Answers

Answers ANSWER KEYS: EXERCISES: CHAPTER 2 4) a) b) c) d) e) f) g) h) i) j) k) l) m) n) o)

Continuous Ordinal Continuous Discrete Continuous Nominal Ordinal Ordinal Continuous Nominal Binary Ordinal Discrete Ordinal Binary

ANSWER KEYS: EXERCISES: CHAPTER 3 6) Boxplot 7) Bar chart—qualitative and quantitative variables Scatter plot—quantitative variables 8) Bar (horizontal and vertical), pie, and Pareto charts 9) Cars sold

Fi

Fri (%)

Fac

Frac (%)

5 6 7 8 9 10 11

4 5 4 6 4 4 3

13.33 16.67 13.33 20 13.33 13.33 10

4 9 13 19 23 27 30

13.33 30 43.33 63.33 76.67 90 100

Sum

30

100

10) Class

Fi

Fri (%)

Fac

Frac (%)

54.7 ├ 61.7

4

61.7 ├ 68.7

4

13.33

4

8

13.33

8

68.7 ├ 75.7

16

10

33.33

18

36

75.7 ├ 82.7

17

56.67

35

70

82.7 ├ 89.7

6

20

41

82

1065

1066

Answers

89.7 ├ 96.7

7

23.33

48

96

96.7 ├ 103.7

2

6.67

50

100

Sum

50

100

11) Type of defect

Fi

Fri (%)

Fac

Frac (%)

Lack of Alignment Scratches Deformation Discoloration Oxygenation

98 67 45 28 12

39.2 26.8 18 11.2 4.8

98 165 210 238 250

39.2 66 84 95.2 100

Total

250

100

12) a) b) c) d) e) f) g)

X ¼ 9:27, Md ¼ 8.685, Mo ¼ 5.12 (there is more than one mode). Q1 ¼ 6.8425, Q3 ¼ 11.16. Observations 63 (19.32) and 83 (23.37) may be outliers. P10 ¼ 5.168, P90 ¼ 14.088. D3 ¼ 7.122, D6 ¼ 9.502. A ¼ 19.44, Dm ¼ 2.698, S2 ¼ 11.958, S ¼ 3.458, SX ¼ 0:3458, CV ¼ 37.3%. Positive asymmetrical. k ¼ 0.242 (leptokurtic).

13) Mean Median Mode Variance Standard Deviation Standard error Q1 Q3 g1 g2

Service 1

Service 2

Service 3

7.56 7.5 2a 13.435 3.665 0.518 4.75 10.25 0.083 1.092

9.66 9 4 20.760 4.556 0.644 6 14 0.183 1.157

11.68 12 5a 21.365 4.622 0.654 8 15 0.191 1.011

a

More than one mode.

c) Services 1, 2, and 3: there are no outliers. d) Services 1, 2, and 3: positive asymmetrical distribution, platykurtic curve. 14) a) b) c) d) e)

X ¼ 39:129, Md ¼ 40, Mo ¼ 40. Q1 ¼ 35, Q3 ¼ 42, D4 ¼ 38, P61 ¼ 41.4, and P84 ¼ 43. There are no outliers. A ¼ 20, S2 ¼ 20.560, S ¼ 4.534, SX ¼ 0:414. g1 ¼ 0.101, g2 ¼ 0.279. Negative asymmetrical distribution and platykurtic curve.

15) a) b) c) d) e)

X ¼ 133:560, Md ¼ 136.098, Mo ¼ 137.826. Q1 ¼ 106.463, Q3 ¼ 163.611, D2 ¼ 97.317, P13 ¼ 82.241, and P95 ¼ 198.636. There are no outliers. A ¼ 180, S2 ¼ 1595.508, S ¼ 39.944, SX ¼ 2:526. AS1 ¼  0.107, k ¼ 0.253. Negative asymmetrical distribution and leptokurtic curve.

Answers

1067

16) a) XA ¼ 28:167, MdA ¼ 28, MoA ¼ 24. XB ¼ 29, MdB ¼ 28, MoB ¼ 28. b) AA ¼ 20, S2A ¼ 27.275, SA ¼ 5.223, SX ¼ 1:066. A AB ¼ 18, S2B ¼ 16.757, SB ¼ 4.118, SX ¼ 0:841. A c) Stock A—the 18th observation (42) may be an outlier. Stock B—the 14th observation (16) may be an outlier. d) Stock A—positive asymmetrical distribution and elongated curve (leptokurtic). Stock B—negative asymmetrical distribution and elongated curve (leptokurtic). 17) a) X ¼ 52, S ¼ 60.69. b) Possible outliers: 8th observation (200) and 13th observation (180). c) X ¼ 30:77, S ¼ 24.863, with no outliers.

ANSWER KEYS: EXERCISES: CHAPTER 4 6) a) Age group * Default Crosstabulation Default

Age group

<¼20

21–30

31–40

41–50

51–60

>60

Count Expected Count % within Age group % within Default % of Total Count Expected Count % within Age group % within Default % of Total Count Expected Count % within Age group % within Default % of Total Count Expected Count % within Age group % within Default % of Total Count Expected Count % within Age group % within Default % of Total Count Expected Count % within Age group % within Default % of Total

Do not have debts

Little indebted

Relatively indebted

Heavily indebted

Total

6 1.1 75.0%

2 1.7 25.0%

0 2.3 .0%

0 3.0 .0%

8 8.0 100.0%

22.2% 3.0% 0 3.8 .0%

4.8% 1.0% 6 5.9 21.4%

.0% .0% 13 8.0 46.4%

.0% .0% 9 10.4 32.1%

4.0% 4.0% 28 28.0 100.0%

.0% .0% 0 7.3 .0%

14.3% 3.0% 0 11.3 .0%

22.8% 6.5% 5 15.4 9.3%

12.2% 4.5% 49 20.0 90.7%

14.0% 14.0% 54 54.0 100.0%

.0% .0% 0 5.4 .0%

.0% .0% 0 8.4 .0%

8.8% 2.5% 24 11.4 60.0%

66.2% 24.5% 16 14.8 40.0%

27.0% 27.0% 40 40.0 100.0%

.0% .0% 5 6.3 10.6%

.0% .0% 27 9.9 57.4%

42.1% 12.0% 15 13.4 31.9%

21.6% 8.0% 0 17.4 .0%

20.0% 20.0% 47 47.0 100.0%

18.5% 2.5% 16 3.1 69.6%

64.3% 13.5% 7 4.8 30.4%

26.3% 7.5% 0 6.6 .0%

.0% .0% 0 8.5 .0%

23.5% 23.5% 23 23.0 100.0%

59.3% 8.0%

16.7% 3.5%

.0% .0%

.0% .0%

11.5% 11.5%

1068

Answers

Total

Count Expected Count % within Age group % within Default % of Total

27 27.0 13.5%

42 42.0 21.0%

57 57.0 28.5%

74 74.0 37.0%

200 200.0 100.0%

100.0% 13.5%

100.0% 21.0%

100.0% 28.5%

100.0% 37.0%

100.0% 100.0%

b) 27% c) 37% d) 3% e) 30.4% f) 42.1% g) Yes. h) 247.642 with sig. ¼ 0.000 (there is association between the variables). i) Coefficient

Value

Sig.

Phi Cramer’s V Contingency

1.113 0.642 0.744

0.000 0.000 0.000

7) a) Company * Motivation Crosstabulation Motivation

Company Petrobras

Bradesco

Fiat

Vivo

Pa˜o de Ac¸u´car

Total

Count Expected Count % within Company % within Motivation % of Total Count Expected Count % within Company % within Motivation % of Total Count Expected Count % within Company % within Motivation % of Total Count Expected Count % within Company % within Motivation % of Total Count Expected Count % within Company % within Motivation % of Total Count Expected Count % within Company % within Motivation % of Total

Very demotivated

Little Demotivated motivated

Very Motivated motivated

Total

36 9.2 72.0% 78.3% 14.4% 0 9.2 .0% .0% .0% 0 9.2 .0% .0% .0% 10 9.2 20.0% 21.7% 4.0% 0 9.2 .0% .0% .0% 46 46.0 18.4% 100.0% 18.4%

8 9.8 16.0% 16.3% 3.2% 0 9.8 .0% .0% .0% 8 9.8 16.0% 16.3% 3.2% 33 9.8 66.0% 67.3% 13.2% 0 9.8 .0% .0% .0% 49 49.0 19.6% 100.0% 19.6%

0 11.2 .0% .0% .0% 16 11.2 32.0% 28.6% 6.4% 9 11.2 18.0% 16.1% 3.6% 0 11.2 .0% .0% .0% 31 11.2 62.0% 55.4% 12.4% 56 56.0 22.4% 100.0% 22.4%

50 50.0 100.0% 20.0% 20.0% 50 50.0 100.0% 20.0% 20.0% 50 50.0 100.0% 20.0% 20.0% 50 50.0 100.0% 20.0% 20.0% 50 50.0 100.0% 20.0% 20.0% 250 250.0 100.0% 100.0% 100.0%

6 11.8 12.0% 10.2% 2.4% 3 11.8 6.0% 5.1% 1.2% 32 11.8 64.0% 54.2% 12.8% 7 11.8 14.0% 11.9% 2.8% 11 11.8 22.0% 18.6% 4.4% 59 59.0 23.6% 100.0% 23.6%

0 8.0 .0% .0% .0% 31 8.0 62.0% 77.5% 12.4% 1 8.0 2.0% 2.5% .4% 0 8.0 .0% .0% .0% 8 8.0 16.0% 20.0% 3.2% 40 40.0 16.0% 100.0% 16.0%

Answers

b) 18.4% c) 78.3% d) 0% e) 64% f) 77.5% g) Yes. h) 375.066 with sig. ¼ 0.000. i) Yes. Coefficient

Value

Sig.

Phi Cramer’s V Contingency

1.225 0.612 0.775

0.000 0.000 0.000

8) a) Strong positive correlation. r ¼ 0.794 with sig. ¼ 0.000.

1069

1070

Answers

b) Positive correlation. r ¼ 0.689 with sig 0.000

Answers

c) Strong positive correlation. r ¼ 0.962 with sig ¼ 0.000.

1071

1072

Answers

9) Case 1: Considering all 30 Brazilian supermarket chains.

Answers

1073

From the scatter plots, there seems to be dependence between the variables, which can be confirmed by the following table. Correlations Revenue ($ Millions) Revenue ($ millions)

Number of stores

Number of employees

Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) N

1 30 .944a .000 30 .988a .000 30

Number of Stores a

.944 .000 30 1

30 .965a .000 30

a

Correlation is significant at the 0.01 level (2-tailed).

We can conclude that there is a strong correlation between all the variable combinations. Case 2: Excluding the four largest Brazilian supermarket chains.

Number of Employees .988a .000 30 .965a .000 30 1 30

1074

Answers

There seems to be correlation between the variables in the second case. This hypothesis can be confirmed by the following table. Correlations

Revenue ($ millions)

Number of stores

Number of employees

a

Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) N Pearson Correlation Sig. (2-tailed) N

Correlation is significant at the 0.01 level (2-tailed). Correlation is significant at the 0.05 level (2-tailed).

b

Revenue ($ Millions)

Number of Stores

Number of Employees

1

.009 .964 26 1

.599a .001 26 .442b .024 26 1

26 .009 .964 26 .599a .001 26

26 .442b .024 26

26

Answers

1075

ANSWER KEYS: EXERCISES: CHAPTER 5 1) S ¼ {00, 10, 01} 2) Mutually exclusive events—do not have elements in common (may not occur simultaneously). Independent events—the probability of one of them occurring is not conditional to the probability of the other occurring. 3) a) 1/4 b) 1/13 c) 3/13 d) 10/13 4) 95% 5) a) S ¼ {1, 2, …, 30} b) 1/3 c) 1/5 d) 7/15 e) ½ f) 1/5 g) 2/3 h) 1/10 6)

a)

b) c) d) e) f) g)

9 8 ð1, 1Þ, ð1, 2Þ, ð1, 3Þ, ð1, 4Þ, ð1, 5Þ, ð1, 6Þ > > > > > > > ð2, 1Þ, ð2, 2Þ, ð2, 3Þ, ð2, 4Þ, ð2, 5Þ, ð2, 6Þ > > > > > > > > = < ð3, 1Þ, ð3, 2Þ, ð3, 3Þ, ð3, 4Þ, ð3, 5Þ, ð3, 6Þ > S¼ > > ð4, 1Þ, ð4, 2Þ, ð4, 3Þ, ð4, 4Þ, ð4, 5Þ, ð4, 6Þ > > > > > > > > > > ð 5, 1 Þ, ð 5, 2 Þ, ð 5, 3 Þ, ð 5, 4 Þ, ð 5, 5 Þ, ð 5, 6 Þ > > > > ; : ð6, 1Þ, ð6, 2Þ, ð6, 3Þ, ð6, 4Þ, ð6, 5Þ, ð6, 6Þ 1/4 1/12 1/9 7/36 2/3 1/12

ANSWER KEYS: EXERCISES: CHAPTER 6 1)



       150 150 150 0 150 1 149 2 148 Pð X  2Þ ¼  0:02  0:98 +  0:02  0:98 + ¼ 0:42  0:02  0:98 0 1 2 E(X) ¼ 150  0.02 ¼ 3 Var(X) ¼ 150  0.02  0.98 ¼ 2.94 2)    10 Pð X ¼ 1Þ ¼  0:12  0:889 ¼ 0:38 1 3) P(X ¼ 5) ¼ 0.125  0.8754 ¼ 0.073 E(X) ¼ 8 Var(X) ¼ 56 4)   32 PðX ¼ 33Þ ¼  0:9530  0:053 ¼ 1:33% 29

1076

Answers

E(X) ¼ 31.6 ffi 32 5) P(X ¼ 4) ¼ 16.8% 6) a) P(X  12) ¼ P(Z  0.67) ¼ 1  P(Z > 0.67) ¼ 0.75 b) P(X < 5) ¼ P(Z <  0.5) ¼ P(Z > 0.5) ¼ 0.3085 c) P(X > 2) ¼ P(Z >  1) ¼ P(Z < 1) ¼ 1  P(Z > 1) ¼ 0.8413 d) P(6 < X  11) ¼ P(0.33 < Z  0.5) ¼ [1  P(Z > 0.5)]  P(Z > 0.33) ¼ 0.3208 7) zc ¼  0.84 8) a) m ¼ np ¼ 40  0.5 ¼ 20 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s ¼ npð1  pÞ ¼ 40  0:5  0:5 ¼ 3:16 P(X ¼ 22) ffi P(21.5 < X < 22.5) ¼ P(0.474 < Z < 0.791) ¼ 0.103 b) P(X > 25.5) ¼ P(Z > 1.74) ¼ 4.09% 9) a) P(X > 120) ¼ e0.028120 ¼ 0.0347 b) P(X > 60) ¼ e0.02860 ¼ 0.1864 10) 220

a) PðX > 220Þ ¼ e 180 ¼ 0:2946 150

b) PðX  150Þ ¼ 1  e 180 ¼ 0:5654 11)

a) P(X > 0.5) ¼ e1.80.5 ¼ 0.4066 b) P(X  1.5) ¼ 1  e1.81.5 ¼ 0.9328

12)

a) P(X > 2) ¼ e0.332 ¼ 0.5134 b) P(X  2.5) ¼ 1  e0.332.5 ¼ 0.5654

13) 6.304 14) a) P(X > 25) ¼ 0.07 b) P(X  32) ¼ 0.99 c) P(25 < X  32) ¼ P(X > 25)  P(X > 32) ¼ 0.06 d) 28.845 e) 6.908 15) a) 2.086 b) E(T) ¼ 0 c) Var(T) ¼ 1.111 16) a) b) c) d) e)

P(T > 3) ¼ 0.0048 P(T  2) ¼ 1  P(T > 2) ¼ 1  0.0344 ¼ 0.9656 P(1.5 < T  2) ¼ P(T > 1.5)  P(T > 2) ¼ 0.0814  0.3444 ¼ 0.0469 1.345 2.145

17) a) P(X > 3) ¼ 0.05 b) 3.73

Answers

1077

c) 4.77 d) E(X) ¼ 1.14 e) Var(X) ¼ 0.98

ANSWER KEYS: EXERCISES: CHAPTER 7 5) Simple random sampling without replacements. 6) Systematic sampling. 7) Stratified sampling. 8) Stratified sampling. 9) Two-stage cluster sampling. 10) By using Expression (7.8) (SRS to estimate the proportion of a finite population), we have n ¼ 262. 11) By using Expression (7.9) (stratified sampling to estimate the mean of an infinite population), we have n ¼ 1,255. 12) By using Expression (7.20) (one-stage cluster sampling to estimate the proportion of an infinite population), we have m ¼ 35.

