 2 = 3 which is the intersection of constraints 0 and 3 + 5x The objective function values is 15 which can also be found with Excel Solver. 6

7 x x 2 15 x x 2 14 x Problem 2: Putte the Pig Name Day Party Putte the Pig is arranging his name day party, again. He is expecting 100 guests to his party and he has to make delicacies for all of them. Putte has decided to make only three kind of delicacies: cake, cookies and buns. In one cake there is enough for 10 persons, one cookie for one person and one batch of buns for 20 persons. Unfortunately there is only 10 h to the party and everything has to be ready by then. Baking a cake takes 30 min, one cookie takes 1 min on average and one batch of buns takes 40 min. The yearly party is expensive so Putte tries to minimize the costs. The ingredients for a cake cost 5 e, for a batch of buns 10 e and for a cookie 1 e. How many of each delicacies should Putte make? Also notice that Putte thinks that there should be enough for at least 20 persons of every delicacy and there should be cake for at least twice as many persons as there are cookies for. Formulate as LP problem and solve with Excel Solver. Solution The decision variables are the quantities of delicacies: := x 2 := x 3 := quantity of cakes (piece) quantity of cookies (piece) quantity of bun batches (piece) Putte the Pig tries to minimize the costs: min 5 + x x 3. 7

8 There are 100 persons and one cake is enough for 10 persons, a cookie for one person and batch of buns for 20 persons: 10 + x x There has to be enough for at least 20 persons of every delicacy: 10 20, x 2 20 and 20x Putte also wants that the ratio between the cake and the cookies is at least two to one:: 10 2 x 2. Putte has 600 min to prepare the delicacies and a cake takes 30 min, a cookie 1 min and batch of bun 40 min: 30 + x x The quantities of delicacies cannot be negative: 0, x 2 0 and x 3 0. Solve the problem with Excel Solver. Putte should optimally make 6 cakes, 20 cookies and one batch of buns. Total costs are 60 e. The solution indicates that the amount of guests and required quantities of cookies and bun batches are the active constraints. If you could decrease the constant term of these constraints the objective function value would decrease. Problem 3: Planning of residential area A construction company owns 800 hectares (ha) of land and is about to build onefamily, two-family and three-family houses. The company estimates that a onefamily house will profit the company e, it requires area of 1 ha, building costs are e, and the water consumption is l per day. Same key figure for two-family house are e, 1,5 ha, eand l per day, and for three-family house e, 2 ha, e and l per day. At least half of the houses have to be one-family houses. The water consumption for the area cannot surpass l per day. For every 200 families there has to be one recreational area. It requires 0,5 ha of area. Building costs are 700 e and water consumption is l per day. Streets etc. require 15 % of the total area. Formulate LP problem and find out how many and what kind of houses should the company build to maximize its profits? Note. In Excel the linear problem can also be solved as integer problem if you constraint the variables to be integers with Solver s kok (int) constraint. 8

9 Solution The decision variables are the quantities of houses and recreational areas: := x 2 := x 3 := x 4 := quantity of one-family houses (piece) quantity of two-family houses (piece) quantity of three-family houses (piece) quantity of recreational areas (piece) The construction company maximizes its profits (income - expenses): max x x x x x 4 At least half of the houses have to one-family houses: 1 2 ( + x 2 + x 3 ) x x 3 0. The water consumption of the residential are cannot surpass l per day and one-family house uses l per day, two-family house uses l per day, three-family house 3200 l per day and recreational area 2500 l per day: x x x For every 200 families there has to be one recreational area: x ( + 2x 2 + 3x 3 ) x x 3 + x 4 0. The construction company has to use 15 % (120 ha) of the total 800 ha of building area for the streets. The remaining 680 ha can be used for the houses and one-family house requires 1 ha of area, two-family house 1,5 ha, three-family house 2 ha and recreational area 0,5 ha: + 1, 5x 2 + 2x 3 + 0, 5x The quantities cannot be negative: 0, x 2 0, x 3 0 and x 4 0. Solve the problem with Excel Solver. The optimal solution is that the company build around 180 one-family houses, 180 two-family houses, no three-family houses and three recreational areas. This solution gives 24 million eprofits. The variables can be set as integers in Excel Solver by adding kok (int) constraint in the Reunaehto (Constraint) section and referring to the decision variable cells. Now the solution is 182 one-family houses, 177 two-family houses, no three-family houses and 3 recreational areas. Problem 4: Solving LP problem in 3D Solve the problem graphically. Check your answer with Excel Solver. 9

