lecture slides - topic one: production line profit ... · a p(n) cost ($) p(n*(a )) ... convert the...
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Topic one: Production line profit maximization subject to a production rate constraint
c⃝2010 Chuan Shi — Topic one: Line optimization : 22/79
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Production line profit maximization
The profit maximization problem
max N
�(N) = �� (N) − �−1∑
�=1
���� − �−1∑
�=1
�� ̄��(N)
s.t. � (N) ≥ �̂ ,
�� ≥ �min, ∀� = 1, ⋅ ⋅ ⋅ , � − 1.
where � (N) = production rate, parts/time unit �̂ = required production rate, parts/time unit � = profit coefficient, $/part
�̄�(N) = average inventory of buffer �, � = 1, ⋅ ⋅ ⋅ , � − 1 �� = buffer cost coefficient, $/part/time unit �� = inventory cost coefficient, $/part/time unit
c⃝2010 Chuan Shi — Topic one: Line optimization : Constrained and unconstrained problems 23/79
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An example about the research goal
0 10 20 30 40 50 60 70 80 90 100 0 10
20 30
40 50
60 70
80 90
100
1040 1060 1080 1100 1120 1140 1160 1180 1200 1220
J(N)
"data.txt""feasible.txt"
1205120011901180116011401120110010801060
Optimalboundary
N1
N2
J(N)
Figure 2: �(N) vs. �1 and �2
c⃝2010 Chuan Shi — Topic one: Line optimization : Constrained and unconstrained problems 24/79
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Two problems
Original constrained problem
max N
�(N) = �� (N) − �−1∑
�=1
���� − �−1∑
�=1
�� ̄��(N)
s.t. � (N) ≥ �̂ ,
�� ≥ �min, ∀� = 1, ⋅ ⋅ ⋅ , � − 1.
Simpler unconstrained problem (Schor’s problem)
max N
�(N) = �� (N) − �−1∑
�=1
���� − �−1∑
�=1
�� ̄��(N)
s.t. �� ≥ �min, ∀� = 1, ⋅ ⋅ ⋅ , � − 1.
c⃝2010 Chuan Shi — Topic one: Line optimization : Constrained and unconstrained problems 25/79
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An example for algorithm derivation
Data �1 = .1, �1 = .01, �2 = .11, �2 = .01, �3 = .1, �3 = .009, �̂ = .88 Cost function �(N) = 2000� (N) − �1 − �2 − �̄1(N) − �̄2(N)
0 10 20 30 40 50 60 70 80 90 100 0 10
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40 50
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80 90
100
1500 1520 1540 1560 1580 1600 1620 1640 1660
J(N)
N1
N2
J(N)
10
20
30
40
50
60
70
80
90
100
20 30 40 50 60 70 80 90 100
N2
N1
P(N1,N2)=P
P(N1,N2)>P
P(N1,N2)< P
Figure 3: �(N) vs. �1 and �2 Figure 4: � (N)
c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 26/79
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An example for algorithm derivation
0 10 20 30 40 50 60 70 80 90 100 0 10
20 30
40 50
60 70
80 90
100
1500 1520 1540 1560 1580 1600 1620 1640 1660
J(N)
N1
N2
J(N)
Figure 5: �(N) vs. �1 and �2
c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 27/79
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Algorithm derivation
Two cases Case 1 The solution of the unconstrained problem is N� s.t. � (N�) ≥ �̂ . In this case, the solution of the constrained problem is the same as the solution of the unconstrained problem. We are done.
Unconstrained problem
max N
�(N) = �� (N) − �−1∑
�=1
����
− �−1∑
�=1
�� ̄��(N)
s.t. �� ≥ �min, ∀� = 1, ⋅ ⋅ ⋅ , � − 1.
( N1 , N2 ) u u
P(N1,N2) > P
10
20
30
40
50
60
70
80
90
100
20 30 40 50 60 70 80 90 100
N2
N1
c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 28/79
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Algorithm derivation
Two cases (continued) Case 2 N� satisfies � (N�) < �̂ . This is not the solution of the constrained problem.
