3QA3 Practice Problems Solutions - Ch 2. Formulate and solve LP problems PDF

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3QA3 Practice Problems Solutions - Ch 2. Formulate and solve LP problems...

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Practice Problems Solutions – Chapter 2. Formulate and Solve LP Problems

Problem 2-3 (p. 56) Under what condition is it possible for an LP problem to have more than one optimal solution? Solution A problem can have alternative optimal solutions if the level profit or level cost line runs parallel to one of the problem’s binding constraints.

Problem 2-4 (p. 57) Under what condition is it possible for an LP problem to have an unbounded solution? Solution A problem can be unbounded if one or more constraints are missing, such that the objective value can be made infinitely larger or smaller without violating any constraints.

Problem 2-7 (p. 57) Consider the following problem. Maximize: Profit = 4𝑋1 + 3𝑋1𝑋2 + 8𝑋2 + 5𝑋3 Subject to: 2𝑋1 + 𝑋2 + 2𝑋3 ≤ 50 𝑋1 − 4𝑋2 ≥ 6 1.5𝑋12 + 6𝑋2 + 3𝑋3 ≥ 21 19𝑋2 − 0.33𝑋3 = 17 5𝑋1 + 4𝑋2 + 3√𝑋3 ≤ 80 Which equations are invalid for use in an LP problem? Why? Solution The objective function is not linear because it contains the product of X1 and X2. The first, second, and fourth constraints are okay as is. The third constraint is not linear because it contains X12. The last constraint is not linear because it contains √X3.

Problem 2-10 (p. 57) What are the components of defining a problem in Excel so that it can be solved using Solver? Solution The three components are: target cell (objective function), changing cells (decision variables), and constraints.

Practice Problems Solutions – Ch 2 … page 1

Problem 2-11 (p. 57) How is the slack (or surplus) calculated for a constraint? How is it interpreted? Solution Slack is defined as the RHS minus the LHS value for a  constraint. It may be interpreted as the amount of unused resource described by the constraint. Surplus is defined as the LHS minus the RHS value for a ≥ constraint. It may be interpreted as the amount of over satisfaction of the constraint.

Problem 2-12 (p. 57) What is an unbounded solution? How does Solver indicate that a problem solution is unbounded? Solution An unbounded solution occurs when the objective of an LP problem can go to infinity (negative infinity for a minimization problem) while satisfying all constraints. Solver indicates an unbounded solution by the message “The Set Cell values do not converge”.

Practice Problems Solutions – Ch 2 … page 2

Problem 2-13 (p. 57) Solve the following LP problem (a) using the graphical procedure and (b) using Excel. Maximize: Profit = 2𝑋 + 𝑌 (0) Subject to: 3𝑋 + 6𝑌 ≤ 32 (1) Constraint (3) is tricky because one of coefficients is negative. The approach is the same; we need to find 2 points on the line 3X – Y = 3. The values of X and Y need 7𝑋 + 𝑌 ≤ 20 (2) to be positive because we need to graph them. 3𝑋 − 𝑌 ≥ 3 (3) So use a little common sense: if X=1 then Y=0 satisfies 3X-Y=3 (3X-Y=31-0=3); 𝑋, 𝑌 ≥ 0 similarly if X=4 then Y=9 satisfies 3X-Y=3 (3X-Y=34-9=3) Solution

See the Excel workbook ‘Practice Problems Solutions – Ch 2.Formulate and Solve LP Problems.xls’. (a) In the graphical procedure the optimal solution occurs at the corner point where constraints (2) and (3) intersect. We determine the optimal solution algebraically as follows: (2) 7𝑋 + 1𝑌 = 20 (3) 3𝑋 − 1𝑌 = 3 (2) + (3) 10𝑋 + 0𝑌 = 23 → 𝑋 = 2.30 Substitute 𝑋 = 2.30 into (2) or (3): (2) → 7 × 2.30 + 1𝑌 = 20 → 𝑌 = 3.90 Substitute 𝑋 = 2.30 and 𝑌 = 3.90 into (0): 𝑃 = 2 × 2.30 + 1 × 3.90 → 𝑃 = 8.50 (b) The optimal solution from Excel is X = 2.30, Y = 3.90, and Profit = 8.50 Practice Problems Solutions – Ch 2 … page 3

Problem 2-17 (p. 57) Solve the following LP problem (a) using the graphical procedure and (b) using Excel. Minimize: Cost = 3𝑋 + 7𝑌 (0) Subject to: 9𝑋 + 3𝑌 ≥ 36 (1) 4𝑋 + 5𝑌 ≥ 40 (2) 𝑋−𝑌 ≤ 0 (3) 2𝑋 ≤ 6 (4) 𝑋, 𝑌 ≥ 0 Solution

Constraint (3) is tricky because one of coefficients is negative. The approach is the same; we need to find 2 points on the line X – Y = 0. The values of X and Y need to be positive because we need to graph them. So use a little common sense: if X=0 then Y=0 satisfies X-Y=0 (X-Y=0-0=0); similarly if X=2 then Y=2 satisfies X-Y=0 (X-Y=2-2=0) Now we must decide whether the feasibile region is above or below the line. Because of the negative coefficient it’s not obvious. So we just pick any point not on the line and see what it tells us. Try the point X=1, Y=3. Then X-Y=1-3=-2; this is  0 so this point is in the feasible region. It’s not necessary but we can try another point, say X=5, Y=3. X-Y=5-3=2. This is not  0 so this point is not in the feasible region. Therefore the feasible region is on and above the line.

