1 You would like to make a nutritious meal of eggs, edamame, and elbow macaroni. The meal should provide at least 40 g of carbohydrates, at least 20 g of protein, and no more than 50 g of fat. An egg (one serving) contains 2 g of carbohydrates, 17 g of protein, and 14 g of fat. A serving of edamame contains 12 g of carbohydrates, 12 g of protein, and 6 g of fat. A serving of elbow macaroni contains 43 g of carbohydrates, 8 g of protein, and 1 g of fat. An egg costs $2, a serving of edamame costs $ 5, and a serving of elbow macaroni costs $3. Formulate a linear optimization model that could be used to determine the number of servings of egg, edamame, and elbow macaroni that should be in the meal in order to meet the nutrition requirements at minimal cost. (You don't need to find the optimal solution to the model you formulate. Biketech manufactures road bikes and mountain bikes. They want to determine the number of each type of bike to produce in order to maximize profit over the next four-week planning horizon. Constraints affecting the production quantities are the production capacities in three departments: welding, assembly, and testing. For the four-week planning period, 260 hours of welding time, 340 hours of assembly time, and 320 hours of testing time are available. The time required to manufacture the two types of bicycles is shown in the following table: Labor (Hours) Road Welding 6 Assembly Testing Mountain 8 4 2 2 1 Road bikes provide a profit of $60 per unit and mountain bikes provide a profit of $50 per unit. Formulate a linear optimization model that will maximize Biketech's profits and satisfy the constraints. (You don't need to find the optimal solution to the model you formulate.) Dunder Mifflin supplies paper to corporate clients. An important client needs 1500 reams of paper. Dunder Mifflin can procure the paper from several suppliers. The characteristics of the suppliers are as follows Cost per ream Delivery time Supplier A 3.50 5 days Supplier B 2.00 10 days Supplier C 6.50 15 days Reams available 200 600 600 Supplier D Supplier E 5.00 4.00 4 days 6 days 200 200 When procuring the paper, Dunder Mifflin needs to ensure the following conditions are satisfied: (a) The client requires that at least 500 of the 1500 reams be delivered within the next 7 days. (b) In order to maintain good relations with Supplier C, Dunder Mifflin must order at least 100 reams from them. (c) Because of a rivalry between supplier B and D, Dunder Mifflin must order at least as many reams from supplier D as they do from supplier B. (d) Each supplier has a limited number of reams of paper available, as shown in the table. Formulate a linear optimization model to help Dunder Mifflin procure the paper at the lowest cost. (You don't need to find the optimal solution to the model you formulate.) Consider the linear optimization model Maximize x + y Subject to 2x + y 8 2x + 3y 12 x, y 0 Graph the constraints and identify the feasible region. (b) Draw the isoquant representing all combinations of x and y that make the objective function equal to 2. (c) Determine the optimal values of x and y. (d) Label the optimal solution on you graph. (e) Calculate the optimal value of the objective function. Consider the linear optimization model Maximize 3x 2y