Transcript decision analysis - University of Redlands

Lecture
2
MGMT 650
Linear Programming Applications
Chapter 4
1
Possible Outcomes of a LP
(Section 2.6)
 A LP is either
Infeasible – there exists no solution which satisfies
all constraints and optimizes the objective function
 or, Unbounded – increase/decrease objective
function as much as you like without violating any
constraint
 or, Has an Optimal Solution



Optimal values of decision variables
Optimal objective function value
2
Infeasible LP – An Example

minimize
4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+16
x33+5x34
Subject to
 x11+x12+x13+x14=100
 x21+x22+x23+x24=200
 x31+x32+x33+x34=150


x11+x21+x31=80
x12+x22+x32=90
x13+x23+x33=120
x14+x24+x34=170

xij>=0, i=1,2,3; j=1,2,3,4


Total demand exceeds total supply
3
Unbounded LP – An Example
maximize 2x1 + x2
subject to
-x1 +
x2  1
x1 - 2x2  2
x1 , x 2  0
x2 can be increased indefinitely without violating any constraint
=> Objective function value can be increased indefinitely
4
Multiple Optima – An Example
maximize x1 + 0.5 x2
subject to
2x1 + x2  4
x1 + 2x2  3
x1 , x2  0
• x1= 2, x2=0, objective function = 2
• x1= 5/3, x2=2/3, objective function = 2
5
Marketing Application: Media Selection
Advertising Media
# of potential
customers reached
Cost ($) per
advertisement
Max times available
per month
Exposure Quality
Units
Day TV
1000
1500
15
65
Evening TV
2000
3000
10
90
Daily newspaper
1500
400
25
40
Sunday newspaper
2500
1000
4
60
Radio
300
100
30
20





Advertising budget for first month = $30000
At least 10 TV commercials must be used
At least 50000 customers must be reached
Spend no more than $18000 on TV adverts
Determine optimal media selection plan
6
Media Selection Formulation

Step 1: Define decision variables






Step 2: Write the objective in terms of the decision variables


DTV = # of day time TV adverts
ETV = # of evening TV adverts
DN = # of daily newspaper adverts
SN = # of Sunday newspaper adverts
R = # of radio adverts
Maximize 65DTV+90ETV+40DN+60SN+20R
Step 3: Write the constraints in terms of the decision variables
DTV
ETV
DN
SN
R
+
25
<=
4
<=
30
<=
30000
0
DN
25
SN
2
R
30
Exposure = 2370
units
Availability of
Media
Budget
+
ETV
>=
10
1500DTV
+
3000ETV
<=
18000
TV Constraints
1000DTV
+
2000ETV
>=
50000
Customers reached
DTV, ETV, DN, SN, R >= 0
+
100R
<=
ETV
DTV
2500SN
+
10
10
3000ETV
+
1000SN
<=
DTV
+
1500DN
+
15
Value
1500DTV
+
400DN
<=
Variable
300R
7
Production-Inventory Model

Nike produces footballs and must decide how many footballs to
produce each month over the next 6 months
Month 1
Month 2
Month 3
Month 4
Month 5
Month 6
Demand
10000
15000
30000
35000
25000
10000
Unit cost ($)
12.50
12.55
12.70
12.80
12.85
12.95






Starting inventory = 5000
Production capacity each month = 30000 footballs
Storage capacity = 10000 footballs
Inventory holding cost of a month = 5% of production cost of that
month
Determine production schedule that minimizes production and
holding cost
Assume for simplicity


Production occurs continuously
Demand occurs at month end
8
Production-Inventory Model Formulation

Step 1: define decision variables



Step 2: formulate objective function is terms of decision
variables


Pj = production quantity in month j
Ij = end-of-month inventory in month j
Sum of production cost + inventory holding cost
Step 3: formulate objective function is terms of decision
variables
 Ij-1



