Nonlinear Optimization Models
Chapter 10
Nonlinear Optimization Models
Solutions:
1.
a.
With $2000 being spent on radio and $1000 being spent on direct mail we can simply substitute
those values into the sales function (remembering that the variables are defined as thousands of
dollars).
S  2 R 2  10M 2  8RM  18R  34M
 2(22 )  10(12 )  8(2)(1)  18(2)  34(1)
 36
Sales of $36,000 will be realized with this allocation of the media budget.
b.
We simply add a budget constraint and nonnegativity constraint to the sales function that is to be
maximized.
max  2 R 2  10M 2  8RM  18R  34M
s.t.
RM 3
R, M  0
The solution is R = $2,500 and M = $500 with Sales of $37,000. The spreadsheet model is:
10 - 1
Nonlinear Optimization Models
2.
a.
The optimization model is
Max 5 L.25C .75
s.t.
25L  75C  75000
L, C  0
10 - 2
Nonlinear Optimization Models
b. The optimal solution is L= 750, C=750 with output of 3,750 units. The spreadsheet model is:
10 - 3
Nonlinear Optimization Models
10 - 4
Nonlinear Optimization Models
3.
a.
The optimization model is
min 50L  100C
s.t.
20 L0.30 C 0.70  50000
L, C  0
b. The optimal solution is L = 2244.265 and C = 2618.336 with a cost of $374,046.85. The spreadsheet
model is:
10 - 5
Nonlinear Optimization Models
10 - 6
Nonlinear Optimization Models
4.
Let OT be the number of overtime hours scheduled. Then the optimization model is
max  3 x12  42 x1  3 x2 2  48 x2  700  5OT
s.t.
4 x1  6 x2  24  OT
x1 , x2 , OT  0
The optimal solution is: x1 = 3.667 and x2 = 3.00 with a profit of $887.333. The spreadsheet
model is:
10 - 7
Nonlinear Optimization Models
10 - 8
Nonlinear Optimization Models
5.
The revenue function is: PsDs + PHDH = Ps( 222 - 0.6Ps + 0.35PH ) + PH(270 + 0.1Ps - 0.64PH).
This is an example of an unconstrained optimization problem because no constraints are required
here. The optimal solution is: Ps = $304.21 and PH = $317.89 with an optimal revenue of
$76,681.48.
The spreadsheet model is:
10 - 9
Nonlinear Optimization Models
6. a.
If X is the weekly production volume in thousands of units at the Dayton plant and Y is the weekly
production volume in thousands of units at the Hamilton plant, then the optimization model is
min X 2  X  5  Y 2  2Y  3
s.t.
X Y  8
X ,Y  0
b.
The optimal solution is X = 4.75 and Y = 3.25 for an optimal objective function value of 42.875.
These are all in thousands. The spreadsheet model is:
10 - 10
Nonlinear Optimization Models
10 - 11
Nonlinear Optimization Models
7.
Min = (y2 - 𝑦̂2)2 + (y3 - 𝑦̂3)2 + (y4 - 𝑦̂4)2 + (y5 - 𝑦̂5)2 + (y6 - 𝑦̂6)2 + (y7 - 𝑦̂7)2
+ (y8 - 𝑦̂8)2 + (y9 - 𝑦̂9)2 + (y10 - 𝑦̂10)2 + (y11 - 𝑦̂11)2 + (y12 - 𝑦̂12)2
s.t.
𝑦̂1 = 𝑦1
𝑦̂2 = 𝑦̂1 + 𝛼(𝑦1 − 𝑦̂1 )
𝑦̂3 = 𝑦̂2 + 𝛼(𝑦2 − 𝑦̂2 )
𝑦̂4 = 𝑦̂3 + 𝛼(𝑦3 − 𝑦̂3 )
𝑦̂5 = 𝑦̂4 + 𝛼(𝑦4 − 𝑦̂4 )
𝑦̂6 = 𝑦̂5 + 𝛼(𝑦5 − 𝑦̂5 )
𝑦̂7 = 𝑦̂6 + 𝛼(𝑦6 − 𝑦̂6 )
𝑦̂8 = 𝑦̂7 + 𝛼(𝑦7 − 𝑦̂7 )
𝑦̂9 = 𝑦̂8 + 𝛼(𝑦8 − 𝑦̂8 )
𝑦̂10 = 𝑦̂9 + 𝛼(𝑦9 − 𝑦̂9 )
𝑦̂11 = 𝑦̂10 + 𝛼(𝑦10 − 𝑦̂10 )
𝑦̂12 = 𝑦̂11 + 𝛼(𝑦11 − 𝑦̂11 )
y1 = 17
y2 = 21
y3 = 19
y4 = 23
y5 = 18
y6 = 16
y7 = 20
y8 = 18
y9 = 22
y10 = 20
y11 = 15
y12 = 22
The optimal solution is  = 0.1744. The spreadsheet model follows.
10 - 12
Nonlinear Optimization Models
10 - 13
Nonlinear Optimization Models
8.
