$\begin{array}{l}{F}_{\mathrm{obj}}=minz=\sum _{i=1}^m\sum _{j=1}^n\sum _{t=1}^T\left(c_{\mathit{ijt}}{x}_{\mathit{ijt}}+i_{\mathit{ijt}}{I}_{\mathit{ijt}}+\sum _{k=1}^p{y}_{\mathit{ijkt}}{z}_{\mathit{ijkt}}\right)\\ \mathrm{s}.\mathrm{t}.\hfill \\ \sum _{k=1}^p{D}_{\mathit{ikt}}{z}_{\mathit{ijkt}}+I_{\mathit{ijt}}=I_{\mathit{ij},t-1}+x_{\mathit{ijt}},i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\left(1\right)\\ \sum _{j=1}^n{z}_{\mathit{ijkt}}=1,k=1,\dots ,p;\left(2\right)\\ x_{\mathit{ijt}}\le x_{\mathit{ijt}}^{\max },i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\left(3\right)\\ I_{\mathit{ijt}}\le I_{\mathit{ijt}}^{\max },i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\left(4\right)\\ z_{\mathit{ijkt}}\in \left\{0,1\right\},i=1,\dots ,m;j=1,\dots ,n;k=1,\dots ,p;t=1,\dots ,T\left(5\right)\\ x_{\mathit{ijt}},I_{\mathit{ijt}}\ge 0i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\end{array}$

Ex. 17

Decision variables:

x_ijt = quantity of product i to be manufactured in facility j in period t

I_ijt = final stock of product i in facility j in period t

Y_ijkt = quantity of product i to be transported from facility j to retailer k in period t

$z_{\mathit{ijt}}=\left\{\begin{array}{l}1\text{if the manufacturing of product}i\text{in period}t\text{occurs in facility}j\\ 0\text{otherwise}\end{array}\right.$

Model parameters:

D_ikt = demand of product i by retailer k in period t

c_ijt = unit production cost of product i in facility j in period t

i_ijt = unit storage cost of product i in facility j in period t

y_ijkt = unit transportation cost of product i from facility j to retailer k in period t

x_ijt^max = maximum production capacity of product i in facility j in period t

I_ijt^max = maximum storage capacity of product i in facility j in period t

General formulation

$\begin{array}{l}minz=\sum _{i=1}^m\sum _{j=1}^n\sum _{t=1}^T\left(c_{\mathit{ijt}}{x}_{\mathit{ijt}}+i_{\mathit{ijt}}{I}_{\mathit{ijt}}+\sum _{k=1}^p{y}_{\mathit{ijkt}}{Y}_{\mathit{ijkt}}\right)\\ \mathrm{s}.\mathrm{t}.\\ I_{\mathit{ijt}}=I_{\mathit{ij},t-1}+x_{\mathit{ijt}}-\sum _{k=1}^p{Y}_{\mathit{ijkt}},i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\left(1\right)\\ \sum _{j=1}^n{Y}_{\mathit{ijkt}}=D_{\mathit{ikt}},i=1,\dots ,m;k=1,\dots ,p;t=1,\dots ,T\left(2\right)\\ x_{\mathit{ijt}}\le \sum _{k=1}^p{D}_{\mathit{ikt}}{z}_{\mathit{ijt}}i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\left(3\right)\\ x_{\mathit{ijt}}\le x_{\mathit{ijt}}^{\max },i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\left(4\right)\\ I_{\mathit{ijt}}\le I_{\mathit{ijt}}^{\max },i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\left(5\right)\\ z_{\mathit{ijt}}\in \left\{0,1\right\},i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\left(6\right)\\ x_{\mathit{ijt}},I_{\mathit{ijt}},Y_{\mathit{ijt}}\ge 0i=1,\dots ,m;j=1,\dots ,n;t=1,\dots ,T\end{array}$

Ex. 18

Time frame with T = 6 periods, t = 1, …, 6 (Jan., Feb., March, April, May, Jun.).

P_t = production in period t (kg)

S_t = production with outsourced labor in period t (kg)

NR_t = number of regular employees in period t

NC_t = number of employees hired from period t − 1 to period t

ND_t = number of employees fired from period t − 1 to period t

HE_t = total amount of overtime in period t

I_t = final stock in period t (kg)

$\begin{array}{l}minz=1.5P_1+2S_1+600{\mathit{NR}}_1+1,000{\mathit{NC}}_1+900{\mathit{ND}}_1+7{\mathit{HE}}_1+1I_1+\\ 1.5P_2+2S_2+600{\mathit{NR}}_2+1,000{\mathit{NC}}_2+900{\mathit{ND}}_2+7{\mathit{HE}}_2+1I_2+\\ ⋮⋮\\ 1.5P_6+2S_6+600{\mathit{NR}}_6+1,000{\mathit{NC}}_6+900{\mathit{ND}}_6+7{\mathit{HE}}_6+1I_6\end{array}$

