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Math - College, section 1, Fall 2019
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Anderson et al - Quantitative Methods 13/e (Homework)

James Finch

Math - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 27 / 68

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving... ()

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Question

Points

1	2	3	4	5	6	7	8	9
1/2	4/5	14/16	1/5	0/3	1/1	0/2	0/10	6/24

Total

27/68 (39.7%)

Instructions

Provide a conceptual understanding of the critical role of quantitative methods in decision-making with Quantitative Methods for Business, 13th edition. Written for the non-mathematician, a unique applications orientation introduces quantitative methods, how they work, and how decision makers can apply and interpret data. A strong managerial emphasis highlights real examples while a problem-scenario approach helps readers apply mathematical concepts. This edition explains how to construct a spreadsheet simulation model using only native Excel functions, as well as how to use Excel add-ins for more sophisticated simulation analyses.

Question 1 includes an interactive applet to explore the union of two events.

Question 2 asks students to derive a formula and then use that formula to calculate probabilities.

Question 3 is a randomized multi-mode question that allows students to determine numerical values and are then guided through the interpretation of those values.

Question 4 contains a Master It.

Question 5 is a randomized multipart question that utilizes the graphing tool, which allows the student to actively sketch a line and corresponding region.

Question 6 is a randomized question that demonstrates how images can be randomized in addition to the standard practice of randomized numbers.

Question 7 lets a student enter coordinate points in a natural (x, y) form.

Question 8 accepts alternative optima as an ordered list that satisfies constraints developed by the student.

Question 9 contains an image with randomized values for a seamless experience for the student. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

Assignment Submission

For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer.

Assignment Scoring

Your last submission is used for your score.

1. 1/2 points | Previous Answers ASWQuant13 2.AQ.501a. My Notes

Use the applet "Probability of the Union of Two Sets" to answer the following questions.

NOTE: If you cannot obtain the exact values listed in the question when clicking and dragging the diagram, select the next closest value and use approximation to find the correct answer.

(a)

Click on the "A" in the left (orange) circle and drag until its probability is

P(A) = 0.164.

Click on the "B" in the right (blue) circle and drag until its probability is

P(B) = 0.07.

Make sure the circles do not overlap.

What is the value of

P(A ∪ B)?

0.125 0.095 correct

0.234 0.199

(b)

Leaving

P(A) = 0.164

and

P(B) = 0.07,

move B until exactly half of its area intersects with A. Now what is the value of

P(A ∪ B)?

0.125 0.095 0.234 correct

0.199

Solution or Explanation

(a)

Click on the "A" in the left (orange) circle and drag until its probability is

P(A) = 0.164.

Click on the "B" in the right (blue) circle and drag until its probability is

P(B) = 0.07.

Make sure the circles do not overlap. The value of

P(A ∪ B)

will be 0.234. Notice that

P(A) + P(B) = 0.164 + 0.07 = 0.234,

so the probability of A OR B is equal to the sum of

P(A)

and

P(B)

when A and B do not intersect (overlap), i.e.

P(A ∩ B) = 0.

(b)

Leaving

P(A) = 0.164

and

P(B) = 0.07,

move B until exactly half of its area intersects with A. Now the value of

P(A ∪ B)

is 0.199. Notice the half of

P(B)

0.07

= 0.035,

and

P(A) + P(B) −

P(B) = 0.164 + 0.07 − 0.035 = 0.199,

so the probability of A OR B is

P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

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2. 4/5 points | Previous Answers ASWQuant13 3.E.029. My Notes

Consider the following exponential probability density function.

f(x) =

e^−x/5 for x ≥ 0

(a)

Write the formula for

P(x ≤ x₀).

1−e−x05

$webMathematica generated answer key$

(b)

Find

P(x ≤ 2).

(Round your answer to four decimal places.)

0.3297

(c)

Find

P(x ≥ 5).

(Round your answer to four decimal places.)

0.3679

(d)

Find

P(x ≤ 6).

(Round your answer to four decimal places.)

0.6988

(e)

Find

P(2 ≤ x ≤ 6).

(Round your answer to four decimal places.)

0.3691

Solution or Explanation

Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.

(a)

P(x ≤ x₀) = 1 − e^−x₀/5

(b)

P(x ≤ 2) = 1 − e^−2/5 = 1 − 0.6703 = 0.3297

(c)

P(x ≥ 5) = 1 − P(x ≤ 5) = 1 − (1 − e^−5/5) = e⁻¹ = 0.3679

(d)

P(x ≤ 6) = 1 − e^−6/5 = 1 − 0.3012 = 0.6988

(e)

P(2 ≤ x ≤ 6) = P(x ≤ 6) − P(x ≤ 2) = 0.6988 − 0.3297 = 0.3691

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3. 14/16 points | Previous Answers ASWQuant13 5.E.015. My Notes

In a gambling game, Player A and Player B both have a $5 and a $10 bill. Each player selects one of the bills without the other player knowing the bill selected. Simultaneously they both reveal the bills selected. If the bills do not match, Player A wins Player B's bill. If the bills match, Player B wins Player A's bill.