ANSWER KEYS: EXERCISES: CHAPTER 8

  ffiffiffiffiffiffi < m < 51 + 1:645 p18 ffiffiffiffiffiffi ¼ 90% 1) P 51  1:645 p18 120 120   ffiffiffiffi < m < 5, 400 + 2:030 p200 ffiffiffiffi ¼ 95% 2) P 5, 400  2:030 p200 36 36 qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi   0:76  0:76 3) P 0:24  1:96 0:24500 < p < 0:24 + 1:96 0:24500 ¼ 95% 60  8

60  8 4) P 83:298 < s2 < 40:482 ¼ 95%

ANSWER KEYS: EXERCISES: CHAPTER 9 7) For the K-S and S-W tests, we have p ¼ 0.200 and 0.151, respectively. Therefore, since P > 0.05, the distribution of data is normal. 8) The data follow a normal distribution (P ¼ 0.200 > 0.05). 9) The variances are homogeneous (P ¼ 0.876 > 0.05—Levene’s test). 10) Since s is unknown, the most suitable test is Student’s t: 6560 pffiffiffiffi ¼ 8:571; tc ¼ 2.030; since Tcal > tc ! we reject H0 (m 6¼ 60). Tcal ¼ 3:5= 36 11) Tcal ¼ 6.921 and P-value ¼ 0.000 < 0.005 ! we reject H0 (m1 6¼ m2). 12) Tcal ¼ 11.953 and P-value ¼ 0.000 < 0.025 ! we reject H0 (mbefore 6¼ mafter), i.e., there was improvement after the treatment. 13) Fcal ¼ 2.476 and P-value ¼ 0.1 > 0.05 ! we do not reject H0 (there is no difference between the population means).

ANSWER KEYS: EXERCISES: CHAPTER 10 4) Sign test. 5) By applying the binomial test for small samples, since P ¼ 0.503 > 0.05, we do not reject H0, concluding that there is no difference in consumers’ preferences. 6) By applying the chi-square test, since w2cal > w2c (6.100 > 5.991) or P < a (0.047 < 0.05), we reject H0, concluding that there are differences in readers’ preferences. 7) By applying the Wilcoxon test, since zcal <  zc (3.135 < 1.645) or P < a (0.0085 < 0.05), we reject H0, concluding that the diet resulted in weight loss. 8) By applying the Mann-Whitney U test (the data do not follow a normal distribution), since zcal >  zc (0.129 > 1.96) or P > a (0.897 > 0.05), we do not reject H0, concluding that the samples come from populations with equal medians. 9) By applying Cochran’s Q test, since Qcal > Qc (8.727 > 7.378) or P < a (0.013 < 0.025), we reject H0, concluding that the proportion of students with high learning levels is not the same in each subject. 10) By applying the Friedman test, since F0 cal > Fc (9.190 > 5.991) or P < a (0.010 < 0.05), we reject H0, concluding that there are differences between the three services.

1078

Answers

ANSWER KEYS: EXERCISES: CHAPTER 11 1) a) Agglomeration Schedule Cluster Combined

Stage Cluster First Appears

Stage

Cluster 1

Cluster 2

Coefficients

Cluster 1

Cluster 2

Next Stage

77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

5 40 25 30 38 1 2 6 4 30 5 29 31 2 1 5 2 1 5 2 1 1 1

13 56 58 55 48 15 14 83 7 42 39 40 38 3 30 25 31 4 6 29 5 2 9

.006 .014 .014 .014 .014 .024 .024 .024 .024 .038 .038 .055 .075 .075 .153 .209 .246 .246 .723 .760 2.764 8.466 173.124

39 56 0 62 75 71 72 74 76 80 77 65 69 83 82 87 90 91 92 93 94 97 98

64 53 26 61 36 55 58 0 68 0 70 78 81 73 86 79 89 85 84 88 95 96 0

87 88 92 86 89 91 90 95 94 91 92 96 93 93 94 95 96 97 97 98 98 99 0

From the agglomeration schedule, it is possible to verify that a big Euclidian distance leap occurs from the 98th stage (when only two clusters remain) to the 99th stage. Analyzing the dendrogram also helps in this interpretation.

b)

In fact, the solution with two clusters is highly advisable at this moment.

1080

Answers

c) Yes. From the agglomeration schedule, it is possible to verify that observation 9 (Antonio) had not clustered in until the moment exactly before the last stage. From the dendrogram, it is also possible to verify that this student differs from the others considerably, which, in this case, results in the generation of only two clusters. d) Agglomeration Schedule Cluster Combined

Stage Cluster First Appears

Stage

Cluster 1

Cluster 2

Coefficients

Cluster 1

Cluster 2

Next Stage

77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98

13 27 1 41 6 30 5 16 1 13 2 14 2 13 1 5 9 1 1 5 1 1

34 29 4 46 82 55 74 57 38 39 15 16 28 30 27 6 13 41 2 14 5 9

.537 .537 .537 .754 1.103 1.103 1.584 1.584 1.584 1.584 2.045 2.149 2.149 3.091 3.091 4.411 4.835 7.134 10.292 12.374 18.848 26.325

67 62 63 0 72 58 68 55 79 77 74 61 87 86 85 83 75 91 94 92 95 97

0 60 69 0 0 53 0 73 66 64 76 84 71 82 78 81 90 80 89 88 96 93

86 91 85 94 92 90 92 88 91 90 89 96 95 93 94 96 98 95 97 97 98 0

Answers

1081

Yes, the new results show that there is one cluster rearrangement in the absence of observation Antonio. e) The existence of an outlier may cause other observations, not so similar to one another, to be allocated in the same cluster because they are extremely different from the first one. Therefore, reapplying the technique, with the exclusion or maintenance of outliers, makes the new clusters better structured, and makes them be generated with higher internal homogeneity.

1082

2) a)

Answers

Answers

1083

b) Agglomeration Schedule Cluster Combined

Stage Cluster First Appears

Stage

Cluster 1

Cluster 2

Coefficients

Cluster 1

Cluster 2

Next Stage

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

8 4 1 4 4 1 1 1 1 11 6 15 7 7 6 6 1

18 16 10 9 8 5 4 3 2 12 11 17 15 14 13 7 6

2.000 2.000 2.000 2.000 2.000 2.000 2.000 2.828 6.633 12.329 14.697 23.409 24.495 32.802 35.665 40.497 78.256

0 0 0 2 4 3 6 7 8 0 0 0 0 13 11 15 9

0 0 0 0 1 0 5 0 0 0 10 0 12 0 0 14 16

5 4 6 5 7 7 8 9 17 11 15 13 14 16 16 17 0

From the agglomeration schedule, it is possible to verify that a big Euclidian distance leap occurs from the 16th stage (when only two clusters remain) to the 17th stage. Analyzing the dendrogram also helps in this interpretation. c) Dendrogram using single linkage

Y

0 Regional 3

8

Regional 3

18

Regional 3

4

Regional 3

16

Regional 3

9

Regional 3

1

Regional 3

10

Regional 3

5

Regional 3

3

Regional 3

2

Regional 1

15

Regional 1

17

Regional 1

7

Regional 1

14

Regional 2

11

Regional 2

12

Regional 2

6

Regional 2

13

5

In fact, there are indications of two clusters of stores.

Rescaled distance cluster combine 10 15

20

25

1084

Answers

d)

Derived stimulus configuration Euclidean distance model 1.5

Store06 Store12

1.0

Store11

Dimension 2

0.5 Store13 Store02 Store03 Store04 Store05 Store01

0.0 Store17 Store15

–0.5

Store07 –1.0 Store14

–1.5 –2

–1

0

1

Dimension 1

The two-dimensional chart generated through the multidimensional scaling allows us to see these two clusters and that one is more homogeneous than the other. e) ANOVA Cluster

Customers’ average evaluation of services rendered (0 to 100) Customers’ average evaluation of the variety of goods (0 to 100) Customers’ average evaluation of the organization (0 to 100)

Error

Mean Square

df

Mean Square

df

F

Sig.

10802.178

1

99.600

16

108.456

.000

12626.178

1

199.100

16

63.416

.000

18547.378

1

314.900

16

58.899

.000

The F tests should be used only for descriptive purposes because the clusters have been chosen to maximize the differences among cases in different clusters. The observed significance levels are not corrected for this and thus cannot be interpreted as tests of the hypothesis that the cluster means are equal.

It is possible to state that both clusters formed present statistically different means for the three variables considered in the study, at a significance level of 0.05 (Prob. F < 0.05). Among the groups, the variable considered most discriminating is the one with the highest F statistic, that is, the variable services rendered (F ¼ 108.456).

1085

Answers

f) Single Linkage* Cluster Number of Case Crosstabulation Count Cluster Number of Case

Single Linkage

1 2

Total

1

2

Total

10 0 10

0 8 8

10 8 18

Yes, there is correspondence between the allocations of the observations in the groups obtained through the hierarchical and k-means methods. g) Yes, based on the dendrogram generated, it is possible to verify that all the stores that belong to regional center 3 form cluster 1, which has the lowest means for all the variables. This fact may determine some specific management action at these stores. After preparing a new cluster analysis, without the stores from cluster 1 (regional center 3), the new agglomeration schedule and its corresponding dendrogram are obtained. From it, we can see the differences between the stores from regional centers 1 and 2 more clearly. Agglomeration Schedule Cluster Combined

Stage Cluster First Appears

Stage

Cluster 1

Cluster 2

Coefficients

Cluster 1

Cluster 2

Next Stage

1 2 3 4 5 6 7

11 6 15 7 7 6 6

12 11 17 15 14 13 7

12.329 14.697 23.409 24.495 32.802 35.665 40.497

0 0 0 0 4 2 6

0 1 0 3 0 0 5

2 6 4 5 7 7 0

1086

Answers

Dendrogram using single linkage 0 11

Regional 2

12

Regional 2

6

Regional 2

13

Regional 1

15

Regional 1

17

Regional 1

7

Regional 1

14

Rescaled distance cluster combine 10 15

20

25

Y

Regional 2

5

3) a) Agglomeration Schedule Cluster Combined Stage

Cluster 1

Cluster 2

1 2 3 4 5 6 7 8 9

18 19 17 16 20 23 17 18 21

33 34 32 31 35 27 19 26 23

Stage Cluster First Appears Coefficients 1.000 .980 .980 .980 .960 .880 .880 .860 .860

Cluster 1

Cluster 2

Next Stage

0 0 0 0 0 0 3 1 0

0 0 0 0 0 0 2 0 6

8 7 7 21 17 9 20 11 18

Answers

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

11 15 13 22 2 4 6 12 11 2 17 3 1 2 9 17 4 9 2 7 1 7 4 2 1

14 18 30 29 13 5 24 20 21 15 25 16 10 3 11 22 8 12 6 28 9 17 7 4 2

.860 .853 .840 .840 .820 .820 .800 .800 .797 .793 .790 .790 .780 .770 .768 .764 .750 .749 .742 .740 .728 .727 .703 .513 .484

0 0 0 0 0 0 0 0 10 14 7 0 0 19 0 20 15 24 23 0 22 29 26 28 30

0 8 0 0 12 0 0 5 9 11 0 4 0 21 18 13 0 17 16 0 27 25 31 32 33

1087

18 19 14 25 19 26 28 27 24 23 25 23 30 28 27 31 32 30 33 31 34 32 33 34 0

Since it is a similarity measure, the values of the coefficients are in descending order in the agglomeration schedule. From this table, it is possible to verify that a considerable leap, in relation to the others, occurs from the 32nd stage (when three clusters are formed) to the 33rd clustering stage. Analyzing the dendrogram also helps in this interpretation.

1088

Answers

b) Dendrogram using average linkage (Between groups) 0 35

18

34

33

33

26

32

15

31

13

30

30

29

2

28

16

27

31

26

3

25

6

24

24

23

4

22

5

21

8

20

22

19

29

18

19

17

34

16

17

15

32

14

25

13

7

12

28

11

1

10

10

9

20

8

35

7

12

6

23

5

27

4

21

3

11

2

14

1

9

5

Rescaled distance cluster combine 10 15

20

25

Answers

1089

In fact, the solution with three clusters is highly advisable. c) Average Linkage (Between Groups)

sector

Health Education Transport

1

2

3

Count

Count

Count

11 0 0

0 12 0

0 0 12

Yes, there is correspondence between the industries and the allocations of companies in the clusters. That is, for the sample under analysis, we can state that companies from the same industry have similarities in relation to how their operations and decision-making processes are carried out. At least as regards the managers’ perception.

