Solved Problems

  1. Solved Problem 7-1 Personal Mini Warehouses is planning to expand its successful Orlando business into Tampa. In doing so, the company must determine how many storage rooms of each size to build. Its objective and constraints follow:

    Maximize monthly earnings=50X1+20X2subject to20X1+40X2≤4,000(advertising budget available)100X1+50X2≤8,000(square footage required)X1≤60(rental limit expected)X1,X2≥0

    where

    X1= number of large spaces developed

    X2= number of small spaces developed

    Solution

    An evaluation of the five corner points of the accompanying graph indicates that corner point C produces the greatest earnings. Refer to the graph and table.

    CORNER POINT VALUES OF X1,X2 OBJECTIVE FUNCTION VALUE ($)
    A (0, 0) 0
    B (60, 0) 3,000
    C (60, 40) 3,800
    D (40, 80) 3,600
    E (0, 100) 2,000
    A graph with three lines for problem 7.1 is shown.

  2. Solved Problem 7-2 The solution obtained with QM for Windows for Solved Problem 7-1 is given in the following program. Use this to answer the following questions.

    1. For the optimal solution, how much of the advertising budget is spent?

    2. For the optimal solution, how much square footage will be used?

    3. Would the solution change if the budget were only $3,000 instead of $4,000?

    4. What would the optimal solution be if the profit on the large spaces were reduced from $50 to $45?

    5. How much would earnings increase if the square footage requirement were increased from 8,000 to 9,000?

    Screenshots of the linear programming results window and the ranging window are seen.

    Solution

    1. In the optimal solution, X1=60 and X2=40. Using these values in the first constraint gives us

      20X1+40X2=20(60)+40(40)=2,800

      Another way to find this is by looking at the slack:

      Slack for constraint 1=1,200, so the amount used is 4,000−1,200=2,800

    2. For the second constraint, we have

      100X1+50X2=100(60)+50(40)=8,000 square feet

      Instead of computing this, you may simply observe that the slack is 0, so all of the 8,000 square feet will be used.

    3. No, the solution would not change. The dual price is 0, and there is slack available. The value 3,000 is between the lower bound of 2,800 and the upper bound of infinity. Only the slack for this constraint would change.

    4. Since the new coefficient for X1 is between the lower bound (40) and the upper bound (infinity), the current corner point remains optimal. So X1=60 and X2=40, and only the monthly earnings change.

      Earnings=45(60)+20(40)=$3,500

    5. The dual price for this constraint is 0.4, and the upper bound is 9,500. The increase of 1,000 units will result in an increase in earnings of 1,000(0.4 per unit)=$400.

  3. Solved Problem 7-3 Solve the following LP formulation graphically, using the isocost line approach:

    Minimize costs=24X1+28X2subject to5X1+4X2≤2,000X1≥80X1+X2≥300X2≥100X1,X2≥0

    Solution

    A graph of the four constraints follows. The arrows indicate the direction of feasibility for each constraint. The next graph illustrates the feasible solution region and plots of two possible objective function cost lines. The first, $10,000, was selected arbitrarily as a starting point. To find the optimal corner point, we need to move the cost line in the direction of lower cost—that is, down and to the left. The last point where a cost line touches the feasible region as it moves toward the origin is corner point D. Thus D, which represents X1=200, X2=100, and a cost of $7,600, is optimal.

    A graph of four constraints for is shown.A graph illustrates feasible region, optimal cost line, and optimal solution.

  4. Solved Problem 7-4 Solve the following problem, using the corner point method. For the optimal solution, how much slack or surplus is there for each?

    Minimize profit=30X1+40X2subject to4X1+2X2≤162X1−X2≥2X2≤2X1,X2≥0

    Solution

    The graph appears next with the feasible region shaded.

    CORNER POINT COORDINATES PROFIT ($)
    A X1=1, X2=0 30
    B X1=4, X2=0 120
    C X1=3, X2=2 170
    D X1=2, X2=2 140

    The optimal solution is (3, 2). For this point,

    4X1+2X2=4(3)+2(2)=16

    Therefore, slack = 0 for constraint 1. Also,

    2X1−1X2=2(3)−1(2)=4>2

    Therefore, surplus=4−2=2 for constraint 2. Also,

    X2=2

    Therefore, slack=0 for constraint 3.

    A graph illustrates the feasible region of a problem.

    The optimal profit of $170 is at corner point C.

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