(a)

Develop the game theory table for this game. The values should be expressed as the gains (or losses) for Player A.

		Player B
		$5	$10
Player A	$5	-5	10
Player A	$10	5	-10

(b)

Is there a pure strategy? Why or why not?

. Since the maximum of the row minimums is

-5

and the minimum of the column maximums is

there is no pure strategy

(c)

Determine the optimal strategies and the value of this game.

probability Player A selects $5 =

1/2

probability Player A selects $10 =

1/2

probability Player B selects $5 =

2/3

probability Player B selects $10 =

1/3

Does the game favor one player over the other?

Yes

(d)

Suppose Player B decides to deviate from the optimal strategy and begins playing each bill 50% of the time. What should Player A do to improve Player A's winnings?

If Player B begins playing each bill 50% of the time, Player A should instead select $5 with probability

and select $10 with probability

Comment on why it is important to follow an optimal game theory strategy.

Following the optimal strategy

prevents

other players from taking advantage of the strategy you're playing, since they cannot improve their expected payout by not playing the optimal strategy.

Solution or Explanation

(a)

		Player B
		$5	$10
Player A	$5	−5	10
Player A	$10	5	−10

(b)

		Player B
		$5	$10	Minimum
Player A	$5	−5	10	−5
	$10	5	−10	−10
	Maximum	5	10

Since the maximum of the row minimums is −5 and the minimum of the column maximums is 5, this game has no pure strategy.

(c)

For Player A, let p = probability of $5 and

(1 − p) = probability

of $10

b₁ = $5, EV = −5p + 5(1 − p)

b₂ = $10, EV = 10p − 10(1 − p)

−5p + 5(1 − p)

10p − 10(1 − p)

−5p + 5 − 5p

10p − 10 + 10p

30p

(1 − p) = 1 −

For Player B, let q = probability of $5 and

(1 − q) = probability

of $10

a₁ = $5, EV = −5q + 10(1 − q)

a₂ = $10, EV = 5q − 10(1 − q)

−5q + 10(1 − q)

5q − 10(1 − q)

−5q + 10 − 10q

5q − 10 + 10q

30q

(1 − q) = 1 −

or q =

(1 − q) =

Value of game using Player A

EV = −5p + 5(1 − p)

−5

+ 5

This is a fair game. Neither player is favored.

(d)

If Player A realizes Player B is using a 50/50 strategy, we can use an expected value with these probabilities to show:

EV(a₁ = $5)

−5

+ 10

= 2.50

EV(a₂ = $10)

− 10

= −2.50

Player A should see that the expected value of a₁ is now larger than the expected value of a₂.

If Player A believes Player B will continue with a 50/50 strategy, then Player A should always play strategy a₁: reveal $5. But, if Player A does this, Player B will catch on and begin revealing a $5 bill all the time. The only way for a player to protect against the opponent taking advantage is to play the optimal strategy all the time.

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4. 1/5 points | Previous Answers ASWQuant13 6.E.004.MI. My Notes

Consider the following time series data.

Month	1	2	3	4	5	6	7
Value	24	13	20	12	19	23	15

(a)

Compute MSE using the most recent value as the forecast for the next period.

MSE =

60.5

What is the forecast for month 8?

(b)

Compute MSE using the average of all the data available as the forecast for the next period. (Round your answer to two decimal places.)

MSE =

36.12

What is the forecast for month 8?

(c)

Which method appears to provide the better forecast?

The most recent value provides a better forecast, because its MSE is smaller than the MSE using the average of all the previous values. The most recent value provides a better forecast, because its forecast for month 8 is smaller than the month 8 forecast using the average of all the previous values. correct

The average of all the previous values provides a better forecast, because its MSE is smaller than the MSE using the most recent value. The average of all the previous values provides a better forecast, because its forecast for month 8 is larger than the month 8 forecast using the most recent value.

Solution or Explanation

Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.

(a)

Month	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	24
2	13	24	−11	121
3	20	13	7	49
4	12	20	−8	64
5	19	12	7	49
6	23	19	4	16
7	15	23	−8	64
Total				363

MSE =

363

= 60.5

forecast for month 8 = 15

(b)

Month	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	24
2	13	24.00	−11.00	121.00
3	20	18.50	1.50	2.25
4	12	19.00	−7.00	49.00
5	19	17.25	1.75	3.06
6	23	17.60	5.40	29.16
7	15	18.50	−3.50	12.25
Total				216.72

MSE =

216.72

= 36.12

forecast for month 8 =

(24 + 13 + 20 + 12 + 19 + 23 + 15)

= 18

(c)

The average of all the previous values is better because its MSE is smaller.

