Report of Analytic Skill
Name: Zhao Mingming
1
Major: Marketing Analytics
ID number: 24202207
Word count: 5,490
2
Introduction
According to Anderson et al (2008), management science is an approach to
managerial decision making based on extensive use of quantitative analysis. Linear
programming plays a key role in management science. It is problem-solving
approaches which can help managers make decisions (ibid). The programming needs
to formulation the problem to develop a model including decision variables,
constraints and objective function (Ignizio, 1968). The applications of linear
programming include production scheduling, media selection, financial planning,
capital budgeting, transportation, product mix and staffing (Hillier and Lieberman,
1995). This report will utilize linear programming to firstly solve a financial planning
problem for Nori & Leets Co. and then find a solution for Alabama Atlantic to solve its
transportation problem. Moreover, all the results are accurate to 2 decimal digits.
Finally, it will examine a report about application of linear programming for
dormitory development plan at Petra Christian University.
Q1
The Nori & Leets Co. needs to reduce the emissions of three main types of pollutants
which are particulate, sculpture oxides and hydrocarbons. The policy standard of
required reduction in annual emission rate is 60 million pounds particulates, 150
million pounds sulfur oxides, and 125 million pounds hydrocarbons respectively. The
two sources of pollution are blast furnaces for making pig iron and opening hearth
furnaces for changing iron into steel. There are three effective abatement methods:
(1) increasing height of the smokestacks, (2) using filter devices, and (3) including
better fuels for the furnaces. However, these methods have technological limit that
emissions has a abatement capacity as Table 2.1 shows.
3
Pollutant Taller smokestacks Filters Better fuels
Blast
Furnaces
Open-Hearth
Furnaces
Blast
Furnaces
Open-Hearth
Furnaces
Blast
Furnaces
Open-Hearth
Furnaces
Particulates 12 9 25 20 17 13
Sulfur oxides 35 42 18 31 56 49
Hydrocarbons 37 53 28 24 29 20
Table 1.1 (millions of pounds)
Furthermore, emission reduction by each method is independent. Engineers also
concluded that it should use some combination of methods. However, each method
can lead to a huge cost such as opportunity revenue loss, operating and maintenance
expenses and start-up cost. Therefore, the company has estimated total annual cost
given in the table below for using the methods at full abatement capacities.
Abatement Method Blast Furnaces Open-Hearth Furnace
Taller Smokestacks $8 million 10 million
Filters 7 million 6 million
Better Fuels 11 million 9 million
Table 1.2
The company intends to minimize the annual cost of achieving the required emission
reduction rates for three pollutants by drawing up a plan which specifies which types
of abatement methods will be used and at what fractions of their abatement
capacities for the blast furnaces and the open-hearth furnaces.
aIt is a deterministic model and there is a linear relationship between total cost and
fraction of capacities. Moreover, the variables are independent. Therefore, it will use
linear programming to solve the financial planning problem.
4
The first step is to define variables. Letting X ij (i= taller smokestacks, filters, better
fuels; j=blast furnaces, open-hearth furnaces) denotes the fraction of total annual
cost from the maximum feasible use of each abatement method on both two sources
of pollutant. The following table shows the variables clearly.
Xij Blast Furnaces Open-Hearth Furnace
Taller Smokestacks Xtb Xto
Filters Xfb Xfo
Better Fuels Xbb Xbo
Table 1.3
The next step is to determine the coefficients of variables, C ij. As the total annual cost
(Z) is the sum of the actual total annual cost of each method, C ij are the total annual
cost from maximum feasible use of each abatement method as shown in Table 1.2.
Hence, objective function is to minimize Z= 8Xtb + 10Xto + 7Xfb + 6Xfo + 11Xbb + 9Xbo.
Furthermore, it needs to set constraints. There are two kinds of constraints: the total
emission reduction of these three pollutants should be more than or equal to
required reduction in annual emission rate (mentioned above), and the X ij should be
less than or equal to 1 and more than 0 as it is fraction.
Total emission reduction
The total emission reduction of each pollutant should be the sum of the fraction of
abatement capacity of every method timing the abatement capacity from maximum
feasible use of that method which is given in table 1.1 above. It is determined that
cost of a method is roughly proportional to the fraction of the abatement capacity.