$\begin{array}{l}\mathrm{s}.\mathrm{t}.\\ I_1=600+P_1-9,600\\ I_2=I_1+P_2-10,600\\ ⋮⋮\\ I_6=I_5+P_6-10,430\end{array}$

Answer Keys: Exercises: Chapter 17

Section 17.2.1 (ex.2)

1. a) Optimal solution: x₁ = 2, x₂ = 1 and z = 10
2. b) Optimal solution: x₁ = 1, x₂ = 4 and z = 14
3. c) Optimal solution: x₁ = 10, x₂ = 6 and z = 52

Section 17.2.1 (ex.4)

1. a) yes
2. b) no
3. c) yes
4. d) no
5. e) yes
6. f) yes
7. g) no
8. h) no
9. i) yes

Section 17.2.2 (ex.2)

1. a) Optimal solution: x₁ = 12, x₂ = 2 and z = 26
2. b) Optimal solution: x₁ = 18, x₂ = 8 and z = 28
3. c) Optimal solution: x₁ = 10, x₂ = 10 and z = 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)

1. a) Any point of the segment CD (C (10, 30);  D (0, 45)).
2. 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 = 5, x₂ = 20 and z = 55

Section 17.3 (ex.2)

a) Ten basic solutions.

c) Optimal solution: x₁ = 7, x₂ = 11, x₃ = 0 and z = 61

Section 17.4.2 (ex.1)

a) Optimal solution: x₁ = 1, x₂ = 17, x₃ = 5 and z = 104

Section 17.4.3 (ex.2)

1. a) Optimal solution: x₁ = 3, x₂ = 3 and z = 15
2. b) Optimal solution: x₁ = 2, x₂ = 4, x₃ = 0 and z = 20
3. c) Optimal solution: x₁ = 4, x₂ = 0, x₃ = 12 and z = 36

Section 17.4.4 (ex.1)

1. a) Optimal solution: x₁ = 0, x₂ = 4 and z = − 4
2. b) Optimal solution: x₁ = 1, x₂ = 7 and z = − 37
3. c) Optimal solution: x₁ = 0, x₂ = 10, x₃ = 35/2 and z = − 55/2
4. d) Optimal solution: x₁ = 100/3, x₂ = 0, x₃ = 40/3 and z = − 140/3

Section 17.4.5.1 (ex.1)

b) Solution 1: x₁ = 115/2, x₂ = 0 and z = 230

Solution 2: x₁ = 60, x₂ = 10 and z = 230

Section 17.4.5.1 (ex.2)

b) Solution 1: x₁ = 310, x₂ = 0 and z = 930

Solution 2: x₁ = 30, x₂ = 140 and z = 930

Section 17.4.5.2 (ex.2)

Solution 1: x₁ = 10, x₂ = 30

Solution 2: x₁ = 30, x₂ = 0

Section 17.4.5 (ex.1)

1. a) Multiple optimal solutions.
2. b) Unlimited objective function z.
3. c) Multiple optimal solutions/degenerate optimal solution.

Section 17.4.5 (ex.2)

1. a) No.
2. b) Unfeasible solution.
3. c) Degenerate optimal solution.
4. d) Multiple optimal solutions.
5. e) Unlimited objective function z.

Section 17.5.2 (ex.1)

b) Optimal solution: x₁ = 70, x₂ = 30, x₃ = 35 and z = 363, 000.

Section 17.5.2 (ex.2)

b) Optimal solution: x₁ = 24,960, x₂ = 17,040 and z = 19,296.

Section 17.5.2 (ex.3)

b) Optimal solution: x₁ = 475, x₂ = 50, x₃ = 50, x₄ = 50, x₅ = 75 and z = 32,475.

Section 17.5.2 (ex.4)

b) Optimal solution:    x₁₁ = 3,600,    x₂₁ = 0,            x₃₁ = 0,          x₄₁ = 8,400,

x₁₂ = 0,           x₂₂ = 10,000    x₃₂ = 0,          x₄₂ = 0,

x₁₃ = 0,           x₂₃ = 0,            x₃₃ = 3,200,   x₄₃ = 4,800 and

z = 5,160

Section 17.5.2 (ex.5)

b) Optimal solution: x₁ = 1, x₂ = 0, x₃ = 1, x₄ = 0, x₅ = 1 and z = 810,548 ($810,548.00).

Section 17.5.2 (ex.6)

b) Optimal solution: x₁ = 20%, x₇ = 20%, x₉ = 20%, x₁₀ = 40%, x₂, x₃, x₄, x₅, x₆, x₈, x₁₁ = 0% and z = 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 = 16.90% per year.

Section 17.5.2 (ex.8)

b) Optimal solution: z = 126,590 ($126,590.00).

Section 17.6.1 (ex.1)

a) x₁ = 60, x₂ = 20 with z = 520

b) 1.333

c) 0.8

d) No.

e) The basic solution remains optimal.

