Mathematics for Management Science Notes 07
prepared by Professor Jenny Baglivo
© Jenny A. Baglivo 2002. All rights reserved.
Calculus and nonlinear programming (NLP):
In nonlinear programming (NLP), either the objective function or the constraints or both
are nonlinear functions of the decision variables.
Solver will use a sequence of approximations to search for a solution. There are several
important things to note about Solver implementations:
•
•
•
•
You need to supply a valid feasible point as a starting point of the search. (In all of
the previous cases, initializing all the changing cells with zero was fine. In fact, the
algorithms never looked at the initial values in the changing cells.)
The algorithm then takes “steps” in its search for an optimal solution. At each point,
the direction of the search is the one judged to give the maximum improvement in the
objective function value.
At the point where the algorithm judges that no improvement is possible from the
current position, it declares victory and reports the results.
You may not get the optimum solution.
(1) You may be at a local optimum instead of a global optimum.
(2) The step size may not be set correctly to judge whether or not searching
further would improve the optimum value.
•
To counteract problem (1), you can start the search at many different initial values,
using points that are of the same magnitude as where you expect the solution to be.
To counteract problem (2), you can use an “automatic scaling” option in the Solver
options box.
Ideas from calculus are central to NLP problems. So, we will review one-variable
calculus and introduce ideas from two-variable calculus.
Example 1: Consider the one-variable function
f(x) = –3 x4 + 4 x3 + 12 x2 + 40
when 0 ≤ x ≤ 3,
whose graph is at the top of the next page. Notice that
•
•
The maximum value is f(2)=72, which occurs at an interior point of the interval and
The minimum value is f(3)=13, which occurs at an endpoint of the interval.
page 1 of 17
We will consider the slope of the curve at various points in the interval. For example, the
slope at the point (1,f(1)) = (1,53) is obtained as the limit of a sequence of
approximations, as illustrated below:
x
0.000
0.500
0.900
0.990
0.999
f(x)
40.0000
43.3125
50.6677
52.7606
52.9760
Slope = (f(x)-f(1))/(x-1)
13.0000
19.3750
23.3230
23.9392
23.9940
The slope of the curve when x=1 is the limit of the approximations: f’(1) = 24.
page 2 of 17
Rules for finding the derivative function (f’(x) ) in simple cases:
(1) The derivative of a constant function is 0:
d(C)/dx = 0
where C is a constant.
(2) The derivative of a linear function is its slope:
d(m x + b)/dx = m
where m and b are constants.
(3) The derivative of a simple power function is as follows:
d(xn)/dx = n xn-1
where n is a constant power.
(4) The derivative of the sum of two functions is the sum of the derivatives:
d(f(x)+g(x))/dx = d(f(x))/dx + d(g(x))/dx = f’(x) + g’(x).
(5) The derivative of a constant multiple of a function is the constant times the derivative
of the function.
d(C f(x))/dx = C d(f(x))/dx = C f’(x).
When a function has a derivative, it is said to be differentiable.
Example 1, continued: Fill-in the blanks
f’(x) = _______________________________________________________
f’(1) = _____________________
f’(2) = _____________________
f’(3) = _____________________
page 3 of 17
Example 2: Consider the function
f(x) =
2500/x + 2 + 0.00125 x2
when 10 ≤ x ≤ 400.
Fill-in the blanks:
f’(x) = _______________________________________________________
f’(100) = _____________________
f’(200) = _____________________
f’(300) = _____________________
page 4 of 17
Local and global optimum points:
• f(x) has a local maximum at x=C (or at (x,y)=(C,f(C))) when f(C) is the largest
value of the function for all x near C.
•
•
•
f(x) has a local minimum at x=C (or at (x,y)=(C,f(C))) when f(C) is the smallest
value of the function for all x near C.
f(x) has a global maximum at x=C (or at (x,y)=(C,f(C))) when f(C) is the largest
value of the function for all x under consideration.
f(x) has a global minimum at x=C (or at (x,y)=(C,f(C))) when f(C) is the smallest
value of the function for all x under consideration.