As a result, the fraction of abatement capacity is roughly equal to X ij. Hence, the
constraints can be displayed as:
Particulates: 12Xtb + 9Xto + 25Xfb + 20Xfo + 17Xbb + 13Xbo ≥ 60
Sulfur oxides: 35Xtb + 42Xto + 18Xfb + 31Xfo + 56Xbb + 49Xbo≥ 1505
Hydrocarbons: 37Xtb + 53Xto + 28Xfb + 24Xfo + 29Xbb + 20Xbo≥ 125
0≤Xij≤1
The problem is calculated by Excel Solver (The process is expressed in Appendix 1).
The optimal solution is Z= $32.15 million and the plan is displayed in Table 1.4.
Taller Smokestacks Filters Better Fuels
Blast
Furnace
s
Open-
Hearth
Furnaces
Blast
Furnac
es
Open-
Hearth
Furnaces
Blast
Furnac
es
Open-
Hearth
Furnaces
Fractio
n1 0.62 0.34 1 0.05 1
Table 1.4
Recommendation:
It is recommended that:
All the three types of abatement methods are used but the fractions of their
abatement capacities are different. Among them, taller smokestacks for blast
furnaces, Filters for open-hearth furnaces, and better fuels for open-hearth furnaces
should be maximum feasible used. Taller Smokestacks for open-hearth furnaces
should be used at 62% of its abatement capacity. Filters for blast furnaces should be
34% of its abatement capacity and Better Fuels for Blast Furnaces is used at 5% of its
abatement capacity.
Sensitivity analysisAccording to Fabrycky and Mize (1982), sensitivity analysis in linear programming is
the study of how the changes in coefficients impact optimal solution. It is important
because in real-life, the problems will seldom satisfy all of these assumptions. If data
used have errors, the system will be modified. For this case, after designing the plan,
management needs to conduct sensitivity analysis. The cost in table 1.2 is estimated
6
and each one may easily be off by as much as 10 percent in either direction.
Moreover, values in table 1.1 are uncertain as well but the uncertainty is less than
cost. The policy standards are constants. However, company and government achieve
the agreement that each 10% increase in the policy standards over the current values
will result in $3.5 million tax reduction for the company. Furthermore, there is a
debate between relative values of the policy standards for three pollutants as well.
The next part is to make senility analysis based on the sensitivity report in Appendix
1.
bObjective function coefficients (OFC) and right-hand side values of constraints are
sensitive parameters in linear programming. As mentioned above, OFCs are cost
parameters and RHS values are required reduction of pollutant. However, for this
case, the constraints Right-Hand Sides values of constraints are constant because
they are policy standards. Hence, only cost parameters which are given in Table 1.2
are sensitive parameters in this case.
To find which parameters should be estimated more closely, it needs to identify the
range of optimality for each coefficient of variables. Range of optimality supplies the
range of values over which the current optimal solution will not change. Moreover, it
is assumed that one coefficient change but others are constant. The formula of range
of optimality is shown as:
current coefficient value−allowabledecrease≤C≤current coefficient value+allowable increase
Allowable increase means that the maximized value the coefficient can increase to
remain the optimal solution. Conversely, allowable decrease refers to the maximized
value the coefficient can decrease to keep optimal solution. Allowable increase and
allowable decrease for each objective function coefficient is given in sensitivity
report in Appendix 1. In terms of allowable decrease, the data of three coefficients is 7
1E+30 which means the number is nearly infinite. This applies that no matter how
much this coefficient decrease, the optimal solution will not change. However,
because the coefficient refers to cost in real world, it cannot be negative and the
lowest value is 0. Hence, the range of optimality (in million dollars) for each OFC is:
Taller Smokestacks for Blast Furnaces: 0≤Ctb≤8.34
Taller Smokestacks for Open-Hearth furnaces: 9.33≤C¿≤10.43
Filters for Blast Furnaces: 5≤C fb≤7.38
Filters for Open-Hearth furnaces: 0≤C fo≤7.82
Better Fuels for Blast Furnaces: 10.96≤Cbb≤13.98
Better Fuels for Open-Hearth furnaces: 0≤Cbo≤9.04
The area value of the range (upper value minus lower value) displays the probability
that estimated value can remain the optimal solution. If the range area are is great,
this implies that the estimate value has a large probability to be located in the range
of optimality. In contrast, if the range area is small, the estimated coefficient will have
a large probability to be out of range of optimality and then the optimal solution
changes. The area value of the range for each coefficient is given as in table 1.5.
