CHAPTER 3
LINEAR PROGRAMMING:
FORMULATION AND APPLICATIONS
SOLUTION TO SOLVED PROBLEMS
3.S1 Farm Management
Dwight and Hattie have run the family farm for over thirty years. They are currently planning
the mix of crops to plant on their 120-acre farm for the upcoming season. The table below gives
the labor hours and fertilizer required per acre, as well as the total expected profit per acre for
each of the potential crops under consideration. Dwight, Hattie, and their children can work at
most 6,500 total hours during the upcoming season. They have 20 tons of fertilizer available.
What mix of crops should be planted to maximize the family’s total profit. Formulate and solve a
linear programming model in a spreadsheet.
Crop
Labor Required
(hours per acre)
Fertilizer Required
(tons per acre)
Expected Profit
(per acre)
Oats
50
1.5
$500
Wheat
60
2
$600
Corn
105
4
$950
This is a resource-allocation problem. The activities are the planting of the three crops and the
limited resources are land, labor, and fertilizer. We will therefore build a spreadsheet following
the template for resource-allocation problems in Figure 3.4. Start by entering the data. The data
for this problem are the labor required, fertilizer required, and expected profit for each crop (per
acre). Following the template, the data in the spreadsheet would be entered as displayed below,
where range names of ProfitPerAcre (C4:E4) and TotalAvailable (H7:H9) are assigned to the
corresponding data cells.
B
3
4
5
6
7
8
9
Profit (per acre)
Resources
Land (acres)
Labor (hours)
Fertilizer (tons)
C
Oats
$500
D
Wheat
$600
E
Corn
$950
Resources Used per Acre
1
1
1
50
60
105
1.5
2
4
1
F
G
H
Total
Available
120
6500
200
The decisions to be made in this problem are how many acres of each crop to plant. Therefore,
we add three changing cells in C12:E12 with range name AcresPlanted. The values in
AcresPlanted (C12:E12) will eventually be determined by the Solver. For now, an arbitrary value
of 1 is entered for each crop.
B
3
4
5
6
7
8
9
10
11
12
Profit (per acre)
Resources
Land (acres)
Labor (hours)
Fertilizer (tons)
Acres Planted
C
Oats
$500
D
Wheat
$600
E
Corn
$950
F
G
Total
Available
120
6500
200
Resources Used per Acre
1
1
1
50
60
105
1.5
2
4
1
1
H
1
The goal is to maximize the family’s total profit. Thus, the target cell should calculate the total
profit. In this case, the total profit will be
Total Profit = ($500)(acres of oats) + ($600)(acres of wheat) + ($950)(acres of corn)
or
Total Profit = SUMPRODUCT(ProfitPerAcre, AcresPlanted).
This formula is entered into cell H12. With 1 acre of each crop planted, the total cost would be
($500)(1) + ($600)(1) + ($950)(1) = $2,050.
B
3
4
5
6
7
8
9
10
11
12
Profit (per acre)
Resources
Land (acres)
Labor (hours)
Fertilizer (tons)
Acres Planted
C
Oats
$500
D
Wheat
$600
E
Corn
$950
Resources Used per Acre
1
1
1
50
60
105
1.5
2
4
1
1
1
11
12
2
F
G
H
Total
Available
120
6500
200
Total Profit
$2,050
H
Total Profit
=SUMPRODUCT(ProfitPerAcre,AcresPlanted)
The functional constraints in this problem involve the limited resources of land, labor, and
fertilizer. Given the AcresPlanted (the changing cells in C12:E12), we calculate the total
resources used in TotalUsed (cells F7:F9). For land, this will be =SUMPRODUCT(C7:E7,
AcresPlanted). Using a range name or an absolute reference for the acres planted, this formula
can be copied into cells F8:F9 to calculate the amount of labor and fertilizer used. The total
resources used must be <= TotalAvailable (H7:H9), as indicated by the <= in G7:G9.
B
3
4
5
6
7
8
9
10
11
12
Profit (per acre)
Resources
Land (acres)
Labor (hours)
Fertilizer (tons)
Acres Planted
C
Oats
$500
D
Wheat
$600
E
Corn
$950
Resources Used per Acre
1
1
1
50
60
105
1.5
2
4
1
1
1
5
6
7
8
9
3
F
Total
Used
3
215
8
G
H
<=
<=
<=
Total
Available
120
6500
200
Total Profit
$2,050
F
Total
Used
=SUMPRODUCT(C7:E7,AcresPlanted)
=SUMPRODUCT(C8:E8,AcresPlanted)
=SUMPRODUCT(C9:E9,AcresPlanted)
The Solver information and solved spreadsheet are shown below.
