Optimization II
Outline
Optimization Extensions
• Multiperiod Models
– Operations Planning: Sailboats
• Network Flow Models
– Transportation Model: Beer Distribution
– Assignment Model: Contract Bidding
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
A
1 objective:
2
3 decision variables:
4
5
6 constraints:
7
8
9
10
11
12
B
(in 100s)
Molding
Cutting
Van
Demand
Nonnegativity (snowboards)
Nonnegativity (skis)
Operations Management -- Prof. Juran
C
6000
D
E
F
4000 $ 154,800 = profit
G
skis
snowboards
18.6
10.8
3
1
2
0
1
0
2
3
1
1
0
1
77.4
51
48
10.8
18.6
10.8
<=
<=
<=
<=
>=
>=
115.5
51
48
16
0
0
3
©The McGraw-Hill Companies, Inc., 2004
A
B
C
D
1 Microsoft Excel 10.0 Answer Report
2 Worksheet: [Book1]downhill
3 Report Created: 3/27/2006 6:28:16 AM
4
5
6 Target Cell (Max)
7
Cell
Name
Original Value
8
$E$1 objective:
$
10,000
9
10
11 Adjustable Cells
12
Cell
Name
Original Value
13
$C$4 skis
1
14
$D$4 snowboards
1
15
16
17 Constraints
18
Cell
Name
Cell Value
19
$E$11 Nonnegativity (snowboards)
18.6
20
$E$12 Nonnegativity (skis)
10.8
21
$E$7 Molding
77.4
22
$E$8 Cutting
51
23
$E$9 Van
48
24
$E$10 Demand
10.8
Operations Management -- Prof. Juran
E
F
G
Final Value
$
154,800
Final Value
18.6
10.8
Formula
$E$11>=$G$11
$E$12>=$G$12
$E$7<=$G$7
$E$8<=$G$8
$E$9<=$G$9
$E$10<=$G$10
Status
Slack
Not Binding 18.6
Not Binding 10.8
Not Binding 38.1
Binding
0
Binding
0
Not Binding
5.2
4
©The McGraw-Hill Companies, Inc., 2004
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
A
B
C
D
Microsoft Excel 10.0 Sensitivity Report
Worksheet: [s-downhill.xls]downhill
Report Created: 3/27/2006 6:31:19 AM
E
F
G
H
Adjustable Cells
Final Reduced Objective
Value
Cost
Coefficient
18.6
0
6000
10.8
0
4000
Allowable
Increase
2000
14000
Allowable
Decrease
4666.67
1000
Final Shadow Constraint
Name
Value Price
R.H. Side
Nonnegativity (snowboards)
18.6
0
0
Nonnegativity (skis)
10.8
0
0
Molding
77.4
0
115.5
Cutting
51
400
51
Van
48
2800
48
Demand
10.8
0
16
Allowable
Increase
18.6
10.8
1E+30
13
27.2
1E+30
Allowable
Decrease
1E+30
1E+30
38.1
27
26
5.2
Cell
Name
$C$4 skis
$D$4 snowboards
Constraints
Cell
$E$11
$E$12
$E$7
$E$8
$E$9
$E$10
Most important number: Shadow Price
The change in the objective function that would result from
a one-unit increase in the right-hand side of a constraint
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Sailboat Problem
• Sailco must determine how many sailboats to produce during
each of the next four quarters.
• At the beginning of the first quarter, Sailco has an inventory of
10 sailboats.
• Sailco must meet demand on time. The demand during each of
the next four quarters is as follows:
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
40
60
75
25
Operations Management -- Prof. Juran
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Sailboat Problem
• Assume that sailboats made during a quarter can be used to
meet demand for that quarter.
• During each quarter, Sailco can produce up to 50 sailboats with
regular-time employees, at a labor cost of $400 per sailboat.
• By having employees work overtime during a quarter, Sailco
can produce unlimited additional sailboats with overtime labor
at a cost of $450 per sailboat.