( N1 , N2 ) u u
P(N1,N2) > P
10
20
30
40
50
60
70
80
90
100
20 30 40 50 60 70 80 90 100
N2
N1
c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 29/79
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Algorithm derivation
Two cases (continued)
Case 2 (continued)
In this case, we consider the following unconstrained problem:
�−1 �−1∑ ∑ max �(N) = �′� (N) − ���� − ���̄�(N)N
�=1 �=1
s.t. �� ≥ �min, ∀� = 1, ⋅ ⋅ ⋅ , � − 1.
in which � is replaced by �′ . Let N★(�′) be the solution to this problem and � ★(�′) = � (N★(�′)).
c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 30/79
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Assertion
The constrained problem �−1 �−1∑ ∑
max �(N) = �′� (N) − ���� − ���̄�(N) N
�=1 �=1
s.t. � (N) � , ≥ ˆ
�� ≥ �min, ∀� = 1, ⋅ ⋅ ⋅ , � − 1.
has the same solution for all �′ in which the solution of the corresponding unconstrained problem ∑�−1 �−1∑
max �(N) = �′� (N) − ���� − ���̄�(N)N
�=1 �=1
s.t. �� ≥ �min, ∀� = 1, ⋅ ⋅ ⋅ , � − 1.
has � ★(�′) ≤ �̂ . c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 31/79
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⋅ ⋅ ⋅ ⋅ ⋅ ⋅
s.t.
s.t.
� (�
Interpretation of the assertion
We claim If the optimal solution of the unconstrained problem is not that of the constrained
−1 ★�
★problem, then the solution of the constrained problem, (�
−1 ★�
, � ), satisfies 1 ,
) = �̂ . ★ , � 1 ,
400
410
420
430
440
450
460
470
480
0 5 10 15 20 25 30 35 40 0
10
20
30
40
50
60
70
80
A’P
(N)
Cos
t ($)
NP(N*(A’)) < P
N*(A’)
max �(�) = 500� (�) − � − �̄(�) N
s.t. � (�) ≥ �̂
� ≥ �min
max �(N) = 500�̂ − � − �̄(�) N
s.t. � (� ) ≥ �̂ ⇒ � (�) = �̂
� ≥ �min
We formally prove this by the Karush-Kuhn-Tucker (KKT) conditions of nonlin-ear programming.
c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 32/79
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⋅ ⋅ ⋅ ⋅ ⋅ ⋅
s.t.
s.t.
� (�
Interpretation of the assertion
We claim If the optimal solution of the unconstrained problem is not that of the constrained
−1 ★�
★problem, then the solution of the constrained problem, (�
−1 ★�
, � ), satisfies 1 ,
) = �̂ . ★ , � 1 ,
400
410
420
430
440
450
460
470
480
0 5 10 15 20 25 30 35 40 0
10
20
30
40
50
60
70
80
A’P
(N)
Cos
t ($)
NP(N*(A’)) < P
N*(A’)
max �(�) = 500� (�) − � − �̄(�) N
s.t. � (�) ≥ �̂
� ≥ �min
max �(N) = 500�̂ − � − �̄(�) N
s.t. � (� ) ≥ �̂ ⇒ � (�) = �̂
� ≥ �min
We formally prove this by the Karush-Kuhn-Tucker (KKT) conditions of nonlin-ear programming.
c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 32/79
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⋅ ⋅ ⋅ ⋅ ⋅ ⋅
s.t.
s.t.
� (�
Interpretation of the assertion
We claim If the optimal solution of the unconstrained problem is not that of the constrained
−1 ★�
★problem, then the solution of the constrained problem, (�
−1 ★�
, � ), satisfies 1 ,
) = �̂ . ★ , � 1 ,
400
410
420
430
440
450
460
470
480
0 5 10 15 20 25 30 35 40 0
10
20
30
40
50
60
70
80
A’P
(N)
Cos
t ($)
NP(N*(A’)) < P
N*(A’)
max �(�) = 500� (�) − � − �̄(�) N
s.t. � (�) ≥ �̂
� ≥ �min
max �(N) = 500�̂ − � − �̄(�) N
s.t. � (� ) ≥ �̂ ⇒ � (�) = �̂
� ≥ �min
We formally prove this by the Karush-Kuhn-Tucker (KKT) conditions of nonlin-ear programming.