Subject to 5x + 2y 20 2x + 3y 12 x, y 0 (a) Graph the constraints and identify the feasible region. (b) Draw the isoquant representing all combinations of x and y that make the objective function equal to 12. (c) Determine the optimal values of x and y. (d) Label the optimal solution on you graph. (e) Calculate the optimal value of the objective function Consider the linear optimization model Maximize 2x + y Subject to x + 2y 8 xy5 x, y 0 (a) Graph the constraints and identify the feasible region. (b) Choose a value and draw the isoquant representing all combinations of x and y that make the objective function equal to that value. (c) Determine the optimal values of x and y. (d) Label the optimal solution on you graph. (e) Calculate the optimal value of the objective function. LP-Exer #1. Definition of Variables: EG= number of servings of eggs in optimal meal ED= number of serving of edamame in optimal meal EM= number of servings of elbow macaroni in optimal meal C= cost function LP Formulation: Minimize: C = 2*EG + 5*ED + 3*EM Subject to: 2*EG + 12*ED +43*EM 40 (carbohydrates) 17*EG + 12*ED + 8*EM 20 (protein) 14*EG + 6*ED + 1*EM 50 (fats) EG 0 ED 0 EM 0 #2. Definition of Variables: R= number of road bikes to produce M= number of mountain bikes to produce P= profit function LP Formulation: Maximize: P = 60*R + 50*M Subject to: 6*R + 8*M 260 (welding) 4*R + 2*M 340 (assembly) 2*R + M 320 (testing) R0 M0 (N.B.: This is a linear programming problem (LP) if we are allowed to sell fractional units of a bike. If optimal solution are required to be integers, then integer programming (IP) has to be used. However, an optimal solution may be obtained by LP if if an LP method for finding an optimal solution already yields integer solutions, i.e., the optimal value based on the IP solutions should be equal to the LP optimal value. ) #3. Definition of Variables: Let A, B, C, D, and E denote the number of reams from suppliers A, B,C, D, and E, respectively. S= denote the number of reams to be delivered within 7 days CF= cost function LP Formulation: Miniimize: CF = 3.5*A + 2*B + 6.5*C + 5*D + 4*E + 0*S Subject to: A + B + C +D + E = 1500 S 500 (condition (a), 500 reams delivered w/n 7 days) A + D + E S (only suppliers A, D, and meet w/n 7-days delivery limit) C 100 (condition (b)) D B (condition (c)) A 200 B 600 C 600 D 200 E 200 A, B, C, D, E 0 (N.B.: This LP doesn't seem to have a solution. Condition ( c) limits the supplies from Suppliers B and D to at most 400 (=200 max for D and D B). However, the maximum supplies from Suppliers A, C, and E is only 200 + 600 + 200 = 1000. All told, these give us a total of 1400 reams only. We will be short by 100 reams.) Page 1 LP-Exer #4 a) Constraints are (boundaries in intercept-intercept form): line1: x/4 + y/8 = 1 (obtained by div both sides of given by 8) line2: x/6 + y/4 = 1 (obtained by div both sides of given by 12) x-axis: y = 0 y-axis: x = 0 (see little arrows for half-planes consistent w/ constraint) Plot of constraints line1 10 9 (0,8) 8 7 y line2 6 5 4 (0,4) 3 (3,2) 2 R 1 (6,0) (4,0) 0 -1-1 0 1 2 3 4 5 6 7 8 9 10 x Feasible region is R with vertices (0,0), (4,0), (3,2) and (0,4) Cramer's Rule for line1 line2: 2 1 2 3 8 1 12 3 D= D(x) = 2 = 8 2 D(y) = (2)(3) - (2)(1) = 6 - 2 = 4 = 12 (8)(3)-(12)(1) = 