• At the end of each quarter (after production has occurred and
the current quarter’s demand has been satisfied), a holding cost
of $20 per sailboat is incurred.
• Problem: Determine a production schedule to minimize the
sum of production and inventory holding costs during the next
four quarters.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Managerial Formulation
Decision Variables
We need to decide on production quantities, both regular
and overtime, for four quarters (eight decisions).
Note that on-hand inventory levels at the end of each
quarter are also being decided, but those decisions will be
implied by the production decisions.
Operations Management -- Prof. Juran
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Managerial Formulation
Objective Function
We’re trying to minimize the total labor cost of
production, including both regular and overtime labor.
Operations Management -- Prof. Juran
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Managerial Formulation
Constraints
There is an upper limit on the number of boats built
with regular labor in each quarter.
No backorders are allowed. This is equivalent to
saying that inventory at the end of each quarter must
be at least zero.
Production quantities must be non-negative.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Managerial Formulation
Note that there is also an accounting constraint: Ending Inventory
for each period is defined to be:
Beginning Inventory + Production – Demand
This is not a constraint in the usual Solver sense, but useful to link
the quarters together in this multi-period model.
Operations Management -- Prof. Juran
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Mathematical Formulation
Decision Variables
Pij = Production of type i in period j.
Let i index labor type; 0 is regular and 1 is overtime.
Let j index quarters; 1 through 4
Operations Management -- Prof. Juran
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Mathematical Formulation
Objective Function
1
4
4
Minimize Z   C i Pij   HI j
i 0 j 1
j 1
Ci = Production Cost; $400 for regular, $450 for overtime
H = Holding Cost; $20 per boat per period
Define Dj to be demand in period j
Define Ij to be ending inventory for period j
1
I j  I j1   Pij  D j
i 0
Operations Management -- Prof. Juran
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Mathematical Formulation
Constraints
P0 j
 50
For all j
Ij
0
For all j
Pij
0
For all i, j
Operations Management -- Prof. Juran
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Solution Methodology
A
1 Production schedule
2 Month
3 Regular time production
4
5 Upper bound
6
7 Overtime production
8
9 Total Production
10
11 Demand
12
13 Ending inventory
14
15
B
C
D
E
I
-$5,700
H
G
F
Total cost
2
1
1
1
<= <=
50 50
1
2
1
2
40 60
3
1
<=
50
4
1
<=
50
1
1
2
75
2
25
-28 -86 -159 -182
>=
>= >= >=
0
0
0
0
Operations Management -- Prof. Juran
=SUM(B3:E3)
Month
Regular time unit cost
Overtime unit cost
Unit holding cost
=SUM(B7:E7)
4
4
8
200
1
400
450
20
Initial inventory
=SUM(B9:E9)
10
Regular time cost
Overtime cost
=SUM(B11:E11)
Holding cost
$1,600
$1,800
-$9,100
J
K
L
4
3
2
400 400 400
450 450 450
20 20 20
=D13+E9-E11
15
©The McGraw-Hill Companies, Inc., 2004
Solution Methodology
A
1 Production schedule
2 Month
3 Regular time production
4
5 Upper bound
6
7 Overtime production
8
9 Total Production
10
11 Demand
12
13 Ending inventory
14
15
B
C
D
E
F
G
H
Total cost
1
2
1
1
<= <=
50 50
3
1
<=
50
4
1
<=
50
4
1
1
1
1
4
2
2
2
2
8
40 60
75
25
200
I
-$5,700
Month
Regular time unit cost
Overtime unit cost
Unit holding cost
1
400
450
20
Initial inventory
10
Regular time cost
Overtime cost
Holding cost
J
K
L
M
=SUM(I10:I12)
N
O
2
3
4
400 400 400
450 450 450
20 20 20
=SUMPRODUCT(I4:L4,B3:E3)
=SUMPRODUCT(I5:L5,B7:E7)
=SUMPRODUCT(I6:L6,B13:E13)
$1,600
$1,800
-$9,100
-28 -86 -159 -182
>= >= >=
>=
0
0
0
0
Operations Management -- Prof. Juran
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Solution Methodology
Operations Management -- Prof. Juran
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Solution Methodology
A
1 Production schedule
2 Month
3 Regular time production
4
5 Upper bound
6
7 Overtime production
8
9 Total Production
10
11 Demand
12
13 Ending inventory
14
15
B
C
D
E
F
G
H
I
$77,350
Total cost
1
50
<=
50
2
50
<=
50
3
50
<=
50
4
25 175
<=
50 200
0
0
15
0
Month
Regular time unit cost
Overtime unit cost
Unit holding cost
1
400
450
20
Initial inventory
10
J
K
L
2
3
4
400 400 400
450 450 450
20 20 20
15
50 50 65 25 190
40 60 75 25 200
Regular time cost
Overtime cost
Holding cost
$70,000
$6,750
$600
20 10 0 0
>= >= >= >=
0 0 0 0
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Solution Methodology
It is optimal to have 15 boats produced on overtime in
the third quarter.