c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 32/79
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Interpretation of the assertion
0 10 20 30 40 50 60 70 80 90 100 0 10
20 30
40 50
60 70
80 90
100
1040 1060 1080 1100 1120 1140 1160 1180 1200 1220
J(N)
"infeasible.txt""feasible.txt"
1205120011901180116011401120110010801060
Optimalboundary
N1
N2
J(N)
A = 1500 (Original Problem)
0 10 20 30 40 50 60 70 80 90 100 0 10
20 30
40 50
60 70
80 90
100 1880 1900 1920 1940 1960 1980 2000 2020 2040 2060 2080
J(N)
"infeasible.txt""feasible.txt"
20602055
2049.82040202020001980195019301900
Optimalboundary
N1
N2
J(N)
A’ = 2500
0 10 20 30 40 50 60 70 80 90 100 0 10
20 30
40 50
60 70
80 90
100
2650
2700
2750
2800
2850
2900
2950
J(N)
"infeasible.txt""feasible.txt"
29212919.7
29102900288028602830280027702730
Optimalboundary
N1
N2
J(N)
A’ = 3500
0 10 20 30 40 50 60 70 80 90 100 0 10
20 30
40 50
60 70
80 90
100 3450 3500 3550 3600 3650 3700 3750 3800 3850
J(N)
"infeasible.txt""feasible.txt"
3819381638133818380037703730370036603620358035403500
Optimalboundary
N1
N2
J(N)
A’ = 4535.82 (Final A’)
c⃝2010 Chuan Shi — Topic one: Line optimization : Algorithm derivation 33/79
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⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅ ⋅
⋅ ⋅ ⋅
Karush-Kuhn-Tucker (KKT) conditions Let �★ be a local minimum of the problem
min �(�) s.t. ℎ1(�) = 0, , ℎ�(�) = 0,
�1(�) ≤ 0, , ��(�) ≤ 0,
where � , ℎ�, and �� are continuously differentiable functions from ℜ�
to Then there exist unique Lagrange multipliers �★ , �★ andℜ. 1, � �★ 1, , �★
� , satisfying the following conditions:
∇��(�★, �★, �★) = 0,
�★� ≥ 0, � = 1, , �,
�★� �� (�
★) = 0, � = 1, , �. ∑� ∑�where �(�, �, �) = �(�) + �=1 ��ℎ�(�) + �=1 �� ��(�) is called the Lagrangian function.
c⃝2010 Chuan Shi — Topic one: Line optimization : Proofs of the algorithm by KKT conditions 34/79
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Convert the constrained problem to minimization form
Minimization form
The constrained problem
min N
−�(N) = −�� (N) + �−1∑
�=1
���� + �−1∑
�=1
�� ̄��(N)
s.t. �̂ − � (N) ≤ 0
�min − �� ≤ 0, ∀� = 1, ⋅ ⋅ ⋅ , � − 1
We have argued that we treat �� as continuous variables, and � (�) and �(�) as continuously differentiable functions.
c⃝2010 Chuan Shi — Topic one: Line optimization : Proofs of the algorithm by KKT conditions 35/79
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⋅ ⋅ ⋅
Applying KKT conditions
The Slater constraint qualification for convex inequalities guarantees the existence of Lagrange multipliers for our problem. So, there exist unique Lagrange multipliers �★
� , � = 0, , � − 1 for the constrained problem to satisfy the KKT conditions:
�−1∑ −∇�(N★) + �★
0∇(�̂ − � (N★)) + ��★∇(�min − ��) = 0 (1)
�=1
or ⎛ ⎛⎞ ⎞⎛⎞⎛⎞⎛
⎞ ∂�(N★) ∂� (N★) ⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎟⎟⎟⎟⎟⎟⎟⎟⎠
− �★ 0
⎜⎜⎜⎜⎜⎜⎜⎜⎝
⎟⎟⎟⎟⎟⎟⎟⎟⎠
∂�1 ∂�(N★) ∂�2 ...
∂�(N★)
∂�1 ∂� (N★ 1 0 0
) ⎜⎜⎜⎝
0 ...
⎟⎟⎟⎠−⋅ ⋅ ⋅− �★ �−1
⎜⎜⎜⎝
0 ...
⎟⎟⎟⎠ =
⎜⎜⎜⎝
0 ...
⎟⎟⎟⎠ ∂�2 ...