The objective is to minimize total production cost. To minimize total production cost, we minimize
the production cost at Aynor plus the production cost at Spartanburg subject to the constraint that total
production of kitchen chairs is equal to 40. The model is:
Min (75Q1  5Q12  100)  (25Q2  2.5Q2 2  150)
s.t. Q1  Q2  40
Q1 , Q2  0
The optimal solution to this model is to produce 10 chairs at Aynor for a production cost of $1350 and
30 chairs at Spartanburg for a production cost of $3150. The total cost is $4500. The spreadsheet
model follows.
10 - 14
Nonlinear Optimization Models
10 - 15
Nonlinear Optimization Models
10 - 16
Nonlinear Optimization Models
9.
The optimal solution is Q1 = 52.223, Q2 = 70.065, Q3 = 37.689 with a total cost of $25,830.
10 - 17
Nonlinear Optimization Models
10 - 18
Nonlinear Optimization Models
10.
Max 1.2712 LN(XA) + 17.414 + 0.3970 LN(XB) + 16.109
s.t.
XA + XB ≤ 500
XA ≥ 0
XB ≥ 0
The optimal solution is XA = 381.009 and XB = 118.006 (these are in thousands of dollars) with a
profit of $42.975 million. The spreadsheet model follows.
10 - 19
Nonlinear Optimization Models
10 - 20
Nonlinear Optimization Models
11.
The demand-weighted objective is:
Min ( 12 ( X  1) 2  (Y  4) 2  24 ( X  1) 2  (Y  2) 2  13 ( X  2.5) 2  (Y  2) 2
 7 ( X  3) 2  (Y  5) 2  17 ( X  4) 2  (Y  4) 2 )
The solution to the un-weighted model is X = 2.23 and Y = 3.349 (from section 10.3)
The solution to the demand-weighted model is: X = 1.909 and Y = 2.721
The solutions are shown in the charts below. The size of the bubble indicates the demand. The
demand weighted solution shifts the optimal location towards the Paint station.
10 - 21
Nonlinear Optimization Models
12.
Let
X = the horizontal coordinate of the tower.
Y = the vertical coordinate of the tower.
Min ( ( X  10) 2  (Y  10) 2  ( X  12) 2  (Y  16) 2  ( X  16) 2  (Y  18) 2 
( X  12) 2  (Y  22) 2 )
s.t.
( X  10) 2  (Y  10) 2  10
( X  12) 2  (Y  16) 2 10
( X  16) 2  (Y  18) 2 10
( X  12) 2  (Y  22) 2 10
The optimal solution is X = 12, Y = 16, with an objective function value of 16.797.
10 - 22
Nonlinear Optimization Models
Let d = the maximum distance and X and Y as defined in part a.
Min d
s.t.
d 
( X  10) 2  (Y  10) 2
d  ( X  12) 2  (Y  16) 2
d  ( X  16) 2  (Y  18) 2
d  ( X  12) 2  (Y  22) 2
The optimal solution is X = 11.006 and Y = 15.999 with a maximum distance of 6.083.
10 - 23
Nonlinear Optimization Models
10 - 24
Nonlinear Optimization Models
Note that the decision variable d is located in cell B21 and the objective function is cell B23. This approach
is necessary because Excel Solver will not let the objective function and a decision variable be the same
cell.
10 - 25
Nonlinear Optimization Models
13.
Let
X = the latitude of the optimal wedding location.
Y = the longitude of the optimal wedding location.
15
Min  Ri ( 69 ( X  lati )2  (Y  longi ) 2 )
i 1
The optimal solution is X = 40.204, Y = -75.214, with an objective function value of 67,444.286.
10 - 26
Nonlinear Optimization Models
10 - 27
Nonlinear Optimization Models
14.
Let
X = the fraction of the portfolio to invest in AAPL
Y = the fraction of the portfolio to invest in AMD
Z = the fraction of the portfolio to invest in ORCL
8
Min 18  ( Rs  R ) 2
s 1
s.t.
39.8 X  42.5Y  10.2 Z