All other demand should be met on regular time.
Total labor cost will be $76,750.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Sensitivity Analysis
Investigate changes in the holding cost, and determine
if Sailco would ever find it optimal to eliminate all
overtime.
Make a graph showing optimal overtime costs as a
function of the holding cost.
Operations Management -- Prof. Juran
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Sensitivity Analysis
Operations Management -- Prof. Juran
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Sensitivity Analysis
A
32 Unit holding cost
33
34
35
36
37
38
39
40
41
42
43
44
45
Operations Management -- Prof. Juran
0
5
10
15
20
25
30
35
40
45
50
55
60
B
C
Overtime cost Holding cost
$6,750
$0
$6,750
$150
$6,750
$300
$6,750
$450
$6,750
$600
$6,750
$750
$11,250
$300
$11,250
$350
$11,250
$400
$11,250
$450
$11,250
$500
$15,750
$0
$15,750
$0
22
©The McGraw-Hill Companies, Inc., 2004
Sensitivity Analysis
$18,000
$800
$16,000
$700
Overtime cost
Holding cost
$600
$12,000
$500
$10,000
$400
$8,000
$300
$6,000
Total Holding Cost
Total Overtime Cost
$14,000
$200
$4,000
$100
$2,000
$0
$0
0
5
10
15
20
25
30
35
40
45
50
55
60
Unit Holding Cost
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Sensitivity Analysis
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Sensitivity Analysis
A
B
16
17 Unit holding cost Total cost
18
0
$76,750
19
5
$76,900
20
10
$77,050
21
15
$77,200
22
20
$77,350
23
25
$77,500
24
30
$77,550
25
35
$77,600
26
40
$77,650
27
45
$77,700
28
50
$77,750
29
55
$77,750
30
60
$77,750
Operations Management -- Prof. Juran
C
1
50
50
50
50
50
50
40
40
40
40
40
30
30
D
E F
G
H I J K L
Regular
Overtime
2 3 4 Total 1 2 3 4 Total
50 50 25 175 15 0 0 0
15
50 50 25 175 0 0 15 0
15
50 50 25 175 0 0 15 0
15
50 50 25 175 0 0 15 0
15
50 50 25 175 0 0 15 0
15
50 50 25 175 0 0 15 0
15
50 50 25 165 0 0 25 0
25
50 50 25 165 0 0 25 0
25
50 50 25 165 0 0 25 0
25
50 50 25 165 0 0 25 0
25
50 50 25 165 0 0 25 0
25
50 50 25 155 0 10 25 0
35
50 50 25 155 0 10 25 0
35
25
©The McGraw-Hill Companies, Inc., 2004
Sensitivity Analysis
Conclusions:
It is never optimal to completely eliminate overtime.
In general, as holding costs increase, Sailco will decide
to reduce inventories and therefore produce more boats
on overtime.