∂� (N★)
★ − �− ,1
0 1 0
∂��−1 ∂��−1
(2)
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⋅ ⋅ ⋅
haha
haha⋅ ⋅ ⋅
⋅ ⋅ ⋅
�
Applying KKT conditions
★ �
and ≥ 0, ∀� = 0, , � − 1, (3)
�0 (�̂ − � (N★)) = 0, ★ (4)
where N★ is the optimal solution of the constrained problem. Assume
★��
★that ��★��
★�(�min − � ) = 0, ∀� = 1, , � − 1, (5)
> �min for all �. In this case, by equation (5), we know that = 0, ∀� = 1, , � − 1.
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Applying KKT conditions
The KKT conditions are simplified to ⎛ ⎞ ⎛ ∂�(N★) ∂� (N★) ⎞⎛
⎞
−
⎜⎜⎜⎜⎜⎜⎜⎜⎝∂�1
∂�(N★) ∂�2 . . .
∂�(N★)
⎟⎟⎟⎟⎟⎟⎟⎟⎠
⎜⎜⎜⎜⎜⎜⎜⎜⎝
∂�1 ∂� (N★) ∂�2 . . .
∂� (N★)
⎟⎟⎟⎟⎟⎟⎟⎟⎠
0
= ⎜⎜⎜⎝
⎟⎟⎟⎠
0 .. .
− �★ 0 (6),
0
∂��−1 ∂��−1
�★ 0(�̂
− � (N★)) = 0, (7)
where �★ 0 ≥ 0. Since N★ is not the optimal solution of the unconstrained
problem, ∇�(N★) =∕ 0. Thus, �★ =∕ 0 since otherwise condition (6) 0 would be violated. By condition (7), the optimal solution N★ satisfies � (N★) = �̂ .
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Applying KKT conditions
The KKT conditions are simplified to ⎛ ⎞ ⎛ ∂�(N★) ∂� (N★) ⎞⎛
⎞
−
⎜⎜⎜⎜⎜⎜⎜⎜⎝∂�1
∂�(N★) ∂�2 . . .
∂�(N★)
⎟⎟⎟⎟⎟⎟⎟⎟⎠
⎜⎜⎜⎜⎜⎜⎜⎜⎝
∂�1 ∂� (N★) ∂�2 . . .
∂� (N★)
⎟⎟⎟⎟⎟⎟⎟⎟⎠
0
= ⎜⎜⎜⎝
⎟⎟⎟⎠
0 .. .
− �★ 0 ,
0
∂��−1 ∂��−1
�★ 0(�̂
− � (N★)) = 0,
In addition, conditions (6) and (7) reveal how we could find �★ 0 and N★ .
For every �★ 0, condition (6) determines N★ since there are � − 1 equations
and � − 1 unknowns. Therefore, we can think of N★ = N★(�0 ★). We
search for a value of �★ 0 such that � (N★(�★
0)) = �̂ . As we indicate in the following, this is exactly what the algorithm does.
c⃝2010 Chuan Shi — Topic one: Line optimization : Proofs of the algorithm by KKT conditions 39/79
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Applying KKT conditions
Replacing �★ 0 by �0 > 0 in constraint (6) gives ⎛ ⎞ ⎛
∂�(N�) ∂� (N�) ⎞⎛
⎞
−
⎜⎜⎜⎜⎜⎜⎜⎜⎝∂�1
∂�(N�) ∂�2 . . .
∂�(N�)
⎟⎟⎟⎟⎟⎟⎟⎟⎠
⎜⎜⎜⎜⎜⎜⎜⎜⎝
∂�1 ∂� (N�) ∂�2 . . .
∂� (N�)
⎟⎟⎟⎟⎟⎟⎟⎟⎠
0
=⎜⎜⎜⎝
⎟⎟⎟⎠
0 .. .
− �0 (8),
0
∂��−1 ∂��−1
where N� is the unique solution of (8). Note that N� is the solution of the following optimization problem:
min N
− �̄(N) = −�(N) + �0( �̂ − � (N))
(9)
s.t. �min − �� ≤ 0, ∀� = 1, ⋅ ⋅ ⋅ , � − 1.