1090 Answers

4) a) Proximity Matrix Correlation between Vectors of Values Case

1:1

1:1

1.000

2:2 .866

-1.000

3:3

4:4 .000

5:5 .998

6:6 .945

-.996

7:7

8:8 .000

1.000

9:9

10:10 .971

-1.000

11:11

12:12 -.500

13:13 .999

14:14 .997

-1.000

15:15

16:16 .327

2:2

.866

1.000

-.866

-.500

.896

.655

-.908

-.500

.866

.721

-.856

-.866

.891

.822

-.881

-.189

3:3

-1.000

-.866

1.000

.000

-.998

-.945

.996

.000

-1.000

-.971

1.000

.500

-.999

-.997

1.000

-.327

4:4

.000

-.500

.000

1.000

-.064

.327

.091

1.000

.000

.240

-.020

.866

-.052

.082

.030

.945

5:5

.998

.896

-.998

-.064

1.000

.922

-1.000

-.064

.998

.953

-.996

-.554

1.000

.989

-.999

.266

6:6

.945

.655

-.945

.327

.922

1.000

-.911

.327

.945

.996

-.951

-.189

.926

.969

-.935

.619

7:7

-.996

-.908

.996

.091

-1.000

-.911

1.000

.091

-.996

-.945

.994

.577

-.999

-.985

.998

-.240

8:8

.000

-.500

.000

1.000

-.064

.327

.091

1.000

.000

.240

-.020

.866

-.052

.082

.030

.945

9:9

1.000

.866

-1.000

.000

.998

.945

-.996

.000

1.000

.971

-1.000

-.500

.999

.997

-1.000

.327

10:10

.971

.721

-.971

.240

.953

.996

-.945

.240

.971

1.000

-.975

-.277

.957

.987

-.963

.545

11:11

-1.000

-.856

1.000

-.020

-.996

-.951

.994

-.020

-1.000

-.975

1.000

.483

-.997

-.998

.999

-.346

12:12

-.500

-.866

.500

.866

-.554

-.189

.577

.866

-.500

-.277

.483

1.000

-.545

-.427

.526

.655

13:13

.999

.891

-.999

-.052

1.000

.926

-.999

-.052

.999

.957

-.997

-.545

1.000

.991

-1.000

.277

14:14

.997

.822

-.997

.082

.989

.969

-.985

.082

.997

.987

-.998

-.427

.991

1.000

-.994

.404

15:15

-1.000

-.881

1.000

.030

-.999

-.935

.998

.030

-1.000

-.963

.999

.526

-1.000

-.994

1.000

-.298

16:16

.327

-.189

-.327

.945

.266

.619

-.240

.945

.327

.545

-.346

.655

.277

.404

-.298

1.000

This is a similarity matrix

Answers

1091

b) Agglomeration Schedule Cluster Combined

Stage Cluster First Appears

Stage

Cluster 1

Cluster 2

Coefficients

Cluster 1

Cluster 2

Next Stage

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1 4 5 3 3 1 3 1 6 1 4 1 4 1 1

9 8 13 11 15 5 7 14 10 6 16 2 12 4 3

1.000 1.000 1.000 1.000 1.000 .999 .998 .997 .996 .987 .945 .896 .866 .619 .577

0 0 0 0 4 1 5 6 0 8 2 10 11 12 14

0 0 0 0 0 3 0 0 0 9 0 0 0 13 7

6 11 6 5 7 8 15 10 10 12 13 14 14 15 0

Since Pearson’s correlation is being used as a similarity measure between observations, the values of the coefficients are in descending order in the agglomeration schedule. From this Table, it is possible to verify that a relevant leap, in relation to the others, occurs from the 13th stage (when three clusters with weekly periods are formed) to the 14th clustering stage. Analyzing the dendrogram also helps in this interpretation.

1092

Answers

c) Dendrogram using single linkage 0 1

1

1

9

1

5

1

13

2

14

2

6

2

10

2

2

4

4

4

8

4

16

4

12

3

3

3

11

3

15

3

7

Rescaled distance cluster combine 10 15

5

20

25

In fact, the solution with three-week clusters is highly advisable at this moment. Moreover, it is possible to verify that the second and third clusters are formed exclusively by the periods related to the third and fourth weeks of each month, respectively. This may offer subsidies to prove that there is recurrence of the joint behavior of banana, orange, and apple sales in these periods, for the data in this example. The following table shows the association between the variable week_month and the allocation of each observation in a certain cluster. Single Linkage

week_month

1 2 3 4

1

2

3

Count

Count

Count

4 4 0 0

0 0 4 0

0 0 0 4

Answers

1093

ANSWER KEYS: EXERCISES: CHAPTER 12 1) a) For each factor, we have the following eigenvalues: Factor 1: (0.917)2 + (0.874)2 + (0.844)2 + (0.031)2 ¼ 2.318 Factor 2: (0.047)2 + (0.077)2 + (0.197)2 + (0.979)2 ¼ 1.005 b) The proportions of variance shared by all the variables to form each factor are: Factor 1: 2:318 4 ¼ 0:580 ð58:00%Þ 1:005 Factor 2: 4 ¼ 0:251 ð25:10%Þ The total proportion of variance lost by the four variables to extract these two factors is: 1  0:580  0:251 ¼ 0:169 ð16:90%Þ c) The proportions of variance shared to form both factors (communalities) are: communalityage ¼ ð0:917Þ2 + ð0:047Þ2 ¼ 0:843 communalityfixedif ¼ ð0:874Þ2 + ð0:077Þ2 ¼ 0:770 communalityvariableif ¼ ð0:844Þ2 + ð0:197Þ2 ¼ 0:751 communalitypeople ¼ ð0:031Þ2 + ð0:979Þ2 ¼ 0:959 d) Based on the two factors extracted, the expressions of each standardized variable are: Zagei ¼ 0:917  F1i + 0:047  F2i + ui , R2 ¼ 0:843 Zfixedif i ¼ 0:874  F1i + 0:077  F2i + ui , R2 ¼ 0:770 Zvariableif i ¼ 0:844  F1i + 0:197  F2i + ui , R2 ¼ 0:751 Zpeoplei ¼ 0:031  F1i + 0:979  F2i + ui , R2 ¼ 0:959 e) 1

people

0.5

variableif fixedif

age

0

–0.5

–1 –1

–0.8

–0.6

–0.4

–0.2

0

0.2

0.4

0.6

0.8

1

f) While variables age, fixedif, and variableif have a high correlation with the first factor (X-axis), variable people has a strong correlation with the second factor (Y-axis). This phenomenon can be a result of the fact that older customers, since they do not like taking risks, invest a great deal more in fixed-income funds, such as, savings accounts or CDB (Bank Deposit Certificates). On the other hand, even though variable variableif has a high correlation with the first factor, the absolute factor loading is negative. This shows that younger customers invest a great deal more in variable-income funds, such as, stocks. Finally, the number of people who live in the household (variable people) has a low correlation with the other variables. Thus, it ends up having a high factor loading with the second factor.

1094

Answers

2) a) YEAR 1 KMO and Bartlett’s Test Kaiser-Meyer-Olkin Measure of Sampling Adequacy.

.719

Bartlett’s Test of Sphericity

Approx. Chi-Square df Sig.

89.637 6 .000

YEAR 2 KMO and Bartlett’s Test Kaiser-Meyer-Olkin Measure of Sampling Adequacy.

.718

Bartlett’s Test of Sphericity

Approx. Chi-Square df Sig.

86.483 6 .000

Based on the KMO statistics, we can state that the overall adequacy of the factor analysis is considered average for each of the years of study (KMO ¼ 0.719 for the first year, and KMO ¼ 0.718 for the second year). In both periods, w2Bartlett statistics allow us to reject, at a significance level of 0.05 and based on the hypothesis of Bartlett’s test of sphericity, that the correlation matrices are statistically equal to the identity matrix with the same dimension. Since w2Bartlett ¼ 89.637 (Sig. w2Bartlett < 0.05 for 6 degrees of freedom) for the first year, and w2Bartlett ¼ 86.483 (Sig. w2Bartlett < 0.05 for 6 degrees of freedom) for the second year. Therefore, the principal component analysis is adequate for each of the years of study. b) YEAR 1 Total Variance Explained Initial Eigenvalues

Extraction Sums of Squared Loadings

Component

Total

% of Variance

Cumulative %

Total

% of Variance

Cumulative %

1 2 3 4

2.589 .730 .536 .146

64.718 18.247 13.391 3.643

64.718 82.965 96.357 100.000

2.589

64.718

64.718

Extraction Method: Principal Component Analysis.

YEAR 2 Total Variance Explained Initial Eigenvalues

Extraction Sums of Squared Loadings

Component

Total

% of Variance

Cumulative %

Total

% of Variance

Cumulative %

1 2 3 4

2.566 .737 .543 .154

64.149 18.435 13.577 3.838

64.149 82.584 96.162 100.000

2.566

64.149

64.149

Extraction Method: Principal Component Analysis.

Answers

1095

Based on the latent root criterion, only one factor is extracted in each of the years, with their respective eigenvalue: Year 1: 2.589 Year 2: 2.566 The proportion of variance shared by all the variables to form the factor each year is: Year 1: 64.718% Year 2: 64.149% c) YEAR 1 Component Matrixa Component 1 Corruption Perception Index - year 1 (Transparency International) Number of murders per 100,000 inhabitants: year 1 (OMS, UNODC and GIMD) Per capita GDP - year 1, using 2000 as the base year (in US$ adjusted for inflation) (World Bank) Average number of years in school per person over 25 years of age - year 1 (IHME)

.900 .614 .911 .755

a 1 component extracted. Extraction Method: Principal Component Analysis.

YEAR 1 Communalities

Corruption Perception Index - year 1 (Transparency International) Number of murders per 100,000 inhabitants: year 1 (OMS, UNODC and GIMD) Per capita GDP - year 1, using 2000 as the base year (in US$ adjusted for inflation) (World Bank) Average number of years in school per person over 25 years of age - year 1 (IHME)

Initial

Extraction

1.000 1.000 1.000 1.000

.810 .378 .830 .571

Extraction Method: Principal Component Analysis.

YEAR 2 Component Matrixa Component 1 Corruption Perception Index - year 2 (Transparency International) Number of murders per 100,000 inhabitants: year 2 (OMS, UNODC and GIMD) Per capita GDP - year 2, using 2000 as the base year (in US$ adjusted for inflation) (World Bank) Average number of years in school per person over 25 years of age - year 2 (IHME) a 1 component extracted. Extraction Method: Principal Component Analysis.

.899 .608 .908 .750

1096

Answers

YEAR 2 Communalities

Corruption Perception Index - year 2 (Transparency International) Number of murders per 100,000 inhabitants: year 2 (OMS, UNODC and GIMD) Per capita GDP - year 2, using 2000 as the base year (in US$ adjusted for inflation) (World Bank) Average number of years in school per person over 25 years of age - year 2 (IHME)

Initial

Extraction

1.000 1.000 1.000 1.000

.808 .370 .825 .563

Extraction Method: Principal Component Analysis.

We can see that slight reductions occurred in the communalities of all the variables from the first to the second year. d) YEAR 1 Component Score Coefficient Matrix Component 1 Corruption Perception Index - year 1 (Transparency International) Number of murders per 100,000 inhabitants: year 1 (OMS, UNODC and GIMD) Per capita GDP - year 1, using 2000 as the base year (in US$ adjusted for inflation) (World Bank) Average number of years in school per person over 25 years of age - year 1 (IHME)

.348 .237 .352 .292

Extraction Method: Principal Component Analysis. Component Scores.

YEAR 2 Component Score Coefficient Matrix Component 1 Corruption Perception Index - year 2 (Transparency International) Number of murders per 100,000 inhabitants: year 2 (OMS, UNODC and GIMD) Per capita GDP - year 2, using 2000 as the base year (in US$ adjusted for inflation) (World Bank) Average number of years in school per person over 25 years of age - year 2 (IHME)

.350 .237 .354 .292

Extraction Method: Principal Component Analysis. Component Scores.

Based on the standardized variables, the expression of the factor extracted each year is: Year 1: Fi ¼ 0:348  Zcpi1i  0:237  Zviolence1i + 0:352  Zcapita_gdp1i + 0:292  Zschool1i Year 2:

Fi ¼ 0:350  Zcpi2i  0:237  Zviolence2i + 0:354  Zcapita_gdp2i + 0:292  Zschool2i

Even if small changes occurred in the factor scores from one year to the next, this only reinforces the importance of reapplying the technique to obtain factors with more precise and updated scores, mainly when they are used to create indexes and rankings.

1097

Answers

e) Year 1

Year 2

Country

Index

Ranking

Country

Index

Ranking

Switzerland Norway Denmark Sweden Japan United States Canada United Kingdom Netherlands Australia Germany Austria Ireland New Zealand Singapore Belgium Israel France Cyprus United Arab Emirates Czech Rep. Italy Poland Spain Chile Greece Kuwait Portugal Romania Oman Saudi Arabia Serbia Argentina Turkey Ukraine Kazakhstan Malaysia Lebanon Russia Mexico China Egypt Thailand Indonesia India Brazil Philippines Venezuela South Africa Colombia

1.6923 1.6794 1.4327 1.4040 1.3806 1.3723 1.3430 1.1560 1.1086 1.0607 1.0297 0.9865 0.9439 0.9269 0.8781 0.8175 0.6322 0.5545 0.5099 0.3157 0.2244 0.0859 0.0373 0.0303 0.0517 0.1432 0.2276 0.2980 0.3028 0.4742 0.5111 0.5407 0.5556 0.6476 0.7109 0.7423 0.7459 0.7966 0.8534 0.8803 0.8840 0.9792 1.0632 1.2245 1.2272 1.3294 1.3466 1.3916 1.8215 1.8534

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

Norway Switzerland Sweden Denmark Japan Canada United States United Kingdom Netherlands Australia Germany Austria Ireland Singapore New Zealand Belgium Israel France Cyprus United Arab Emirates Czech Rep. Poland Spain Chile Italy Kuwait Greece Portugal Romania Saudi Arabia Oman Argentina Serbia Malaysia Turkey Ukraine Kazakhstan Lebanon Russia China Mexico Egypt Thailand Indonesia India Brazil Philippines Venezuela Colombia South Africa

1.6885 1.6594 1.4388 1.4225 1.3848 1.3844 1.3026 1.1321 1.1007 1.0660 1.0401 0.9903 0.9411 0.9184 0.9063 0.8265 0.6444 0.5448 0.4606 0.2849 0.1857 0.0868 0.0334 0.0170 0.0064 0.1462 0.2247 0.2794 0.3150 0.4321 0.5034 0.5342 0.5544 0.6098 0.6401 0.6807 0.6970 0.8060 0.8513 0.8982 0.9323 0.9485 1.0800 1.2431 1.2533 1.3468 1.3885 1.4149 1.7697 1.9173

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

From the first to the second year, there were some changes in the relative positions of the countries in the ranking.