Even if holding costs are reduced to zero, Sailco will
need to produce at least 15 boats on overtime. Demand
for the first three quarters exceeds the total capacity of
regular time production.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Gribbin Brewing
Regional brewer Andrew Gribbin distributes kegs of
his famous beer through three warehouses in the
greater News York City area, with current supplies as
shown:
Warehouses
Hoboken
Bronx
Brooklyn
Operations Management -- Prof. Juran
Supply
80
145
120
27
©The McGraw-Hill Companies, Inc., 2004
On a Thursday morning, he has his usual
weekly orders from his four loyal
customers, as shown :
Bars
Der Ratkeller
McGoldrick's Pub
Night Train Bar & Grill
Stern Business School
Operations Management -- Prof. Juran
Demand
80
65
70
85
28
©The McGraw-Hill Companies, Inc., 2004
Tracy Chapman, Gribbin’s shipping manager, needs to
determine the most cost-efficient plan to deliver beer to
these four customers, knowing that the costs per keg are
different for each possible combination of warehouse
and customer:
Hoboken
Bronx
Brooklyn
Ratkeller
$4.64
$3.52
$9.95
McGoldrick's
$5.13
$4.16
$6.82
Operations Management -- Prof. Juran
Night Train
$6.54
$6.90
$3.88
Stern
$8.67
$7.91
$6.85
29
©The McGraw-Hill Companies, Inc., 2004
a)
b)
c)
d)
What is the optimal shipping plan?
How much will it cost to fill these four orders?
Where does Gribbin have surplus inventory?
If Gribbin could have one additional keg at one of the three
warehouses, what would be the most beneficial location, in
terms of reduced shipping costs?
e) Gribbin has an offer from Lu Leng Felicia, who would like to
sublet some of Gribbin’s Brooklyn warehouse space for her
tattoo parlor. She only needs 240 square feet, which is
equivalent to the area required to store 40 kegs of beer, and
has offered Gribbin $0.25 per week per square foot. Is this a
good deal for Gribbin? What should Gribbin’s response be to
Lu Leng?
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Managerial Problem Formulation
Decision Variables
Numbers of kegs shipped from each of three warehouses
to each of four customers (12 decisions).
Objective
Minimize total cost.
Constraints
Each warehouse has limited supply.
Each customer has a minimum demand.
Kegs can’t be divided; numbers shipped must be integers.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Mathematical Formulation
Decision Variables
Define Xij = Number of kegs shipped from warehouse i to customer j.
Define Cij = Cost per keg to ship from warehouse i to customer j.
i = warehouses 1-3, j = customers 1-4
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Mathematical Formulation
Objective
3
Minimize Z =
4
 X C
i 1 j 1
ij
ij
Constraints
Define Si = Number of kegs available at warehouse i.
4
X
j 1
ij
 Si
Define Dj = Number of kegs ordered by customer j.
3
X
i 1
ij
 Dj
Do we need a constraint to ensure that all of the Xij are integers?
