ISDS 514
Solutions to Homework 1: Setting up basic linear programs
1.1
A linear constraint means that the left hand side corresponds to decision variables multiplied by
constant numbers and then added with each other and the left hand side is a constant
number.
a. linear
b. nonlinear (the square root term)
c. linear, can be rewritten as: 4 X1 – .3333 X2 = 75
d. not linear in the form it is, as decision variables are divided by each other.
Side note (we will see more of this later): This constraint can be linearized if we transform it. If we assume that
the decision variables are non-negative, the denominator would be non-negative. We can multiply both
sides by the denominator to get:
3𝑋𝑋1 + 2𝑋𝑋2 − 3𝑋𝑋3 ≤ 0.9(𝑋𝑋1 + 𝑋𝑋2 + 𝑋𝑋3 )
Brining all decision variables to the left hand side and leaving a constant number on the right hand side, we get
2.1 X1 + 1.1 X2 – 3.9 X3 ≤ 0 which is a linear constraint.
e. nonlinear – squared value of decision variables.
1.2 9.
The constraint 3𝑋𝑋1 + 3𝑋𝑋2 ≤ 480 is redundant as it does not limit the feasible area any more than the
other constraints too. If we remove this constraint, the problem does not change at all.
X1 = 50 and X2 = 50 is a feasible solution as it falls on one of the sides of the feaible area.
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To get the optimal solution, notice that the optimal solution would be one of the corner points. The
corner points here are (0,0), (80,0), (60,40), (0,100). Calculate the objective function for
each of these points. Since we are maximizing, the point with the highest value of objective
is the optimal solution.
1.3
a) objective function
functional constraints
nonnegativity constraints
Z = x1 + 2x2
x1 + x2 ≤ 5
x1 + 3x2 ≤ 9
x1 ≥ 0, x2 ≥ 0
b & e)
c) Yes. The point (3,1) satisfies all the constraints in the problem.
d) No. The point (1,3) does not satisfy the constraint x1 + 3x2 ≤ 9.
1.4
X1 = number of propane grills to produce, X2 = number of electric grills to produce
MAX
100 X1 + 80 X2
ST
2 X1 + 1 X2 ≤ 2400
4 X1 + 5 X2 ≤ 6000
2 X1 + 3 X2 ≤ 3300
1 X1 + 1 X2 ≤ 1500
X1, X2 ≥ 0
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See the Excel solution file for part d.
1.5
a) To build an algebraic model for this problem, start by defining the decision variables. In this case, the
two decisions are how many Collegiates to produce and how many Minis to produce. These
variables are defined below:
C = Number of Collegiates to produce,
M = Number of Minis to produce.
Next determine the goal of the problem. In this case, the goal is to produce the number of each type
of backpack to achieve the highest possible total profit. Each Collegiate yields a unit profit of $32
while each Mini yields a unit profit of $24. The objective function is therefore
Maximize Total Profit = $32C + $24M.
The first set of constraints in this problem involve the limited resources (nylon and labor hours).
Given the number of backpacks produced, C and M, and the required nylon and labor hours for
each, the total resources used can be calculated. These total resources used need to be less than or
equal to the amount available. Since the labor available is in units of hours, the labor required for
each backpack needs to be in units of hours (3/4 hour and 2/3 hour) rather than minutes (45
minutes and 40 minutes). These constraints are as follows:
Nylon:
Labor Hours:
3C + 2M ≤ 5400 square feet,
(3/4)C + (2/3)M ≤ 1400 hours.
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The final constraint is that they should not produce more of each backpack than the sales forecast.
Therefore,
Sales Forecast:
C ≤ 1000
M ≤ 1200.
After adding nonnegativity constraints, the complete algebraic formulation is given below:
Let
C = Number of Collegiates to produce,
M = Number of Minis to produce.
Maximize
Total Profit = $32C + $24M,
subject to
Nylon:
3C + 2M ≤ 5400 square feet,
Labor Hours:
(3/4)C + (2/3)M ≤ 1400 hours,
Sales Forecast:
C ≤ 1000
M ≤ 1200.
and C ≥ 0, M ≥ 0.
b) To build a spreadsheet model for this problem, start by entering the data. The data for this
problem are the unit profit of each type of backpack, the resource requirements (square feet of
nylon and labor hours required), the availability of each resource, 5400 square feet of nylon and (35
laborers)(40 hours/laborer) = 1400 labor hours, and the sales forecast for each type of backpack
(1000 Collegiates and 1200 Minis). In order to keep the units consistent in row 8 (hours), the labor
required for each backpack (in cells C8 and D8) are converted from minutes to hours (0.75 hours =
45 minutes, 0.667 hours = 40 minutes). The range names UnitProfit (C4:D4), Available (G7:G8), and
SalesForecast (C13:D13) are added for these data.