A
1
2
3
4
5
6
7
8
9
10
11
12
B
C
D
E
Oats
$500
Wheat
$600
Corn
$950
G
H
<=
<=
<=
Total
Available
120
6500
200
Farm Management
Profit (per acre)
Resources
Land (acres)
Labor (hours)
Fertilizer (tons)
Acres Planted
Resources Used per Acre
1
1
1
50
60
105
1.5
2
4
80
40
F
Total
Used
=SUMPRODUCT(C7:E7,AcresPlanted)
=SUMPRODUCT(C8:E8,AcresPlanted)
=SUMPRODUCT(C9:E9,AcresPlanted)
Total
Used
120
6,400
200
Total Profit
$64,000
0
Range Nam e
AcresPla nted
Profi tPerAcre
TotalAvai labl e
TotalProfit
TotalUsed
5
6
7
8
9
F
11
12
Cells
C12:E1 2
C4:E4
H7:H9
H12
F7:F9
H
Total Profit
=SUMPRODUCT(ProfitPerAcre,AcresPlanted)
Thus, oats should be planted on 80 acres and wheat on 40 acres, while not planting any corn,
with a resulting total profit of $64,000.
4
3.S2 Diet Problem
The kitchen manager for Sing Sing Prison is trying to decide what to feed its prisoners. She
would like to offer some combination of milk, beans, and oranges. The goal is to minimize cost,
subject to meeting the minimum nutritional requirements imposed by law. The cost and
nutritional content of each food, along with the minimum nutritional requirements, are shown
below. What diet should be fed to each prisoner? Formulate and solve a linear programming
model in a spreadsheet.
Milk
(gallons)
3.2
1.12
32.0
2.00
Niacin (mg)
Thiamin (mg)
Vitamin C (mg)
Cost ($)
Navy
Beans
(cups)
4.9
1.3
0.0
0.20
Oranges
(large Calif.
Valencia)
0.8
0.19
93.0
0.25
Minimum
Daily
Requirement
13.0
1.5
45.0
This is a cost-benefit-trade-off problem. The activities are the quantities of food to feed each
prisoner and the required benefits are the minimum nutritional requirements. We will therefore
build a spreadsheet following the template for cost-benefit-trade-off problems in Figure 3.6. Start
by entering the data. The data for this problem are the nutrient content of each food, the
minimum daily requirement for each nutrient, and the cost of each food. Following the template,
the data in the spreadsheet would be entered as displayed below, where range names of UnitCost
(C5:E5), NutritionalContents (C9:E11), and MinimumRequirement (H9:H11) are assigned to the
corresponding data cells.
B
3
4
5
6
7
8
9
10
11
Unit Cost
Niacin
Thiamin
Vitamin C
C
Milk
(gal.)
$2.00
D
Beans
(cups)
$0.20
E
F
G
H
Oranges
$0.25
Nutritional Contents (mg)
3.2
4.9
0.8
1.12
1.3
0.19
32
0
93
5
Minimum
Requirement
13
1.5
45
The decisions to be made in this problem are how much of each food type should be fed to each
prisoner. Therefore, we add three changing cells in C13:E13, with range name Quantity. The
values in Quantity (C13:E13) will eventually be determined by the Solver. For now, an arbitrary
value of 1 is entered for each food type.
The goal is to minimize the total cost per prisoner. Thus, the target cell should calculate this cost:
Total Cost = ($2)(gallons of milk) + ($0.20)(cups of beans) + ($0.25)(number of oranges)
or
Total Cost = SUMPRODUCT(UnitCost, Quantity).
This formula is entered into cell H14.
B
3
4
5
6
7
8
9
10
11
12
13
14
Unit Cost
Niacin
Thiamin
Vitamin C
Quantity
(per prisoner)
C
Milk
(gal.)