Section 17.6.1 (ex.2)

1. a) x₁ = 15, x₂ = 0 with z = 120
2. b) c₁ ≥ 2.4 or c₁ ≥ c₁⁰ − 5.6
3. c) c₂ ≤ 20 or c₂ ≤ c₂⁰ + 14

Section 17.6.1 (ex.3)

1. a) x₁ = 0, x₂ = 17 with z = 102
2. b) Unlimited objective function z.
3. c) c₁ ≥ 3 or c₁ ≥ c₁⁰ − 5
4. d) 0 ≤ c₂ ≤ 16 or c₂⁰ − 6 ≤ c₂ ≤ c₂⁰ + 10

Section 17.6.1 (ex.4)

1. a) $0.133\le \frac{c_1}{c_2}\le 0.25$
2. b) The basic solution remains optimal with z = 1,700.
3. c) 8 ≤ c₁ ≤ 15 or c₁⁰ − 4 ≤ c₁ ≤ c₁⁰ + 3
4. d) 48 ≤ c₂ ≤ 90 or c₂⁰ − 12 ≤ c₂ ≤ c₂⁰ + 30
5. e) The basic solution remains optimal with z = 1,830.
6. f) The basic solution remains optimal with z = 2,440.
7. g) 13.333 ≤ c₁ ≤ 25

Section 17.6.2 (ex.1)

1. a) P₁ = 0, P₂ = 34.286, P₃ = 85.714
2. b) b₁ ≥ b₁⁰ − 8.5
   b₂⁰ − 5.95 ≤ b₂ ≤ b₂⁰ + 6.125
   b₃⁰ − 3.267 ≤ b₃ ≤ b₃⁰ + 2.164
3. c) 0
4. d) $137.14 (z = 1,902.86), x₁ = 115.71 and x₂ = 8.57

Section 17.6.2 (ex.2)

1. a) P₁ = 0, P₂ = 1.222, P₃ = 0.444 (2^nd operation)
2. b) b₁ ≥ b₁⁰ − 20
   b₂⁰ − 180 ≤ b₂ ≤ b₂⁰ + 22.5
   b₃⁰ − 36 ≤ b₃ ≤ b₃⁰ + 180
3. c) $27.50
4. d) $16.00

Section 17.6.3 (ex.1)

b) z₁¹ = 3, $z_2^1=\frac{6}{5}$, z₁^∗ = 3 and z₂^∗ = 2

Section 17.6.3 (ex.2)

b) z₁^∗ = − 4 and z₂^∗ = − 2

Section 17.6.4 (ex.3)

1. a) Degenerate optimal solution.
2. b) Multiple optimal solutions.
3. c) Degenerate optimal solution.
4. d) Multiple optimal solutions.
5. e) Multiple optimal solutions.
6. f) Degenerate optimal solution.
7. g) Degenerate optimal solution.

Answer Keys: Exercises: Chapter 18

Ex.1

1. a) N = {1, 2, 3, 4, 5, 6}
2. b) A = {(1, 2), (1, 3), (2, 3), (3, 4), (3, 5), (4, 2), (4, 5), (4, 6), (5, 6)}
3. c) Directed network.
4. d) 1 → 2 → 3 → 4 → 2
5. e) 1 → 3 → 5 → 4
6. f) 1 → 3 → 4 → 6
7. g) 2 → 3 → 4 → 2
8. h) 3 → 4 → 5 → 3

Ex.2

1. a) N = {1, 2, 3, 4, 5, 6}
2. b) A = {(1, 2), (1, 3), (2, 3), (2, 4), (3, 5), (4, 6), (5, 2), (5, 4), (6, 5)}
3. c) Directed network.
4. d) 2 → 3 → 5 → 4 → 6 → 5
5. e) 1 → 2 → 5 → 4 → 6 → 5
6. f) 1 → 3 → 5 → 4
7. g) 2 → 3 → 5 → 2
8. h) 1 → 2 → 3 → 1

Ex.3

1. a) Tree

1. b) Cover tree

Ex.4

Ex.5

Classic transportation problem:

Optimal FBS: x₁₁ = 40, x₁₄ = 30, x₂₂ = 60, x₂₄ = 20, x₃₃ = 50 with z = 1, 110.

Ex.6

Maximum flow problem:

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

Ex.7

Shortest route problem:

Optimal FBS: x₁₃ = 1, x₃₆ = 1, x₆₈ = 1 (1 − 3 − 6 − 8) with z = 11.

Ex.8

x₁₁ = 50, x₂₂ = 10, x₂₃ = 20, x₃₃ = 20.

Ex.9

x₁₁ = 80, x₁₃ = 70, x₂₂ = 50, x₂₃ = 80 with z = 4,590.

Ex.10

x₁₃ = 150, x₂₁ = 80, x₂₂ = 50 with z = 4,110.