1098

Answers

3) a) Correlation Matrix

Perception of the variety of goods (0 to 10) Correlation Perception of the variety of goods (0 to 10)

Perception of the quality and speed of inventory replacement (0 to 10)

Perception of the store’s layout (0 to 10)

Perception of Perception of the store’s Perception of Perception of thermal, Perception of prices the quality of the the store’s acoustic and the store’s compared to the services general visual comfort inside the store cleanliness (0 to rendered (0 to competition (0 to discount policy (0 to 10) 10) 10) 10) (0 to 10)

1.000

.753

.898

.733

.640

.193

.084

.053

Perception of the quality and speed of inventory replacement (0 to 10)

.753

1.000

.429

.633

.548

.208

-.449

-.367

Perception of the store’s layout (0 to 10)

.898

.429

1.000

.641

.567

.142

.413

.318

Perception of thermal, acoustic and visual comfort inside the store (0 to 10)

.733

.633

.641

1.000

.864

.227

.235

.174

Perception of the store’s general cleanliness (0 to 10)

.640

.548

.567

.864

1.000

.194

.220

.173

Perception of the quality of the services rendered (0 to 10)

.193

.208

.142

.227

.194

1.000

.137

.113

Perception of the store’s prices compared to the competition (0 to 10)

.084

-.449

.413

.235

.220

.137

1.000

.906

Perception of the store’s discount policy (0 to 10)

.053

-.367

.318

.174

.173

.113

.906

1.000

Yes. Based on the magnitude of some Pearson’s correlation coefficients, it is possible to identify a first indication that the factor analysis may group the variables into factors. b) KMO and Bartlett’s Test Kaiser-Meyer-Olkin Measure of Sampling Adequacy.

.610

Bartlett’s Test of Sphericity

13752.938 28 .000

Approx. Chi-Square df Sig.

Yes. From the result of the w2Bartlett statistic, it is possible to reject that the correlation matrix is statistically equal to the identity matrix with the same dimension, at a significance level of 0.05 and based on the hypothesis of Bartlett’s test of sphericity, since w2Bartlett ¼ 13,752.938 (Sig. w2Bartlett < 0.05 for 28 degrees of freedom). Therefore, the principal component analysis can be considered adequate. c) Total Variance Explained Initial Eigenvalues

Extraction Sums of Squared Loadings

Component

Total

% of Variance

Cumulative %

Total

% of Variance

Cumulative %

1 2 3 4 5 6 7 8

3.825 2.254 .944 .597 .214 .126 .025 .016

47.812 28.174 11.794 7.458 2.679 1.570 .313 .201

47.812 75.986 87.780 95.238 97.917 99.486 99.799 100.000

3.825 2.254

47.812 28.174

47.812 75.986

Extraction Method: Principal Component Analysis.

Answers

1099

Considering the latent root criterion, two factors are extracted, with the respective eigenvalues: Factor 1: 3.825 Factor 2: 2.254 The proportion of variance shared by all the variables to form each factor is: Factor 1: 47.812% Factor 2: 28.174% Thus, the total proportion of variance shared by all the variables to form both factors is equal to 75.986%. d) The total proportion of variance lost by all the variables to extract these two factors is: 1  0:75986 ¼ 0:24014 ð24:014%Þ e)

Communalities

Perception of the variety of goods (0 to 10) Perception of the quality and speed of inventory replacement (0 to 10) Perception of the store’s layout (0 to 10) Perception of thermal, acoustic and visual comfort inside the store (0 to 10) Perception of the store’s general cleanliness (0 to 10) Perception of the quality of the services rendered (0 to 10) Perception of the store’s prices compared to the competition (0 to 10) Perception of the store’s discount policy (0 to 10)

Initial

Extraction

1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

.873 .914 .766 .827 .721 .101 .978 .900

Extraction Method: Principal Component Analysis.

Note that the loadings and communality of the variable services rendered are considerably low. This may demonstrate the need to extract a third factor, which decharacterizes the latent root criterion.

1100

Answers

f)

Communalities

Perception of the variety of goods (0 to 10) Perception of the quality and speed of inventory replacement (0 to 10) Perception of the store’s layout (0 to 10) Perception of thermal, acoustic and visual comfort inside the store (0 to 10) Perception of the store’s general cleanliness (0 to 10) Perception of the quality of the services rendered (0 to 10) Perception of the store’s prices compared to the competition (0 to 10) Perception of the store’s discount policy (0 to 10)

Initial

Extraction

1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

.887 .917 .804 .828 .722 .987 .978 .900

Extraction Method: Principal Component Analysis.

Yes, it is possible to confirm the construct of the questionnaire proposed by the store’s general manager, because variables variety of goods, replacement, layout, comfort, and cleanliness have a stronger correlation with a specific factor, variables price and discounts, with another factor, and, finally, variable services rendered, with a third factor. g) To the detriment of the extraction based on the latent root criterion, the decision to extract three factors increases the communalities of the variables, highlighting the variable services rendered, now, more strongly correlated with the third factor. h)

Answers

1101

Varimax rotation redistributes the variable loadings in each factor, which facilitates the confirmation of the construct proposed by the store’s general manager. i) Component plot in rotated space

1.0

Components 2

Discounts Prices 0.5

Layout Cleanliness

0.0

Assortment

Comfort

Services Replacement

–0.5

–1.0 –1.0

–0.5

Com

0.0

pon

0.5 1 .0 –1.0

ents

3

–0.5

0.0

0.5

Components

1

1.0

1102

Answers

Component plot in rotated space

1.0

Components 2

Discounts Prices 0.5 Layout Services

0.0

Cleanliness Comfort Assortment

–0.5 Replacement –1.0 –1.0

–0.5

Com

0.0

pon

0.5 1.0 1.0

ents

1

0.5

0.0

Components

ANSWER KEYS: EXERCISES: CHAPTER 13 1) a) b) c) d) e) f) 2)

–0.5

Y^ ¼ 3:8563 + 0:3872  X R2 ¼ 0.9250 Yes (P-value of t ¼ 0.000 < 0.05). 9.9595 billion dollars (we must make Y ¼ 0 and solve the equation). 3.8563% (we must make X ¼ 0). 0.4024% (mean) 1.2505% (minimum) 2.0554% (maximum)

3

–1.0

Answers

1103

a) Yes, since the P-value of the F statistic < 0.05, we can state that at least one of the explanatory variables is statistically significant to explain the behavior of variable cpi, at a significance level of 0.05. b) Yes, since the P-value of both t statistics < 0.05, we can state that their parameters are statistically different from zero, at a significance level of 0.05. Therefore, the Stepwise procedure would not exclude any of the explanatory variables of the final model. c) c^ pii ¼ 15:1589 + 0:0700  agei  0:4245  hoursi d) R2 ¼ 0.3177 e) By analyzing the signs of the final model’s coefficients, for this cross-section, we can state that countries with billionaires with lower average ages have lower cpi indexes. That is, there is a higher corruption perception from society. Besides, on average, a greater number of hours worked per week has a negative relationship with variable cpi. That is, countries with a higher corruption perception (lower cpi´s) have a higher workload per week. It is important to mention that countries with lower cpi´s are those considered emerging countries. f)

By using the Shapiro-Francia test, the most suitable for the size of this sample, we can see that the residuals follow a normal distribution, at a significance level of 0.05. We would have arrived at the same conclusion if the test used had been the Shapiro-Wilk. g)

From the Breusch-Pagan/Cook-Weisberg test, it is possible to verify if there is homoskedasticity in the model proposed. h)

Since the final model obtained does not have very high VIF statistics (1 – Tolerance ¼ 0.058), we may consider that there are no multicollinearity problems.

1104

Answers

3)

a) The difference between the average cpi value for emerging and for developed countries is 3.6318. That is, while emerging countries have an average cpi ¼ 4.0968, developed countries have an average cpi ¼ 7.7286. This is exactly the value of the cpi regression intercept based on variable emerging, since the dummy emerging for developed countries ¼ 0. Yes, this difference is statistically significant, at a significance level of 0.05, since the P-value of t statistic < 0.05 for the variable emerging. b)

c^ pii ¼ 13:1701  0:1734  hoursi  3:2238  emergingi c)

c^ pi ¼ 13:1701  0:1734  ð37Þ  3:2238  ð1Þ ¼ 3:5305

Answers

d)

1105

c^ pimin ¼ 8:0092  0:3369  ð37Þ  4:0309  ð1Þ ¼ 8:4870 c^ pimax ¼ 18:3310  0:0099  ð37Þ  2:4168  ð1Þ ¼ 15:5479

Obviously, the confidence interval is extremely broad and makes no sense. This happened because the value of R2 is not so high. e)

c^ pii ¼ 27:4049  5:7138  ln ðhoursi Þ  3:2133  emergingi 2

f) Since the adjusted R is slightly higher in the model with nonlinear functional form (logarithmic functional form for variable hours) than in the model with linear functional form, we choose the nonlinear estimated model seen in item (e). Since, in both cases, neither the number of variables nor the sample size used changes, such analysis could be carried out directly based on the values of R2. 4) a)

chole^ sterolt ¼ 136:7161 + 1:9947  bmit  5:1635  sportt b) We can see that the body mass index has a positive relationship with the LDL cholesterol index, such that, every time the index increases by one unit, on average, there is an increase of almost 2 mg/dL of the cholesterol commonly known as bad cholesterol, ceteris paribus. Analogously, the increase in the frequency of physical activities per week by one unit makes the LDL cholesterol index drop, on average, more than 5 mg/dL, ceteris paribus. Therefore, maintaining one’s weight, or even losing weight, plus establishing a routine of weekly physical activities, may contribute to the establishment of a healthier life.

1106

Answers

c)

Since we have, at a significance level of 0.05 and for a model with 3 parameters and 48 observations, 0.938 < dL ¼ 1.45, we can state that there is a positive first-order autocorrelation between the error terms. d)

By analyzing the Breusch-Godfrey test, we can see that, besides the first-order autocorrelation between the error terms, there are also autocorrelation problems between the 3rd, 4th, and 12th order residuals. This shows the existing seasonality in the executive’s behavior regarding his body mass and engagement to doing physical activities.

ANSWER KEYS: EXERCISES: CHAPTER 14 1)

a) Yes. Since the P-value of the w2 statistic < 0.05, we can state that at least one of the explanatory variables is statistically significant to explain the probability of default, at a significance level of 0.05. b) Yes. Since the P-value of all Wald z statistics < 0.05, we can state that their respective parameters are statistically different from zero, at a significance level of 0.05. Therefore, no explanatory variable will be excluded from the final model. c) pi ¼

1 1 + eð2:975070:02433  agei + 0:74149  genderi 0:00025  incomei Þ

d) Yes. Since the parameter estimated for the variable gender is positive, on average, male individuals (dummy ¼ 1) have higher probabilities of default than female individuals, as long as the other conditions are kept constant. The chances of the event occurring will be multiplied by a factor greater than 1.

Answers

1107

e) No. Older people, on average, tend to have smaller probabilities of default, maintaining the remaining conditions constant, since the parameter of the variable age is negative, that is, the chances of the event occurring is multiplied by a factor less than 1, as the age increases. f) p ¼

1 1 + e½2:975070:02433  ð37Þ + 0:74149  ð1Þ0:00025  ð6, 850Þ

¼ 0:7432

The average probability of default estimated for this individual is 74.32%. g)

The chance of being default as the income increases by one unit is, on average and maintaining the remaining conditions constant, multiplied by a factor of 0.99974 (a chance 0.026% lower). h)

While the overall model efficiency is 77.40%, the sensitivity is 93.80% and the specificity is 30.23% (for a cutoff of 0.5).

1108

Answers

2) a)

Only the category bad of variable price was not statistically significant, at a significance level of 0.05, to explain the probability of the event occurring - the event we are interested in. That is, there are no differences that would change the probability of someone becoming loyal to the retailer when they answer terrible or bad on their perception of the prices, maintaining the remaining conditions constant. b)

Answers

c)

For a cutoff of 0.5, the overall model efficiency is 86.00%. d)

Sensitivity/specificity

1.00

0.75

0.50

0.25

0.00 0.00

0.25

0.50 Probability cutoff Sensitivity

0.75 Specificity

1.00

1109

1110

Answers

The cutoff from which the specificity becomes slightly higher than the sensitivity is equal to 0.57.

e) On average, the chance of becoming loyal to the establishment is multiplied by a factor of 5.39 when their perception of services rendered changes from terrible to bad. Whereas, from terrible to regular, this chance is multiplied by a factor of 6.17. From terrible to good, it is multiplied by a factor of 27.78, and, finally, from terrible to excellent, by a factor of 75.60. These answers will only be valid if the other conditions are kept constant. f) On average, the chance of becoming loyal to the establishment is multiplied by a factor of 6.43 when their perception of variety of goods changes from terrible to bad. Whereas, from terrible to regular, this chance is multiplied by a factor of 7.83. From terrible to good, it is multiplied by a factor of 28.09, and, finally, from terrible to excellent, by a factor of 381.88. Conversely, for the variable accessibility, on average, the chance of becoming loyal to the establishment is multiplied by a factor of 10.49 when their perception changes from terrible to bad. From terrible to regular, this chance is multiplied by a factor of 18.55. From terrible to good, it is multiplied by a factor of 127.40, and, finally, from terrible to excellent, by a factor of 213.26. Finally, for the variable price, on average, the chance of becoming loyal to the establishment is multiplied by a factor of 18.47 when their perception changes from terrible or bad to regular. From terrible or bad to good, this chance is multiplied by a factor of 20.82. Lastly, from terrible or bad to excellent, the chance of becoming loyal to the establishment is multiplied by a factor of 49.87. These answers will only be valid if the other conditions are kept constant in each case. g) Based on the analysis of these chances, if the establishment wishes to invest in a single perceptual variable to increase the probability of consumers becoming loyal, such that, they leave their terrible perceptions behind and begin, with higher frequency, to have excellent perceptions of this issue, it must invest in the variable variety of goods, since this variable is the one that shows the highest odds ratio (381.88). In other words, the chances of becoming loyal to the establishment, when their perception changes from terrible variety of goods to excellent, are, on average, multiplied by a factor of 381.88 (38,088% higher), maintaining the remaining conditions constant.