Need Help?

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5. 0/3 points | Previous Answers ASWQuant13 7.E.002. My Notes

6. 1/1 points | Previous Answers ASWQuant13 7.E.007. My Notes

Identify the feasible region for the following set of constraints.

0.5A + 0.25B	≥	35
1A + 5B	≥	250
0.25A + 0.5B	≤	55
A, B ≥ 0

The A B-coordinate plane is given. There are 3 lines and a shaded region on the graph.
The first line enters the window at B = 140 on the positive B-axis, goes down and right, passes through the point (20, 100) crossing the third line, passes through the point (50, 40) crossing the second line, and exits the window at A = 70 on the positive A-axis.
The second line enters the window at B = 50 on the positive B-axis, goes down and right, passes through the point (50, 40) crossing the first line, passes through the point (200, 10) crossing the third line, crosses the A-axis at A = 250, and exits the window in the fourth quadrant.
The third line enters the window at B = 110 on the positive B-axis, goes down and right, passes through the point (20, 100) crossing the first line, passes through the point (200, 10) crossing the second line, and exits the window at A = 220 on the positive A-axis.
The region is below the first line, above the second line, and below the third line.

Solution or Explanation

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7. 0/2 points | Previous Answers ASWQuant13 7.E.010. My Notes

For the linear program

Max

2A + 3B

s.t.

1A	+	2B ≤ 4
5A	+	3B ≤ 15
	A,	B ≥ 0

find the optimal solution using the graphical solution procedure. What is the value of the objective function at the optimal solution?

$webMathematica generated answer key$ at (A, B) =

$webMathematica generated answer key$

Solution or Explanation

The A B-coordinate plane is given. There are 2 lines and a shaded region on the graph.
The first line labeled A + 2B = 4 enters the window at B = 2 on the positive B-axis, goes down and right, passes through the approximate point (2.57, 0.71) crossing the second line labeled 5A + 3B = 15, and exits the window at A = 4 on the positive A-axis.
The second line labeled 5A + 3B = 15 enters the window at B = 5 on the positive B-axis, goes down and right, passes through the approximate point (2.57, 0.71) crossing the first line labeled A + 2B = 4, and exits the window at A = 3 on the positive A-axis.
The region is below the first line labeled A + 2B = 4 and below the second line labeled 5A + 3B = 15.
The intersection of the lines is labeled Optimal Solution.
The value of the objective function is labeled 2(18⁄7) + 3(5⁄7) = 51⁄7.

(1)

(2)

(1) ✕ 5

10B

(3)

(2) − (3)

−

−5

From (1),

A = 4 − 2

= 4 −

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8. 0/10 points | Previous Answers ASWQuant13 9.E.009. My Notes

A linear programming computer package is needed.

Epsilon Airlines services predominately the eastern and southeastern United States. A vast majority of Epsilon's customers make reservations through Epsilon's website, but a small percentage of customers make reservations via phone. Epsilon employs call-center personnel to handle these reservations along with any problems with the website reservation system and for the rebooking of flights for customers if their plans change or their travel is disrupted. Staffing the call center appropriately is a challenge for Epsilon's management team. Having too many employees on hand is a waste of money, but having too few results in very poor customer service and the potential loss of customers.

Epsilon analysts have estimated the minimum number of call-center employees needed by day of week for the upcoming vacation season (June, July, and the first two weeks of August). These estimates are given in the following table.

Day	Minimum Number of Employees Needed
Monday	85
Tuesday	50
Wednesday	45
Thursday	75
Friday	100
Saturday	85
Sunday	60

The call-center employees work five consecutive days and then have two consecutive days off. An employee may start work any day of the week. Each call-center employee receives the same salary. Assume that the schedule cycles and ignore start-up and stopping of the schedule. Develop a model that will minimize the total number of call-center employees needed to meet the minimum requirements. (Let

X_i

= the number of call-center employees who start work on day i where

i = 1 = Monday,

i = 2 = Tuesday,

etc).

Min

$webMathematica generated answer key$

s.t. Monday

$webMathematica generated answer key$ Tuesday

$webMathematica generated answer key$ Wednesday

$webMathematica generated answer key$ Thursday

$webMathematica generated answer key$ Friday

$webMathematica generated answer key$ Saturday

$webMathematica generated answer key$ Sunday

$webMathematica generated answer key$

X₁, X₂, X₃, X₄, X₅, X₆, X₇ ≥ 0

Find the optimal solution.