$2.00
D
Beans
(cups)
$0.20
E
G
H
Oranges
$0.25
Nutritional Contents (mg)
3.2
4.9
0.8
1.12
1.3
0.19
32
0
93
1
F
1
1
13
14
6
Minimum
Requirement
13
1.5
45
Total Cost
$2.45
H
Total Cost
=SUMPRODUCT(UnitCost,Quantity)
The functional constraints in this problem involve the minimum daily requirement of each
nutrient. Given the amount of food fed each prisoner (the changing cells in C13:E13), we
calculate the total resources used in F9:F11. For niacin, this will be =SUMPRODUCT(C9:E9,
$C$13:$E$13). Using an absolute reference for the acres planted, this formula can be copied into
cells F10-F11 to calculate the thiamin and vitamin C. The benefit achieved (total of each
nutrient) must be >= the minimum needed (H9:H11), as indicated by the >= in G9:G11.
B
3
4
5
6
7
8
9
10
11
12
13
14
Unit Cost
Niacin
Thiamin
Vitamin C
Quantity
(per prisoner)
C
Milk
(gal.)
$2.00
D
Beans
(cups)
$0.20
E
G
H
Oranges
$0.25
Nutritional Contents (mg)
3.2
4.9
0.8
1.12
1.3
0.19
32
0
93
1
F
1
1
7
8
9
10
11
7
Total
Nutrients
8.9
>=
2.610
>=
125
>=
Minimum
Requirement
13
1.5
45
Total Cost
$2.45
F
Total
Nutrients
=SUMPRODUCT(C9:E9,Quantity)
=SUMPRODUCT(C10:E10,Quantity)
=SUMPRODUCT(C11:E11,Quantity)
The Solver information and solved spreadsheet are shown below.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
B
C
D
E
Beans
(cups)
$0.20
Oranges
$0.25
G
H
The Prison Diet Problem
Unit Cost
Niacin
Thiamin
Vitamin C
Quantity
(per prisoner)
Milk
(gal.)
$2.00
Nutritional Contents (mg)
3.2
4.9
0.8
1.12
1.3
0.19
32
0
93
0
2.574
F
Total
Nutrients
=SUMPRODUCT(C9:E9,Quantity)
=SUMPRODUCT(C10:E10,Quantity)
=SUMPRODUCT(C11:E11,Quantity)
Total
Nutrients
13
>=
3.438
>=
45
>=
0.484
Range Name
Mini mumRequirem ent
Nutri tionalContents
Quantity
TotalCos t
TotalNutrients
UnitCost
7
8
9
10
11
F
13
14
Minimum
Requirement
13
1.5
45
Total Cost
$0.64
Cells
H9:H11
C9:E11
C13:E13
H14
F9:F11
C5:E5
H
Total Cost
=SUMPRODUCT(UnitCost,Quantity)
Thus, each prisoner should be fed a daily average of 2.574 cups of beans and 0.484 oranges for a
total cost of $0.64.
8
3.S3 Cutting Stock Problem
Decora Accessories manufactures a variety of bathroom accessories, including decorative towel
rods and shower curtain rods. Each of the accessories includes a rod made out of stainless steel.
However, many different lengths are needed: 12”, 18”, 24”, 40”, and 60”. Decora purchases
60” rods from an outside supplier and then cuts the rods as needed for their products. Each 60”
rod can be used to make a number of smaller rods. For example, a 60” rod could be used to
make a 40” and an 18” rod (with 2” of waste), or 5 12” rods (with no waste). For the next
production period, Decora needs 25 12” rods, 52 18” rods, 45 24” rods, 30 40” rods, and 12
60” rods. What is the fewest number of 60” rods that can be purchased to meet their production
needs? Formulate and solve an integer programming model in a spreadsheet.
This is a cost-benefit-trade-off problem. For each length of rod, the required benefit is obtaining
the minimum number needed of those rods. Each possible way of cutting a 60” rod will represent
an activity. For example, a 60” rod can be cut into 5 12” rods, or 3 12” rods and an 18” rod (with
6” scrap), or 3 12” rods and a 24” rod, etc. We will build a spreadsheet following the template for
cost-benefit-trade-off problems in Figure 3.6. Start by entering the data. The data for this
problem are the total length of the uncut rods, the minimum number of each rod length required
and, for each possible pattern of cutting a 60” rod, the number of each length rod created. There
are 11 possible patterns of cutting a 60” rod that either leave no scrap or an unusable piece of
scrap (i.e., less than 12”). Following the template, the data in the spreadsheet would be entered as
displayed, where range names of UncutRodLength (F3), MinimumRequirement (P7:P11), and
Waste (C12:M12) are assigned to the corresponding data cells.