Answers

1111

3) a)

b)

Yes. Since the P-value of the w2 statistic < 0.05, we can reject the null hypothesis that all parameters bjm (j ¼ 1, 2; m ¼ 1, 2, 3, 4) are statistically equal to zero at a significance level of 0.05. That is, at least one of the explanatory variables is statistically significant to form the occurrence probability expression of at least one of the classifications proposed for the LDL cholesterol index. c) Since all the parameters are statistically significant for all the logits (Wald z tests at a significance level of 0.05), the final equations estimated for the average occurrence probabilities of the classifications proposed for the LDL cholesterol index can be written the following way:

1112

Answers

Probability of an individual i having a very high LDL cholesterol index:

pi ¼

1 … 1 + eð0:420:31  cigarettei + 0:16  sporti Þ + eð2:620:41  cigarettei + 1:01  sporti Þ … ð2:461:41  cigarette + 1:13  sport Þ ð2:861:67  cigarette + 1:16  sport Þ i i +e i i +e

Probability of an individual i having a high LDL cholesterol index:

pi ¼

eð0:420:31  cigarettei + 0:16  sporti Þ … 1 + eð0:420:31  cigarettei + 0:16  sporti Þ + eð2:620:41  cigarettei + 1:01  sporti Þ … ð2:461:41  cigarette + 1:13  sport Þ ð2:861:67  cigarette + 1:16  sport Þ i i +e i i +e

Probability of an individual i having a borderline LDL cholesterol index:

pi ¼

eð2:620:41  cigarettei + 1:01  sporti Þ … 1 + eð0:420:31  cigarettei + 0:16  sporti Þ + eð2:620:41  cigarettei + 1:01  sporti Þ … ð2:461:41  cigarette + 1:13  sport Þ ð2:861:67  cigarette + 1:16  sport Þ i i +e i i +e

Probability of an individual i having a near optimal LDL cholesterol index:

pi ¼

eð2:461:41  cigarettei + 1:13  sporti Þ … + eð2:620:41  cigarettei + 1:01  sporti Þ

1 + eð0:420:31  cigarettei + 0:16  sporti Þ …

+eð2:461:41  cigarettei + 1:13  sporti Þ + eð2:861:67  cigarettei + 1:16  sporti Þ

Probability of an individual i having an optimal LDL cholesterol index:

pi ¼

eð2:861:67  cigarettei + 1:16  sporti Þ … + eð2:620:41  cigarettei + 1:01  sporti Þ

1 + eð0:420:31  cigarettei + 0:16  sporti Þ …

+eð2:461:41  cigarettei + 1:13  sporti Þ + eð2:861:67  cigarettei + 1:16  sporti Þ

d) For an individual who does not smoke and only practices sports once a week, we have: Probability of having a very high LDL cholesterol index ¼ 41.32%. Probability of having a high LDL cholesterol index ¼ 31.99%. Probability of having a borderline LDL cholesterol index ¼ 8.23%. Probability of having a near optimal LDL cholesterol index ¼ 10.92%. Probability of having an optimal LDL cholesterol index ¼ 7.54%.

Answers

1113

e)

If people start practicing sports twice a week, they will considerably increase their probability of having near-optimal or optimal levels of LDL cholesterol.

f) The chances of having a high cholesterol index, in comparison to a level considered very high, are, on average, multiplied by a factor of 1.1745 (17.45% higher), when we increase the number of times physical activities are done weekly by one unit and maintaining the remaining conditions constant. g) The chances of having an optimal cholesterol index, on average and in comparison to a level considered near optimal, are multiplied by a factor of 1.2995 (0.2450047 / 0.1885317), when people stop smoking and maintaining the remaining conditions constant. That is, the chances are 29.95% higher.

1114

Answers

Tip: For those who are in doubt about this procedure, you just need to change the reference category of variable cigarette (now, smokes ¼ 0) and estimate the model with the category near optimal of the dependent variable as the reference category. h) and i)

ANSWER KEYS: EXERCISES: CHAPTER 15 1) a) Statistic Mean Variance

1.020 1.125

Even if in a preliminary way, we can see that the mean and variance of the variable purchases are quite close.

Answers

1115

b)

Since the P-value of the t-test that corresponds to the b parameter of lambda is greater than 0.05, we can state that the data of the dependent variable purchases do not present overdispersion. So, the Poisson regression model estimated is suitable due to the presence of equidispersion in the data. c)

The result of the w2 test suggests that there is quality in the adjustment of the Poisson regression model estimated. That is, there are no statistically significant differences, at a significance level of 0.05, between the observed and the predicted probability distributions of annual use incidence of closed-end credit. d) Since all the zcal values < 1.96 or > 1.96, the P-values of the Wald z statistics < 0.05 for all the parameters estimated, thus, we arrive at final Poisson regression model. Therefore, the final expression for the estimated average number of annual use of closed-end credit financing when purchasing durable goods, for a consumer i, is: purchasesi ¼ eð7:0480:001  incomei 0:086  agei Þ e) purchases ¼ e[7.0480.001(2,600)0.086(47)] ¼ 1.06 We recommend that this calculation be carried out with a larger number of decimal places.

f) The annual use incidence rate of closed-end credit financing when there is an increase in the customer’s monthly income of US$1,00 is, on average and as long as the other conditions are kept constant, multiplied by a factor of 0.9988 (0.1124% lower). Consequently, at each increase of US$100,00 in the customer’s monthly income, we expect

1116

Answers

the annual use incidence rate of closed-end credit financing to be 11.24% lower, on average and provided the other conditions are kept constant. g) The annual use incidence rate of closed-end credit financing when there is an increase of 1 year in consumers’ average age is, on average and as long as the other conditions are kept constant, multiplied by a factor of 0.9171 (8.29% lower). h)

In the constructed chart, it is possible to see that higher monthly incomes lead to a decrease in the expected annual use of closed-end credit financing when purchasing durable goods, with an average reduction rate of 12.0% at each increase of US$100.00 in income. i)

Answers

1117

j) Young people and with lower monthly income. 2) a) Statistic Mean Variance

2.760 8.467

Even if in a preliminary way, there are indications of overdispersion in the data of the variable property, since its variance is extremely higher than its mean.

1118

Answers

b)

Since the P-value of the t-test that corresponds to the b parameter of lambda is lower than 0.05, we can state that the data of the dependent variable property present overdispersion, making the Poisson regression model estimated not suitable.

Furthermore, the result of the w2 test suggests the inexistence of adjustment quality in the Poisson regression model estimated. That is, there are statistically significant differences, at a significance level of 0.05, between the probability distributions observed and predicted for the number of real estate properties for sale per square. c)

d) Since the confidence interval for f (alpha in Stata) does not include zero, we can state that, at a 95% confidence level, f is statistically different from zero and has an estimated value equal to 0.230. The result of the likelihood-ratio test for parameter f (alpha) itself suggests that the null hypothesis that this parameter is statistically equal to zero can be rejected at a significance level of 0.05. This proves that there is overdispersion in the data and, therefore, we must choose the estimation of the negative binomial model.

Answers

1119

e) Since all the zcal values < 1.96 or > 1.96, the P-values of the Wald z statistics < 0.05 for all the parameters estimated and, thus, we arrive at the final negative binomial regression model. Therefore, the expression for the estimated average number of real estate properties for sale in a certain square ij is: propertyij ¼ eð0:608 + 0:001  distparkij 0:687  mallij Þ f) property ¼ e[0.608+0.001(820)0.687(0)] ¼ 5.07 We recommend that this calculation be carried out with a larger number of decimal places.

g) The number of real estate properties for sale per square is multiplied, on average and provided the other conditions are kept constant, by a factor of 1.0012 at each 1 meter further away from the municipal park. Hence, when there is an approximation of 1 meter from the park, we must divide the average amount of real estate properties for sale per square by this same factor. That is, the number will be multiplied by a factor of 0.9987 (0.1237% lower). Thus, at each approximation of 100 meters from the park, we expect that the average amount of real estate properties for sale to be, on average and as long as the other conditions are kept constant, 12.37% lower. h) The expected number of real estate properties for sale when a commercial center or mall is built in the microregion (square) is, as long as the other conditions are kept constant, multiplied by a factor of 0.5031. That is, on average, it becomes 49.69% lower. i)

1120

Answers

j)

k) Yes, we can state that proximity to parks and green spaces and the existence of malls and commercial centers in the microregion make the number of real estate properties for sale go down. That is, these features may be helping reduce the intention of selling residential real estate. l)

Answers

1121

m)

We can see that the adjustment of the negative binomial regression model is better than the adjustment of the Poisson regression model, since: – the maximum difference between the probabilities observed and the ones predicted is lower for the negative binomial model; – Pearson’s total value is also lower for the negative binomial regression model.

1122

Answers

n)

ANSWER KEYS: EXERCISES: CHAPTER 16 Ex. 3 max x1 + x2 s:t: ¼ 10 2x1  5x2 a) x1 + 2x2 + x3 ¼ 50 x1 ,x2 , x3  0

ð1Þ ð 2Þ ð3Þ

min 24x1 + 12x2 s:t: ¼4 ð1Þ 3x1 + 2x2  x3 b) + x4 ¼ 26 ð2Þ 2x1  4x2  x5 ¼ 3 ð 3Þ x2 x1 , x2 , x3 ,x4 ,x5  0 ð4Þ

Answers

max 10x1  x2 s:t: 6x1 + x2 + x3 ¼ 10 c) x2  x4 ¼ 6 x1 ,x2 ,x3 , x4  0

ð 1Þ ð2Þ ð 3Þ

max 3x1 + 3x2  2x3 s:t: 6x1 + 3x2  x3 + x4 ¼ 10 d) x2 + x3  x5 ¼ 20 4 x1 , x2 , x3 ,x4 ,x5  0

ð 1Þ ð 2Þ ð3Þ

Ex. 4 max x1 + x2 s:t: 2x1  5x2  10 a)  2x1 + 5x2  10 x1 + 2x2  50 x1 ,x2  0

ð 1Þ ð2Þ ð 3Þ ð4Þ

min 24x1 + 12x2 s:t: 3x1 + 2x2  4 b)  2x1 + 4x2  26 x2  3 x1 ,x2  0

ð 1Þ ð 2Þ ð 3Þ ð 4Þ

max 10x1  x2 s:t: 6x1 + x2  10 c)  x2  6 x1 ,x2  0

ð 1Þ ð 2Þ ð3Þ

max 3x1 + 3x2  2x3 s:t: ð 1Þ 6x1 + 3x2  x3  10 d) x2   x3  20 ð2Þ 4 ð 3Þ x1 , x 2 , x 3  0 Ex. 5 a) min  z ¼  10x1 + x2 b) min  z ¼  3x1  3x2 + 2x3 Ex. 7 xi ¼ number of vehicles of model i to be manufactured per week, i ¼ 1, 2, 3. x1 ¼ number of vehicles of model Arlington to be manufactured per week. x2 ¼ number of vehicles of model Marilandy to be manufactured per week. x3 ¼ number of vehicles of model Lagoinha to be manufactured per week. Fobj ¼ max z ¼ 2, 500x1 + 3, 000x2 + 2,800x3 subject to 3x1 + 4x2 + 3x3  480 ðminutes  machine=week available for injectionÞ 5x1 + 5x2 + 4x3  640 ðminutes  machine=week available for foundryÞ 2x1 + 4x2 + 4x3  400 ðminutes  machine=week available for machiningÞ 4x1 + 5x2 + 5x3  640 ðminutes  machine=week available for upholsteryÞ 2x1 + 3x2 + 3x3  320 ðminutes  machine=week available for final assemblyÞ  50 ðminimum sales potential of the Arlington modelÞ x1 x2  30 ðminimum sales potential of the Marilandy modelÞ x3  30 ðminimum sales potential of the Lagoinha modelÞ x1 ,x2 , x3  0

1123

1124

Answers

Ex. 8 xi ¼ liters of product i to be manufactured per month, i ¼ 1, 2 x1 ¼ liters of beer to be manufactured per month. x2 ¼ liters of soft drink to be manufactured per month. Fobj ¼ max z ¼ 0:5x1 + 0:4x2 subject to 2x1  57,600 ðminutes=month available to extract beer maltÞ 4x1  115,200 ðminutes=month available to process wortÞ 3x1  96,000 ðminutes=month available to ferment beerÞ 4x1  115,200 ðminutes=month available to process beerÞ 5x1  96,000 ðminutes=month available to bottle beerÞ 1x2  57,600 ðminutes=month available to prepare simple syrupÞ 3x2  67,200 ðminutes=month available to prepare compound syrupÞ 4x2  76,800 ðminutes=month available to dilute soft drinksÞ 5x2  96,000 ðminutes=month available to carbonate soft drinksÞ 2x2  48,000 ðminutes=month available to bottle soft drinksÞ x1 + x2  42, 000 ðmaximum demand of beer and soft drinksÞ x1 ,x2  0 Ex. 9 xi ¼ quantity of product i to be manufactured per week, i ¼ 1, 2, …, 5. x1 ¼ number of refrigerators to be manufactured per week. x2 ¼ number of freezers to be manufactured per week. x3 ¼ number of stoves to be manufactured per week. x4 ¼ number of dishwashers to be manufactured per week. x5 ¼ number of microwave ovens to be manufactured per week. Fobj ¼ max z ¼ 52x1 + 37x2 + 35x3 + 40x4 + 29x5 subject to 0:2x1 + 0:2x2 + 0:4x3 + 0:4x4 + 0:3x5  400 ðh  machine=week pressingÞ 0:2x1 + 0:3x2 + 0:3x3 + 0:3x4 + 0:2x5  350 ðh  machine=week paintingÞ 0:4x1 + 0:3x2 + 0:3x3 + 0:3x4 + 0:2x5  250 ðh  machine=week moldingÞ 0:2x1 + 0:4x2 + 0:4x3 + 0:4x4 + 0:4x5  200 ðh  machine=week assemblyÞ 0:1x1 + 0:2x2 + 0:2x3 + 0:2x4 + 0:3x5  200 ðh  machine=week packagingÞ 0:5x1 + 0:4x2 + 0:5x3 + 0:4x4 + 0:2x5  480 ðh  employee=week pressingÞ 0:3x1 + 0:4x2 + 0:4x3 + 0:4x4 + 0:3x5  400 ðh  employee=week paintingÞ 0:5x1 + 0:5x2 + 0:3x3 + 0:4x4 + 0:3x5  320 ðh  employee=week moldingÞ 0:6x1 + 0:5x2 + 0:4x3 + 0:5x4 + 0:6x5  400 ðh  employee=week assemblyÞ 0:4x1 + 0:4x2 + 0:4x3 + 0:3x4 + 0:2x5  1, 280 ðh  employee=week packagingÞ 200  x1  1,000 ð min :demand; max :capac:refrigeratorÞ ð min :demand; max :capac:freezerÞ 50  x2  800 50  x3  500 ð min :demand; max :capac:stoveÞ ð min :demand; max :capac:dishwasherÞ 50  x4  500 ð min :demand; max :capac:microwaveÞ 40  x5  200 Ex. 10 xij ¼ liters of type i petroleum used daily to produce gasoline j, i ¼ 1, 2, 3, 4; j ¼ 1, 2, 3. x11 ¼ liters of petroleum 1 used daily to produce regular gasoline. ⋮ x41 ¼ liters of petroleum 4 used daily to produce regular gasoline. x12 ¼ liters of petroleum 1 used daily to produce green gasoline. ⋮ x42 ¼ liters of petroleum 4 used daily to produce green gasoline. x13 ¼ liters of petroleum 1 used daily to produce yellow gasoline. ⋮ x43 ¼ liters of petroleum 4 used daily to produce yellow gasoline.

