# answer the questions

**Problems**

**•1-1**

**•1-5 (hint: use words for this, not math)**

**•1-7 (hint: invert refers to shape not necessary number placement)**

**•1-8**

1-1. In the tickets example, identify a fourth feasible alternative.

1-5. In a baseball game, Jim is the pitcher and Joe is the batter. Suppose that Jim can throw either a fast or a curve ball at random. If Joe correctly predicts a curve ball, he can maintain a .500 batting average, else if Jim throws a curve ball and Joe prepares for a fastball, his batting average is kept down to .200. On the other hand, if Joe correctly predicts a fastball, he gets a .300 batting average, else his batting average is only .100.

Define the alternatives for this situation.

Define the objective function for the problem, and discuss how it differs from the familiar optimization (maximization or minimization) of a criterion.

1-7. A (two-dimensional) pyramid is constructed in four layers: The bottom layer consists of (equally spaced) dots 1, 2, 3, and 4; the next layer includes dots 5, 6, and 7; the following layer has dots 8 and 9; and the top layer has dot 10. You want to invert the pyramid (i.e., bottom layer has one dot and top layer has four) by moving the dots around.

Identify two feasible solutions.

Determine the smallest number of moves needed to invert the pyramid.

1-8. You have four chains, each consisting of three solid links. You need to make a bracelet by connecting all four chains. It costs 2 cents to break a link and 3 cents to resolder it.

Identify two feasible solutions and evaluate them.

Determine the cheapest cost for making the bracelet.

1.2 Operations Research Models

Consider the following tickets purchasing problem D. A businessperson has a 5-week commitment

traveling between Fayetteville (FYV) and Denver (DEN). Weekly departure from Fayetteville occurs on

Mondays for return on Wednesdays. A regular roundtrip ticket costs $400, but a 20% discount is granted

if the roundtrip dates span a weekend. A one-way ticket in either direction costs 75% of the regular price.

How should the tickets be bought for the 5-week period?

We can look at the situation as a decision-making problem whose solution requires answering three

questions:

1. What are the decision alternatives D?

2. Under what restrictions D is the decision made?

3. What is an appropriate objective criterion D for evaluating the alternatives?

Three plausible alternatives come to mind:

1. Buy five regular FYV-DEN-FYV for departure on Monday and return on Wednesday of the same

week.

2. Buy one FYV-DEN, four DEN-FYV-DEN that span weekends, and one DEN-FYV.

3. Buy one FYV-DEN-FYV to cover Monday of the first week and Wednesday of the last week and

four DEN-FYV-DEN to cover the remaining legs. All tickets in this alternative span at least one

weekend.

The restriction on these options is that the businessperson should be able to leave FYV on Monday and

return on Wednesday of the same week.

An obvious objective criterion for evaluating the proposed alternatives is the price of the tickets. The

alternative that yields the smallest cost is the best. Specifically, we have:

=

Alternative 1 cost

Alternative 2 cost

Alternative 3 cost

=

5 x $400 = $2000

.75 $400 + 4 x 0.8 $400) +.75 * $ 400 = $1880

5 ~ (.8 x $400) = $1600

–

Alternative 3 is the cheapest.

Though the preceding example illustrates the three main components of an OR model – alternatives,

objective criterion, and constraints-situations differ in the details of how each component is developed,

and how the resulting model is solved. To illustrate this point, consider the following garden problem :

A home owner is in the process of starting a backyard vegetable garden. The garden must take on a

rectangular shape to facilitate row irrigation. To keep critters out, the garden must be fenced. The owner

has enough material to build a fence of length L = 100 ft. The goal is to fence the largest possible

rectangular area.

In contrast with the tickets example, where the number of alternatives is finite, the number of alternatives

in the present example is infinite; that is, the width and height of the rectangle can each assume

(theoretically) infinity of values between 0 and L. In this case, the width and the height are continuous

variables D.

=

Because the variables of the problem are continuous, it is impossible to find the solution by exhaustive

enumeration. However, we can sense the trend toward the best value of the garden area by fielding

increasing values of width (and hence decreasing values of height). For example, for L 100 ft , the

combinations (width, height) = (10, 40), (20, 30), (25, 25), (30, 20), and (40, 10) respectively yield (area) =

(400, 600, 625, 600, and 400), which demonstrates, but not proves, that the largest area occurs when

width = height = L/4 = 25 ft. Clearly, this is no way to compute the optimum, particularly for

situations with several decision variables. For this reason, it is important to express the problem

mathematically in terms of its unknowns, in which case the best solution is found by applying appropriate

solution methods.

To demonstrate how the garden problem is expressed mathematically in terms of its two unknowns, width

and height, define

W =

width of the rectangle in feet

height of the rectangle in feet

h

Based on these definitions, the restrictions of the situation can be expressed verbally as

1. Width of rectangle + Height of rectangle = Half the length of the garden fence

2. Width and height cannot be negative

These restrictions are translated algebraically as

1.2(w+h) = L

2. W > 0, h > 0

The only remaining component now is the objective of the problem; namely, maximization of the area of

the rectangle. Let z be the area of the rectangle, then the complete model becomes

Maximize z= wh

subject to

2(w+h) = L

w, h > 0

Actually, this model can be simplified further by eliminating one of the variables in the objective function

using the constraint equation; that is,

L

W =

2

-h

The result is

Lh

z = wh

et = -1) –

h2

2

The maximization of z is achieved by using differential calculus (Chapter 20 C), which yields the best

L

L

solution as h

25 ft. Back substitution in the constraint equation then yields w = – 25 ft.

4

4

Thus the solution calls for constructing a square-shaped garden.

=

–

Based on the preceding two examples, the general OR model can be organized in the following general

format:

Maximize or minimize Objective Function

subject to

Constraints

A solution is feasible if it satisfies all the constraints. It is optimal if, in addition to being feasible, it

yields the best (maximum or minimum) value of the objective function. In the ticket purchasing problem,

the problem considers three feasible alternatives, with the third alternative being optimal. In the garden

L

problem, a feasible alternative must satisfy the condition w+h- with w and h > 0, that is,

2

nonnegative variables D. This definition leads to an infinite number of feasible solutions and, unlike the

ticket purchasing problem, which uses simple price comparisons, the optimum solution is determined

using differential calculus.

9

Though OR models are designed to optimize a specific objective criterion subject to a set of constraints,

the quality of the resulting solution depends on the degree of completeness of the model in representing

the real system. Take, for example, the ticket purchasing model. If all the dominant alternatives for

purchasing the tickets are not identified, then the resulting solution is optimum only relative to the

alternatives represented in the model. To be specific, if for some reason alternative 3 is left out of the

model, the resulting “optimum” solution would call for purchasing the tickets for $1880, which is a

suboptimal solution. The conclusion is that “the” optimum solution of a model is best only for that

model. If the model happens to represent the real system reasonably well, then its solution is optimum

also for the real situation.