+ Pj = Dj + Ij
Pj < = 30000
Ij <= 10000
Pj, Ij >=0
9
Production-Inventory Formulation in LINDO


min 12.50p1+12.55p2+12.70p3+12.80p4+12.85p5+12.95p6+0.625i1+0.6275i2+0.635i3+0.64i4+0.6425i5+0.6475i6
st
p1-i1=5000
Production-inventory constraint for month 1

p2+i1-i2=15000
Production-inventory constraint for month 2

p3+i2-i3=30000
Production-inventory constraint for month 3

p4+i3-i4=35000
Production-inventory constraint for month 4

p5+i4-i5=25000
Production-inventory constraint for month 5

p6+i5-i6=10000
Production-inventory constraint for month 6

p1<=30000
p2<=30000
p3<=30000
p4<=30000
p5<=30000
p6<=30000











i1<=10000
i2<=10000
i3<=10000
i4<=10000
i5<=10000
i6<=10000
Month
1
Month
2
Month
3
Month
4
Month
5
Month
6
Demand
10000
15000
30000
35000
25000
10000
Production
5000
20000
30000
30000
25000
10000
Inventory
0
5000
5000
0
0
0
Production capacity constraints
Inventory storage constraints
Cost = 1,535,562.00
10
Blending Problem – Self-study



Ferdinand Feed Company receives four raw grains from
which it blends its dry pet food.
The pet food advertises that each 8-ounce packet meets
the minimum daily requirements for vitamin C, protein
and iron.
The cost of each raw grain as well as the vitamin C,
protein, and iron units per pound of each grain are as
follows:
Vitamin C Protein
Iron
Grain
1
2
3
4

Units/lb
9
16
8
10
Units/lb
12
10
10
8
Units/lb
0
14
15
7
Cost/lb
0.75
0.9
0.8
0.7
Ferdinand is interested in producing the 8-ounce mixture
at minimum cost while meeting the minimum daily
requirements of 6 units of vitamin C, 5 units of protein,
and 5 units of iron.
11
Blending Problem Formulation

Define the decision variables
xj = the pounds of grain j (j = 1,2,3,4)
used in the 8-ounce mixture
 Define the objective function in terms of decision
variables
Minimize the total cost for an 8-ounce
mixture:
MIN .75x1 + .90x2 + .80x3 + .70x4
12
Blending Problem - Constraints

Define the constraints
Total weight of the mix is 8-ounces (.5 pounds):
(1) x1 + x2 + x3 + x4 = .5
Total amount of Vitamin C in the mix is at least 6 units:
(2) 9x1 + 16x2 + 8x3 + 10x4 >= 6
Total amount of protein in the mix is at least 5 units:
(3) 12x1 + 10x2 + 10x3 + 8x4 >= 5
Total amount of iron in the mix is at least 5 units:
(4) 14x2 + 15x3 + 7x4 >= 5
Nonnegativity of variables: xj > 0 for all j
13
Blending Problem – Optimal Solution
OBJECTIVE FUNCTION VALUE = 0.406





VARIABLE
VALUE
REDUCED COSTS
X1
0.099
0.000
X2
0.213
0.000
X3
0.088
0.000
X4
0.099
0.000
Thus, the optimal blend is about
.10 lb. of grain 1,
.21 lb. of grain 2,
.09 lb. of grain 3, and
.10 lb. of grain 4.
The mixture costs 40.6 cents.
14
Transportation Problem – Chapter 7

Objective:






determination of a transportation plan of a single commodity
from a number of sources
to a number of destinations,
such that total cost of transportation is minimized
Sources may be plants, destinations may be warehouses
Question:





how many units to transport
from source i
to destination j
such that supply and demand constraints are met, and
total transportation cost is minimized
15
A Transportation Table
1
Factory
Warehouse
3
2
4
4
7
7
1
100
1
3
12
8
8
200
2
10
8
16
Factory 1
can
supply
100
units per
period
5
150
3
450
Demand
80
90
120
160
450
Warehouse B’s demand is 90
units per period
Total demand
per period
Total
supply
capacity
per
period
16
LP Formulation of Transportation Problem

minimize
4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+
16x33+5x34
Subject to
Minimize total cost of transportation
 x11+x12+x13+x14=100
Supply constraint for factories
 x21+x22+x23+x24=200
 x31+x32+x33+x34=150
 x11+x21+x31=80
 x12+x22+x32=90
Demand constraint of warehouses
 x13+x23+x33=120
 x14+x24+x34=160
 xij>=0, i=1,2,3; j=1,2,3,4
17
Solution in Management Scientist
Total transportation cost = 4(80) + 7(0) + 7(10)+ 1(10)
+ 12(0) + 3(90) + 8(110) + 8(0) + 8(0) +10(0) + 16(0)
+5 (150) = $2300
18
Assignment Problem – Chapter 7