Answers

Fobj ¼ max z ¼ ð0:40  0:20Þx11 + ð0:40  0:25Þx21 + ð0:40  0:30Þx31 + ð0:40  0:30Þx41 + ð0:45  0:20Þx12 + ð0:45  0:25Þx22 + ð0:45  0:30Þx32 + ð0:45  0:30Þx42 + ð0:50  0:20Þx13 + ð0:50  0:25Þx23 + ð0:50  0:30Þx33 + ð0:50  0:30Þx43 subject to 0:10x21  0:05x31 + 0:20x41  0 0:07x11 + 0:02x21  0:12x31  0:03x41  0  0:05x12 + 0:05x22  0:10x32  0:15x42  0 + 0:10x32  0:05x42  0 0:05x12  0:15x33 + 0:10x43  0  0:10x13 0:03x13  0:02x23 + 0:08x33  0:07x43  0 x11 + x21 + x31 + x41  12, 000 x12 + x22 + x32 + x42  10, 000 x13 + x23 + x33 + x43  8, 000 x11 + x12 + x13  15, 000 x21 + x22 + x23  15, 000 x31 + x32 + x33  15, 000 x41 + x42 + x43  15, 000 x11 + x21 + x31 + x41 + x12 + x22 + x32 + x42 + x13 + x23 + x33 + x43  60,000 x11 , x21 ,x31 , x41 ,x12 , x22 ,x32 , x42 ,x13 , x23 ,x33 , x43  0 Ex. 12 xi ¼ 1 if the company invests in project i 0 otherwise x1 ¼ if the company invests in the development of new products or not. x2 ¼ if the company invests in capacity building or not. x3 ¼ if the company invests in Information Technology or not. x4 ¼ if the company invests in expanding the factory or not. x5 ¼ if the company invests in expanding the depot or not. Fobj ¼ max z ¼ 355:627x1 + 110:113x2 + 213:088x3 + 257:190x4 + 241:833x5 subject to : 360x1 + 240x2 + 180x3 + 480x4 + 320x5  1,000 ðBudget constraintÞ 0 ðProject 2 depends on 3Þ x2  x3 ðMutually excluding projectsÞ x4 + x 5  1 xi ¼ 0 or 1 Ex. 13 xi ¼ percentage of stock i to be allocated in the portfolio, i ¼ 1, …, 10. x1 ¼ percentage of stock 1 from the banking sector to be allocated in the portfolio. x2 ¼ percentage of stock 2 from the banking sector to be allocated in the portfolio. ⋮ x10 ¼ percentage of stock 10 from the electrical sector to be allocated in the portfolio. Fobj ¼ 0:0439x1 + 0:0453x2 + 0:0455x3 + 0:0439x4 + 0:0402x5 + 0:0462x6 + 0:0421x7 + 0:0473x8 + 0:0233x9 + 0:0221x10 s:t: x1 + x2 + ⋯ + x10 ¼ 1 ð 1Þ 0:0122x1 + 0:0121x2 + ⋯ + 0:0148x10  0:008 ð2Þ ð 3Þ 0:0541x1 + 0:0528x2 + ⋯ + 0:0267x10  0:05 ð 4Þ x1 + x2 + x3 + x4 + x5  0:50 ð 5Þ x1 + x2 + x3 + x4  0:20 ð 6Þ x6 + x7 + x8  0:20 ð 7Þ x9 + x10  0:20 ð 8Þ 0  x1 ,x2 ,⋯,x10  0:40 Ex. 16 Decision variables: xijt ¼ quantity of product i to be manufactured in facility j in period t Iijt ¼ final  stock of product i in facility j in period t 1 if product i is delivered by facility j to retailer k in period t zijkt ¼ 0 otherwise

1125

1126

Answers

Model parameters: Dikt ¼ demand of product i by retailer k in period t cijt ¼ unit production cost of product i in facility j in period t iijt ¼ unit storage cost of product i in facility j in period t yijkt ¼ total transportation cost of product i from facility j to retailer k in period t xmax ijt ¼ maximum production capacity of product i in facility j in period t Imax ijt ¼ maximum storage capacity of product i in facility j in period t General formulation ! p m X n X T X X Fobj ¼ min z ¼ cijt xijt + iijt Iijt + yijkt zijkt i¼1 j¼1 t¼1

s:t:

k¼1

p X Dikt zijkt + Iijt ¼ Iij, t1 + xijt , k¼1

n X zijkt ¼ 1,

i ¼ 1, …, m; j ¼ 1,…, n; t ¼ 1, …, T

ð 1Þ

k ¼ 1, …,p;

ð2Þ

j¼1

xijt  xmax ijt , max Iijt  Iijt , zijkt 2 f0, 1g, xijt , Iijt  0

i ¼ 1, …,m; i ¼ 1, …,m; i ¼ 1, …,m; i ¼ 1, …,m;

j ¼ 1, …,n; j ¼ 1, …, n; j ¼ 1, …, n; j ¼ 1, …, n;

t ¼ 1, …, T t ¼ 1, …, T k ¼ 1, …,p; t ¼ 1, …,T t ¼ 1, …, T

Ex. 17 Decision variables: xijt ¼ quantity of product i to be manufactured in facility j in period t Iijt ¼ final stock of product i in facility j in period t Yijkt ¼quantity of product i to be transported from facility j to retailer k in period t 1 if the manufacturing of product i in period t occurs in facility j zijt ¼ 0 otherwise Model parameters: Dikt ¼ demand of product i by retailer k in period t cijt ¼ unit production cost of product i in facility j in period t iijt ¼ unit storage cost of product i in facility j in period t yijkt ¼ unit transportation cost of product i from facility j to retailer k in period t xmax ijt ¼ maximum production capacity of product i in facility j in period t Imax ijt ¼ maximum storage capacity of product i in facility j in period t General formulation ! p m X n X T X X cijt xijt + iijt Iijt + yijkt Yijkt minz ¼ i¼1 j¼1 t¼1

s:t: Iijt ¼ Iij, t1 + xijt  n X Yijkt ¼ Dikt ,

k¼1 p X

Yijkt , i ¼ 1,…, m; j ¼ 1,…, n; t ¼ 1,…, T

ð 1Þ

i ¼ 1, …,m; k ¼ 1, …,p; t ¼ 1,…, T

ð 2Þ

k¼1

j¼1

xijt 

p X Dikt zijt

k¼1 xijt  xmax ijt , max Iijt  Iijt ,

zijt 2 f0, 1g, xijt , Iijt , Yijt  0

i ¼ 1, …,m; j ¼ 1, …, n; t ¼ 1, …,T

ð 3Þ

i ¼ 1, …,m; j ¼ 1, …,n; i ¼ 1,…, m; j ¼ 1,…, n; i ¼ 1,…, m; j ¼ 1, …,n; i ¼ 1,…, m; j ¼ 1,…,n;

ð 4Þ ð 5Þ ð6Þ

Ex. 18 Time frame with T ¼ 6 periods, t ¼ 1, …, 6 (Jan., Feb., March, April, May, Jun.). Pt ¼ production in period t (kg) St ¼ production with outsourced labor in period t (kg)

t ¼ 1, …, T t ¼ 1, …,T t ¼ 1, …,T t ¼ 1, …,T

ð 3Þ ð 4Þ ð 5Þ

Answers

NRt ¼ number of regular employees in period t NCt ¼ number of employees hired from period t  1 to period t NDt ¼ number of employees fired from period t  1 to period t HEt ¼ total amount of overtime in period t It ¼ final stock in period t (kg) minz ¼ 1:5P1 + 2S1 + 600NR1 + 1, 000NC1 + 900ND1 + 7HE1 + 1I1 + 1:5P2 + 2S2 + 600NR2 + 1,000NC2 + 900ND2 + 7HE2 + 1I2 + ⋮ ⋮ 1:5P6 + 2S6 + 600NR6 + 1, 000NC6 + 900ND6 + 7HE6 + 1I6 s:t: I1 ¼ 600 + P1  9, 600 I2 ¼ I1 + P2  10, 600 ⋮ ⋮ I6 ¼ I5 + P6  10, 430

ANSWER KEYS: EXERCISES: CHAPTER 17 Section 17.2.1 (ex.2) a) Optimal solution: x1 ¼ 2, x2 ¼ 1 and z ¼ 10 b) Optimal solution: x1 ¼ 1, x2 ¼ 4 and z ¼ 14 c) Optimal solution: x1 ¼ 10, x2 ¼ 6 and z ¼ 52 Section 17.2.1 (ex.4) a) yes b) no c) yes d) no e) yes f) yes g) no h) no i) yes Section 17.2.2 (ex.2) a) Optimal solution: x1 ¼ 12, x2 ¼ 2 and z ¼ 26 b) Optimal solution: x1 ¼ 18, x2 ¼ 8 and z ¼ 28 c) Optimal solution: x1 ¼ 10, x2 ¼ 10 and z ¼ 100 Section 17.2.3 (ex.1) e) Multiple optimal solutions. f) There is no optimal solution. g) Unlimited objective function z. h) Multiple optimal solutions. i) Degenerate optimal solution. j) There is no optimal solution. Section 17.2.3 (ex.2) a) Any point of the segment CD (C (10, 30); D (0, 45)). b) Any point of the segment AB (A (8, 0); B (7/2, 3)). Section 17.3 (ex.1) a) Six basic solutions. c) Optimal solution: x1 ¼ 5, x2 ¼ 20 and z ¼ 55 Section 17.3 (ex.2) a) Ten basic solutions. c) Optimal solution: x1 ¼ 7, x2 ¼ 11, x3 ¼ 0 and z ¼ 61 Section 17.4.2 (ex.1) a) Optimal solution: x1 ¼ 1, x2 ¼ 17, x3 ¼ 5 and z ¼ 104

1127

1128

Answers

Section 17.4.3 (ex.2) a) Optimal solution: x1 ¼ 3, x2 ¼ 3 and z ¼ 15 b) Optimal solution: x1 ¼ 2, x2 ¼ 4, x3 ¼ 0 and z ¼ 20 c) Optimal solution: x1 ¼ 4, x2 ¼ 0, x3 ¼ 12 and z ¼ 36 Section 17.4.4 (ex.1) a) Optimal solution: b) Optimal solution: c) Optimal solution: d) Optimal solution:

x1 ¼ 0, x2 ¼ 4 and z ¼  4 x1 ¼ 1, x2 ¼ 7 and z ¼  37 x1 ¼ 0, x2 ¼ 10, x3 ¼ 35/2 and z ¼  55/2 x1 ¼ 100/3, x2 ¼ 0, x3 ¼ 40/3 and z ¼  140/3

Section 17.4.5.1 (ex.1) b) Solution 1: x1 ¼ 115/2, x2 ¼ 0 and z ¼ 230 Solution 2: x1 ¼ 60, x2 ¼ 10 and z ¼ 230 Section 17.4.5.1 (ex.2) b) Solution 1: x1 ¼ 310, x2 ¼ 0 and z ¼ 930 Solution 2: x1 ¼ 30, x2 ¼ 140 and z ¼ 930 Section 17.4.5.2 (ex.2) Solution 1: x1 ¼ 10, x2 ¼ 30 Solution 2: x1 ¼ 30, x2 ¼ 0 Section 17.4.5 (ex.1) a) Multiple optimal solutions. b) Unlimited objective function z. c) Multiple optimal solutions/degenerate optimal solution. Section 17.4.5 (ex.2) a) No. b) Unfeasible solution. c) Degenerate optimal solution. d) Multiple optimal solutions. e) Unlimited objective function z. Section 17.5.2 (ex.1) b) Optimal solution: x1 ¼ 70, x2 ¼ 30, x3 ¼ 35 and z ¼ 363, 000. Section 17.5.2 (ex.2) b) Optimal solution: x1 ¼ 24,960, x2 ¼ 17,040 and z ¼ 19,296. Section 17.5.2 (ex.3) b) Optimal solution: x1 ¼ 475, x2 ¼ 50, x3 ¼ 50, x4 ¼ 50, x5 ¼ 75 and z ¼ 32,475. Section 17.5.2 (ex.4) b) Optimal solution:

x11 ¼ 3,600, x21 ¼ 0, x22 ¼ 10,000 x12 ¼ 0, x23 ¼ 0, x13 ¼ 0, z ¼ 5,160

x31 ¼ 0, x41 ¼ 8,400, x32 ¼ 0, x42 ¼ 0, x33 ¼ 3,200, x43 ¼ 4,800 and

Section 17.5.2 (ex.5) b) Optimal solution: x1 ¼ 1, x2 ¼ 0, x3 ¼ 1, x4 ¼ 0, x5 ¼ 1 and z ¼ 810,548 ($810,548.00). Section 17.5.2 (ex.6) b) Optimal solution: x1 ¼ 20%, x7 ¼ 20%, x9 ¼ 20%, x10 ¼ 40%, x2, x3, x4, x5, x6, x8, x11 ¼ 0% and z ¼ 3.07%. Section 17.5.2 (ex.7) b) Optimal solution: 50% ($250,000.00) in the RF_C fund 25% ($125,000.00) in the Petrobras stock fund 25% ($125,000.00) in the Vale stock fund Objective function z ¼ 16.90% per year.

Answers

1129

Section 17.5.2 (ex.8) b) Optimal solution: z ¼ 126,590 ($126,590.00). Solution

Jan.

Feb.

Mar.

Apr.

May

Jun.

Pt St NRt NCt NDt HEt It

9600 0 5 0 5 0 600

10,000 0 5 0 0 28.57 0

12,800 0 6 1 0 91.43 0

11,520 0 6 0 0 0 870

10,770 0 5 0 1 83.57 0

10,430 0 5 0 0 59.29 0

Section 17.6.1 (ex.1) a) x1 ¼ 60, x2 ¼ 20 with z ¼ 520 b) 1.333 c) 0.8 d) No. e) The basic solution remains optimal. Section 17.6.1 (ex.2) a) x1 ¼ 15, x2 ¼ 0 with z ¼ 120 b) c1 2.4 or c1 c01  5.6 c) c2  20 or c2  c02 + 14 Section 17.6.1 (ex.3) a) x1 ¼ 0, x2 ¼ 17 with z ¼ 102 b) Unlimited objective function z. c) c1 3 or c1 c01  5 d) 0  c2  16 or c02  6  c2  c02 + 10 Section 17.6.1 (ex.4) a) 0:133  cc1  0:25 2 b) The basic solution remains optimal with z ¼ 1,700. c) 8  c1  15 or c01  4  c1  c01 + 3 d) 48  c2  90 or c02  12  c2  c02 + 30 e) The basic solution remains optimal with z ¼ 1,830. f) The basic solution remains optimal with z ¼ 2,440. g) 13.333  c1  25 Section 17.6.2 (ex.1) a) P1 ¼ 0, P2 ¼ 34.286, P3 ¼ 85.714 b) b1 b01  8.5 b02  5.95  b2  b02 + 6.125 b03  3.267  b3  b03 + 2.164 c) 0 d) $137.14 (z¼ 1,902.86), x1 ¼ 115.71 and x2 ¼ 8.57 Section 17.6.2 (ex.2) a) P1 ¼ 0, P2 ¼ 1.222, P3 ¼ 0.444 (2nd operation) b) b1 b01  20 b02  180  b2  b02 + 22.5 b03  36  b3  b03 + 180 c) $27.50 d) $16.00 Section 17.6.3 (ex.1) b) z11 ¼ 3, z12 ¼ 65, z∗1 ¼ 3 and z∗2 ¼ 2 Section 17.6.3 (ex.2) b) z∗1 ¼  4 and z∗2 ¼  2

1130

Answers

Section 17.6.4 (ex.3) a) Degenerate optimal solution. b) Multiple optimal solutions. c) Degenerate optimal solution. d) Multiple optimal solutions. e) Multiple optimal solutions. f) Degenerate optimal solution. g) Degenerate optimal solution.