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
A
Total Cost
1
2
3 Shipping Plan
4
Hoboken
5
Bronx
6
Brooklyn
7
8
9
10
11
Costs
12
Hoboken
13
Bronx
14
Brooklyn
B
74.97
C
D
E
=SUMPRODUCT(B4:E6,B12:E14)
Ratkeller McGoldrick's
1
1
1
1
1
1
3
3
=
=
80
65
$ 4.64
$ 3.52
$ 9.95
$
$
$
5.13
4.16
6.82
Operations Management -- Prof. Juran
Night Train
1
1
1
3
=
70
$
$
$
6.54
6.90
3.88
Stern
1
1
1
3
=
85
F
G
H
I
=SUM(B4:E4)
4 <= 80
4 <= 145
4 <= 120
=SUM(E4:E6)
$ 8.67
$ 7.91
$ 6.85
34
©The McGraw-Hill Companies, Inc., 2004
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Operations Management -- Prof. Juran
36
©The McGraw-Hill Companies, Inc., 2004
A
Total Cost
1
2
3 Shipping Plan
4
Hoboken
5
Bronx
6
Brooklyn
7
8
9
10
11
Costs
12
Hoboken
13
Bronx
14
Brooklyn
B
1469.55
C
Ratkeller McGoldrick's
0
0
80
65
0
0
80
65
=
=
80
65
$ 4.64
$ 3.52
$ 9.95
$
$
$
Operations Management -- Prof. Juran
5.13
4.16
6.82
D
E
Night Train
0
0
70
70
=
70
Stern
35
0
50
85
=
85
$
$
$
6.54
6.90
3.88
F
G
H
35 <= 80
145 <= 145
120 <= 120
$ 8.67
$ 7.91
$ 6.85
37
©The McGraw-Hill Companies, Inc., 2004
A
B
C
6 Adjustable Cells
7
8
Cell
Name
9
$B$8 Hoboken to Der Ratkeller
10
$C$8 Hoboken to McGoldrick's Pub
11
$D$8 Hoboken to Night Train Bar & Grill
12
$E$8 Hoboken to Stern Business School
13
$B$9 Bronx to Der Ratkeller
14
$C$9 Bronx to McGoldrick's Pub
15
$D$9 Bronx to Night Train Bar & Grill
16
$E$9 Bronx to Stern Business School
17
$B$10 Brooklyn to Der Ratkeller
18
$C$10 Brooklyn to McGoldrick's Pub
19
$D$10 Brooklyn to Night Train Bar & Grill
20
$E$10 Brooklyn to Stern Business School
21
22 Constraints
23
24
Cell
Name
25
$B$11 Der Ratkeller Demand
26
$C$11 McGoldrick's Pub Demand
27
$D$11 Night Train Bar & Grill Demand
28
$E$11 Stern Business School Demand
29
$F$8 Hoboken Supply
30
$F$9 Bronx Supply
31
$F$10 Brooklyn Supply
Operations Management -- Prof. Juran
D
E
F
G
H
Final Reduced Objective Allowable Allowable
Value
Cost
Coefficient Increase Decrease
0
0.360
4.640
1E+30
0.360
0
0.210
5.130
1E+30
0.210
0
0.840
6.540
1E+30
0.840
35
0.000
8.670
0.210
0.760
80
0.000
3.520
0.360
1E+30
65
0.000
4.160
0.210
1E+30
0
1.960
6.900
1E+30
1.960
0
0.000
7.910
0.760
0.210
0
7.490
9.950
1E+30
7.490
0
3.720
6.820
1E+30
3.720
70
0.000
3.880
0.840
1E+30
50
0.000
6.850
1.82
0.840
Final Shadow Constraint Allowable Allowable
Value Price
R.H. Side Increase Decrease
80
4.280
80
0
35
65
4.920
65
0
35
70
5.700
70
45
35
85
8.670
85
45
35
35
0.000
80
1E+30
45
145
-0.760
145
35
0
120
-1.820
120
35
45
38
©The McGraw-Hill Companies, Inc., 2004
Where does Gribbin have surplus inventory?
The only supply constraint that is not binding is the
Hoboken constraint. It would appear that Gribbin has
45 extra kegs in Hoboken.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
If Gribbin could have one additional keg at one of the
three warehouses, what would be the most beneficial
location, in terms of reduced shipping costs?
Operations Management -- Prof. Juran
40
©The McGraw-Hill Companies, Inc., 2004
According to the sensitivity report,
•One more keg in Hoboken is worthless.
•One more keg in the Bronx would have
reduced overall costs by $0.76.
•One more keg in Brooklyn would have reduced
overall costs by $1.82.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Gribbin has an offer from Lu Leng Felicia, who would like
to sublet some of Gribbin’s Brooklyn warehouse space for
her tattoo parlor. She only needs 240 square feet, which is
equivalent to the area required to store 40 kegs of beer,
and has offered Gribbin $0.25 per week per square foot.