If f(x) is continuous on a ≤ x ≤ b, then f assumes a global maximum and a global
minimum on that interval.
Example 1 (the interval is 0 ≤ x ≤ 3): Fill-in the blanks
Local maximum at ___________________________
Local minimum at ___________________________
Global maximum at __________________________
Global minimum at __________________________
Example 2 (the interval is 10 ≤ x ≤ 400): Fill-in the blanks
Local maximum at ___________________________
Local minimum at ___________________________
Global maximum at __________________________
Global minimum at __________________________
page 5 of 17
The second derivative test for finding local maxima and local minima:
Suppose that f’(C) = 0. Then C is said to be a critical value of the function and the
point (C,f(C)) is said to be a critical point of the function.
Recall that the second derivative function (f’’(x)) is the derivative of the derivative
function. The second derivative is related to the concavity of the function:
1. If f’’(x) > 0, then the curve is concave up (or cupped up) at (x, f(x)).
2. If f’’(x) < 0, then the curve is concave down (cupped down) at (x, f(x)).
The second derivative test can be used to determine local maxima and local minima when
f’(C) = 0 and f’’(C) ≠ 0.
• If f’(C) = 0 and f’’(C) > 0, then there is a local minimum at x = C.
• If f’(C) = 0 and f’’(C) < 0, then there is a local maximum at x = C.
Example 2, continued: Use the second derivative test to check that (100,f(100)) is a local
minimum point.
page 6 of 17
Example 3: Consider the following function
f(x) = –3 x4 + 4 x3 + 12 x2 + 40 for all real numbers x.
(a) Find all critical values and critical points of f(x).
(b) For each critical point, use the second derivative to determine if that point is a local
maximum or a local minimum for the function.
page 7 of 17
Example 4: A manufacturer has been selling 1000 TV sets a week at $450 each. A market
survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will
increase by 100 per week.
Let x equal the number of TV sets sold in a given week.
(a) Write weekly revenue as a function of x.
(b) Use calculus to determine how large a rebate the company should offer in order to maximize
its revenue.
(c) If the company's weekly cost is C(x) = 68000 + 150 x, how should it set the size of the rebate
in order to maximize its profit?
page 8 of 17
Application type: economic order quantity (EOQ) model
Assume that demand for a certain product is constant over the year and that each new
order is delivered in full when the inventory level reaches zero. The problem is to
determine the optimal number of units of a product to purchase whenever an order is
placed.
Standard notation:
D
C
H
R
Total demand during the year
Unit purchase cost
Cost of holding one unit in inventory
(often given as a percent of C)
Cost of placing an order (reorder cost)
Let x equal the number of items ordered at one time (the order quantity). Then the total
cost can be written as follows:
Total Cost = Purchasing Cost + Inventory Cost + Ordering Cost
f(x) = D C + H (x/2) + R (D/x).
The basic model is: Minimize f(x) subject to 1 ≤ x ≤ D.
Notes:
(1) Since the demand is assumed to be constant, the average number of units on hand is x/2.
(2) If the total demand is D and you order x at a time, then you place D/x orders.
(3) You may have further limitations to worry about.
page 9 of 17
Example 5: Alan Wang is responsible for purchasing the paper used in all the copy machines
and laser printers at the corporate headquarters of MetroBank. Alan projects that in the coming
year he will need to purchase a total of 24,000 boxes of paper, which will be used at a fairly
steady rate throughout the year. Each box of paper costs $35. Alan estimates that it costs $50
each time an order is placed (this includes the cost of placing an order and the related costs of
shipping and receiving). MetroBank assigns a cost of 18% to funds allocated to supplies and
inventories because such funds are the lifeline of the bank and could be lent out to credit card
customers who are willing to pay this rate on money borrowed from the bank. Alan has been
placing paper orders once a quarter, but he wants to determine if another ordering pattern would
be better. He wants to determine the most economical order quantity to use in purchasing paper.