(X₁, X₂, X₃, X₄, X₅, X₆, X₇) =

$webMathematica generated answer key$

Give the number of call-center employees that exceed the minimum required.

(M, Tu, W, Th, F, Sa, Su) =

$webMathematica generated answer key$

Solution or Explanation

Let X_i = the number of call center employees who start work on day i.

(i = 1 = Monday, i = 2 = Tuesday,

)

Min

X₁

X₂

X₃

X₄

X₅

X₆

X₇

s.t.

X₁

X₄

X₅

X₆

X₇

≥

X₁

X₂

X₅

X₆

X₇

≥

X₁

X₂

X₃

X₆

X₇

≥

X₁

X₂

X₃

X₄

X₇

≥

X₁

X₂

X₃

X₄

X₅

≥

100

X₂

X₃

X₄

X₅

X₆

≥

X₃

X₄

X₅

X₆

X₇

≥

X₁, X₂, X₃, X₄, X₅, X₆, X₇ ≥ 0

Solution: X₁ = 20, X₂ = 20, X₃ = 0, X₄ = 55, X₅ = 5, X₆ = 5, X₇ = 0.

Total Number of Employees = 105.

Excess employees: Thursday = 20, Sunday = 5, all others = 0.

Note: There are alternative optima to this problem (Number of employees may differ from above, but will have objective function value = 105).

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9. 6/24 points | Previous Answers ASWQuant13 10.E.002. My Notes

Question Part

Points

Submissions Used

0/1

1/1

0/1

1/1

0/1

1/100

Total

6/24

Consider the following network representation of a transportation problem.

A network diagram between five locations is shown. The left-hand side of the graph is labeled "Supplies" and lists Jefferson City and Omaha. The right-hand side of the graph is labeled "Demands" and lists Des Moines, Kansas City, and St. Louis. Lines are shown between various locations. The following list contains the numbers placed on the graph.

Jefferson City: 40

Omaha: 20

Des Moines: 20

Kansas City: 10

St. Louis: 30

Jefferson City–Des Moines: 14

Jefferson City–Kansas City: 9

Jefferson City–St. Louis: 8

Omaha–Des Moines: 8

Omaha–Kansas City: 11

Omaha–St. Louis: 25

The supplies, demands, and transportation costs per unit are shown on the network.

(a)

Develop a linear programming model for this problem; be sure to define the variables in your model.

Let

x₁₁ = amount shipped from Jefferson City to Des Moines
x₁₂ = amount shipped from Jefferson City to Kansas City
x₁₃ = amount shipped from Jefferson City to St. Louis
x₂₁ = amount shipped from Omaha to Des Moines
x₂₂ = amount shipped from Omaha to Kansas City
x₂₃ = amount shipped from Omaha to St. Louis

Min

x₁₁

x₁₂

x₁₃

x₂₁

x₂₂

x₂₃

s.t.

x₁₁

x₁₂

x₁₃

≤

x₂₁

x₂₂

x₂₃

≤

x₁₁

x₂₁

x₁₂

x₂₂

x₁₃

x₂₃

x₁₁, x₁₂, x₁₃, x₂₁, x₂₂, x₂₃ ≥ 0

(b)

Solve the linear program to determine the optimal solution.

	Amount	Cost
Jefferson City–Des Moines	0	0
Jefferson City–Kansas City	10	90
Jefferson City–St. Louis	30	240
Omaha–Des Moines	20	160
Omaha–Kansas City	0	0
Omaha–St. Louis	0	0
Total		490

Solution or Explanation

(a)

Let

x₁₁ = amount shipped from Jefferson City to Des Moines
x₁₂ = amount shipped from Jefferson City to Kansas City
x₁₃ = amount shipped from Jefferson City to St. Louis
x₂₁ = amount shipped from Omaha to Des Moines
x₂₂ = amount shipped from Omaha to Kansas City
x₂₃ = amount shipped from Omaha to St. Louis

Min

14x₁₁

9x₁₂

8x₁₃

8x₂₁

11x₂₂

25x₂₃

s.t.

x₁₁

x₁₂

x₁₃

≤

x₂₁

x₂₂

x₂₃

≤

x₁₁

x₂₁

x₁₂

x₂₂

x₁₃

x₂₃

x₁₁, x₁₂, x₁₃, x₂₁, x₂₂, x₂₃ ≥ 0

(b)

Optimal Solution

	Amount	Cost
Jefferson City–Des Moines	0	0
Jefferson City–Kansas City	10	90
Jefferson City–St. Louis	30	240
Omaha–Des Moines	20	160
Omaha–Kansas City	0	0
Omaha–St. Louis	0	0
Total		490

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Anderson et al - Quantitative Methods 13/e (Homework)

Current Score : 27 / 68

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving... ()

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