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⋅ ⋅ ⋅
⋅ ⋅ ⋅
⋅ ⋅ ⋅
Applying KKT conditions The problem above is equivalent to
max �̄(N) = �(N) − �0(�̂ − � (N)) N
(10)
s.t. �min − �� ≤ 0, ∀� = 1, , � − 1.
or
�−1 �−1∑ ∑ max �̄(N) = �� (N) − ���� − ���̄� − �0(�̂ − � (N)) N
�=1 �=1 (11)
s.t. �min − �� ≤ 0, ∀� = 1, , � − 1.
or �−1 �−1∑ ∑
max �̄(N) = (� + �0)� (N) − ���� − ���̄� N (12)
�=1 �=1 s.t. �� ≥ �min, ∀� = 1, , � − 1.
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⋅ ⋅ ⋅
⋅ ⋅ ⋅
⋅ ⋅ ⋅
Applying KKT conditions The problem above is equivalent to
max �̄(N) = �(N) − �0(�̂ − � (N)) N
(10)
s.t. �min − �� ≤ 0, ∀� = 1, , � − 1.
or
�−1 �−1∑ ∑ max �̄(N) = �� (N) − ���� − ���̄� − �0(�̂ − � (N)) N
�=1 �=1 (11)
s.t. �min − �� ≤ 0, ∀� = 1, , � − 1.
or �−1 �−1∑ ∑
max �̄(N) = (� + �0)� (N) − ���� − ���̄� N (12)
�=1 �=1 s.t. �� ≥ �min, ∀� = 1, , � − 1.
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⋅ ⋅ ⋅
⋅ ⋅ ⋅
⋅ ⋅ ⋅
Applying KKT conditions The problem above is equivalent to
max �̄(N) = �(N) − �0(�̂ − � (N)) N
(10)
s.t. �min − �� ≤ 0, ∀� = 1, , � − 1.
or
�−1 �−1∑ ∑ max �̄(N) = �� (N) − ���� − ���̄� − �0(�̂ − � (N)) N
�=1 �=1 (11)
s.t. �min − �� ≤ 0, ∀� = 1, , � − 1.
or �−1 �−1∑ ∑
max �̄(N) = (� + �0)� (N) − ���� − ���̄� N (12)
�=1 �=1 s.t. �� ≥ �min, ∀� = 1, , � − 1.
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⋅ ⋅ ⋅
Applying KKT conditions
or, finally,
�−1 �−1∑ ∑ max �̄(N) = �′� (N) − ���� − ���̄� N
�=1 �=1 (13)
s.t. �� ≥ �min, ∀� = 1, , � − 1.
where �′ = � + �0. This is exactly the unconstrained problem, and N�
is its optimal solution. Note that �0 > 0 indicates that �′ > �.
In addition, the KKT conditions indicate that the optimal solution of the constrained problem N★ satisfies � (N★) = �̂ . This means that, for every �′ > � (or �0 > 0), we can find the corresponding optimal solution N�
satisfying condition (8) by solving problem (13). We need to find the �′ such that the solution to problem (13), denoted as N★(�′), satisfies � (N★(�′)) = �̂ .
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⋅ ⋅ ⋅
Applying KKT conditions
or, finally,
�−1 �−1∑ ∑ max �̄(N) = �′� (N) − ���� − ���̄� N
�=1 �=1 (13)
s.t. �� ≥ �min, ∀� = 1, , � − 1.
where �′ = � + �0. This is exactly the unconstrained problem, and N�
is its optimal solution. Note that �0 > 0 indicates that �′ > �.
In addition, the KKT conditions indicate that the optimal solution of the constrained problem N★ satisfies � (N★) = �̂ . This means that, for every �′ > � (or �0 > 0), we can find the corresponding optimal solution N�
satisfying condition (8) by solving problem (13). We need to find the �′ such that the solution to problem (13), denoted as N★(�′), satisfies � (N★(�′)) = �̂ .
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d
Applying KKT conditions
Then, �0 = �′ − � and N★(�′) satisfy conditions (6) and (7):
−∇�(N★(�′)) + �★ 0∇(�̂ − � (N★(�′))) = 0,
�★ 0(�̂ − � (N★(�′))) = 0.
Hence, �★ = �′ −� is exactly the Lagrange multiplier satisfying the KKT 0 conditions of the constrained problem, and N★ = N★(�′) is the optimal solution of the constrained problem.