• Use calculus to solve Alan's problem.
page 10 of 17
Example 5 solution and formulas sheets:
A
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
Metro
B
C
D
E
F
Bank
Annual demand
Holding cost as % of unit cost
Cost per box
Reorder cost
24000
18%
35
50
Minimum
Reorder bounds
Maximum
1
24000
Minimize B20
By Changing B15 (initial value 100)
Subject to
B24 >= D24
B25 <= D25
MODEL
Decision
Variable
Order Quantity
#Boxes
617.2133995
Purchase cost:
Inventory cost:
Reorder cost:
840000
1944.222208
1944.222211
843888.4444
Minimize total cost:
Options:
Use Automatic Scaling
Assume non-negative
Subject to
LHS
Minimum ordered
Maximum ordered
RHS
617.2133995
617.2133995
>=
<=
A
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
Metro
1
24000
B
C
D
Bank
Annual demand
Holding cost as % of unit cost
Cost per box
Reorder cost
24000
0.18
35
50
Reorder bounds
Minimum
1
Maximum
24000
MODEL
Decision
Variable
Order Quantity
#Boxes 100
Purchase cost: =B3*B5
Inventory cost: =(B4*B5)*(B15/2)
Reorder cost: =B6*(B3/B15)
Minimize total cost: =SUM(B17:B19)
Subject to
LHS
Minimum ordered
Maximum ordered
=B15
=B15
RHS
>=
<=
page 11 of 17
=B9
=C9
G
Example 6: Keith Shoe Store carries a basic black dress shoe for men that sells at an
approximate constant rate of 500 pairs of shoes every three months. Keith's current buying policy
is to order 500 pairs each time an order is placed. It costs Keith $150 to place an order. The
annual holding cost rate is 25%. With the order quantity of 500, Keith obtains shoes at the lowest
possible unit cost of $32 per pair. Other quantity discounts offered by the manufacturer are as
follows:
Order Quantity:
1-149
150-299
300 or more
Price per pair:
$36
$34
$32
(a) What is the minimum cost order quantity for the shoes?
(b) What are the annual savings of the minimum cost plan over the policy currently being used
by Keith?
page 12 of 17
Example 6 solution sheet:
A
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
51
52
53
54
55
56
Keith's
Shoe
B
Annual demand
Holding cost as % of unit cost
Reorder cost
(1)
C
D
E
F
G
Store
2000
25%
150
Quantities 1-149
MODEL 1:
Cost per pair
Reorder minimum
Reorder maximum
36
1
149
Decision
Purchase cost:
Inventory cost:
Reorder cost:
Minimize total cost:
Subject to
RHS
149
149
>=
<=
1
149
Quantities 150-299
MODEL
Cost per pair
Reorder minimum
Reorder maximum
2:
34
150
299
Decision
Variable
Order Quantity
# P a i r s : 265.6845
Purchase cost:
Inventory cost:
Reorder cost:
Minimize total cost:
Subject to
68000
1129.159
1129.159
70258.32
LHS
Minimum ordered
Maximum ordered
(3)
72000
670.5
2013.423
74683.92
LHS
Minimum ordered
Maximum ordered
(2)
Variable
Order Quantity
#Pairs:
149
RHS
265.6845
265.6845
>=
<=
150
299
Quantities 300+
MODEL
Cost per pair
Reorder minimum
Reorder maximum
3:
32
300
2000
Decision
Variable
Order Quantity
#Pairs:
300
Purchase cost:
Inventory cost:
Reorder cost:
Minimize total cost:
Subject to
Minimum ordered
Maximum ordered
page 13 of 17
64000
1200
1000
66200
LHS
300
300
RHS
>=
<=
300
2000
Properties of the total cost function in the EOQ model:
•
•
•
The function f(x) is concave up throughout its domain and has a unique critical point.
Thus, that critical point corresponds to the global minimum.
At the global minimum, the inventory and order costs are equal.
