Transcript Nonlinear Programming and Inventory Control (Multiple Items)

Nonlinear Programming
and Inventory Control
(Multiple Items)
Multiple Items
• Materials management involves many items and
transactions
• Uneconomical to apply detailed inventory control
analysis to all items
• Because a small percentage of inventory items
accounts for most of the inventory value
• Must focus on important items
• Isolate those items requiring precise control
• ABC analysis indicates where managers should
concentrate
ABC Analysis
• Divides inventory into three classes according to dollar
volume (dollar volume=Annual demand *unit purchase
cost)
• A class is high value items whose dollar volume
accounts for 75-80% of the total inventory value, while
representing 15-20% of the inventory items
• B class is lesser value items whose dollar volume
accounts for 10-15% of the total inventory value, while
representing 20-25% of the inventory items
• C class is low value items whose volume accounts for 510% of the total inventory value, while representing 6065% of the inventory items
• Thus, the same degree of control is not justified for all
items
Example
Item
Annual usage
cost
Annual dollar usage
Rank
G-1
40,000
0.07
2,800
5
G-2
195,000
0.11
21,450
1
G-3
4000
0.10
400
9
M-1
100,000
0.05
5000
3
M-2
2,000
0.14
280
10
M-3
240,000
0.07
16,800
2
M-4
16,000
0.08
1280
6
P-1
80,000
0.06
4800
4
P-2
10,000
0.07
700
7
P-3
5,000
0.09
450
8
53,960
$A:75-80
%A:15-20
$C:5-10
Example-Cont. $B:10-15
%B:20-25 %C:60-65
Item
Annual
dollar usage
Cumulative
dollar usage
%
Cumulative
%
class
G-2
21,450
21,450
39.8
39.8
A
M-3
16,800
38,250
31.1
70.9
A
M-1
5000
43250
9.3%
80.2
B
P-1
4800
48050
8.8%
89.0
B
G-1
2800
50850
5.2%
94.2
B
M-4
1280
52130
2.4%
96.6
C
P-2
700
52,830
1.3%
97.9
C
P-3
450
53,280
0.8%
98.7
C
G-3
400
…
…
C
M-2
280
…
100.0
C
53,960
53,960
EOQ and Multiple Items
• Most inventory systems stock many items
• May treat each item individually and then
add them up
• Restrictions may be imposed (limited
warehouse capacity, upper limit on the
maximum dollar investment, number of
orders per year, and so on)
• Will be covered after covering NLP
Overview of NLP
• Many realistic problems have nonlinear functions
• When LP problems contain nonlinear functions, they are
referred NLP
• Have a separate name, because they are solved
differently
• In LP, solutions are found at the intersections
• In NLP, there may no corner point
• Solution space can be undulating line or surface
• Like a mountain range with many peaks and valleys
• Optimal point at the top of any peak or at the bottom of
any valley
• In NLP, we have local and global points
Local and Global Optimal Point
• Solution techniques
generally search for high
points or low points
• Difficulty of NLP is to
determine whether the
identified is a local or
global optimal point
• Global optimal point can
be found by the very
complex mathematical
techniques
Global optimal point
Local optimal points
Constrained and Unconstrained
Optimization
• Constrained optimization
• Profit function: Z= vp-cf-vcv,
v=1500-24.6p, cf=$10,000
and cv=$8
• vp creates a curvilinear
relationship
• Results in quadratic
function:
profit
Z= 1696.8p-24.6p2-22,000
•
δZ/δp=0, 1696.8-49.2p=0, p=34.49
• Called classical unconstrained
optimization
price
Constrained Optimization
Max Z= 1696.8p-24.6p2-22,000
s.t.
p<=20
• Referred NLP
• Solution is on the boundary
formed by constraint
• Change RHS from 20 to 40
• Solution is no longer on the
boundary
• Makes process of finding optimal
solution difficult, particularly, with
more variables and constraints
Feasible
space
P=20
price
Feasible
space
P=40
price
Single Facility Location Problem
• Locate a centralized facility
that serves several
customers
• Minimize the total miles
traveled between the
facility and all customers
• Locations of the cities and
the number of trips are:
(20,20,75), (10,35,105),
(25,9,135), (32,15, 60),
(18,8,90)
• d=[(xi-x)2+(yi-y)2]1/2
• Min Σditi
city1
city3
city5
city2
city4
NLP with Multiple Constraints
• Consider a problem with two constraints
• A company produces two products and the production is subject to
resource constraints
• Demand of each product is dependent on the price (x1=1500-24.6p1
and x2=2700-63.8p2)
• Cost of producing x1 and x2 are 12 and 9
• Production resources are (2x1+2.7 x2<=6000, 3.6x1+2.9x2<=8500,
7.2x1+8.5 x2<=15,000)
• Model: Max z=(p1-12)x1+(p2-9)x2
s.t.
2x1+2.7 x2<=6000
3.6x1+2.9x2<=8500
7.2x1+8.5 x2<=15,000
• Where x1=1500-24.6p1 and x2=2700-63.8p2
• Decision variables are?
Solution Techniques
• Very complex
• Two method
– Least or Substitution method
• Transferring a constraint optimization to an
unconstraint optimization
– Lagrange multiplier
Substitution Method
• Restricted to models containing only
equality constraints
• Involves solving the constraint for one
variable for another
• New expression will be substituted into the
objective function to eliminate it completely
Example
• Max Z= vp-cf-vcv
s.t.
v=15,00 -24.6 p
• cf=$10,000 and cv=$8
• vp creates a curvilinear relationship
• Constraint has been solved v for p
• Substitute it in objective function
Z= 1500p-24.6p2-cf-1500cv+24.6 pcv
Z= 1696.8p-24.6p2-22,000
• Differentiating and setting it equal to zero
• 0=1696.8-49.2p
• p=34.49
Example
• Consider the Furniture Company
• Assume contribution of each declines as the quantity
increases
• Relationship for x1: $4-0.1x1
• Relation for x2: $5-0.2 x2
• Profit earned form each:($4-0.1x1)x1 and ($5-0.2 x2)x2
• Total profit: z=4x1+5x2-0.1x12-0.2x22
• Consider just one constraint: x1+2x2=40 or x1=40-2x2
• z=4(40-2x2)+5x2-0.1 (40-2x2)-0.2x22
• z=13x2-0.6x22
• x1=18.4, z=$70.42
Limitation of Substitution
• Highest order of decision variable was a
power of two
• Dealt only with two decision variables and
single constraint
Lagrange Multiplier
• Used for constraint optimization consisting of nonlinear
objective function and constraints
• Transform objective function into a Lagrangian function
• Constraints as multiples of Lagrange multipliers are
subtracted from the objective function
• Consider an example
Max z=4x1+5x2-0.1x12-0.2x22
s.t.
x1+2x2=40
• Forming Lagrangian function
L=4x1+5x2-0.1x12-0.2x22 - (x1+2x2-40)
Lagrange Multiplier-Cont.
• Partial derivatives of L with respect to each
of variables
• δL/δx1=0, δL/δx2=0, δL/δ=0
• 4-0.2x1=0, 5-0.4-2=0, x1-2x2+40=0
• X1=18.3, x2=10.8, =0.33, z=70.42
Sensitivity Analysis
• Lagrangian multiplier, , is analogous to dual
variables in LP
• Shows changes in objective function value by
changing the RHS
• Positive value of  shows the increase in
objective function
• Example
Max z=4x1+5x2-0.1x12-0.2x22
s.t.
x1+2x2=41
• x1=18.8, x2=11.2, =0.27, z=70.75
Mathematical Notations
•
•
•
•
Di=Annual demand for item i in units
Ci=Unit purchase cost of item i in dollar
Ai= ordering cost of item i in dollar
fi=Required storage space for item i in square
foot
• F=Maximum total storage space available
• TC=Total average annual costs in dollar
• Hi= Annual holding cost per unit i per year in
dollar
Application of NLP and
Inventory Modeling
Example-Limited Working Capital
Demand
Item cost
Order cost
Item 1
Item 2
Item 3
EOQ QC
1000
50
20
1000
20
50
2000
80
50
100
158
112
5000
3160
8960
Total=17120
Available budget=15,000
Inventory carrying charge=0.2
Solution
• Solved by the method of
Lagrange-multiplier
• Before applying the method, we should
solve the total cost function by ignoring the
constraint
• Otherwise, we form the Lagrangian
expression
Lagrangian Expression
• Total budget=17120> available budget=15000
• Form the Lagrangian Expression
L
n
n
n
D C   A D /Q  
i
i 1
i
i
i
i
i 1
i 1
n
H i Qi
  ( Qi Ci  B)
2
i 1