24 - 12 = 12 (2)(12) - (2)(8) = 24 - 16 = 8 = x = D(x)/D = 12/4 =3 y = D(y)/D = 8/4 = 2 b) isoquant line: x + y =2 Plot of constraints line1 10 9 (0,8) 8 7 y line2 6 5 4 Optimal solution (3,2); for letter d); P = 5 (0,4) 3 (3,2) 2 R 1 -1-1 0 (6,0) (4,0) 0 1 2 3 4 5 6 7 8 9 10 x isoquant line: x + y = 2 (red line); intercepts are (2,0) and (0,2) c), d), e): optimal solution, value: Compare the values of the obj. function P at the vertices of the feasible region R (use Extreme Value Theorem) Vertices (0,0) (4,0) (3,2) (0,4) P=x+y 0+0=0 4+0=4 3+2=5 0+4=4 MAX Page 2 (see figure in b) for plot) LP-Exer #5 a) Constraints are (in intercept-intercept form): line1: x/4 + y/10 = 1 (obtained by div both sides of given by 20) line2: x/6 + y/4 = 1 (obtained by div both sides of given by 12) x-axis: y = 0 y-axis: x = 0 (see little arrows for half-planes consistent w/ constraint) Plot of constraint 12 line1 10 (0,10) 8 y 6 (0,4) 4 line2 (36/11, 20/11) 2 (6,0) R 0 0 1 3 (4,0) 4 2 5 6 7 x Feasible region is R with verices (4,00, (6,0) and (36/11, 20/11) Cramer's Rule for line1 line2: 5 2 12 3 5 20 2 D(y) = 3 20 D(x) = 2 2 D= 12 = (5)(3) - (2)(2) = 15 - 4 = 11 = (20)(3) - (12)(2) = 60 - 24 = 36 = (5)(12) - (2)(20) = 60 - 40 = 20 x = D(x)/D = 36/11 y = D(y)/D = 20/11 b) isoquant line: 3x - 2y = 12 or y = (3x-12)/2 x y = (3x - 12)/2 4 0 4.5 0.75 12 5 1.5 line1 10 6 3 7 4.5 Plot of constraint (0,10) y 8 6 3x - 2y =12 (isoquant line; passes thru (4,0), (4.5, 0.75), (5, 1.5), (6,3), and (7,4.5)) (0,4) 4 line2 (36/11, 20/11) 2 (6,0) R 0 0 1 2 3 (4,0) 4 5 x 6 7 Optimal solution (4,0) (for letter d); P = 18 isoquant line: 3x - 2y = 12 (red line) c), d), e): optimal solution, value: Compare the values of the obj. function P at the vertices of the feasible region R (use Extreme Value Theorem) Vertices (4,0) (6,0) P = 3x - 2y 2(4) - 3(0) = 8 3(6) - 2(0) = 18 (36/11, 20/11) 3(36/11) - 2(20)/11 = (108 - 40)/11 = 68/11 = 6.1818... Page 3 LP-Exer #6 a) Constraints are (in intercept-intercept form): line1: x/8 + y/4 = 1 (obtained by div both sides of given by 8) line2: x/5 - y/5 = 1 (obtained by div both sides of given by 5) x-axis: y = 0 y-axis: x = 0 (see little arrows for half-planes consistent w/ constraint) Plot of constraints line1 5 (0,4) 4 3 2 R y 1 (6,1) (8,0) 0 -1 0 1 2 3 (5,0) 5 6 4 7 8 9 -2 -3 -4 -5 (0,-5) -6 line2 x Feasible region is R with vertices (0,0), (5,0), (6,1) and (0,4) Cramer's Rule for line1 line2: 1 2 5 -1 1 8 1 D(y) = -1 8 D(x) = 2 1 D= 5 = = = (1)(-1) - (2)(1) = -1 - 2 = -3 (8)(-1) - (5)(2) = --8 - 10 = -18 (1)(5) - (1)(8) = 5 - 8 = -3 x = D(x)/D = (-18)/(-3) = 6 y = D(y)/D = (-3)/(-3) = 1 b) isoquant line: 2x + y = 2 (the isoquant 2x + y = 4 is also easy to plot below) Plot of constraints line1 5 4 (0,4) Optimal solution (6,1); for letter d); max P = 13 3 2 R y 1 (6,1) (8,0) 0 -1 0 1 2 3 (5,0) 5 6 4 7 8 9 -2 -3 -4 -5 (0,-5) -6 line2 Isoquant line: 2x + y = 2 w/ intercepts (0,2) and (1,0) x isoquant line: 2x + y = 2 (red line); intercepts are (1,0) and (0,2) c), d), e): optimal solution, value: Compare the values of the obj. function P at the vertices of the feasible region R (use Extreme Value Theorem) Vertices (0,0) (5,0) (6,1) (0,4) P = 2x + y 2(0) + 0 = 0 2(5) + 0 = 10 2(6) + 1 = 13 2(0) + 4 = 4 MAX (see figure in b) for plot) Page 4