Ex.11

1. a) Optimal FBS: x₁₂ = 100, x₁₃ = 100, x₂₃ = 100, x₃₁ = 150, x₃₂ = 50 with z = 6,800.
2. b) Optimal FBS: x₁₃ = 50, x₃₁ = 100, x₄₁ = 20, x₄₂ = 150, x₄₃ = 30 with z = 1,250.
3. c) Optimal FBS: x₁₂ = 20, x₁₄ = 30, x₂₁ = 20, x₂₄ = 10, x₃₂ = 20, x₃₃ = 60 with z = 1,490.
   Alternative solution: x₁₁ = 20, x₁₂ = 20, x₁₄ = 10, x₂₄ = 30, x₃₂ = 20, x₃₃ = 60 with z = 1,490.

Ex.12

Indexes:

Suppliers i ∈ I

Consolidating centers j ∈ J

Factory k ∈ K

Products p ∈ P

Model parameters:

Model’s decision variables:

The problem can be formulated as follows:

$min\sum _p\sum _i\sum _j{c}_{\mathit{pij}}{x}_{\mathit{pij}}+\sum _p\sum _j\sum _k{c}_{\mathit{pjk}}{y}_{\mathit{pjk}}+\sum _p\sum _i\sum _k{c}_{\mathit{pik}}{z}_{\mathit{pik}}$

s.t.:

$\sum _j{y}_{\mathit{pjk}}+\sum _i{z}_{\mathit{pik}}=D_{\mathit{pk}},\forall p,k$

$\sum _p\sum _i{x}_{\mathit{pij}}\le C_{max,j},\forall j$

$\sum _j{x}_{\mathit{pij}}+\sum _k{z}_{\mathit{pik}}\le S_{\mathit{ip}},\forall i,p$

$\sum _i{x}_{\mathit{pij}}=\sum _k{y}_{\mathit{pjk}},\forall p,j$

$x_{\mathit{pij}},y_{\mathit{pjk}},z_{\mathit{pik}}\ge 0,\forall p,i,j,k$

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.

Ex.13

$x_{\mathit{ij}}=\left\{\begin{array}{l}1\text{if}\text{task}i\text{is designated to machine}j,i=1,\dots ,4,j=1,\dots ,4\\ 0\text{otherwise}\end{array}\right.$

1. a) Optimal FBS: x₁₂ = 1, x₂₄ = 1, x₃₃ = 1, x₄₁ = 1 with z = 37.
2. b) Optimal FBS: x₁₃ = 1, x₂₄ = 1, x₃₃ = 1, x₄₁ = 1 with z = 35.

Ex.14

$x_{\mathit{ij}}=\left\{\begin{array}{l}1\text{if}\text{route}\left(i,j\right)\text{is included in the shortest route},\forall i,j\\ 0\text{otherwise}\end{array}\right.$

$\begin{array}{l}min6x_{12}+9x_{13}+4x_{23}+4x_{24}+7x_{25}+6x_{35}+2x_{45}+7x_{46}+3x_{56}\\ \mathrm{s}.\mathrm{t}.\\ x_{12}+x_{13}=1\\ x_{46}+x_{56}=1\\ x_{12}-x_{23}-x_{24}-x_{25}=0\\ x_{13}+x_{23}-x_{35}=0\\ x_{24}-x_{45}-x_{46}=0\\ x_{25}+x_{35}+x_{45}-x_{56}=0\\ x_{\mathit{ij}}\in \left\{0,1\right\}\text{or}{x}_{\mathit{ij}}\ge 0\end{array}$

Optimal FBS: x₁₂ = 1, x₂₄ = 1, x₄₅ = 1, x₅₆ = 1 (1 − 2 − 4 − 5 − 6) with z = 15.

Ex.15

Optimal FBS: x_AB = 1, x_BD = 1, x_DE = 1 (A − B − D − E) with z = 64.

Ex.16

x₁₂ = 6, x₁₃ = 4, x₂₃ = 0, x₂₄ = 6, x₃₄ = 1, x₃₅ = 3, x₄₅ = 0, x₄₆ = 7, x₅₆ = 3 with z = 10.

Answer Keys: Exercises: Chapter 19

Section 19.1 (ex.1)

1. a) BP
2. b) MIP
3. c) IP
4. d) BIP
5. e) BP
6. f) MBP
7. g) MIP

Section 19.2 (ex.1)

1. a) No
2. b) Yes (x₁ = 10,x₂ = 0 with z = 20)
3. c) No
4. d) Yes (x₁ = 0, x₂ = 4 with z = 32)
5. e) Yes (x₁ = 1,x₂ = 0 with z = 4)
6. f) No
7. g) Yes (x₁ = 6,x₂ = 5 with z = 58)

Section 19.2 (ex.2)

b) SF = $\left\{\begin{array}{l}\left(0,0\right);\left(0,1\right);\left(0,2\right);\left(0,3\right);\left(0,4\right);\left(1,0\right);\left(1,1\right);\left(1,2\right);\left(1,3\right);\left(2,0\right);\left(2,1\right);\\ \left(2,2\right);\left(2,3\right);\left(3,0\right);\left(3,1\right);\left(3,2\right);\left(4,0\right);\left(4,1\right);\left(4,2\right);\left(5,0\right);\left(5,1\right);\left(6,0\right)\end{array}\right\}$

c) Optimal solution: x₁ = 4 and x₂ = 2 with z = 14.