• Derivative with respect to Qi and lambda
L
 0   Ai Di / Qi2  H i / 2  Ci
Qi
L
0

n
C Q  B  0
i
i 1
i
Final Result
• From the first equation
2 Ai Di
Ci (i  2 )
Qi* 
• From the second equation
n
C Q
i
i
B
i 1
n
C
i
i 1
n

i 1
2 Ai Di Ci
B
Ci (i  2 )
2 Ai Di Ci
B
(i  2 )
Problem Solution
HW Assignment #1
1
2
3
4
Demand/year
1000
5000
10000
8000
Ordering cost ($)
6
10
10
8
Cost/unit ($)
10
3
5
2
Floor space required
sq.ft
5
1
1
1.5
•Assume that the annual inventory carrying charge is 10%
and that 15000 sq.ft of floor space are available.
•What is the optimal inventory policy for these items.
•Determine the cost of having only 15,000 sq.ft of floor space?
HW#2
Item
Annual Usage
Cost
($)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
10,000
4,000
15,000
20,000
3,000
40,000
25,000
12,000
30,000
35,000
26,000
38,000
16,000
27,000
24,000
10
15
5
3.5
18
2
4.5
6
3
1.8
2
0.5
6
1
7
Classify these items into A, B, and C.
HW#3
• The Furniture Company has developed the
following NLP model to determine the optimal
number of chairs and tables to produce daily.
Max z=7x1-0.3x12 +8x2-0.4x22
S.t.
4x1+5x2=100 hr
• Determine the optimal solution to this NLP using
the substitution method.
• Determine the optimal solution to this NLP using
the Lagrange multipliers.
HW#4
• Consider the following NLP model to
determine solution.
Max z=30x1-2x12 +25x2-0.5x22
S.t.
3x1+6x2=300
• Determine the optimal solution to this NLP
using the substitution method.
• Determine the optimal solution to this NLP
using the Lagrange multipliers.
HW#5
• Section 11.2c, Problems 1 and 4