See the Excel workbook ‘Practice Problems Solutions – Ch 2.Formulate and Solve LP Problems.xls’. (a) In the graphical procedure the optimal solution occurs at the corner point where constraints (2) and (4) intersect. We determine the optimal solution algebraically as follows: (2) 4𝑋 + 5𝑌 = 40 (4) 2𝑋 =6 From (4) 2𝑋 = 6 → 𝑋 = 3.00

Substitute 𝑋 = 3.00 into (2): (2) → 4 × 3.00 + 5𝑌 = 40 → 𝑌 = 5.60 Substitute 𝑋 = 3.00 and 𝑌 = 5.60 into (0): 𝐶 = 3 × 3.00 + 7 × 5.60 → 𝑃 = 48.20

(b) The optimal solution from Excel is X = 3.00, Y = 5.60, and Cost = 48.20. Practice Problems Solutions – Ch 2 … page 4

Problem 2-27 (p. 59) An investor is interested in the following two stocks: Carolina Solar Power South West Steel Short-term appreciation $0.46 $0.26 Long-term appreciation $1.72 $1.93 Dividend income 9% 13% The investor wants to: (1) Grow his investment by at least $6,000 in the long term (over three years) and at least $900 in the short term. (2) Earn a dividend of at least $300 per year. Assuming these data are indicative of what will happen over the next three years, what is the smallest investment, in dollars, that the investor needs to make to achieve his investment goals? Solution:

LP Problem Minimize: Subject to:

Cost = 𝑋 + 𝑌 0.46𝑋 + 0.26𝑌 ≥ 900 1.72𝑋 + 1.93𝑌 ≥ 6,000 0.09𝑋 + 0.13𝑌 ≥ 300 𝑋, 𝑌 ≥ 0

(0) (1) (2) (3)

See the Excel workbook ‘Practice Problems Solutions – Ch 2.Formulate and Solve LP Problems.xls’. Practice Problems Solutions – Ch 2 … page 5

Problem 2-29 (p. 59) An airline makes lattes in the galley and sells them to passengers. A regular latte contains a shot of expresso, I cup of 2% milk, frothed, and 0.5 cup of whipped cream. The low-fat latte contains a shot of expresso, 1.25 cups of skim milk, frothed, and no whipped cream. An airplane is stocked with 100 shots of expresso, 60 cups of skim milk, 60 cups of 2% milk, and 30 cups of whipped cream. The airline makes a profit of $1.58 on each regular latte and $1.65 on each low-fat latte. Assuming that all lattes that are made can be sold, what would be the mix of regular and low-fat lattes that maximizes the profit for the airline? Solution

LP Problem Maximize: Subject to:

Profit = 1.58𝑅 + 1.65𝐿 1𝑅 + 1𝐿 ≤ 100 1.25𝐿 ≤ 60 1𝑅 ≤ 60 . 5𝑅 ≤ 30 𝑅, 𝐿 ≥ 0

(0) (1) shots of expresso (2) cups of skim milk (3) cups of 2% milk (4) cups of whipped cream

See the Excel workbook ‘Practice Problems Solutions – Ch 2.Formulate and Solve LP Problems.xls’. Practice Problems Solutions – Ch 2 … page 6

Problem 2-43 (p. 61) An investor is considering three different television network stocks for her portfolio: British Broadcasting Company (BBC), Canadian Broadcasting Company (CBC), and Australian Broadcasting Company (ABC). Her broker has gathered the following information: Short-term growth Intermediate-term growth Dividend rate per $ invested per $ invested BBC $0.39 $1.59 8% CBC $0.26 $1.70 4% ABC $0.42 $1.45 6% The investor’s goals are: (1) The investment should yield short-term growth of at least $1,000, (2) The investment should yield intermediate-term growth of at least $6,000, and (3) The dividends should be at least $250 per year. Determine the least amount the investor can invest and how the investment should be allocated between the three stocks. Solution

LP Problem Minimize: Subject to:

Investment = 𝐵 + 𝐶 + 𝐴 0.39𝐵 + 0.26𝐶 + 0.42𝐴 ≥ 1,000 1.59𝐵 + 1.70𝐶 + 1.45𝐴 ≥ 6,000 0.08𝐵 + 0.04𝐶 + 0.06𝐴 ≥ 250 𝐵, 𝐶, 𝐴 ≥ 0

(0) (1) (2) (3)

See the Excel workbook ‘Practice Problems Solutions – Ch 2.Formulate and Solve LP Problems.xls’. Practice Problems Solutions – Ch 2 … page 7...