10 max 4x + 2y + 11z s.e. x + 2y + 3z 15 x + z 6 z 2 x 0, y 0, z 0 Solution Draw the constraint planes into x, y, z-space to perceive the feasible solutions. Draw level surface of the objective function and move in the direction of plane s perpendicular until there are only one feasible solution left, the optimal solution. The optimal solution is x = 4, y = 2, 5 and z = 2 and the objective function value is

the right hand side. Unravel the meaning of the values in the reports, especially Varjohintaa (Shadow price). Tip.

11 Problem 5: Sensitivity analysis In this exercise you get know the Sensitivity Report of the Excel Solver. Solve the Exercise 2 with Excel Solver and when the Ratkaisimen tulokset (Solver results) window opens select the reports (Herkkyysanalyysi (Sensitivity), Vastaus (Answer) and Rajoitus (Limits)) from the right hand side. Unravel the meaning of the values in the reports, especially Varjohintaa (Shadow price). Tip. Try how the optimum solution changes if you, for example, change one of the constant terms in the constraints and solve the problem again. Solution Solve the Exercise 2 and select the reports (Herkkyysanalyysi (Sensitivity), Vastaus (Answer) and Rajoitus (Limits)) from the right hand side of the Ratkaisimen tulokset (Solver results) window. This opens three new spreadsheets in the Excel. On the first row of the Answer spreadsheet there are the initial and final values of the objective function. Next there are the initial and final values of the decision variables. From Kokonaisluku (Integer) section you can see if the variable has real or integer value. From Reunaehdot (Boundary condition) section you can see the values of the left side of the constraints and the formulas of the constraints. Tila (State) section tells if the constraint is binding (equality applies) or not. Liukuma (Slide) section tells how much the constraint value can change before the equality applies. In the first section of the Sensitivity spreadsheet there are final values. Vähentyneet kustannukset (Reduced costs) section tells how much can the coefficient in the objective function change before the decision variable enter the base (value is zero for base variables). Tavoitekerroin (Target coefficient) indicates how much the objective function value changes if value of certain decision variable is changed by one. Sallittu lisäys (Allowed increase) and vähennys (decrease) tell how much can the 11

12 objective function coefficients of the decision variables increase or decrease before the variable values in the optimal solution differ. In the second part of the spreadsheet Varjohinta (Shadow price) indicates how much the objective function increases if the value of the constant term in the constraint increases with one. Oikean puolen reunaehto (Right side boundary condition tells the value of constraint s constant term and Sallittu lisäys (Allowed increase) and vähennys (decrease) indicate how much the constant term can increase or decrease without changing the values of decision variables in the optimal solution. First on the Limits spreadsheets is the objective function value. Second section tells the final value, the minimum value and the maximum value of decision variables and the values of objective function with these decision variable values. Note. There are two optimal solutions in which = 4 and x 3 = 2 or = 5 and x 3 = 1 which affects the report values. 12

LINEAR PROGRAMMING: A GEOMETRIC APPROACH. Copyright Cengage Learning. All rights reserved.

3 LINEAR PROGRAMMING: A GEOMETRIC APPROACH Copyright Cengage Learning. All rights reserved. 3.4 Sensitivity Analysis Copyright Cengage Learning. All rights reserved. Sensitivity Analysis In this section,