Section 19.2 (ex.3)

b) SF = {(0, 0); (0, 1); (0, 2); (1, 0); (1, 1); (2, 0); (2, 1); (3, 0)}

c) Optimal solution: x₁ = 2 and x₂ = 1 with z = 4.

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 = 3 and x₂ = 2 with z = 13.

Section 19.3 (ex.1)

Optimal FBS = {x₃ = 1, x₄ = 1, x₆ = 1, x₈ = 1} with z = 172.

Section 19.4 (ex.1)

$\begin{array}{l}maxz=7x_1+12x_2+8x_3+10x_4+7x_5+6x_6\\ \mathrm{s}.\mathrm{t}.\\ 4x_1+7x_2+5x_3+6x_4+4x_5+3x_6\le 20\\ x_5+x_6\le 1\\ x_3-x_2\le 0\\ x_1,x_2,x_3,x_4,x_5,x_6\in \left\{0,1\right\}\end{array}$

Optimal solution: x₁ = 1, x₂ = 1, x₃ = 0, x₄ = 1, x₅ = 0, x₆ = 1 with z = 35.

Section 19.5 (ex.1)

Indexes

i, j = 1, .., n that represent the customers (index 0 represents the depot)

v = 1, …, NV that represent the vehicles

Parameters

C_max,v = maximum capacity of vehicle ν

d_i = demand of client i

c_ij = travel cost from client i to client j

Decision variables

$x_{\mathit{ij}}^v=\left\{\begin{array}{l}1\text{if the}\mathrm{arc}\text{from}i\text{to}j\text{is}\text{traveled}\mathrm{by}\text{vehicle}v\\ 0\text{otherwise}\end{array}\right.$

$y_i^v=\left\{\begin{array}{l}1\text{if order of client}i\text{is delivered}\mathrm{by}\text{vehicle}v\\ 0\text{otherwise}\end{array}\right.$

Model formulation

$min\sum _i\sum _j\sum _v{c}_{\mathit{ij}}{x}_{\mathit{ij}}^v$

s.t.

$\sum _v{y}_i^v=1,i=1,\dots ,n$

$\sum _v{y}_i^v=\mathit{NV},i=0$

$\sum _i{d}_i{y}_i^v\le C_{max,v},v=1,\dots ,\mathit{NV}$

$\sum _i{x}_{\mathit{ij}}^v=y_j^v,j=0,\dots ,n,v=1,\dots ,\mathit{NV}$

$\sum _j{x}_{\mathit{ij}}^v=y_i^v,i=0,\dots ,n,v=1,\dots ,\mathit{NV}$

$\sum _{\mathit{ij}\in S}{x}_{\mathit{ij}}^v=x_{\mathit{ij}}^v\le \left|S\right|-1,S\subseteq \left\{1,\dots ,n\right\},2\le \left|S\right|\le n-1,v=1,\dots ,\mathit{NV}$

$x_{\mathit{ij}}^v\in \left\{0,1\right\},i=0,\dots ,nj=0,\dots ,n,v=1,\dots ,\mathit{NV}$

$y_i^v\in \left\{0,1\right\},i=0,\dots ,n,v=1,\dots ,\mathit{NV}$

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 = 0). 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 x_ij^v and y_i^v will be binary.

Section 19.6 (ex.1)

Indexes

i = 1, …, m that represent the distribution centers (DCs)

j = 1, …, n that represent the consumers

Model parameters

f_i = fixed costs to maintain DC i open

c_ij = transportation costs from DC i to consumer j

D_j = demand of customer j

C_{max, i} = maximum capacity of DC i

Decision variables

$y_i=\left\{\begin{array}{l}1\text{if}\mathrm{DC}i\text{opens}\\ 0\text{otherwise}\end{array}\right.$

$x_{\mathit{ij}}=\left\{\begin{array}{l}1\text{if consumer}j\text{is supplied}\mathrm{by}\mathrm{DC}i\\ 0\text{otherwise}\end{array}\right.$

General formulation

$\begin{array}{l}{F}_{\mathrm{obj}}=minz=\sum _{i=1}^m{f}_i{y}_i+\sum _{i=1}^m\sum _{j=1}^n{c}_{\mathit{ij}}{x}_{\mathit{ij}}{D}_j\\ \mathrm{s}.\mathrm{t}.\\ \sum _{j=1}^n{x}_{\mathit{ij}}{D}_j\le C_{\text{max},i}\cdot y_i,i=1,\dots ,m\left(1\right)\\ \sum _{i=1}^m{x}_{\mathit{ij}}=1,j=1,\dots ,n\left(2\right)\\ x_{\mathit{ij}},y_i\in \left\{0,1\right\},i=1,\dots ,m,j=1,\dots ,n\left(3\right)\end{array}$

which corresponds to a binary programming problem.