The curve is very flat in the vicinity of the global minimum. (When you rely on
Solver to get the solution, it may stop far away from the minimum. )
Footnote on the Keith Shoe Problem:
There are three distinct total cost functions. The function for the first range (1-149) has
its global minimum at x=258.2, the function for the second range (150-299) has its global
minimum at x=265.7, and the function for the third range (300 or more) has its global
minimum at the point where x=274. Only in the second case was the global minimum in
the range of interest in the problem.
Footnotes on functions with one critical point:
1. If f''(x) > 0 throughout the domain of interest and f has a unique critical point when
x=a, then (a,f(a)) is the global minimum point.
2. If f''(x) < 0 throughout the domain of interest and f has a unique critical point when
x=a, then (a,f(a)) is the global maximum point.
page 14 of 17
These attachments are for examples in the next set of notes (notes08):
Solution and sensitivity reports for Example 4 of notes08 with an inequality constraint:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Lawn
King,
A
Inc.
B
Max advertising (thous):
C
D
E
MODEL
Thousands dollars:
Maximize
Variables
Direct Mail
1
sales:
1
Solver options:
Assume Non-negative
Use Automatic Scaling
32
Subject to
LHS
Advertising budget
G
Maximize B11
By Changing B9:C9
(Initial values 0.50, 0.50)
Subject to:
B14 <= D14
2
Decision
Radio
F
RHS
2
<=
2
Adjustable Cells
Cell
$B$9
$C$9
Final
Value
Name
Thousands dollars: Radio
Thousands dollars: Direct Mail
Reduced
Gradient
1
1
0
0
Constraints
Cell
Final
Value
Name
$B$14 Advertising budget LHS
page 15 of 17
Lagrange
Multiplier
2
5.9999978
H
Solution and sensitivity sheets for Example 6 of notes08:
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
B
C
D
E
F
G
H
TMC
Probability constant
NPV (in thousands)
Project 1
12
800
Engineers Available
Project 2
Project 3
9
700
Maximize B17
By Changing B12:D12
(Several initial values,including 0,0,0)
Subject to:
B20 <= D20
4
1200
36
MODEL
#Engineers:
Expected P(Success):
Maximize
Expected
NPV
Options:
Assume Non-negative
Use Automatic Scaling
Decision Variables
Project 1
Project 2
Project 3
12.23325
10.63140
13.13536
0.50481
0.54155
0.76656
1702.813
Subject to:
Total Engineers
LHS
RHS
36
<=
36
Adjustable Cells
Cell
Final
Value
Name
$B$12 #Engineers: Project 1 12.23325
$C$12 #Engineers: Project 2 10.63140
$D$12 #Engineers: Project 3 13.13536
Reduced
Gradient
0.00000
0.00000
0.00000
Constraints
Cell
Final
Value
Name
$B$20 Total Engineers LHS
page 16 of 17
36
Lagrange
Multiplier
16.34737
Formulas sheets for Examples 4 and 6 of notes08:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
King, Inc.
Lawn
B
D
Max advertising (thous): 2
MODEL
Decision
Variables
Radio
Direct Mail
Thousands dollars: 0.5
Maximize
sales:
0.5
=-2*B9^2-10*C9^2-8*B9*C9+18*B9+34*C9
Subject to
LHS
Advertising budget
RHS
=B9+C9
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
C
<=
B
C
=B3
D
TMC
Project 1
Probability constant
NPV (in thousands)
12
800
Engineers Available
36
Project 2
9
700
Project 3
4
1200
MODEL
Decision
Project 1
#Engineers: 0
Expected P(Success):
=B12/(B12+B4)
Maximize
Expected
=SUMPRODUCT(B5:D5,B14:D14)
NPV
Subject to:
Total Engineers
LHS
=SUM(B12:D12)
page 17 of 17
Variables
Project 2
Project 3
0
0
=C12/(C12+C4)
=D12/(D12+D4)
RHS
<=
=B7