Is this a good deal for Gribbin?
What should Gribbin’s response be to Lu Leng?
Operations Management -- Prof. Juran
42
©The McGraw-Hill Companies, Inc., 2004
Assuming that the current situation will continue into
the foreseeable future, it would appear that Gribbin
could reduce his inventory in Hoboken without losing
any money (i.e. the shadow price is zero).
However, we need to check the sensitivity report to
make sure that the proposed decrease of 40 kegs is
within the allowable decrease.
This means that he could make a profit by renting space
in the Hoboken warehouse to Lu Leng for $0.01 per
square foot.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Lu Leng wants space in Brooklyn, but Gribbin would need
to charge her more than $1.82 for every six square feet
(about $0.303 per square foot), or else he will lose money
on the deal.
Note that the sensitivity report indicates an allowable
decrease in Brooklyn that is enough to accommodate Lu
Leng.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
As for the Bronx warehouse, note that the allowable
decrease is zero. This means that we would need to rerun the model to find out the total cost of renting Bronx
space to Lu Leng.
A possible response from Gribbin to Lu Leng:
“I can rent you space in Brooklyn, but it will cost you
$0.35 per square foot. How do you feel about
Hoboken?”
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Contract Bidding Example
A company is taking bids on four construction jobs.
Three contractors have placed bids on the jobs. Their
bids (in thousands of dollars) are given in the table
below. (A dash indicates that the contractor did not bid
on the given job.)
Contractor 1 can do only one job, but contractors 2 and 3
can each do up to two jobs.
Contractor 1
Contractor 2
Contractor 3
Job 1
50
51
—
Operations Management -- Prof. Juran
Job 2
46
48
47
Job 3
42
44
45
Job 4
40
—
45
46
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Formulation
Decision Variables
Which contractor gets which job(s).
Objective
Minimize the total cost of the four jobs.
Constraints
Contractor 1 can do no more than one job.
Contractors 2 and 3 can do no more than two jobs each.
Contractor 2 can’t do job 4.
Contractor 3 can’t do job 1.
Every job needs one contractor.
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©The McGraw-Hill Companies, Inc., 2004
Formulation
Decision Variables
Define Xij to be a binary variable representing the
assignment of contractor i to job j. If contractor i ends up
doing job j, then Xij = 1. If contractor i does not end up with
job j, then Xij = 0.
Define Cij to be the cost; i.e. the amount bid by contractor i
for job j.
Objective
Minimize Z =
3
4
X
i 1 j 1
ij
C ij
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Formulation
Constraints
3
X
i 1
ij
4
X
j 1
j 1
for all j.
ij
 1 for i = 1.
ij
2
4
X
1
for i = 2, 3.
X2 , 4  X 3 , 1  0
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Solution Methodology
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Contractor 1
Contractor 2
Contractor 3
B
Job 1
50
51
10000
C
Job 2
46
48
47
D
Job 3
42
44
45
E
Job 4
40
10000
45
F
G
H
=SUM(B8:E8)
Assignment of contractors to jobs
Contractor 1
Contractor 2
Contractor 3
Total
Required
Job 1
0
0
0
0
=
1
Job 2
0
0
0
0
=
1
Job 3
0
0
0
0
=
1
Job 4
Total
Max
0
0 <= 1
0
0 <= 2
0
0 <= 2
0
=SUM(E8:E10)
=
1
=SUMPRODUCT(B2:E4,B8:E10)
Total cost ($1000s)
Operations Management -- Prof. Juran
0
50
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Solution Methodology
Notice the very large values in cells B4 and
E3. These specific values (10,000) aren’t
important; the main thing is to assign these
particular contractor-job combinations costs
so large that they will never be in any
optimal solution.