For this problem, index i corresponds to:

i = 1 (Belem), i = 2 (Palmas), i = 3 (Sao Luis), i = 4 (Teresina), and i = 5 (Fortaleza);

and index j corresponds to:

j = 1 (Belo Horizonte), j = 2 (Vitoria), j = 3 (Rio de Janeiro), j = 4 (Sao Paulo), and j = 5 (Campo Grande).

Optimal FBS: x₂₂ = 1, x₂₄ = 1, x₄₅ = 1, x₅₁ = 1, x₅₃ = 1, y₂ = 1, y₄ = 1, y₅ = 1 with z = 459,400.00.

Section 19.6 (ex.2)

Indexes:

Suppliers i ∈ I

Consolidating centers j ∈ J

Factory k ∈ K

Products p ∈ P

Model parameters:

Model’s decision variables:

The problem can be formulated as follows:

$min\sum _p\sum _i\sum _j{c}_{\mathit{pij}}{x}_{\mathit{pij}}+\sum _p\sum _j\sum _k{c}_{\mathit{pjk}}{y}_{\mathit{pjk}}+\sum _p\sum _i\sum _k{c}_{\mathit{pik}}{z}_{\mathit{pik}}+\sum _j{f}_j{z}_j$

s.t.:

$\sum _j{y}_{\mathit{pjk}}+\sum _i{z}_{\mathit{pik}}=D_{\mathit{pk}},\forall p,k$

$\sum _p\sum _i{x}_{\mathit{pij}}\le C_{max,j}\cdot z_j,\forall j$

$\sum _j{x}_{\mathit{pij}}+\sum _k{z}_{\mathit{pik}}\le S_{\mathit{ip}},\forall i,p$

$\sum _i{x}_{\mathit{pij}}=\sum _k{y}_{\mathit{pjk}},\forall p,j$

$x_{\mathit{pij}},y_{\mathit{pjk}},z_{\mathit{pik}}\ge 0,\forall p,i,j,k$

$z_j\in \left\{0,1\right\},\forall z$

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 z_j is binary.

Section 19.7 (ex.1)

x_i = number of buses that start working in shift i, i = 1, 2, …, 9.

Therefore, we have:

x₁ = number of buses that start working at 6:01.

x₂ = number of buses that start working at 8:01.

x₃ = number of buses that start working at 10:01.

x₄ = number of buses that start working at 12:01.

x₅ = number of buses that start working at 14:01.

x₆ = number of buses that start working at 16:01.

x₇ = number of buses that start working at 18:01.

x₈ = number of buses that start working at 20:01.

x₉ = number of buses that start working at 22:01.

$\begin{array}{l}{F}_{\mathrm{obj}}=minz=x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9\\ \text{subject to}\\ x_1\ge 20\left(6:01-8:00\right)\\ x_1+x_2\ge 24\left(8:01-10:00\right)\\ x_1+x_2+x_3\ge 18\left(10:01-12:00\right)\\ x_1+x_2+x_3+x_4\ge 15\left(12:01-14:00\right)\\ x_2+x_3+x_4+x_5\ge 16\left(14:01-16:00\right)\\ x_3+x_4+x_5+x_6\ge 27\left(16:01-18:00\right)\\ x_4+x_5+x_6+x_7\ge 18\left(18:01-20:00\right)\\ x_5+x_6+x_7+x_8\ge 12\left(20:01-22:00\right)\\ x_6+x_7+x_8+x_9\ge 10\left(22:01-24:00\right)\\ x_7+x_8+x_9\ge 4\left(00:01-02:00\right)\\ x_8+x_9\ge 3\left(02:01-04:00\right)\\ x_9\ge 8\left(04:01-06:00\right)\\ x_i\ge 0,i=1,2,\dots ,9\end{array}$

Optimal solution: x₁ = 24, x₂ = 0, x₃ = 0, x₄ = 0, x₅ = 16, x₆ = 11, x₇ = 0, x₈ = 0, x₉ = 8 with z = 59.

Section 19.7 (ex.2)

x_i = number of employees that start working on day i, i = 1, 2, …, 7.

x₁ = number of employees that start working on Monday.

x₂ = number of employees that start working on Tuesday.

⋮

x₇ = number of employees that start working on Sunday.

$\begin{array}{l}minz=x_1+x_2+x_3+x_4+x_5+x_6+x_7\\ \text{subject to}\\ x_1+x_4+x_5+x_6+x_7\ge 15\left(\text{Monday}\right)\\ x_1+x_2+x_5+x_6+x_7\ge 20\left(\text{Tuesday}\right)\\ x_1+x_2+x_3+x_6+x_7\ge 17\left(\text{Wednesday}\right)\\ x_1+x_2+x_3+x_4+x_7\ge 22\left(\text{Thursday}\right)\\ x_1+x_2+x_3+x_4+x_5\ge 25\left(\text{Friday}\right)\\ x_2+x_3+x_4+x_5+x_6\ge 15\left(\text{Saturday}\right)\\ x_3+x_4+x_5+x_6+x_7\ge 10\left(\text{Sunday}\right)\\ x_i\ge 0,i=1,\dots ,7\end{array}$

Alternative optimal solution: x₁ = 10, x₂ = 6, x₃ = 0, x₄ = 5, x₅ = 4, x₆ = 0, x₇ = 1 with z = 26.