Coefficient Area of range of optimality
C tb 8.34 (8.34−0)
C ¿ 1.1 (10.43−9.33)
C fb 2.38 (7.38−5)
C fo 7.82 (7.82−0)
Cbb 3.02 (13.98−10.96)
Cbo 9.04 (9.04−0)
Table 1.5
Recommendation
As the Table 1.5 show, Cto (Taller Smokestacks for Open-Hearth furnaces) has the
lowest area value and it is the parameter that the company should be estimated
8
more closely to true value without changing optimal solution. In addition, Cfb (Filter
for Blast Furnaces) and Cbb (Better fuels for Blast Furnaces) should also be estimated
a little accurately as their area of range of optimality is not very large.
cAs mentioned above, the true cost parameter may be off by 10% in either direction.
To analyze the effect of inaccuracy in estimating each cost parameter, it needs to
assume that if one parameter change, others are constant as well. The assumption is
valid for every parameter. In terms of the situation when true value is less than
estimated value, the true value and range of optimality is displayed in Table 1.6.
Coefficient Estimated value True value (10% decrease) Range of optimality
C tb 8 7.2 [0, 8.34]
C ¿ 10 9 [9.33, 10.43]
C fb 7 6.3 [5, 7.38]
C fo 6 5.4 [0, 7.82]
Cbb 11 9.9 [10.96, 13.98]
Cbo 9 8.1 [0,9.04]
Table 1.6 (million dollars)
It is clear that the true value of Ctb, Cfb, Cfo and Cbo is located in its range of optimality
respectively. Therefore, true value of these four coefficients does not change the
optimal solution. However, for Cto (Taller Smokestacks for Open-Hearth Furnaces) and
Cbb (Better Fuels for Blast Furnaces), its true value is out of their range of optimality
and the optimal solution will change.
As for the situation when the true value is 10% more than estimated value, the
process is similar to the situation when true value is 10% less than estimated value. It
also assumes that when one parameter changes, other parameters do not change.
The following table shows the true value and range of optimality.
9
Coefficient Estimated value True value (10% decrease) Range of optimality
C tb 8 8.8 [0, 8.34]
C ¿ 10 11 [9.33, 10.43]
C fb 7 7.7 [5, 7.38]
C fo 6 6.6 [0, 7.82]
Cbb 11 12.1 [10.96, 13.98]
Cbo 9 9.9 [0,9.04]
Table 1.7 (million dollars)
It is explicit that if true value of cost parameter is 10% more than estimated value,
nearly all of them is not in their range of optimality except Cbb. This displays that
except Cbb, the optimal solution will change when other cost parameter’s true value
is 10% more than estimated value.
Recommendation:
The company should conduct much effort to make its true cost value lower than its
estimated value. The reason is that the probability of changing optimal solution
when the true value is more than estimated value is really high. As the table 1.7
shows, five of six parameters will change optimal solution. However, the probability
of changing optimal solution when true value is less than estimated value is relatively
low. Two of six parameters will modify optimal solution.
Moreover, the company should put more focus on Cbb and Cbo to estimate more
closely. In perspective of Cbb, even the estimated value is in the optimality, but 11 is
so close to lower limit which is 10.96 in range of optimality. If the true value of Cbb is
a little less than 11, the optimal solution will change probably. Similar to Cbb, Cbo’s
estimated value is too close to upper limit which is 9.04 as well. If the true value of
Cbo is a bit more than 9, the optimal may solution change in a large probability.
10
Therefore， the company should revise Cbb (better fuels for blast furnaces) and Cbo
(better fuels for open-hearth furnaces).
dShadow price in the constraints part of sensitivity report refers to the change in the
value of the solution because of per unit increase in the right-hand side of the
constraints. However, shadow price is only valid when there is fairly small change in
right-hand side. Allowable increase and allowable decrease in sensitivity report
specify how much the right-hand side value can be changed without changing the
value of optimal solution. As mentioned above, the required reduction in the annual
emission rate of pollutant is right-hand side value and the total cost is the value of
optimal solution. Hence, shadow price is a good approach to specify the rate at
which the total cost of an optimal solution would change with any small change in
the required reduction in annual emission rate of each pollutant.