3
4
5
6
7
8
9
10
11
12
12
B
C D
Length of Uncut Rods
Length of
Cut Rod
12
18
24
40
60
Waste
B
Waste
1
5
2
3
1
E
F G
60
3
3
4
2
2
1
0
6
0
0
H
I
Pattern
5 6 7
1 1 1
1
1 2
1
6
0
8
J
K
L
M
8
9
10 11
3
2
1
1
1
6
0
2
1
0
N
O
P
Minimum
Requirement
25
52
45
30
12
C
=UncutRodLength-SUMPRODUCT($B$7:$B$11,C7:C11)
9
The decisions to be made in this problem are how many rods should be cut into each of the
possible patterns. Therefore, we add 11 changing cells in C14:M14, with range name
PatternQuantity. The values in PatternQuantity (C14:M14) will eventually be determined by the
Solver. For now, an arbitrary value of 1 is entered for each pattern.
B
3
4
5
6
7
8
9
10
11
12
13
14
C D E F G
Length of Uncut Rods 60
Length of
Cut Rod
12
18
24
40
60
Waste
Pattern Quantity
1
5
2
3
1
3
3
4
2
2
1
H
I
Pattern
5 6 7
1 1 1
1
1 2
1
J
K
L
M
8
9
10 11
3
2
1
1
N
O
Minimum
Needed
25
52
45
30
12
1
0
6
0
0
6
0
8
6
0
2
1
0
1
1
1
1
1
1
1
1
1
1
1
P
The goal is to minimize the total number of 60” (uncut) rods needed. This is just the sum of the
changing cells:
Total Uncut Rods needed = SUM(PatternQuantity).
This formula is entered into cell P14.
B
3
4
5
6
7
8
9
10
11
12
13
14
C D E F G
Length of Uncut Rods 60
Length of
Cut Rod
12
18
24
40
60
Waste
Pattern Quantity
1
5
2
3
1
3
3
4
2
2
1
H
I
Pattern
5 6 7
1 1 1
1
1 2
1
J
K
L
M
8
9
10 11
3
2
1
1
N
6
0
0
6
0
8
6
0
2
1
0
1
1
1
1
1
1
1
1
1
1
1
Total Uncut Rods Needed
11
13
14
10
P
Minimum
Needed
25
52
45
30
12
1
0
O
P
Total Uncut Rods Needed
=SUM(PatternQuantity)
The functional constraints in this problem require that the number of cut rods of each length must
be at least the minimum number needed. Given PatternQuantity (the changing cells in C14:M14),
we calculate the total number of each rod length produced in TotalProduced (N7:N11). For 12”
rods, this will be =SUMPRODUCT(C7:M7, PatternQuantity). Using a range name or absolute
reference for the quantity of each pattern, this formula can be copied into cells N8:N11 to
calculate the total number of other lengths created. For each rod length, the benefit achieved (the
total number of that rod length cut) must be >= MinimumRequirement (P7:P11), as indicated by
the >= in O7:O11.
3
4
5
6
7
8
9
10
11
12
13
14
B
C D E F G
Length of Uncut Rods 60
Length of
Cut Rod
12
18
24
40
60
Waste
1
5
2
3
1
3
3
4
2
2
1
H
I
Pattern
5 6 7
1 1 1
1
1 2
1
J
K
L
M
8
9
10 11
3
2
1
1
1
0
6
0
0
6
0
8
6
0
2
1
0
Pattern Quantity 1
1
1
1
1
1
1
1
1
1
1
5
6
7
8
9
10
11
11
N
Total
Produced
16
10
5
2
1
O
P
>=
>=
>=
>=
>=
Minimum
Needed
25
52
45
30
12
Total Uncut Rods Needed
11
N
Total
Produced
=SUMPRODUCT(C7:M7,PatternQuantity)
=SUMPRODUCT(C8:M8,PatternQuantity)
=SUMPRODUCT(C9:M9,PatternQuantity)
=SUMPRODUCT(C10:M10,PatternQuantity)
=SUMPRODUCT(C11:M11,PatternQuantity)
Also, an integer number of each pattern must be cut. Therefore, we constrain the changing cells
to be integer. The Solver information and solved spreadsheet are shown below.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
B
C
D
E
F
G
H
I
J
L
M
N
O
P
Cutting Stock Problem
Length of Uncut Rods 60
Length of
Cut Rod
12
18
24
40
60
Waste
1
5
2
3
1
3
3
4
2
2
1
Pattern Quantity
Pattern
5 6 7
1 1 1
1
1 2
1
8
3
0
6
0
0
6
0
8
6
2
0
0
0
0
17
0
0
Range Na me
Mini mumNee ded
PatternQua ntity
To talProduced
To talRod sNeede d
UncutRodL ength
Wa ste
5
6
7
8
9
10
11
12
K
B
Waste
Total
Minimum
10 11 Produced
Needed
27
>=
25
2 1
52
>=
52
1
45
>=
45
1
30
>=
30
1
12
>=
12
0 2 0
Total Uncut Rods Needed
11 30 12
72
9
Cells
P7:P11
C14:M1 4
N7:N11
P14
F3
C12:M1 2
N
Total
Produced
=SUMPRODUCT(C7:M7,PatternQuantity)
=SUMPRODUCT(C8:M8,PatternQuantity)
=SUMPRODUCT(C9:M9,PatternQuantity)
=SUMPRODUCT(C10:M10,PatternQuantity)
=SUMPRODUCT(C11:M11,PatternQuantity)
C
=UncutRodLength-SUMPRODUCT($B$7:$B$11,C7:C11)
13
14
P
Total Uncut Rods Needed
=SUM(PatternQuantity)
Thus, PatternQuantity (C14:M14) show the number of rods that should be cut into the respective
patterns. TotalRodsNeeded (P14) indicates that a total of 72 uncut rods are needed.