Consequently, solving the constrained problem through our algorithm is essentially finding the unique Lagrange multipliers and optimal solution of the problem.
c⃝2010 Chuan Shi — Topic one: Line optimization : Proofs of the algorithm by KKT conditions 43/79
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d
Applying KKT conditions
Then, �0 = �′ − � and N★(�′) satisfy conditions (6) and (7):
−∇�(N★(�′)) + �★ 0∇(�̂ − � (N★(�′))) = 0,
�★ 0(�̂ − � (N★(�′))) = 0.
Hence, �★ = �′ −� is exactly the Lagrange multiplier satisfying the KKT 0 conditions of the constrained problem, and N★ = N★(�′) is the optimal solution of the constrained problem.
Consequently, solving the constrained problem through our algorithm is essentially finding the unique Lagrange multipliers and optimal solution of the problem.
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Algorithm summary for case 2
Solve unconstrained problem
Solve, by a gradient method, the unconstrained prob-lem for fixed �′
max N
�(N) = �′� (N) − �−1∑
�=1
���� − �−1∑
�=1
�� ̄��(N)
s.t. �� ≥ �min, ∀� = 1, ⋅ ⋅ ⋅ , � − 1.
Search
Do a one-dimensional search on �′ > � to find �′
such that the solution of the unconstrained problem, N★(�′), satisfies
� (N★(�′)) = �̂ .
P(N*(A'))=P?
Solve unconstrained problem
Search: Choose A'
Yes
No
Quit
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⋅ ⋅ ⋅
Numerical results
Numerical experiment outline
Experiments on short lines.
Experiments on long lines.
Computation speed.
Method we use to check the algorithm
�̂ surface search in (�1, , ��−1) space. All buffer size allocations, N, such that � (N) = �̂ compose the �̂ surface.
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�̂ surface search
15
20
25
30
35
40 40 45 50 55 60 65 70
70
75
80
85
90
N3
P surfaceThe optimal solution
N1
N2
Figure 6: �̂ Surface search
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Experiment on short lines (4-buffer line)
Line parameters: �̂ = .88
machine �1 �2 �3 �4 �5
� .11 .12 .10 .09 .10 � .008 .01 .01 .01 .01
Machine 4 is the least reliable machine (bottleneck) of the line.
Cost function
4 4∑ ∑ �(N) = 2500� (N) − �� − �̄�(N)
�=1 �=1
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Experiment on short lines (4-buffer line)
Results
Optimal solutions
�̂ Surface Search The algorithm Error Rounded �★
Prod. rate .8800 .8800 .8800 �★
1 28.85 28.8570 0.02% 29.0000 �★
2 58.46 58.5694 0.19% 59.0000 �★
3 92.98 92.9068 0.08% 93.0000 �★
4 87.39 87.4415 0.06% 87.0000 �̄1 19.0682 19.0726 0.02% 19.1791 �̄2 34.3084 34.3835 0.23% 34.7289 �̄3 48.7200 48.6981 0.04% 48.9123 �̄4 31.9894 32.0063 0.05% 31.9485
Profit ($) 1798.2 1798.1 0.006% 1797.4000
The maximal error is 0.23% and appears in �̄2.
Computer time for this experiment is 2.69 seconds.