For particulates, as given in sensitivity report, its shadow price is 0.11. As a result,
when the policy standard of particulates changes by 1 million, the total cost of
optimal solution will alter 0.11 million dollars in same direction. Its allowable
increase is 14.30 that the value of policy standard for particulate can be increased at
most 14.30 million pounds without changing the shadow price. The allowable
decrease is 7.48 which refers to that government can just decrease 7.48 million
pounds in maximum for policy standard for particulate without changing the rate
above.
For sulfur oxides, the shadow price is 0.13 which implies that if government change 1
million pounds for policy standard of sulfur oxides, the total cost will change 0.13
million dollars in same direction simultaneously. Allowable increase for this pollutant
11
is 20.45 that its value of policy standard can go up 20.45 million pounds without
changing shadow price. The policy standard can reduce 1.69 million pounds for sulfur
oxides with remaining shadow price as the allowable decrease is 1.69.
The shadow price of hydrocarbons is 0.07 which implies that 1 million changes in
policy standard of hydrocarbons will result in 0.07 million dollars change in same
direction. Allowable increase and allowable decrease is 2.01 and 21.69 respectively
which means that government can increase 2.01 million pounds and cut down 21.69
million pounds for policy standard for hydrocarbons without changing the rate.
eAs the debate about the relative values of policy standards for the three pollutants,
the company wants to determine how much change in the opposite direction for
sulfur oxides can offset the impact of total cost of optimal solution by each unit
change in policy standard for particulates. Shadow price is the best way to solve it.
As argued above, the total cost of optimal solution will change by 0.11 million dollars
with the same direction change of 1 million pounds for particulates’ policy standard.
To offset the effect on total cost, surfer oxides should change the same cost in
opposite direction by altering the value of policy standard. Letting Q refers to the
quantity change of sulfur oxides and the formula is:
Q * shadow price of sulfur oxides = −¿0.11
Shadow price of sulfur oxides is equal to 0.13 and the result is Q= 0.11/0.13= −¿0.85
million pounds. Hence, if 1 million pounds is increased in values of policy standard
for particulates, the value of policy standard for sulfur oxides should decrease 0.85
million pounds to remain the total cost of the optimal solution. If values of policy
standard for particulates decrease 1 million pounds, 0.85 million pounds should be
increased for policy standards for sulfur oxides.12
However, because the change happens simultaneously, it is necessary to check if the
change is small enough to keep shadow price valid. 100 percent rule is a tool to
examine it. For particulates, 1 million increases should be divided by allowable
increase which is 14.30 million and the result is 7%. Concerning sulfur oxides, the
corresponding decrease is 0.85 and its allowable decrease is 1.69 million. The
percent is 50%. The sum of two percent is 57% which is less than 100%.
Consequently, the shadow price is valid and the corresponding 0.85 million decrease
is available.
Taking the situation when government decreases 1 million for particulates into
consideration, its 100 percent rule calculation is similar to the process above. 1
million decreases should be divided by its allowable decrease which is 1.69 million
and the percent is 59%. For sulfur oxides, 0.85 million increases should be divided by
its allowable decrease which is 20.45 million. The percent is 4%. The sum of two
percent is 63% which less than 100%. As a result, the shadow price is valid and the
0.85 million decrease is available as well.
fAs mentioned above, government concludes that each 10% increase in all the policy
standard for pollutant will result in reducing 3.5 million dollar tax for the company.
Moreover, the percentage cannot be over 50%. The tax reduction can be reflected in
reduction of total cost. Therefore, the objective function and right-hand side
constraints will change. Letting θ denotes the percentage change in all the policy
standard and θ is equal to one of 10%, 20%, 30%, 40% and 50%. Assuming Y means
the total tax reduction and k refers to the relationship between θ and Y. Based on the
information above, Y= k * θ. As mentioned above, when θ equals 10%, Y is 3.5
million. When θ is equal to 20%, Y will be 7 million. Thus, k refers to 35. As a result, 13
the objectives function in original linear programming will change into:
Z= 8Xtb + 10Xto + 7Xfb + 6Xfo + 11Xbb + 9Xbo –Y= 8Xtb + 10Xto + 7Xfb + 6Xfo + 11Xbb + 9Xbo –
35* θ
In terms of constraints, the left-hand side value will not change, however, right-hand
side value changes. The new constraints are shown as:
Particulates: 12Xtb + 9Xto + 25Xfb + 20Xfo + 17Xbb + 13Xbo ≥ 60*(1+θ)
Sulfur oxides: 35Xtb + 42Xto + 18Xfb + 31Xfo + 56Xbb + 49Xbo≥ 150*(1+θ)
Hydrocarbons: 37Xtb + 53Xto + 28Xfb + 24Xfo + 29Xbb + 20Xbo≥ 125*(1+θ)
Other information in original linear programming will not change. When Θ is equal to
0, its result is the same as the total cost of original optimal solution. By using Excel
Solver, the minimized total cost for each θ can be gained and given in Table 1.7.