ANSWER KEYS: EXERCISES: CHAPTER 18 Ex.1 a) b) c) d) e) f) g) h)

N ¼ {1, 2, 3, 4, 5, 6} A ¼ {(1, 2), (1, 3), (2, 3), (3, 4), (3, 5), (4, 2), (4, 5), (4, 6), (5, 6)} Directed network. 1!2!3!4!2 1!3!5!4 1!3!4!6 2!3!4!2 3!4!5!3

Ex.2 a) b) c) d) e) f) g) h)

N ¼ {1, 2, 3, 4, 5, 6} A ¼ {(1, 2), (1, 3), (2, 3), (2, 4), (3, 5), (4, 6), (5, 2), (5, 4), (6, 5)} Directed network. 2!3!5!4!6!5 1!2!5!4!6!5 1!3!5!4 2!3!5!2 1!2!3!1

Ex.3 a) Tree

1

3

5 4

b) Cover tree

1

3

5 2

4

6

Answers

1131

Ex.4

3

5

1

7

2

8 4

6

Ex.5 Classic transportation problem:

4

4

6

60

3

50

4

50

8 8

6

5 7

5

3

50

2

9

2

80

40

8

1

70

1

9

Optimal FBS: x11 ¼ 40, x14 ¼ 30, x22 ¼ 60, x24 ¼ 20, x33 ¼ 50 with z ¼ 1, 110. Ex.6 Maximum flow problem: 8

2 6

6

6

3

7

1

5 6

4

7 4

6 5

3

3

6

7

Optimal solution: x12 ¼ 6, x13 ¼ 2, x14 ¼ 7, x24 ¼ 3, x25 ¼ 3, x34 ¼ 2, x36 ¼ 0, x45 ¼ 3, x46 ¼ 3, x47 ¼ 6, x57 ¼ 6, x67 ¼ 3 with z ¼ 15. Ex.7 Shortest route problem:

3

2 1

4

5

4

1

7

3

5

6

4

4

3

5

2

8

4

4

5

6

Optimal FBS: x13 ¼ 1, x36 ¼ 1, x68 ¼ 1 (1  3  6  8) with z ¼ 11. Ex.8 x11 ¼ 50, x22 ¼ 10, x23 ¼ 20, x33 ¼ 20.

3

1

1132

Answers

Ex.9 x11 ¼ 80, x13 ¼ 70, x22 ¼ 50, x23 ¼ 80 with z ¼ 4,590. Ex.10 x13 ¼ 150, x21 ¼ 80, x22 ¼ 50 with z ¼ 4,110. Ex.11 a) Optimal FBS: x12 ¼ 100, x13 ¼ 100, x23 ¼ 100, x31 ¼ 150, x32 ¼ 50 with z ¼ 6,800. b) Optimal FBS: x13 ¼ 50, x31 ¼ 100, x41 ¼ 20, x42 ¼ 150, x43 ¼ 30 with z ¼ 1,250. c) Optimal FBS: x12 ¼ 20, x14 ¼ 30, x21 ¼ 20, x24 ¼ 10, x32 ¼ 20, x33 ¼ 60 with z ¼ 1,490. Alternative solution: x11 ¼ 20, x12 ¼ 20, x14 ¼ 10, x24 ¼ 30, x32 ¼ 20, x33 ¼ 60 with z ¼ 1,490. Ex.12 Indexes: Suppliers i 2 I Consolidating centers j 2 J Factory k 2 K Products p 2 P Model parameters: Cmax, j Dpk Sip cpij cpjk cpik

maximum capacity of consolidating center j. demand of product p in factory k. capacity of supplier i to produce product p. unit transportation cost of p from supplier i to consolidating center j. unit transportation cost of p from consolidating center j to factory k. unit transportation cost of p from supplier i to factory k.

Model’s decision variables: xpij ypjk zpik

amount of product p transferred from supplier i to consolidating center j. amount of product p transferred from consolidating center j to factory k. amount of product p transferred from supplier i to factory k.

The problem can be formulated as follows: XXX XXX XXX min cpij xpij + cpjk ypjk + cpik zpik p

s.t.:

i

p

j

X j

ypjk +

XX p

X j

i

xpij +

X i

j

X

p

k

zpik ¼ Dpk ,

i

k

8p,k

(1)

8j

(2)

8i,p

(3)

i

xpij  C max , j , X k

xpij ¼

zpik  Sip ,

X

ypjk ,

8p, j

(4)

8p,i, j, k

(5)

k

xpij , ypjk , zpik  0,

In the objective function, the first term represents suppliers’ transportation costs up to the consolidation terminals, the second refers to the transportation costs from the consolidation terminals to the final client (factory in Harbin), and the third represents suppliers’ transportation costs directly to the final client. Constraint (1) ensures that client k’s demand for product p will be met. Constraint (2) refers to the maximum capacity of each consolidation terminal. Constraint (3) represents supplier i’s capacity to supply product p. Whereas constraint (4) refers to the preservation of the input and output flows in each transshipment point. Finally, we have the non-negativity constraints.

Answers

Ex.13

 xij ¼

1 if task i is designated to machine j, i ¼ 1,…, 4, j ¼ 1, …, 4 0 otherwise

a) Optimal FBS: x12 ¼ 1, x24 ¼ 1, x33 ¼ 1, x41 ¼ 1 with z ¼ 37. b) Optimal FBS: x13 ¼ 1, x24 ¼ 1, x33 ¼ 1, x41 ¼ 1 with z ¼ 35. Ex.14

 xij ¼

1 if route ði, jÞ is included in the shortest route, 8i, j 0 otherwise

min 6x12 + 9x13 + 4x23 + 4x24 + 7x25 + 6x35 + 2x45 + 7x46 + 3x56 s:t: x12 + x13 ¼ 1 x46 + x56 ¼ 1 x12  x23  x24  x25 ¼ 0 x13 + x23  x35 ¼ 0 x24  x45  x46 ¼ 0 x25 + x35 + x45  x56 ¼ 0 xij 2 f0, 1g or xij  0 Optimal FBS: x12 ¼ 1, x24 ¼ 1, x45 ¼ 1, x56 ¼ 1 (1  2  4  5  6) with z ¼ 15. Ex.15 Optimal FBS: xAB ¼ 1, xBD ¼ 1, xDE ¼ 1 (A  B  D  E) with z ¼ 64. Ex.16 x12 ¼ 6, x13 ¼ 4, x23 ¼ 0, x24 ¼ 6, x34 ¼ 1, x35 ¼ 3, x45 ¼ 0, x46 ¼ 7, x56 ¼ 3 with z ¼ 10.

ANSWER KEYS: EXERCISES: CHAPTER 19 Section 19.1 (ex.1) a) BP b) MIP c) IP d) BIP e) BP f) MBP g) MIP Section 19.2 (ex.1) a) No b) Yes (x1 ¼ 10,x2 ¼ 0 with z ¼ 20) c) No d) Yes (x1 ¼ 0, x2 ¼ 4 with z ¼ 32) e) Yes (x1 ¼ 1,x2 ¼ 0 with z ¼ 4) f) No g) Yes (x1 ¼ 6,x2 ¼ 5 with z ¼ 58) Section 19.2((ex.2) ) ð0, 0Þ;ð0, 1Þ;ð0, 2Þ;ð0, 3Þ;ð0, 4Þ;ð1, 0Þ;ð1, 1Þ;ð1, 2Þ;ð1, 3Þ;ð2, 0Þ;ð2, 1Þ; b) SF ¼ ð2, 2Þ;ð2, 3Þ;ð3, 0Þ;ð3, 1Þ;ð3, 2Þ;ð4, 0Þ;ð4, 1Þ;ð4, 2Þ;ð5, 0Þ;ð5, 1Þ;ð6, 0Þ c) Optimal solution: x1 ¼ 4 and x2 ¼ 2 with z ¼ 14. Section 19.2 (ex.3) b) SF ¼ {(0, 0); (0, 1); (0, 2); (1, 0); (1, 1); (2, 0); (2, 1); (3, 0)} c) Optimal solution: x1 ¼ 2 and x2 ¼ 1 with z ¼ 4.

1133

1134

Answers

Section 19.2 (ex.4) b) {(0, 0); (0, 1); (0, 2); (1, 0); (1, 1); (1, 2); (2, 0); (2, 1); (2, 2); (3, 0); (3, 1); (3, 2); (4, 0)} c) Optimal solution: x1 ¼ 3 and x2 ¼ 2 with z ¼ 13. Section 19.3 (ex.1) Optimal FBS ¼ {x3 ¼ 1, x4 ¼ 1, x6 ¼ 1, x8 ¼ 1} with z ¼ 172. Section 19.4 (ex.1) max z ¼ 7x1 + 12x2 + 8x3 + 10x4 + 7x5 + 6x6 s:t: 4x1 + 7x2 + 5x3 + 6x4 + 4x5 + 3x6  20 x5 + x6  1 x3  x2  0 x1 , x2 , x3 , x4 ,x5 ,x6 2 f0, 1g Optimal solution: x1 ¼ 1, x2 ¼ 1, x3 ¼ 0, x4 ¼ 1, x5 ¼ 0, x6 ¼ 1 with z ¼ 35. Section 19.5 (ex.1) Indexes i, j ¼ 1, .., n that represent the customers (index 0 represents the depot) v ¼ 1, …, NV that represent the vehicles Parameters Cmax,v ¼ maximum capacity of vehicle n di ¼ demand of client i cij ¼ travel cost from client i to client j Decision variables

 xvij

¼ 

yvi Model formulation min s.t.

¼

PPP i

j

1 if the arc from i to j is traveled by vehicle v 0 otherwise 1 if order of client i is delivered by vehicle v 0 otherwise

v v cij xij

X

yvi ¼ 1,

i ¼ 1,…, n

(1)

v

X

X X

yvi ¼ NV,

i¼0

(2)

v

di yvi  C max , v , v ¼ 1, …, NV

(3)

i

xvij ¼ yvj ,

j ¼ 0,…, n,

v ¼ 1, …, NV

(4)

xvij ¼ yvi ,

i ¼ 0,…, n,

v ¼ 1, …, NV

(5)

i

X X

j

xvij ¼ xvij  jSj  1, S f1, …, ng, 2  jSj  n  1, v ¼ 1, …, NV

(6)

ij2S

xvij 2 f0, 1g, i ¼ 0, …, n j ¼ 0,…,n, v ¼ 1, …, NV

(7)

yvi 2 f0, 1g, i ¼ 0, …,n, v ¼ 1,…, NV

(8)

The main objective of the model is to minimize the total travel costs. Constraint (1) guarantees that each node (client) will be visited by only one vehicle. Whereas constraint (2) guarantees that all the routes will begin and end at the depot (i ¼ 0).

Answers

1135

Constraint (3) guarantees that vehicle capacity will not be exceeded. Constraints (4) and (5) guarantee that vehicles will not interrupt their routes at one client. They are the constraints for the preservation of the input and output flows. Constraint (6) guarantees that subroutes will not be formed. Finally, constraints (7) and (8) guarantee that variables xvij and yvi will be binary. Section 19.6 (ex.1) Indexes i ¼ 1, …, m that represent the distribution centers (DCs) j ¼ 1, …, n that represent the consumers Model parameters fi ¼ fixed costs to maintain DC i open cij ¼ transportation costs from DC i to consumer j Dj ¼ demand of customer j Cmax, i ¼ maximum capacity of DC i Decision variables

( yi ¼ ( xij ¼

1 if DC i opens 0 otherwise

1 if consumer j is supplied by DC i 0 otherwise

General formulation Fobj ¼ min z ¼

m X

fi yi +

i¼1

s:t:

n X

m X n X

cij xij Dj

i¼1 j¼1

xij Dj  Cmax,i  yi ,

i ¼ 1, …,m

ð 1Þ

j ¼ 1,…, n

ð 2Þ

j¼1 m X xij ¼ 1, i¼1

xij ,yi 2 f0, 1g,

i ¼ 1,…, m, j ¼ 1,…, n

ð3Þ

which corresponds to a binary programming problem. For this problem, index i corresponds to: i ¼ 1 (Belem), i ¼ 2 (Palmas), i ¼ 3 (Sao Luis), i ¼ 4 (Teresina), and i ¼ 5 (Fortaleza); and index j corresponds to: j ¼ 1 (Belo Horizonte), j ¼ 2 (Vitoria), j ¼ 3 (Rio de Janeiro), j ¼ 4 (Sao Paulo), and j ¼ 5 (Campo Grande). Optimal FBS: x22 ¼ 1, x24 ¼ 1, x45 ¼ 1, x51 ¼ 1, x53 ¼ 1, y2 ¼ 1, y4 ¼ 1, y5 ¼ 1 with z ¼ 459,400.00. Section 19.6 (ex.2) Indexes: Suppliers i 2 I Consolidating centers j 2 J Factory k 2 K Products p 2 P Model parameters: Cmax, j fj Dpk Sip cpij cpjk cpik

maximum capacity of consolidating center j. fixed costs to open consolidating center j. demand of product p in factory k. capacity of supplier i to produce product p. unit transportation cost of p from supplier i to consolidating center j. unit transportation cost of p from consolidating center j to factory k. unit transportation cost of p from supplier i to factory k.

1136

Answers

Model’s decision variables: xpij ypjk zpik zj

amount of product p transported from supplier i to consolidating center j. amount of product p transported from consolidating center j to factory k. amount of product p transported from supplier i to factory k. binary variable that assumes value 1 if center j operates, and 0 otherwise.

The problem can be formulated as follows: XXX XXX XXX X min cpij xpij + cpjk ypjk + cpik zpik + f j zj p

i

p

j

s.t.:

X

j

ypjk +

j

XX p

X j

X

k

j

8p,k

(1)

8j

(2)

8i, p

(3)

xpij  C max , j  zj ,

xpij +

X k

xpij ¼

i

zpik ¼ Dpk ,

i

i

i

X

p

k

zpik  Sip ,

X

8p, j

(4)

8p, i, j, k

(5)

8z

(6)

ypjk ,

k

xpij , ypjk , zpik  0, zj 2 f0, 1g,

In the objective function, the first term represents suppliers’ transportation costs up to the consolidation terminals, the second refers to the transportation costs from the consolidation terminals to the final client (factory in Harbin), and the third represents suppliers’ transportation costs directly to the final client, and the last one the fixed costs related to the consolidation terminals’ location. Constraint (1) ensures that client k’s demand for product p will be met. Constraint (2) refers to the maximum capacity of each consolidation terminal. Constraint (3) represents supplier i’s capacity to supply product p. Whereas constraint (4) refers to the preservation of the input and output flows in each transshipment point. Finally, we have the non-negativity constraints and that variable zj is binary. Section 19.7 (ex.1) xi ¼ number of buses that start working in shift i, i ¼ 1, 2, …, 9.

