Solving An Allocation Problem

A taxi company limits its drivers to one eight-hour shift per day, with one half-hour break for a meal. A taxi cannot carry more than three passengers, all in the back seat and cannot combine stops; that is, all the passengers must be going to the same destination

Answer

The drivers have an 8 hour shift with a 30 minute-break for lunch, which means they are in work for 8-0.5=7.5 hours a day. Letting S be the length of the shift, a daily shift is S ≤ 8 hours. Also, letting B be the length of the meal break; the length of a meal break is B ≤ 30 minutes.

The back seat is to carry a maximum of 3 passengers per trip. Letting the number of passengers be P, the number of passengers in the backseat is P ≤ 3. Additionally, all the passengers must be heading to the same destination. Thus, the number of stops per taxi ride=1.

This is done to prevent passengers from taking advantage of the route being used by the taxi. This way, the taxi company is able maximize profits since everyone takes their own taxi to their specific destination.

Question 2

ACME Unlimited sells custom machine tools. The sales department must send each order to Engineering for approval. There are no exceptions to this rule, which creates delays.

Answer

For ACME its sales must be confirmed by engineering department. The constraint present is number of sales = number of orders sent to Engineering for approval. This is done to ensure that all sales of the custom machine tools are high quality and are functional as they have been checked by Engineering for efficiency.

Question 3

An online university offers six-week terms, with three two-week modules per term. Each module is supposedly limited to one topic, or set of related topics. Faculty members find it difficult to create interesting courses, and students express frustration with their learning experiences.

Answer

An online university offers a 6-week term. Therefore, with the length of the term being T, then the constraint for the length of the term is T = 6 weeks. There are 3 modules of 2 weeks each. The number of modules per terms (M) can be expressed as M = 3 modules, and the length of each module (TM), can be expressed as TM = 2 weeks.

Each module is limited to one topic; or a set of related topics. Therefore, with T representing the number of topics per module, the constraint for the number of topics is T ≥ 1. This highly rigid schedule makes the learning experience frustrating, due to the constraints present.

Part II: Describing Constraints

Question 1

Patty makes pottery in her home studio. She works a total of six hours (360 mintes) per day. It takes her 10 minutes to make a cup, 15 minutes to make a bowl, and 30 minutes to make a vase. Write a constraint governing the joint production of cups, bowls, and vases in a day.

Variables:

C = number of cups made

B = number of bowls

V = number of vases

Answer

C * (10 minutes) + B * (15 minutes) + V * (30 minutes) = Joint production * (360 minutes)

Question 3

Patty earns the following profit for each product:

Vase: $1.35

Cup: $0.75

Bowl: $2.40

Using the variable labels given in question 1, Part II, write the profit equation for one of Patty’s days.

Answer

Profit equation for 1 day will be:

Profit for 1 day (360 minutes) = C * $0.75 *(10 minutes) + B * $2.40* (15 minutes) + V * $1.35* (30 minutes)

Profit for 1 day (360 minutes) = C * 7.5 + B * 36 + V * 40.5

Question 2

An aid agency is buying generators for storm survivors. The generators will be shipped on a convoy of flatbed trucks having a total usable cargo area of 1,350 square feet. Generator A has a footprint (floor space required) of 8 square feet, and Generator B has a footprint of 12 square feet. Write a constraint governing the number of generators of each type that can be shipped.

Variables:

GenA = number of type A generators shipped

GenB = number of type B generators shipped

Answer:

GenA * 8 ft2 + GenB * 12 ft2

≤ 1350 ft2

Part III: Solving an Allocation Problem

A refinery produces gasoline and fuel oil under the following constraints.

Let:

Gas = number of gallons of gasoline produced per day

Fuel = number of gallons of fuel oil produced per day

Demand constraints

Minimum daily demand for fuel oil = 3 million gallons ( fuel ≥ 3 )

Maximum daily demand for gasoline = 6.4 million gallons ( gas ≤ 6.4 )

Production constraints

Refining one gallon of fuel oil produces at least 0.5 gallons of gasoline

(Fuel ≤ 0.5 * gas, gas ≥ 2 fuel)

Wholesale prices (earned by the refinery)

Gas: $1.90 per gallon

Fuel Oil: $1.5 per gallon

Requirement: Maximize the refinery’s daily profit by determining the optimum mix of fuel, oil, and gasoline that should be produced. The correct answer consists of a number for fuel oil, and a number for gasoline that maximizes the following profit equation:

P = (1.90)*gas + (1.5)*fuel (Answer will be in millions of dollars).

Run at least 10 trials.

Answer