The decision to be made in this problem is how many of each type of backpack to make.
Therefore, we add two changing cells with range name UnitsProduced (C11:D11). The values in
CallsPlaced will eventually be determined by the Solver. For now, arbitrary values of 10 and 10 are
entered.
The goal is to produce backpacks so as to achieve the highest total profit. Thus, the objective cell
should calculate the total profit, where the objective will be to maximize this objective cell. In this
case, the total profit will be
Total Profit = ($32)(# of Collegiates) + ($24)(# of Minis) or
Total Cost = SUMPRODUCT(UnitProfit, UnitsProduced).
This formula is entered into cell G11 and given a range name of TotalProfit. With 10 Collegiates and
10 Minis produced, the total profit would be ($32)(10) + ($24)(10) = $560.
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The first set of constraints in this problem involve the limited available resources (nylon and labor
hours). Given the number of units produced (UnitsProduced in C11:D11), we calculate the total
resources required. For nylon, this will be =SUMPRODUCT(C7:D7, UnitsProduced) in cell E7. By using
a range name or an absolute reference for the units produced, this formula can be copied into cell
E8 to calculate the labor hours required. The total resources used
(TotalResources in E7:E8) must be <= Available (in cells G7:G8), as indicated by the <= in F7:F8.
The final constraint is that it does not make sense to produce more backpacks than can be sold (as
predicted by the sales forecast). Therefore UnitsProduced (C11:D11) should be less-than-orequal-to
the SalesForecast (C13:D13), as indicated by the <= in C12:D12
The Solver information and solved spreadsheet are shown below.
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Solver Parameters
Set Objective Cell: TotalProfit
To: Max
By Changing Variable Cells:
UnitsProduced
Subject to the Constraints:
TotalRequired <= Available
UnitsProduced <= SalesForecast
Solver Options:
Make Variables Nonnegative
Solving Method: Simplex LP
Range Name
Cells
Available
G7:G8
SalesForecast
C13:D13
TotalProfit
G11
TotalRequired
E7:E8
UnitProfit
C4:D4
UnitsProduced
C11:D11
E
5
6
7
8
Total
Required
=SUMPRODUCT(C7:D7,UnitsProduced)
=SUMPRODUCT(C8:D8,UnitsProduced)
G
10
11
Total Profit
=SUMPRODUCT(UnitProfit,UnitsProduced)
Thus, they should produce 1000 Collegiates and 975 Minis to achieve the maximum total profit of
$55,400.
c) Start by plotting a graph with Collegiates (C) on the horizontal axis and Minis (M) on the
vertical axis.
Next, the four constraint boundary lines (where the left-hand-side of the constraint exactly equals
the right-hand-side) need to be plotted. The easiest way to do this is by determining where these
lines intercepts the two axes. For the Nylon constraint boundary line (3C + 2M = 5400), setting M = 0
yields a C-intercept of 1800 while setting C = 0 yields an M-intercept of 2700. For the
Labor constraint boundary line ((3/4)C + (2/3)M = 1400), setting M = 0 yields a C-intercept of
1866.67 while setting C = 0 yields an M-intercept of 2100. The sales forecast constraints are a
horizontal line at M = 1200 and a vertical line at C = 1000. These constraint boundary lines are
plotted below.
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Mini
3000
2500
3C + 2M = 5400
2000
C = 1000
(3/4)C + (2/3)M = 1400
1500
M = 1200
1000
500
500
1000
1500
2000
Collegiate
A feasible solution must be below and/or to the left of all four of these constraints while being
above the Collegiate axis (since C ≥ 0) and to the right of the Mini axis (since M ≥ 0). This yields the
feasible region shown below.
Mini
3000
2500
2000
1500
1000
500
500
1000
1500
2000
Collegiate
To find the optimal solution, an objective function line is plotted by setting the objective function
equal to a value. For example, the objective function line when the value of the objective function is
$48,000 is plotted as a dashed line below.
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Mini
3000
2500
2000
1500
1000
500
500
1000
1500
2000
Collegiate
$32C + $24M = $48,000
All objective function lines will be parallel to this one. To find the feasible solution that maximizes
profit, slide this line out as far as possible while still touching the feasible region. This occurs when
the profit is $55,400, and the objective function line intersect the feasible region at the single point
with (C, M) = (1000, 975) as shown below.