c⃝2010 Chuan Shi — Topic one: Line optimization : Numerical results 48/79
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Experiment on long lines (11-buffer line)
Line parameters: �̂ = .88
machine �1 �2 �3 �4 �5 �6
� .11 .12 .10 .09 .10 .11 � .008 .01 .01 .01 .01 .01
machine �7 �8 �9 �10 �11 �12
� .10 .11 .12 .10 .12 .09 � .009 .01 .009 .008 .01 .009
Cost function
11 11∑ ∑ �(N) = 6000� (N) − �� − �̄�(N)
�=1 �=1
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Experiment on long lines (11-buffer line)
Results
Optimal solutions, buffer sizes:
�̂ Surface Search The algorithm Error Rounded �★
Prod. rate .8800 .8800 .8799 �★
1 29.10 29.1769 0.26% 29.0000 �★
2 59.20 59.2830 0.14% 59.0000 �★
3 97.80 97.7980 0.002% 98.0000 �★
4 107.50 107.4176 0.08% 107.0000 �★
5 84.50 84.4804 0.02% 84.0000 �★
6 70.80 70.6892 0.17% 71.0000 �★
7 63.10 63.1893 0.14% 63.0000 �★
8 53.10 52.9274 0.33% 53.0000 �★
9 47.20 47.2232 0.05% 47.0000 � ★
10 47.90 47.7967 0.22% 48.0000 � ★
11 48.80 48.7716 0.06% 49.0000
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Experiment on long lines (11-buffer line)
Results (continued)
Optimal solutions, average inventories:
�̂ Surface Search The algorithm Error Rounded �★
�̄1 19.2388 19.2986 0.31% 19.1979 �̄2 34.9561 35.0423 0.25% 34.8194 �̄3 52.5423 52.6032 0.12% 52.6833 �̄4 45.1528 45.1840 0.07% 45.0835 �̄5 34.4289 34.4770 0.14% 34.2790 �̄6 30.7073 30.7048 0.01% 30.8229 �̄7 28.0446 28.1299 0.30% 28.0902 �̄8 21.5666 21.5438 0.11% 21.5932 �̄9 21.5059 21.5442 0.18% 21.4299 �̄10 22.6756 22.6496 0.11% 22.7303 �̄11 20.8692 20.8615 0.04% 20.9613
Profit ($) 4239.3 4239.2 0.002% 4239.5000
Computer time is 91.47 seconds.
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Experiments for Tolio, Matta, and Gershwin (2002) model
Consider a 4-machine 3-buffer line with constraints �̂ = .87. In addition, � = 2000 and all �� and �� are 1.
machine �1 �2 �3 �4
��1 .10 .12 .10 .20 ��1 .01 .008 .01 .007 ��2 – .20 – .16 ��2 – .005 – .004
�̂ Surf. Search The algorithm Error � (N★) .8699 .8699 �★
1 29.8600 29.9930 0.45% �★
2 38.2200 38.0206 0.52% �★
3 20.6800 20.7616 0.39% �̄1 17.2779 17.3674 0.52% �̄2 17.2602 17.1792 0.47% �̄3 6.1996 6.2121 0.20%
Profit ($) 1610.3000 1610.3000 0.00%
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Experiments for Levantesi, Matta, and Tolio (2003) model
Consider a 4-machine 3-buffer line with constraints �̂ = .87. In addition, � = 2000 and all �� and �� are 1.
machine �1 �2 �3 �4
�� 1.0 1.02 1.0 1.0 ��1 .10 .12 .10 .20 ��1 .01 .008 .01 .012 ��2 – .20 – .16 ��2 – .005 – .006
� ★ Surf. Search The algorithm Error � (N★) .8699 .8700 �★
1 27.7200 27.9042 0.66% �★
2 38.7900 38.9281 0.34% �★
3 34.0700 34.1574 0.26% �̄1 15.4288 15.5313 0.66% �̄2 19.8787 19.9711 0.46% �̄3 13.8937 13.9426 0.35%
Profit ($) 1590.0000 1589.7000 0.02%
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Computation speed
Experiment
Run the algorithm for a series of experiments for lines having iden-tical machines to see how fast the algorithm could optimize longer lines.
Length of the line varies from 4 machines to 30 machines.
Machine parameters are � = .01 and � = .1.
In all cases, the feasible production rate is �̂ = .88.
The objective function is
�−1 �−1∑ ∑ �(N) = �� (N) − �� − �̄�(N).
�=1 �=1
where � = 500� for the line of length �.
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Computation speed
0
500
1000
1500
2000
2500
3000
5 10 15 20 25 30
Com
pute
r tim
e (s
econ
ds )
Length of production lines
1 min3 mins
10 mins
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Algorithm reliability
We run the algorithm on 739 randomly generated 4-machine 3-buffer lines. 98.92% of these experiments have a maximal error less than 6%.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
50
100
150
200
250
300
350
400
450
Maximal Error
Freq
uen
cy
c⃝2010 Chuan Shi — Topic one: Line optimization : Numerical results 56/79
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Algorithm reliability
Taking a closer look at those 98.92% experiments, we find a more accurate distribution of the maximal error. We find that, out of the total 739 experiments, 83.90% of them have a maximal error less than 2%.
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
10
20
30
40
50
60
Maximal Error
Freq
uen
cy
c⃝2010 Chuan Shi — Topic one: Line optimization : Numerical results 57/79
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2.852 Manufacturing Systems Analysis Spring 2010
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