12
3.S4 Bidding for Classes
In the MBA program at a prestigious university in the Pacific Northwest, students bid for
electives in the second year of their program. Each student has 100 points to bid (total) and must
take two electives. There are four electives available: Management Science, Finance, Operations
Management, and Marketing. Each class is limited to 5 students. The bids submitted for each of
the 10 students are shown in the table below.
Management
Science
60
20
45
50
30
50
70
25
35
60
Student
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
Finance
10
20
45
20
30
50
20
25
15
10
Operations
Management
10
40
5
5
30
0
10
35
35
10
Marketing
20
20
5
25
10
0
0
15
15
20
a. Formulate and solve a spreadsheet model to determine an assignment of students to classes so as
to maximize the total bid points of the assignments.
This is a special kind of assignment problem. The assignees are the students and the tasks are the
classes. Each student is assigned to take two classes. Each class has a limit of five students. Start
by entering the data. The data for this problem are the bid points for each student for each class,
where the corresponding data cells are assigned a range name of Points (C5:F14).
B
3
4
5
6
7
8
9
10
11
12
13
14
Points
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
C
Management
Science
60
20
45
50
30
50
70
25
35
60
D
Finance
10
20
45
20
30
50
20
25
15
10
13
E
Operations
Management
10
40
5
5
30
0
10
35
35
10
F
Marketing
20
20
5
25
10
0
0
15
15
20
The decisions to be made in this problem are whether or not to assign each student to each class.
Therefore, a table of changing cells is created for each student and class combination in C18:F27,
and given a range name of Assignment. The values in Assignment (C18:F27) will eventually be
determined by the Solver. For now, arbitrary values of 0 and 1 are entered.
B
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
Points
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
C
Management
Science
60
20
45
50
30
50
70
25
35
60
Assignment
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
Management
Science
1
0
1
0
1
0
1
0
1
0
D
Finance
10
20
45
20
30
50
20
25
15
10
E
Operations
Management
10
40
5
5
30
0
10
35
35
10
Marketing
20
20
5
25
10
0
0
15
15
20
Finance
0
1
0
1
0
1
0
1
0
1
Operations
Management
1
0
1
0
1
0
1
0
1
0
Marketing
0
1
0
1
0
1
0
1
0
1
14
F
The goal is to maximize the total bid points of the assignments.
Total Bid Points = SUMPRODUCT(Points, Assignment).
This formula is entered into cell I29.
B
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
Points
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
C
Management
Science
60
20
45
50
30
50
70
25
35
60
Assignment
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
Management
Science
1
0
1
0
1
0
1
0
1
0
D
Finance
10
20
45
20
30
50
20
25
15
10
E
Operations
Management
10
40
5
5
30
0
10
35
35
10
F
Marketing
20
20
5
25
10
0
0
15
15
20
Finance
0
1
0
1
0
1
0
1
0
1
Operations
Management
1
0
1
0
1
0
1
0
1
0
Marketing
0
1
0
1
0
1
0
1
0
1
G
H
Total Points
29
I
535
H
I
Total Points =SUMPRODUCT(Points,Assignment)
15
The functional constraints in this problem are that each student must be assigned to two classes,
each class is limited to five students, and each student can take each class at most once. The
number of classes a student is assigned to is just the sum of the row of changing cells for each
student. For example, for George it is =SUM(C18:F18). This formula is copied into G18:G27.
