Decision Support Systems Problems 7.1 to 7.6
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Chapter 7.1 to 7.6 Problems
7-16 A candidate for mayor in a small town has allocated $40,000 for last-minute advertising in
the days preceding the election. Two types of ads will be used: radio and television. Each radio
ad costs $200 and reaches an estimated 3,000 people. Each television ad costs $500 and reaches
an estimated 7,000 people. In planning the advertising campaign, the campaign manager would
like to reach as many people as possible, but she has stipulated that at least 10 ads of each type
must be used. Also, the number of radio ads must be at least as great as the number of television
ads. How many ads of each type should be used? How many people will this reach?
The optimal solution is the solution which maximizes the number of people reached by
different number of radio and television ads. These are 175 radio ads and 10 television
ads.
The maximum number of people reached by advertisements is 595000 people by doing
175 radio ads and 10 television ads for maximizing the number of people covered.
7-19 MSA Computer Corporation manufactures two models of smartphones, the Alpha 4 and the
Beta 5. The firm employs five technicians, working 160 hours each per month, on its assembly
line. Management insists that full employment (i.e., all 160 hours of time) be maintained for each
worker during next month’s operations. It requires 20 labor hours to assemble each Alpha 4
computer and 25 labor hours to assemble each Beta 5 model. MSA wants to see at least 10 Alpha
4s and at least 15 Beta 5s produced during the production period. Alpha 4s generate $1,200 profit
per unit, and Beta 5s yield $1,800 each. Determine the most profitable number of each model of
smartphone to produce during the coming month.
MSA Computer Corporation is faced with the decision of how many Alpha 4 and Beta 5
smartphones to produce in the coming month to maximize profits. The following
constraints and information are given:
Labor Hours Constraint:
There are 5 technicians, each working 160 hours per month, totaling 800 hours available
for production.
It takes 20 labor hours to assemble each Alpha 4 and 25 labor hours to assemble each
Beta 5.
This constraint can be expressed as: 20x + 25y ≤ 800, where x is the number of Alpha 4s
produced and y is the number of Beta 5s produced.
Production Requirements:
MSA wants to produce at least 10 Alpha 4s (x ≥ 10) and at least 15 Beta 5s (y ≥ 15).
Non-negativity Constraints:
Production quantities cannot be negative: x ≥ 0 and y ≥ 0.
The objective is to maximize profit, which is calculated as: Profit (Z) = 1200x + 1800y,
where x is the number of Alpha 4s and y is the number of Beta 5s.
To determine the most profitable production quantities, the profit function is evaluated at
the corner points of the feasible region. The combination of x (Alpha 4s) and y (Beta 5s)
that yields the highest profit at one of these corner points represents the optimal
production plan for maximizing profits in the coming month.
Chapter 7.1 to 7.6 Problems
The solution to the problem provides the specific quantities of Alpha 4 and Beta 5
smartphones to produce to achieve the highest possible profit while meeting all
constraints.
7-25 Woofer Pet Foods produces a low-calorie dog food for overweight dogs. This product is
made from beef products and grain. Each pound of beef costs $0.90, and each pound of grain
costs $0.60. A pound of the dog food must contain at least 9 units of Vitamin 1 and 10 units of
Vitamin 2. A pound of beef contains 10 units of Vitamin 1 and 12 units of Vitamin 2. A pound of
grain contains 6 units of Vitamin 1 and 9 units of Vitamin 2. Formulate this as an LP problem to
minimize the cost of the dog food. How many pounds of beef and grain should be included in
each pound of dog food? What are the cost and vitamin content of the final product?
Minimize Z = $0.90x + $0.60y
10x + 6y >= 9
12x + 9y >= 10
X, Y >= 0
Here is the cost and vitamin content of the final product that can be calculated:
Cost per pound of dog food: 0.90B+0.60G
Vitamin 1 content: 10B+6G
Vitamin 2 content: 12B+9G
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