Θ (%) Total cost (million dollars)
0 32.15
10 32.09
20 32.02
30 31.96
40 31.90
50 32.21
As a result, it will be chosen when Θ equals 40% because its total cost is minimized
which is 31.90 million dollars.
14
Q2
Alabama Atlantic is a lumber company and it has three sources of wood and five
markets to be supplied. Moreover, the delivery of each route is independent. The
annual capacity of wood at sources is 15, 20 and 15 million board feet and the annual
demand at markets is 11, 12, 9, 10 and 8 million board feet. In the past, the company
has shipping wood by rail. However, because of the increasing shipping cost of rail,
the alternative of using ships is being investigated. Now the company has three
options about shipping channels which are exclusively by rail, exclusively by water
and, by rail and water together to distribute wood from sources to market. The
problem for the company is to determine the overall shipping plan that minimizes
the total equivalent uniform annual cost (Z). This is a typical transportation problem
and as Hiller and Lieberman (1995) suggests, the problem can be solved by linear
programming.
Option 1: Ship exclusively by ship
To formulate the model, first thing is to define variables. Letting X ij (i=1,2,3; and j=
1,2,3,4,5) be the number of million board feet of wood from resource i to market j
which can shown in Table 2.1 below.
Xij 1 2 3 4 5
1 X11 X12 X13 X14 X15
2 X21 X22 X23 X24 X25
3 X31 X32 X33 X34 X35
Table 2.1 (millions)
Simultaneously, the objective function coefficient, which is indicated as C ij1
(coefficient of option 1)demonstrates that the equivalent uniform annual cost of per
million board feet by rail from source i to market j. As shipping by rail does not have
15
other cost, its equivalent uniform annual cost is just unit cost by rail. The C ij1 is shown
in table 2.2.
Cij1 Unit Cost by Rail ($1,000s) to market
Source 1 2 3 4 5
1 61 72 45 55 66
2 69 78 60 49 56
3 59 66 63 61 47
Table 2.2
Letting Z1 refers to total equivalent annual cost which is the sum of equivalent annual
cost of each route. Hence, according to the general formula shown in appendix 1, the
objective function is to minimize Z1= 61 X11 +72 X12 +45 X13+ 55 X14+66 X15 +69 X21+78
X22+60 X23 +49 X24 +56 X25 + 59 X31+ 66 X32+ 63 X33+ 61 X34+47 X35.
The next stage is checking the constraints. There are three constraints which are
supply constraints, demand constraints as every source has capacity of wood and the
amount of demand is limited in each market, and all Xij are nonnegative.
Supply constraints:
Source 1: ∑j=1
5
X1 j≤15 (X11 +X12 +X13 +X14 +X15≤15)
Source 2: ∑j=1
5
X2 j≤20 (X21 +X22 +X23 +X24 +X25≤20)
Source 3: ∑j=1
5
X3 j≤15 (X31 +X32 +X33 +X34 +X35≤15)
Demand constraints:
Market 1: ∑i=1
3
Xi1=11 (X11 +X21 +X31¿11)
16
Market 2: ∑i=1
3
Xi2=12 (X12 +X22 +X32 ¿12)
Market 3: ∑i=1
3
Xi3=9 (X13 +X23 +X33¿9)
Market 4: ∑i=1
3
Xi4=10 (X14 +X24 +X34¿10)
Market 5: ∑i=1
3
Xi5=8 (X15 +X25 +X35 ¿8)
Xij ≥0
By using Excel Solver, the minimize cost is $2,816,000 and the optimal solution is
shown in the Table 2.3 and diagram 2.1 below. (The process is shown in Appendix 2)
Xij Optimal distribution design (millions)
Source 1 2 3 4 5
1 6 0 9 0 0
2 2 0 0 10 8
3 3 12 0 0 0
Table 2.3
17
1 1
2
2 3
4
15
20
11
12
10
9
8
10
2
9
3
6
Diagram 2.1
Option 2: Ship exclusively by water
Similar to calculate option 1, letting Xij represents number of million board feet of
wood from resource i to market j as shown in Table 1 above.
