LAGRANGEAN EXAMPLE
Prof. Stephen Graves
Production quantities for metal stamping (from David B. Kletter’s SM thesis,
Determining Production Lot Sizes and Safety Stocks for an Automobile Stamping Plant,
1994) : A metal stamping plant for an automobile manufacturer will produce 100’s of
parts on dozens of stamping lines to satisfy part demand from the assembly plants. In
running the stamping plant, a key planning decision is the assignment of parts to
stamping lines. Once this assignment is made, then inventory control policies need to be
established for the family of parts assigned to each line. A common control policy is to
determine the relative production frequency (and/or batch size) for each part and the
necessary buffer stock to assure a high level of service for the assembly plants.
The following describes how to find the production batch sizes.
n
Si Di H i Qi
min ∑
+
2
i =1 Qi
n
s.t.
Ti Di
=K
∑
i =1 Qi
Qi = production quantity or lot size
Si
= setup cost
Hi = holding cost
Ti
= setup time
Di = annual demand
K
= available capacity for setups
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Introduce Lagrange Multiplier: λ
⎛ n
⎞
n S Di H iQ
⎜
⎟
T
D
i
i
i
i
⎜
L(Q,λ ) = ∑
+
+ λ ⎜∑
− K ⎟⎟
⎜⎜
⎟⎟
Qi
2
⎝ i =1
⎠
i =1 Qi
Approach:
• for given λ, find Qi that minimize L (Q, λ);
• then choose λ such that Qi satisfy original constraint.
Look at first-order conditions:
∂L = 0
∂ Qi
for i = 1, ... n
∂L = 0
∂λ
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∂ L = − Si Di + Hi − λTi Di = 0
∂ Qi
2
Qi2
Qi2
⇒ Qi2 = 2 ( S i + λT i ) Di
Hi
*
Qi
=
2
+
H i ( S i λT i ) Di
Now choose λ such that
∂ L = n T i Di − K = 0
∑
∂λ i=1 Q*i
n
∑
i =1
T i Di
2
+ λT ) D
(
S
i
i
i
Hi
−K =0
Solve by univariate search.
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Note that
∂ L = −λ
∂K
⇒
λ is a shadow price on the capacity, and equals the amount
we would pay for additional capacity, on the margin.
This approach is very useful for removing "troublesome"
constraints.
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Lagrange Example
Limit
Total
Item 1
Item 2
Item 3
25
10000
1
20
707.11
40
20000
1
10
1264.91
50
5000
1
25
707.11
617.73
2679.12
282.84
707.11
158.11
1264.91
176.78
707.11
400
160.77
122.91
116.32
2898.16
822.98
1305.23
769.95
Annual
Setup
Time
Quantity
Item 1
Quantity
Item 2
Quantity
Item 3
total cost
617.73
496.58
250.97
326.39
405.19
359.27
380.09
392.11
398.69
402.11
400.39
399.54
399.96
400.18
400.07
400.02
707.11
948.68
2121.32
1581.14
1224.74
1414.21
1322.88
1274.36
1249.00
1236.12
1242.58
1245.79
1244.19
1243.38
1243.78
1243.99
1264.91
1414.21
2366.43
1897.37
1612.45
1760.68
1688.19
1650.45
1630.95
1621.11
1626.04
1628.50
1627.27
1626.65
1626.96
1627.11
707.11
866.03
1732.05
1322.88
1060.66
1198.96
1131.92
1096.59
1078.19
1068.88
1073.55
1075.87
1074.71
1074.13
1074.42
1074.56
2679.12
2732.34
3710.15
3169.42
2884.89
3026.59
2955.22
2919.57
2901.61
2892.68
2897.14
2899.38
2898.26
2897.70
2897.98
2898.12
Setup cost($)
Demand (per yr)
Hold. cost($/un-yr)
Setup time (hrs)
EOQ
EOQ setup time/year
EOQ Cost
Avail. Setup Time
400
Constrained cost
Lagrange multiplier
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10
5
2.5
3.75
3.125
2.81
2.65
2.57
2.61
2.63
2.62
2.615
2.6175
2.61875
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Microsoft Excel Answer Report
Worksheet: Lagrange example
Target Cell (Min)
Cell
$C$19
Name
Constrained cost Total
Original Value
Final Value
2679.12
2898.16
Original Value
Final Value
707.11
1264.91
707.11
1244.05
1627.16
1074.61
Cell Value
Formula
Adjustable Cells
Cell
$D$4
$E$4
$F$4
Name
Quantity Item 1
Quantity Item 2
Quantity Item 3
Constraint
Cell
$C$17
$D$4
$E$4
$F$4
Name
Avail. Setup Time Total
Quantity Item 1
Quantity Item 2
Quantity Item 3
400
1244.05
1627.16
1074.61
$C$17<=$B$17
$D$4>=0
$E$4>=0
$F$4>=0
Status
Slack
Binding
Not Binding
Not Binding
Not Binding
Microsoft Excel Sensitivity Report
Worksheet: Lagrange example
Changing Cells
Cell
$D$4
$E$4
$F$4
Name
Quantity Item 1
Quantity Item 2
Quantity Item 3
Final
Value
Reduced
Gradient
1244.05
1627.16
1074.61
0.00
0.00
0.00
Final
Value
Lagrange
Multiplier
400
-2.61915
Constraints
Cell
$C$17
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Avail. Setup Time
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Summer 2003
0
1244.05
1627.16
1074.61