R1
10.1X  13.6Y  137.9 Z

R2
124.9 X  56.9Y  170.6 Z

R3
151.8 X  36.7Y  16.6 Z

R4
58.3 X  34.8Y  40.7 Z

R5
14.3 X  67.4Y  30.3Z

R6
41.9 X  183.6Y  15.2 Z

R7

R8

1
R

R
R

X , Y, Z 
25
0
57.1X  6.3Y  .6 Z
X Y  Z
8
1
8
s 1
s
10 - 28
Nonlinear Optimization Models
10 - 29
Nonlinear Optimization Models
10 - 30
Nonlinear Optimization Models
15.
Let:
FS  proportion of portfolio invested in the foreign stock mutual fund
IB  proportion of portfolio invested in the intermediate-term bond fund
LG  proportion of portfolio invested in the large-cap growth fund
LV  proportion of portfolio invested in the large-cap value fund
SG  proportion of portfolio invested in the small-cap growth fund
SV  proportion of portfolio invested in the small-cap value fund
R = the expected return of the portfolio
Rs = the return of the portfolio in year s
Max R
s.t.
10.06 FS  17.64 IB  32.41LG  32.36 LV  33.44 SG  24.56 SV  R1
13.12 FS  3.25 IB  18.71LG  20.61LV  19.40SG  25.32SV  R2
13.47 FS  7.51IB  33.28 LG  12.93LV  3.85SG  6.70 SV  R3
45.42 FS  1.33IB  41.46 LG  7.06 LV  58.68SG  5.43SV  R4
21.93FS  7.36 IB  23.26 LG  5.37 LV  9.02SG  17.31SV  R5
FS 
IB 
LG 
LV 
SG 
SV  1
5
1
5
R
s 1
5
1
5
 (R
s 1
s
 R)2
s
R
 30
FS , I B, LG , LV , SG , SV  0
10 - 31
Nonlinear Optimization Models
10 - 32
Nonlinear Optimization Models
10 - 33
Nonlinear Optimization Models
10 - 34
Nonlinear Optimization Models
16.
Let
X = the fraction of the portfolio to invest in Stock 1
Y = the fraction of the portfolio to invest in Stock 2
Z = the fraction of the portfolio to invest in Stock 3
12
Min 112  ( Rs  R ) 2
s 1
s.t.
.300 X  .225Y  .149 Z