Special case of transportation problem

When # of rows = # of columns in the
transportation tableau
 All supply and demands =1

Objective: Assign n jobs/workers to n machines
such that the total cost of assignment is minimized
 Plenty of practical applications

Job shops
 Hospitals
 Airlines, etc.
19
Cost Table for Assignment Problem
Aircraft (j)
1
2
3
4
1
$1
$4
$6
$3
2
$9
$7
$10
$9
3
$4
$5
$11
$7
4
$8
$7
$8
$5
Pilot (i)
All assignment costs in thousands of $
20
Formulation of Assignment Problem

minimize x11+4x12+6x13+3x14 + 9x21+7x22+10x23+9x24 +
4x31+5x32+11x33+7x34 + 8x41+7x42+8x43+5x44
subject to
Pilot Assigned to
Cost
 x11+x12+x13+x14=1
aircraft # (`000 $)
 x21+x22+x23+x24=1
1
1
1
 x31+x32+x33+x34=1
 x41+x42+x43+x44=1
2
3
10
3
2
5
 x11+x21+x31+x41=1
4
4
5
 x12+x22+x32+x42=1
Optimal Solution:
 x13+x23+x33+x43=1
x11=1; x23=1; x32=1; x44=1; rest=0
 x14+x24+x34+x44=1
Cost of assignment = 1+10+5+5=$21 (`000)

xij = 1, if pilot i is assigned to aircraft j, i=1,2,3,4; j=1,2,3,4
0 otherwise
21
Transshipment Problem – Chapter 7

Transshipment problems are transportation problems in which a shipment may
move through intermediate nodes (transshipment nodes) before reaching a
particular destination node.
s1
c15
Supply
s2
3
c13
1
c37
c14
Sources
c25
6
c46
c47
4
c23
2
c36
c56
c24
5
Demand
7
c57
d1
d2
Destinations
Intermediate Nodes
Network Representation
22
Example: Goodyear Tires

The Detroit (1) and Akron (2) facilities of
Goodyear supply three customers at Memphis,
Pittsburgh, and Newark.
 Distribution is done through warehouses located
at Charlotte (3) and Atlanta (4).
 Current weekly demands by the customers are 50,
60 and 40 units for Memphis (5), Pittsburgh (6),
and Newark (7) respectively.
 Both facilities at Detroit and Akron can supply at
most 75 units per week.
23
Transportation Costs

Network Representation
ZROX
Detroit
75 ARNOLD
5
1
Charlotte
8
4
50
Pittsburgh
HEWES
60
5
8
3
7
75 Akron
Memphis
WASH
Atlanta
BURN
4
4
Newark 40
24
Goodyear Tires Formulation

Define Decision Variables


xij = amount shipped from manufacturer i to warehouse j
xjk = amount shipped from warehouse j to customer k

where




i = 1 (Detroit), i = 2 (Akron),
j = 3 (Charlotte), j = 4 (Atlanta),
k = 5 (Memphis), k = 6 (Pittsburgh), k = 7 (Newark)
Define Objective Function
Minimize Overall Shipping Costs:
Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37 +
3x45 + 4x46 + 4x47
25
Goodyear Tires Formulation

Define Constraints
Amount Out of Detroit:
x13 + x14 < 75
Amount Out of Akron:
x23 + x24 < 75
Amount Through Charlotte: x13 + x23 - x35 - x36 - x37 = 0
Amount Through Atlanta: x14 + x24 - x45 - x46 - x47 = 0
Amount Into Memphis:
x35 + x45 = 50
Amount Into Pittsburgh:
x36 + x46 = 60
Amount Into Newark:
x37 + x47 = 40
Non-negativity of variables: xij , xjk > 0, for all i, j and k.
26
Goodyear Tires Solutions
Objective Function Value =
1150.000
Variable
Value
Reduced Costs
X13
X14
X23
X24
X35
X36
X37
X45
X46
X47
75.000
0.000
0.000
75.000
50.000
25.000
0.000
0.000
35.000
40.000
0.000
2.000
4.000
0.000
0.000
0.000
3.000
3.000
0.000
0.000
27
Goodyear Tires Solutions
ZROX
Detroit
75 ARNOLD
5
1
75
Charlotte
8
4
50
Pittsburgh
HEWES
60
5
8
3 4
7
75 Akron
Memphis
WASH
Atlanta
BURN
4
Newark 40
28