The number of students assigned to a class is the sum of the column of changing cells for each
class. For example, for Management Science it is =SUM(C18:C27). This formula is copied into
C28:F28.
B
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
C
Management
Points
Science
George
60
Fred
20
Ann
45
Eric
50
Susan
30
Liz
50
Ed
70
David
25
Tony
35
Jennifer
60
Management
Science
1
0
1
0
1
0
1
0
1
0
5
<=
Capacity
5
Assignment
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
D
G
Finance
10
20
45
20
30
50
20
25
15
10
E
F
Operations
Management Marketing
10
20
40
20
5
5
5
25
30
10
0
0
10
0
35
15
35
15
10
20
Finance
0
1
0
1
0
1
0
1
0
1
5
<=
5
Operations
Management Marketing
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
5
5
<=
<=
5
5
Total
Classes
2
2
2
2
2
2
2
2
2
2
H
I
=
=
=
=
=
=
=
=
=
=
Classes
to Take
2
2
2
2
2
2
2
2
2
2
Total Points
535
K
16
17
18
19
20
21
22
23
24
25
26
27
28
Total
Classes
=SUM(C18:F18)
=SUM(C19:F19)
=SUM(C20:F20)
=SUM(C21:F21)
=SUM(C22:F22)
=SUM(C23:F23)
=SUM(C24:F24)
=SUM(C25:F25)
=SUM(C26:F26)
=SUM(C27:F27)
B
C
D
E
F
Total in Class =SUM(C18:C27) =SUM(D18:D27) =SUM(E18:E27) =SUM(F18:F27)
16
The Solver information and solved spreadsheet are shown below.
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
B
C
D
E
F
Management
Science
60
20
45
50
30
50
70
25
35
60
Finance
10
20
45
20
30
50
20
25
15
10
Operations
Management
10
40
5
5
30
0
10
35
35
10
Marketing
20
20
5
25
10
0
0
15
15
20
Management
Science
1
0
1
0
0
1
1
0
0
1
5
<=
5
Finance
0
0
1
1
1
1
0
1
0
0
5
<=
5
Operations
Management
0
1
0
0
1
0
1
1
1
0
5
<=
5
Marketing
1
1
0
1
0
0
0
0
1
1
5
<=
5
G
H
I
=
=
=
=
=
=
=
=
=
=
Classes
to Take
2
2
2
2
2
2
2
2
2
2
J
K
Bidding for Classes
Points
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
Assignment
George
Fred
Ann
Eric
Susan
Liz
Ed
David
Tony
Jennifer
Total in Class
Capacity
Range Na me
As sign ment
Capa city
Clas sesToTake
Poi nts
Stu dentPoints
To talCla sses
To talInClass
To talPo ints
G
16
17
18
19
20
21
22
28
Total
Classes
2
2
2
2
2
2
2
2
2
2
Total Points
705
Cells
C18:F27
C30:F30
I1 8:I2 7
C5:F14
K18 :K27
G18:G2 7
C28:F28
I2 9
K
Total
Classes
=SUM(C18:F18)
=SUM(C19:F19)
=SUM(C20:F20)
:
:
16
17
18
19
20
21
22
Student
Points
=SUMPRODUCT(C5:F5,C18:F18)
=SUMPRODUCT(C6:F6,C19:F19)
=SUMPRODUCT(C7:F7,C20:F20)
:
:
B
C
D
E
F
Total in Class =SUM(C18:C27) =SUM(D18:D27) =SUM(E18:E27) =SUM(F18:F27)
29
H
I
Total Points =SUMPRODUCT(Points,Assignment)
17
Student
Points
80
60
90
45
60
100
80
60
50
80
Thus, the 1’s in Assignment (C18:F27) show the assignments that should be made, achieving a
total of 705 points.
b. Does the resulting solution seem like a fair assignment?
No. For example, Eric did not get into Management Science despite bidding 50 points, while
Ann got in with only 45 points. Also, Eric got into classes worth only 45 total bid points to him
while Liz got classes worth 100 bid points to her.
c. Which alternative objectives might lead to a fairer assignment?
Perhaps maximizing the minimum total number of bid points achieved by each student.
18