Therefore, we have: x1 ¼ number of buses x2 ¼ number of buses x3 ¼ number of buses x4 ¼ number of buses x5 ¼ number of buses

that that that that that

start start start start start

working working working working working

at at at at at

Shift

Period

1 2 3 4 5 6 7 8 9

6:01–14:00 8:01–16:00 10:01–18:00 12:01–20:00 14:01–22:00 16:01–24:00 18:01–02:00 20:01–04:00 22:01–06:00

6:01. 8:01. 10:01. 12:01. 14:01.

Answers

x6 ¼ number x7 ¼ number x8 ¼ number x9 ¼ number

of of of of

buses buses buses buses

that that that that

start start start start

working working working working

at at at at

1137

16:01. 18:01. 20:01. 22:01.

Fobj ¼ min z ¼ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 subject to  20 ð6 : 01  8 : 00Þ x1 x1 + x2  24 ð8 : 01  10 : 00Þ  18 ð10 : 01  12 : 00Þ x1 + x 2 + x 3 x1 + x2 + x3 + x4  15 ð12 : 01  14 : 00Þ x2 + x3 + x4 + x5  16 ð14 : 01  16 : 00Þ  27 ð16 : 01  18 : 00Þ x3 + x4 + x5 + x6 x4 + x5 + x6 + x7  18 ð18 : 01  20 : 00Þ  12 ð20 : 01  22 : 00Þ x5 + x6 + x7 + x8 x6 + x7 + x8 + x9  10 ð22 : 01  24 : 00Þ x7 + x8 + x9  4 ð00 : 01  02 : 00Þ  3 ð02 : 01  04 : 00Þ x8 + x 9 x9  8 ð04 : 01  06 : 00Þ  0, i ¼ 1, 2,…, 9 xi Optimal solution: x1 ¼ 24, x2 ¼ 0, x3 ¼ 0, x4 ¼ 0, x5 ¼ 16, x6 ¼ 11, x7 ¼ 0, x8 ¼ 0, x9 ¼ 8 with z ¼ 59. Section 19.7 (ex.2) xi ¼ number of employees that start working on day i, i ¼ 1, 2, …, 7. x1 ¼ number of employees that start working on Monday. x2 ¼ number of employees that start working on Tuesday. ⋮ x7 ¼ number of employees that start working on Sunday. min z ¼ x1 + x2 + x3 + x4 + x5 + x6 + x7 subject to + x4 + x5 + x6 + x7  15 ðMondayÞ x1 + x5 + x6 + x7  20 ðTuesdayÞ x1 + x2 x6 + x7  17 ðWednesdayÞ x1 + x2 + x3 + x7  22 ðThursdayÞ x1 + x2 + x3 + x4 + x1 + x2 + x3 + x4 + x5  25 ðFridayÞ  15 ðSaturdayÞ x2 + x3 + x4 + x5 + x6 x3 + x4 + x5 + x6 + x7  10 ðSundayÞ xi  0, i ¼ 1,…, 7 Alternative optimal solution: x1 ¼ 10, x2 ¼ 6, x3 ¼ 0, x4 ¼ 5, x5 ¼ 4, x6 ¼ 0, x7 ¼ 1 with z ¼ 26.

ANSWER KEYS: EXERCISES: CHAPTER 20 4) P (I < 0) ¼ 15.92% by using the NORM.DIST function in Excel or P (I < 0) ¼ 12.19% analyzing the data generated in the Monte Carlo simulation for variable I. Note: The results can change at each new simulation. 5) P (Index > 0.07) ¼ 22.50% by using the NORM.DIST function in Excel or P (Index > 0.07) ¼ 20.43% analyzing the values generated in the simulation. Note: The results can change at each new simulation.

ANSWER KEYS: EXERCISES: CHAPTER 21 1) Fcal ¼ 2.476 (sig. 0.100), that is, there are no differences in the production of helicopters in the three factories. 2) There are no significant differences between the hardness measures of the different converters. That is, the “Type of Converter” factor does not have a significant effect on the variable “Hardness.” On the other hand, we can conclude

1138

Answers

that there are significant differences in the hardness of the different types of ore. That is, the “Type of Ore” factor has a significant effect on the variable “Hardness.” We can also conclude that there is a significant interaction between the two factors. Tests of Between-Subjects Effects Dependent Variable: Hardness Source Corrected Model Intercept Converter Ore Converter * Ore Error Total Corrected Total

Type III Sum of Squares a

15006.222 2023032.111 66.074 14433.852 506.296 3250.667 2041289.000 18256.889

df

Mean Square

F

Sig.

8 1 2 2 4 72 81 80

1875.778 2023032.111 33.037 7216.926 126.574 45.148

41.547 44808.751 .732 159.850 2.804

.000 .000 .485 .000 .032

R Squared ¼ .822 (Adjusted R Squared ¼ .802).

a

3) There are significant differences between the octane rating indexes of the different types of petroleum and between the octane rating indexes of the different oil refining processes. That is, both factors have a significant effect on the octane rating index. Finally, we can conclude that there is significant interaction between the two factors. Tests of Between-Subjects Effects Dependent Variable: Octane Rating Source Corrected Model Intercept Petroleum Refining Petroleum * Refining Error Total Corrected Total

Type III Sum of Squares a

450.229 399857.521 402.792 31.729 15.708 35.250 400343.000 485.479

R Squared ¼ .927 (Adjusted R Squared ¼ .905).

a

ANSWER KEYS: EXERCISES: CHAPTER 22 1) a) Control charts for X UCL ¼ 17.4035 Average ¼ 16.5318 LCL ¼ 15.6600 Control charts for R UCL ¼ 2.7305 Average ¼ 1.1965 LCL ¼ 0.0000 b) Cp ¼ 0.860 Cpk ¼ 0.842 Cpm ¼ 0.859

df

Mean Square

F

Sig.

11 1 2 3 6 36 48 47

40.930 399857.521 201.396 10.576 2.618 .979

41.801 408365.128 205.681 10.801 2.674

.000 .000 .000 .000 .030

Answers

2) a) Control charts for X UCL ¼ 17.3895 Average ¼ 16.5318 LCL ¼ 15.6740 Control charts for S UCL ¼ 1.1938 Average ¼ 0.5268 LCL ¼ 0.0000 b) Cp ¼ 0.9491 Cpk ¼ 0.9290 3) a) Control charts for X UCL ¼ 6.7113 Average ¼ 6.0625 LCL ¼ 5.4137 Control charts for R UCL ¼ 2.0322 Average ¼ 0.8905 LCL ¼ 0.0000 b) Cp ¼ 0.771 Cpk ¼ 0.722 Cpm ¼ 0.542 4) a) Control charts for X UCL ¼ 6.7162 Average ¼ 6.0625 LCL ¼ 5.4088 Control charts for S UCL ¼ 0.9098 Average ¼ 0.4015 LCL ¼ 0.0000 b) Cp ¼ 0.8302 Cpk ¼ 0.7783 5) P chart UCL ¼ 0.1748 Average ¼ 0.0680 LCL ¼ 0.0000 6) UCL ¼ 8.7403 Average ¼ 3.4000 LCL ¼ 0.0000 7) UCL ¼ 11.9996 Average ¼ 5.1750 LCL ¼ 0.0000

1139

1140

Answers

8) Control chart: defects defects UCL Center = 1.1357 LCL

Fraction of nonconformities

2.0

1.5

1.0

0.5

24.00 23.00 22.00 21.00 20.00 19.00 18.00 17.00 16.00 15.00 14.00 13.00 12.00 11.00 10.00 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 100

0.0

Sigma level:

ANSWER KEYS: EXERCISES: CHAPTER 23 1) a)

3

Answers

In fact, this is a balanced clustered data structure. b)

1141

1142

Answers

c)

d) Yes. Since the estimation of variance component τ00, which corresponds to random intercept u0j, is considerably higher than its standard error, it is possible to verify that there is variability, at a significance level of 0.05, in the score obtained between students from different countries. Statistically, z ¼ 422.619/125.284 ¼ 3.373 > 1.96, where 1.96 is the critical value of the standard normal distribution, which results in a significance level of 0.05. e) Since Sig. w2 ¼ 0.000, it is possible to reject the null hypothesis that the random intercepts are equal to zero (H0: u0j ¼ 0), which makes the estimation of a traditional linear regression model be ruled out for these clustered data. f) rho ¼

τ00 422:619 ¼ 0:974 ¼ τ00 + s2 422:619 + 11:196

which suggests that approximately 97% of the total variance in students’ grades in science is due to differences between the participants’ countries of origin. g)

Answers

1143

h)

i) The parameters estimated of the fixed- and random-effects components are statistically different from zero, at a significance level of 0.05. j)

1144

Answers

k)

l)

The significance level of the test is equal to 1.000 (much greater than 0.05) because the logarithms of both restricted likelihood functions are identical (LLr ¼ 357.501), the model with only random effects in the intercept is favored, since random error terms u1j are statistically equal to zero.

Answers

m)

n) scoreij ¼ 13:22 + 0:0028  incomeij + 0:0008  resdevelj  incomeij + u0j + rij o)

1145

1146

Answers

2) a)

In fact, this is an unbalanced clustered data structure of real estate properties in districts. b)

This is also an unbalanced data panel. c)

Answers

d)

e)

1147

1148

Answers

f)

g) l

Level-2 intraclass correlation:

rhoproperty|district ¼

l

τu000 + τr000 0:1228 + 0:0368 ¼ 0:996 ¼ 2 τu000 + τr000 + s 0:1228 + 0:0368 + 0:0007

Level-3 intraclass correlation: rhodistrict ¼

τu000 0:1228 ¼ ¼ 0:766 2 τu000 + τr000 + s 0:1228 + 0:0368 + 0:0007

The correlation between the natural logarithms of the rental prices per square meter of the properties in the same district is equal to 76.6% (rhodistrict), and the correlation between these annual indexes, for the same property of a certain district, is equal to 99.6% (rhoproperty j district). Thus, we estimate that the real estate and districts random effects form more than 99% of the total variance of the residuals! h) Given the statistical significance of the estimated variances τu000, τr000, and s2 (relationships between values estimated and respective standard errors higher than 1.96, and this is the critical value of the standard normal distribution which results in a significance level of 0.05), we can state that there is variability in the rental price of the commercial properties throughout the period analyzed. Moreover, there is variability in the rental price, throughout time, between real estate properties in the same district and between properties located in different districts.

Answers

1149

i) Since Sig. w2 ¼ 0.000, it is possible to reject the null hypothesis that the random intercepts are equal to zero (H0: u00k ¼ r0jk ¼ 0), which makes the estimation of a traditional linear regression model be ruled out for these data. j)

k) First, we can see that the variable that corresponds to the year (linear trend) with fixed effects is statistically significant, at a significance level of 0.05 (Sig. z ¼ 0.000 < 0.05), which demonstrates that, each year, rental prices of commercial properties increase, on average, 1.10% (e0.011 ¼ 1.011), ceteris paribus. In relation to the random-effects components, it is also possible to verify that there is statistical significance in the variances of u00k, r0jk, and etjk, at a significance level of 0.05, because the estimations of τu000, τr000, and s2 are considerably higher than the respective standard errors.

1150

l)

Answers

Answers

1151

m)

n) l

Level-2 intraclass correlation:

rhoproperty|district ¼ ¼

l

τu000 + τu100 + τr000 + τr100 τu000 + τu100 + τr000 + τr100 + s2 0:142444 + 0:000043 + 0:039638 + 0:000047 ¼ 0:9994 0:142444 + 0:000043 + 0:039638 + 0:000047 + 0:000103

Level-3 intraclass correlation:

rhodistrict ¼ ¼

τu000 + τu100 τu000 + τu100 + τr000 + τr100 + s2 0:142444 + 0:000043 ¼ 0:7817 0:142444 + 0:000043 + 0:039638 + 0:000047 + 0:000103

For this model, we estimate that the real estate and districts random effects form more than 99.9% of the total variance of the residuals!

1152

Answers

o)

Since Sig. w22 ¼ 0.000, we choose the linear trend model with random intercepts and slopes. p)

q) ln ðpÞtjk ¼ 4:134 + 0:015  year jk + 0:231  foodjk + 0:189  space4jk  0:004  valetjk  year jk + u00k + u10k  year jk + r 0jk + r 1jk  year jk + etjk Note: At this moment, we decide to insert the parameter of the variable space4 in the expression, statistically significant at a significance level of 0.10.

Answers

1153

r) Yes, it is possible to state that the natural logarithm of the rental price per square meter of the real estate properties follows a linear trend throughout time. In addition, there is a significant variance in the intercepts and slopes between those located in the same district and between those located in different districts. Yes, the existence of restaurants or food courts in the building, at least four or a higher number of parking spaces, and valet parking in the building where the property is located explain part of the evolution in variability of the natural logarithm of the rental price per square meter of the properties. s)

t) l

Random-effects variance-covariance matrix for level district: 

   u00k 0:037004 0 var ¼ u10k 0 0:000016

1154

l

Answers

Random-effects variance-covariance matrix for level property: 

   r0jk 0:030961 0 ¼ var r1jk 0 0:000044 u)

v) l

Random-effects variance-covariance matrix for level district:  var

   u00k 0:037253 0:000653 ¼ u10k 0:000653 0:000014

Answers

l

1155

Random-effects variance-covariance matrix for level property: 

   r0jk 0:031679 0:000484 ¼ var r1jk 0:000484 0:000046 w)

Since Sig. w22 ¼ 0.000, the structure of the random-terms variance-covariance matrices is considered unstructured, that is, we can conclude that error terms u00k and u10k are correlated (cov(u00k , u10k) 6¼ 0), and that error terms r0jk and r1jk are also correlated (cov(r0jk , r1jk) 6¼ 0). x) ln ðpÞtjk ¼ 3:7807 + 0:0144  year jk + 0:2314  food jk + 0:2071  space4jk + 0:5111  subwayk  0:0031  valetjk  year jk  0:0072  subwayk  year jk + 0:0001  violencek  year jk + u00k + u10k  year jk + r 0jk + r 1jk  year jk + etjk y) Yes, it is possible to state that the existence of subway and the violence index in the district explain part of the variability of the evolution of the natural logarithm of the rental price per square meter between real estate properties located in different districts. z)