Taking objective function coefficient into account, Cij2 (coefficient of option 2)
represents the equivalent uniform annual cost of per million board feet by water
from source i to market j. Different from shipping by rail, shipping by water needs
capital investment in ships and the equivalent uniform annual cost of these
investment is one-tenth the amount of capital investment. Hence, C ij2
is equal to one-
tenth the amount of capital investment of each route plus the unit shipping cost. A ij
denotes the capital investment for ships per million board feet from source i to
market j and the data is displayed in table 2.4. Bij means the shipping cost by water
per million board feet from source i to market j whose number is shown in table 2.5.
Thus, Cij2= 0.1*Aij +Bij. However, C14
2 and C31
2 do not apply to this formula because ship
is not feasible from source 1 to market 4 and source 3 to market 1. The company can
just delivery wood by rail from source 1 to market 4 and source 3 to market 1.
Consequently, C14 2and C31
2 is shipping cost by rail which is 55 and 59 ($1,000s)
respectively. Thus, the result of Cij is expressed in Table 2.6.
Aij
Unit investment for ships ($1,000s) to
market
Souc
e1 2 3 4 5
1 275 303 238－
285
18
3
515
8
12
2 293 318 270 250 265
3 － 283 275 268 240
Table 2.4
Bij Unit Cost by Rail ($1,000s) to market
Source 1 2 3 4 5
1 31 38 24－
35
2 36 43 28 24 31
3－
33 36 32 26
Table 2.5
Cij2 Equivalent uniform annual cost ($1,000s)
Source 1 2 3 4 5
1 58.5 68.3 47.8 55 63.5
2 65.3 74.8 55 49 57.5
3 59 61.3 63.5 58.8 50
Table 2.6
Therefore, the objective function is to minimize Z2= 58.5 X11 +68.3 X12 +47.8 X13+ 55
X14+63.5 X15 +65.3 X21+74.8 X22+55 X23 +49 X24 +57.5 X25 + 59 X31+ 61.3 X32+ 63.5 X33+
58.8 X34+50 X35.
As other conditions do not change, the constraints are the same as option 1 which
are:
Supply constraints:
Source 1: ∑j=1
5
x 1 j≤15 (X11 +X12 +X13 +X14 +X15≤15)
19
Source 2: ∑j=1
5
x 2 j≤20 (X21 +X22 +X23 +X24 +X25≤20)
Source 3: ∑j=1
5
x 3 j ≤15 (X31 +X32 +X33 +X34 +X35≤15)
Demand constraints:
Market 1: ∑i=1
3
xi 1=11 (X11 +X21 +X31¿11)
Market 2: ∑i=1
3
xi 2=12 (X12 +X22 +X32 ¿12)
Market 3: ∑i=1
3
xi 3=9 (X13 +X23 +X33¿9)
Market 4: ∑i=1
3
xi 4=10 (X14 +X24 +X34¿10)
Market 5: ∑i=1
3
xi 5=8 (X15 +X25 +X35 ¿8)
Xij ≥0
The problem is solved by Excel Solver as well and the minimize cost is Z2= $2,770,800
and optimal solution is expressed in Table 2.7 and diagram 2.2 (The process is shown
in Appendix 2).
Xij Optimal distribution design (millions)
Source 1 2 3 4 5
1 6 0 9 0 0
2 5 0 0 10 5
3 0 12 0 0 3
Table 2.7
20
6
Diagram 2.2
Option 3: Ship by either rail or water, depending on which is less expensive for the
particular route
As option 1 and option 2, letting Xij displays number of million board feet of wood
from resource i to market j as shown in Table 1 above.
The objective function coefficient Cij3
(coefficient of option 3) is the less expensive
equivalent uniform annual cost of each route by rail or water. In other words, if the
equivalent uniform annual cost of a route by rail is lower than that by water, C ij3
equals to the equivalent uniform annual cost by rail. On the contrary, if that cost by
rail is higher than that by water, C ij is the equivalent uniform annual cost by water.