Answer Keys: Exercises: Chapter 20

4) P (I < 0) = 15.92% by using the NORM.DIST function in Excel or P (I < 0) = 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) = 22.50% by using the NORM.DIST function in Excel or P (Index > 0.07) = 20.43% analyzing the values generated in the simulation.

Note: The results can change at each new simulation.

Answer Keys: Exercises: Chapter 21

1) F_cal = 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 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.

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.

Answer Keys: Exercises: Chapter 22

1)

a) Control charts for$\overline{\mathit{X}}$

• UCL = 17.4035
• Average = 16.5318
• LCL = 15.6600

Control charts for R

• UCL = 2.7305
• Average = 1.1965
• LCL = 0.0000

b)C_p = 0.860

C_pk = 0.842

C_pm = 0.859

2)

a) Control charts for$\overline{\mathit{X}}$

• UCL = 17.3895
• Average = 16.5318
• LCL = 15.6740

Control charts for S

• UCL = 1.1938
• Average = 0.5268
• LCL = 0.0000

b)C_p = 0.9491

C_pk = 0.9290

3)

a) Control charts for$\overline{\mathit{X}}$

• UCL = 6.7113
• Average = 6.0625
• LCL = 5.4137

Control charts for R

• UCL = 2.0322
• Average = 0.8905
• LCL = 0.0000

b)C_p = 0.771

C_pk = 0.722

C_pm = 0.542

4)

a) Control charts for$\overline{\mathit{X}}$

• UCL = 6.7162
• Average = 6.0625
• LCL = 5.4088

Control charts for S

• UCL = 0.9098
• Average = 0.4015
• LCL = 0.0000

b)C_p = 0.8302

C_pk = 0.7783

5)P chart

• UCL = 0.1748
• Average = 0.0680
• LCL = 0.0000

6)UCL = 8.7403

Average = 3.4000

LCL = 0.0000

7)UCL = 11.9996

Average = 5.1750

LCL = 0.0000

8)

Answer Keys: Exercises: Chapter 23

1)

a)

In fact, this is a balanced clustered data structure.

b)

c)

d)

Yes. Since the estimation of variance component τ₀₀, which corresponds to random intercept u_0j, 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 = 422.619/125.284 = 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. χ² = 0.000, it is possible to reject the null hypothesis that the random intercepts are equal to zero (H₀: u_0j = 0), which makes the estimation of a traditional linear regression model be ruled out for these clustered data.

f)

$\mathit{rho}=\frac{{\mathtt{\tau }}_{00}}{{\mathtt{\tau }}_{00}+{\mathtt{\sigma }}^2}=\frac{422.619}{422.619+11.196}=0.974$

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)

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)

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 (LL_r = − 357.501), the model with only random effects in the intercept is favored, since random error terms u_1j are statistically equal to zero.

m)

n)

${\mathit{score}}_{\mathit{ij}}=13.22+0.0028\cdot {\mathit{income}}_{\mathit{ij}}+0.0008\cdot {\mathit{resdevel}}_j\cdot {\mathit{income}}_{\mathit{ij}}+{\mathit{u}}_{0\mathit{j}}+r_{\mathit{ij}}$

o)

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)

d)

e)

f)

g)

• • Level-2 intraclass correlation:

${\mathit{rho}}_{\mathit{property}\mid \mathit{district}}=\frac{{\mathtt{\tau }}_{u000}+{\mathtt{\tau }}_{r000}}{{\mathtt{\tau }}_{u000}+{\mathtt{\tau }}_{r000}\mathtt{+}{\mathtt{\sigma }}^2}=\frac{0.1228+0.0368}{0.1228+0.0368+0.0007}=0.996$

• • Level-3 intraclass correlation:

${\mathit{rho}}_{\mathit{district}}=\frac{{\mathtt{\tau }}_{u000}}{{\mathtt{\tau }}_{u000}+{\mathtt{\tau }}_{r000}\mathtt{+}{\mathtt{\sigma }}^2}=\frac{0.1228}{0.1228+0.0368+0.0007}=0.766$

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% (rho_district), and the correlation between these annual indexes, for the same property of a certain district, is equal to 99.6% (rho_{property | 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 τ_u000, τ_r000, and σ² (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.

i)

Since Sig. χ² = 0.000, it is possible to reject the null hypothesis that the random intercepts are equal to zero (H₀: u_00k = r_0jk = 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 = 0.000 < 0.05), which demonstrates that, each year, rental prices of commercial properties increase, on average, 1.10% (e^0.011 = 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 u_00k, r_0jk, and e_tjk, at a significance level of 0.05, because the estimations of τ_u000, τ_r000, and σ² are considerably higher than the respective standard errors.