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Solution Methodology
Operations Management -- Prof. Juran
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Solution Methodology
Operations Management -- Prof. Juran
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Optimal Solution
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Contractor 1
Contractor 2
Contractor 3
B
Job 1
50
51
10000
C
Job 2
46
48
47
D
Job 3
42
44
45
E
Job 4
40
10000
45
Job 1
0
1
0
1
=
1
Job 2
0
0
1
1
=
1
Job 3
0
1
0
1
=
1
Job 4
1
0
0
1
=
1
F
G
H
Assignment of contractors to jobs
Contractor 1
Contractor 2
Contractor 3
Total
Required
Total cost ($1000s)
Operations Management -- Prof. Juran
Max
Total
1 <= 1
2 <= 2
1 <= 2
182
54
©The McGraw-Hill Companies, Inc., 2004
Conclusions
The optimal solution is to award Job 4 to
Contractor 1, Jobs 1 and 3 to Contractor 2, and Job
2 to Contractor 3. The total cost is $182,000.
Operations Management -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Sensitivity Analysis
1. What is the “cost” of restricting Contractor 1
to only one job?
2. How much more can Contractor 1 bid for Job
4 and still get the job?
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©The McGraw-Hill Companies, Inc., 2004
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
A
B
C
D
Microsoft Excel 9.0 Sensitivity Report
Worksheet: [03b-02-bids.xls]Sheet1
Report Created: 12/30/01 5:11:36 PM
E
F
G
H
Adjustable Cells
Cell
$B$8
$C$8
$D$8
$E$8
$B$9
$C$9
$D$9
$E$9
$B$10
$C$10
$D$10
$E$10
Name
Contractor 1 Job 1
Contractor 1 Job 2
Contractor 1 Job 3
Contractor 1 Job 4
Contractor 2 Job 1
Contractor 2 Job 2
Contractor 2 Job 3
Contractor 2 Job 4
Contractor 3 Job 1
Contractor 3 Job 2
Contractor 3 Job 3
Contractor 3 Job 4
Final Reduced Objective Allowable Allowable
Value
Cost
Coefficient Increase Decrease
0
1
50
1E+30
1
0
1
46
1E+30
1
0
0
42
1
3
1
0
40
3
1E+30
1
0
51
1
1E+30
0
1
48
1E+30
1
1
0
44
1
1
0
9958
10000
1E+30
9958
0
9949
10000
1E+30
9949
1
0
47
1
1E+30
0
1
45
1E+30
1
0
3
45
1E+30
3
Constraints
Cell
$B$11
$C$11
$D$11
$E$11
$F$8
$F$9
$F$10
Name
Total Job 1
Total Job 2
Total Job 3
Total Job 4
Contractor 1 Total
Contractor 2 Total
Contractor 3 Total
Final Shadow Constraint Allowable Allowable
Value
Price
R.H. Side
Increase Decrease
1
51
1
0
1
1
47
1
1
1
1
44
1
0
1
1
42
1
0
1
1
-2
1
1
0
2
0
2
1E+30
0
1
0
2
1E+30
1
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©The McGraw-Hill Companies, Inc., 2004
Conclusions
The sensitivity report indicates a shadow price
of –2 (cell E29).
(Allowing Contractor 1 to perform one additional job
would reduce the total cost by 2,000.)
The allowable increase in the bid for Job 4 by
Contractor 1 is 3.
(He could have bid any amount up to $43,000 and
still have won that job.)
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Network Representation
Con. 1
Job 1
Con. 2
Job 2
Operations Management -- Prof. Juran
Con. 3
Job 3
Job 4
59
©The McGraw-Hill Companies, Inc., 2004
Optimal Solution
Con. 1
Con. 2
Con. 3
44
51
Job 1
47
Job 2
Operations Management -- Prof. Juran
40
Job 3
Job 4
60
©The McGraw-Hill Companies, Inc., 2004
Summary
• Optimization Extensions
• Multiperiod Models
– Operations Planning: Sailboats
• Network Flow Models
– Transportation Model: Beer Distribution
– Assignment Model: Contract Bidding
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