Mini
3000
2500
2000
1500
1000
Optimal:C(, M) = (1000,975)
500
500
1000
1500
2000
$32C + $24M = $55,400
Collegiate
Therefore, the optimal solution is to produce 1000 Collegiates and 975 Minis, yielding a total profit
of $55,400.
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1.6
b) The decisions to be made are what fraction of each venture to participate. The constraints
on the decisions are that you can’t become more than a full partner in either venture, that
your money is limited to $12,000, and time is limited to 600 hours. In addition, negative
involvement is not possible. The overall measure of performance for the decisions is the
profit to be made.
c) Let x1 be the fraction purchased of the partnership in the first friends venture.
Let x2 be the fraction purchased of the partnership in the second friends venture.
Maximize P = $9,000x1 + $9,000x2,
subject to
x1 ≤ 1
x2 ≤ 1
$10,000x1 + $8,000x2 ≤ $12,000
400x1 + 500x2 ≤ 600 hours
and
x1 ≥ 0, x2 ≥ 0.
d)
Data cells:
Changing cells:
Objective cell:
Output cells:
B2:C2, B5:C6, F5:F6, and B11:C11
B9:C9
F9
D5:D6
e) This is a linear programming model the objective function as well as the left hand side of
constraints involve “linear” functions, meaning that they involve decision variables that are
multiplied by fixed numbers, which are then added to each other. This can also be seen by
the use of SUMPRODUCT function when expressing these functions in Excel.
The constraints are also in form of either <=, >= or = .
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i)
1.7
Optimal solution = (x1, x2) = (0.667, 0.667). P = $12,000.
b) The decisions to be made are how many windows of each type to produce. The constraints
on the decisions are the amounts of glass, aluminum and wood available. In addition,
negative production levels are not possible. The overall measure of performance for the
decisions is the profit to be made.
c) This is the same as the algebraic formulation of the problem.
Let W be the number of wood-framed windows to produce.
Let A be the number of aluminum-framed windows to produce.
The following table gives the data for the problem:
Resource Usage per Unit of Activity
Amount of
Resource
Wood-framed
Aluminum-framed
Resource Available
Glass
6
8
48
Aluminum
0
1
4
Wood
1
0
6
Unit Profit
$60
$30
glass: 6 (#wood-framed) + 8 (# aluminum-framed) ≤ 48
aluminum:
1 (# aluminum-framed) ≤ 4
wood:
1 (#wood-framed) ≤ 6
Nonnegativity:
(#wood-framed) ≥ 0, (# aluminum-framed) ≥ 0
Profit = $60 (#wood-framed) + $30 (# aluminum-framed)
Alternatively and more simply, we can write this in terms of decision variables:
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Maximize P = 60W + 30A
subject to
6W + 8A ≤ 48
W≤6
A≤4
and
W ≥ 0, A ≥ 0.
d)
Data cells:
Changing cells:
Objective cell:
Output cells:
B2:C2, B5:C5, F5, B10:C10
B8:C8
F8
D5, F8
e) This is a linear programming model the objective function as well as the left hand side of
constraints involve “linear” functions, meaning that they involve decision variables that are
multiplied by fixed numbers, which are then added to each other. This can also be seen by
the use of SUMPRODUCT function when expressing these functions in Excel.
The constraints are also in form of either <=, >= or = .
1.8
a.
See the excel solution file
X1 = the number of weekday ads placed
X2 = the number of Sunday ads placed
MAX 30000X1 + 80000X2
≤
26 (Max. daily ads)
S.T.
X1
≤
5 (Max. Sunday ads)
X2
≥
8 (Min. daily ads)
X1
≥
2 (Min. Sunday ads)
X2
≤ 40000
(Budget)
2000X1 + 8000X2
X1, X2 ≥ 0
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The “no interaction” assumption essentially means that the exposure obtained from weekday ads has
no influence or is not impacted by the exposure obtained from weekend ads. This means that the total
exposure from the two types of ads is simply the addition of exposures of obtained from each type of ad
and therefore total exposure is linear in terms of placement of ads.
This assumption may not be fully valid for this problem because there might be people who read the
newspaper each day. Consequently, some of the Sunday readership will include its weekday readership
and the total exposure obtained from weekday and weekend ads may “interact”. The full exposure by
PMA’s advertising decision may not truly be obtained by addition of exposures from weekend and
weekday ads. Place 12 daily ads and 2 Sunday ads. The total exposure = 520,000.
b. A redundant constraint means that the addition and removal of it would not change the optimal
solution. Therefore, all non-binding constraints are “redundant”. In this problem, given the budget, a
maximum of 20 daily ads or 5 Sunday ads are already guaranteed, and the constraints corresponding to
them are non-binding.
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