Hence, comparing the cost for each route in table 2 and table 6, the Cij3 is:
Cij3 Total equivalent uniform annual cost($1,000s)
Source 1 2 3 4 521
1 1
2
2
3
3
4
5
15
20
15
11
12
8
10
9
512
10
5
9
3
1 58.5 (w) 68.3(w) 45(r) 55(r) 63.5(w)
2 65.3(w) 74.8(w) 55(w) 49(r) 56(r)
3 59(r) 61.3(w) 63(w) 58.8(w) 47(r)
Table 2.8
Hence, the objective function is to minimize Z3= 58.5 X11 +68.3 X12 +45 X13+ 55
X14+63.5 X15 +65.3 X21+74.8 X22+55 X23 +49 X24 +56 X25 + 59 X31+ 61.3 X32+ 63 X33+ 58.8
X34+47 X35
The constraints are identical as option 1 and option 3 which are:
Supply constraints:
Source 1: ∑j=1
5
x 1 j≤15 (X11 +X12 +X13 +X14 +X15≤15)
Source 2: ∑j=1
5
x 2 j≤20 (X21 +X22 +X23 +X24 +X25≤20)
Source 3: ∑j=1
5
x 3 j ≤15 (X31 +X32 +X33 +X34 +X35≤15)
Demand constraints:
Market 1: ∑i=1
3
xi 1=11 (X11 +X21 +X31¿11)
Market 2: ∑i=1
3
xi 2=12 (X12 +X22 +X32 ¿12)
Market 3: ∑i=1
3
xi 3=9 (X13 +X23 +X33¿9)
Market 4: ∑i=1
3
xi 4=10 (X14 +X24 +X34¿10)
Market 5: ∑i=1
3
xi 5=8 (X15 +X25 +X35 ¿8)
22
Xij ≥0
The minimized cost gained by Excel Solver is that Z3= $2,729,100 and the optimal
solution is given in Table 2.9 (The process is shown in Appendix 2).
Xij3 Optimal distribution design (millions)
Sourc
e1 2 3 4 5
1 6(w) 0 9(r) 0 0
2 5(w) 0 0 10(r) 5(r)
3 0 12(w) 0 0 3(r)
Table 2.9
Diagram 2.3
Recommendation:
Comparing these options, option 3 is the best as the total equivalent uniform annual
cost is the lowest.
23
1 1
2
2
3
3
4
5
15
20
15
11
12
8
10
9
6 by water
5 by rail
12 by
water
10 by rail
5 by
water
9 by rail
3 by rail
Q3
a
The title of paper is “Application of Linear Programming for Dormitory Development
plan at Petra Christian University”. The authors were Connie Susilawati, Desire L.
Litaay and Andre Parsaulian. Moreover, it was published in Volume 3 and Number 2
of journal “Dimensi Teknik Sipil” in 2001.
b
The paper considered a dormitory problem. A survey to Petra Christian University’s
students showed the required facilities and their financial ability and the author
defined dormitory as students residential with book shop, cafeteria, sport and other
support facilities. Therefore, the investors wanted to know how to allocate scarce
resources to different types of uses to obtain optimum profit. This paper was to solve
the site allocation problem in dormitory that how many rooms and how much area
of each facility should be built to obtain the maximum profit and satisfy students’
needs simultaneously.
c
Linear programming was utilized to solve the problem. To formulate the model, the
study defines the variables as numbers of room and the area of supporting facilities
firstly. However, the dimensions of some facilities depended on the number of
occupants which were based on the number of rooms. According to the survey the
university conducted, four units of two-bed rooms need to be built one three-bed
room. Moreover, three bathrooms are requisite for these 11 occupants. To meet
linear programming, the author transformed the dependent facilities such as
common room, living room and dining room into three bedroom unit (I2). For
instance, when building a three bedroom, it needs 8.8 m2 areas for common room.
Another example is that when building a three bed room, it needs to build four two
24
bed room. The transformation was given in table 3.1. The other six variables are
independent which was shown as X1 ,X2, X3, X4, X5 and X6. The variables represented
the area of kitchen, book shop, mini market, phone booths, sporting facilities and
garden.