R1
.103 X  .290Y  .260 Z

R2
.216 X  .216Y  .419 Z

R3
.046 X  .272Y  .078Z

R4
.071X  .144Y  .169 Z

R5
.056 X  .107Y  .035Z

R6
.038 X  .321Y  .133Z

R7
.089 X  .305Y  .732 Z

R8
.090 X  .195Y  .021Z

R9
.083 X  .390Y  .131Z

R10
.035 X  .072Y  .006 Z

R11
.176 X  .715Y  .908Z

R12
X Y  Z

1
R
s

R
R

.15
X , Y, Z 
0
12
1
12
s 1
10 - 35
Nonlinear Optimization Models
10 - 36
Nonlinear Optimization Models
10 - 37
Nonlinear Optimization Models
17.
Let:
FS  proportion of portfolio invested in the foreign stock mutual fund
IB  proportion of portfolio invested in the intermediate-term bond fund
LG  proportion of portfolio invested in the large-cap growth fund
LV  proportion of portfolio invested in the large-cap value fund
SG  proportion of portfolio invested in the small-cap growth fund
SV  proportion of portfolio invested in the small-cap value fund
Ds = the difference between the portfolio return and the S&P 500 return, year s
Min D12  D2 2  D3 2  D4 2 + D52
s.t.
10.06 FS  17.64 IB  32.41LG  32.36 LV  33.44SG  24.56SV  25.00
 D1
13.12 FS  3.25IB  18.71LG  20.61LV  19.40SG  25.32SV  20.00  D2
13.47 FS  7.51IB  33.28LG  12.93LV  3.85SG 6.70SV  8.00
 D3
45.42 FS  1.33IB  41.46 LG  7.06 LV  58.68SG 5.43SV  30.00
 D4
21.93FS  7.36 IB  23.26 LG  5.37 LV  9.02SG 17.31SV  10.00
 D5
FS 
IB 
LG 
LV 
SG 
SV  1
FS , I B, LG, LV , SG, SV  0
10 - 38
Nonlinear Optimization Models
Note from the dialog box that follows, cells B17 through F17 are not designated as decision
variables. Rather they are simple calculations in the spreadsheet. These cells correspond to the
deviation variables defined in the algebraic model.
10 - 39
Nonlinear Optimization Models
10 - 40
Nonlinear Optimization Models
18.
Let:
FS  proportion of portfolio invested in the foreign stock mutual fund
IB  proportion of portfolio invested in the intermediate-term bond fund
LG  proportion of portfolio invested in the large-cap growth fund
LV  proportion of portfolio invested in the large-cap value fund
SG  proportion of portfolio invested in the small-cap growth fund
SV  proportion of portfolio invested in the small-cap value fund
R = expected return of the portfolio
Rs = the return of the portfolio in year s
ds = the difference between the expected portfolio return and return for year s
Min
1
5
(d12  d 2 2  d32  d 42 + d52 )
s.t.
10.06 FS  17.64 IB  32.41LG  32.36 LV  33.44SG  24.56SV
 R1
13.12 FS  3.25IB  18.71LG  20.61LV  19.40SG  25.32SV  R2
13.47 FS  7.51IB  33.28LG  12.93LV  3.85SG  6.70SV  R3
45.42 FS  1.33IB  41.46 LG  7.06 LV  58.68SG  5.43SV  R4
21.93FS  7.36IB  23.26 LG  5.37 LV  9.02SG  17.31SV  R5
5
1
5
R  R
s 1
s
d1  R - R1
d 2  R - R2
d3  R - R3
d 4  R - R4
FS 
IB 
LG 
LV 
SG 
SV  1
FS , I B, LG, LV , SG, SV  0
10 - 41
Nonlinear Optimization Models
10 - 42
Nonlinear Optimization Models
10 - 43
Nonlinear Optimization Models
10 - 44
Nonlinear Optimization Models
19.
The frontier is shown below. As the maximum variance increases the expected return increases but
at a decreasing rate. In Figure 10.12, as the required expected return increases, the variance
increases at an increasing rate. These curves are known as the efficient frontier and are really the
same curve from two different perspectives. For example, a variance of 40 with an expected
return of 11.835 is on both curves.
10 - 45
Nonlinear Optimization Models
20.
A spreadsheet model follows. Results might vary somewhat form the values in Table 10.4
10 - 46
Nonlinear Optimization Models
10 - 47
Nonlinear Optimization Models
We used the Multistart option with population size of 100, but not requiring bounds. We used a starting
point of m = 149.89, with p = q = 0.
10 - 48