l)

m)

n)

• • Level-2 intraclass correlation:

$\begin{array}{l}{\mathit{rho}}_{\mathit{property}\mid \mathit{district}}=\frac{{\mathtt{\tau }}_{u000}+{\mathtt{\tau }}_{u100}\mathtt{+}{\mathtt{\tau }}_{r000}+{\mathtt{\tau }}_{r100}}{{\mathtt{\tau }}_{u000}+{\mathtt{\tau }}_{u100}\mathtt{+}{\mathtt{\tau }}_{r000}+{\mathtt{\tau }}_{r100}\mathtt{+}{\mathtt{\sigma }}^2}\\ =\frac{0.142444+0.000043+0.039638+0.000047}{0.142444+0.000043+0.039638+0.000047+0.000103}=0.9994\end{array}$

• • Level-3 intraclass correlation:

$\begin{array}{c}{\mathit{rho}}_{\mathit{district}}=\frac{{\mathtt{\tau }}_{u000}+{\mathtt{\tau }}_{u100}}{{\mathtt{\tau }}_{u000}+{\mathtt{\tau }}_{u100}\mathtt{+}{\mathtt{\tau }}_{r000}+{\mathtt{\tau }}_{r100}\mathtt{+}{\mathtt{\sigma }}^2}\\ =\frac{0.142444+0.000043}{0.142444+0.000043+0.039638+0.000047+0.000103}=0.7817\end{array}$

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!

o)

Since Sig. χ₂² = 0.000, we choose the linear trend model with random intercepts and slopes.

p)

q)

$\begin{array}{l}ln{\left(p\right)}_{\mathit{tjk}}=4.134+0.015\cdot {\mathit{year}}_{\mathit{jk}}+0.231\cdot {\mathit{food}}_{\mathit{jk}}+0.189\cdot \mathit{space}{4}_{\mathit{jk}}-0.004\cdot {\mathit{valet}}_{\mathit{jk}}\cdot {\mathit{year}}_{\mathit{jk}}\\ +{\mathit{u}}_{00k}+{\mathit{u}}_{10k}\cdot {\mathit{year}}_{\mathit{jk}}+{\mathit{r}}_{0\mathit{jk}}+{\mathit{r}}_{1\mathit{jk}}\cdot {\mathit{year}}_{\mathit{jk}}+{\mathit{e}}_{\mathit{tjk}}\end{array}$

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.

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)

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

$var\left[\begin{array}{c}{u}_{00k}\\ u_{10k}\end{array}\right]=\left[\begin{array}{cc}0.037004& 0\\ 0& 0.000016\end{array}\right]$

• • Random-effects variance-covariance matrix for level property:

$var\left[\begin{array}{c}{r}_{0\mathit{jk}}\\ r_{1\mathit{jk}}\end{array}\right]=\left[\begin{array}{cc}0.030961& 0\\ 0& 0.000044\end{array}\right]$

u)

v)

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

$var\left[\begin{array}{c}{u}_{00k}\\ u_{10k}\end{array}\right]=\left[\begin{array}{cc}0.037253& -0.000653\\ -0.000653& 0.000014\end{array}\right]$

• • Random-effects variance-covariance matrix for level property:

$var\left[\begin{array}{c}{r}_{0\mathit{jk}}\\ r_{1\mathit{jk}}\end{array}\right]=\left[\begin{array}{cc}0.031679& -0.000484\\ -0.000484& 0.000046\end{array}\right]$

w)

Since Sig. χ₂² = 0.000, the structure of the random-terms variance-covariance matrices is considered unstructured, that is, we can conclude that error terms u_00k and u_10k are correlated (cov(u_00k , u_10k) ≠ 0), and that error terms r_0jk and r_1jk are also correlated (cov(r_0jk , r_1jk) ≠ 0).

x)

$\begin{array}{l}ln{\left(p\right)}_{\mathit{tjk}}=3.7807+0.0144\cdot {\mathit{year}}_{\mathit{jk}}+0.2314\cdot {\mathit{food}}_{\mathit{jk}}+0.2071\cdot \mathit{space}{4}_{\mathit{jk}}+0.5111\cdot {\mathit{subway}}_{\mathit{k}}\\ -0.0031\cdot {\mathit{valet}}_{\mathit{jk}}\cdot {\mathit{year}}_{\mathit{jk}}-0.0072\cdot {\mathit{subway}}_{\mathit{k}}\cdot {\mathit{year}}_{\mathit{jk}}+0.0001\cdot {\mathit{violence}}_{\mathit{k}}\cdot {\mathit{year}}_{\mathit{jk}}\\ +{\mathit{u}}_{00k}+{\mathit{u}}_{10k}\cdot {\mathit{year}}_{\mathit{jk}}+{\mathit{r}}_{0\mathit{jk}}+{\mathit{r}}_{1\mathit{jk}}\cdot {\mathit{year}}_{\mathit{jk}}+{\mathit{e}}_{\mathit{tjk}}\end{array}$

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)