The coefficient of objective function could be calculated by the net present value of
the cash flow for each variable. For I2, its coefficient is composed by the net present
value of cash flow of independent and dependent variables in table 3.1. The result is
34,952,096.51. For other six variables, their coefficients are －576,000, 1,564,218.78,
1,564,218.78, 1,564,218.78, －50,000 and －49,529.5 (Rp) respectively.
d
Hence, the objective function was to maximize Z= 34,952,096.51*I2 － 576,000*X1
+1,564,218.78*X2+1,564,218.78*X3+1,564,218.78*X－ 50,000*X5－ 49,529.5*X6 －
1,385,557,930
The objective function is to maximize the net cash flow which was equal to the sum
of the net cash flow of every variable. Coefficients of kitchen, sport facilities and
garden were negative because they could not gain profit and were referred to cost.
Moreover, it needed to minus 1,385,557,930 which were fixed cost.
There were three constraints in this model which were physical constraints,
regulation constraints and market constraints. The physical constraints consisted of
land area, minimum room capacity and layout. The constraints were shown in Table
3.2. Letting Yi stand for the constraints, Y1 was policy constraints as the policy that the
Building Coverage Ratio in the region was 50%. Y2 and Y3 were physical constraints as 25
it was limited by layout. For example, for Y3, the constraint emerged because second
floor of dormitory could only have living room, kitchen, bedrooms and bathrooms.
Moreover, Y4 was physical constraints as well because it referred to land area.
Additionally, Y7 to Y12 represented the market demand constraints of book shop, mini
market, phone booths and sporting facilities respectively. Moreover, I2 were integer
because it was bedrooms.
Table 3.2 (Source: Susilawati et al, 2001)
f
The result was that the maximum of net cash flow was equal to Rp 392,952,557. The
optimal solution was shown as Table 3.3.
Decision variables Description Value Unit
I2 Bedroom 35.00 Rooms
X1 kitchen 73.15 m2
X2 Book shop 149.33 m2
X3 Mini market 169.29 m2
X4 Phone booths 49.29 m2
X5 Sporting facilities 381.43 m2
X6 Garden 0.00 m2
Table 3.3 (Source: Susilawati et al, 2001)
26
I2 just showed the optimal number of three-bed room and it should be transformed
to other dependent variable. Therefore, the final is result is shown in Table 3.4.
Table 3.4 (Source: Susilawati et al, 2001)
e
However, this model also had some limitations and shortcomings. As the paper
argued, this model ignored the qualitative factors such as social, politic or ethic
issues which could impact the result significantly in real world. Moreover, when the
model transformed dependent variable into I2, the ratio between favorable rooms
was gained by the former survey which might have errors. Additionally, the model
also assumed the ratio constant which is not realistic. Furthermore, because
objective function coefficients were calculated by net present value of the cash flow,
it needed to use discount rate, 17%. Nevertheless, there was risk and uncertainty
about discount rate and the discount rate fluctuation would lead to inaccuracy of the
model.
27
Appendix 1
Sheet 1 for Qa:
Sheet 2 for sensitivity report
28
Sheet 3 for θ=10%
29
Sheet 4 for θ=20%
Sheet 5 for θ=30%
30
Sheet 6 for θ=40%
Sheet 7 for θ=50%
31
Appendix 2
The general linear programming model for the transportation problem is:
Min
∑i=1
n
∑j=1
m
Cij X ij
s.t.
∑j=1
n
X ij≤si, i=1, 2,……m Supply
∑j=1
m
X ij=d j , j= 1, 2,…..n Demand
X ij ≥0 for all i and j
where32
i= index for origins, i=1, 2,…..m
J= index for destinations, j=1, 2,…..n
X ij= number of units shipped from origin I to destination j
C ij= cost per unit shipped from origin I to destination j
si= supply or capacity in units at origin i
d j= demand in units at destination j
Sheet 1 for shipping by rail
Sheet 2 for shipping by water
33
Sheet for shipping by either water and rail
34
Reference:
Anderson,D.R., Sweeney,D.J., Williams, T.A. & Martin R.K. (2008), An introduction to
management, United States: Thomson South-Western
Hillier, F.S. & Lieberman, G.J. (1995), Introduction to operations research, New York:
McGraw-Hill
Ignizio, J.P. (1982) Linear Programming in Single- & Multiple- Objective System,
United States: Prenitice-Hall
Susilawati, C. & Litaay, D.L. & Parsaulian, A. (2001), Application of Linear
Programming for Dormitory Development Plan at Petra Christian University,
“Dimensi Teknik Sipil”, vol.3, no.2, pp. 59-63
35
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