IE homework 2
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Northeastern University *
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Course
6200
Subject
Health Science
Date
Feb 20, 2024
Type
docx
Pages
12
Uploaded by UltraGullPerson674
2
nd
solution:
Given that: the Roanoke Medical Center is partnering with a national religious health association
to establish a free clinic in the region, three potential clinic locations under consideration are:
Cherry Hill Commons, Eastgate, and Kenwood
On an average clinic patient expected to spend $10 per visit
At each location they are expected to get $1 million as donations, & they are expected to
grow 15% annually over the next three years.
Annual cost as salary for each location is $275,000.
Fixed cost for cherry hill is $340,000.
Fixed cost for Eastgate is $410,000.
Fixed cost for Kenwood is $378,000.
Average variable cost at cherry hill $20
Average variable cost at Eastgate is $27
Average variable cost at Kenwood is $24
Expected annual patients 6180, 5925, 6220 for cherry hill, Eastgate, Kenwood
respectively.
Location
Cherry Hill
Eastgate
Kenwood
Fixed cost (FC)
340,000
410,000
378,000
Annual cost as salary
275,000
275,000
275,000
Total fixed cost = (FC + Annual cost of salary)
615,000
685,000
653,000
Variable cost
20*6180=123,600
27*5925= 159,975
24*6220= 149,280
Total cost= FC+VC
738,600
844,975
802,280
Income from visits
10*6180=61,800
10*5925= 59,250
10*6220= 62,200
Income from donation
1,000,000
1,000,000
1,000,000
Total income (income from visits + income from donation)
1,061,800
1,059,250
1,062,200
Profit = (total income-
total cost)
1,061,800-
7386,600=323,200
1,059,250-
844,975=214,275
1,062,200-
802,280=259,920
a.
Which location should be chosen based on total costs in year 1 of operations?
Answer: we can see that Cherryhill
is the least among the calculation with $738,600. So
therefore, Cherryhill is selected based on the year 1 operation.
b.
Which location should be chosen based on total profits in year 1 of operations?
Answer: Cherry Hill is selected because of the highest profits produced in 1
st
year with profit
of $323,200.
Now the assume by the 3
rd
year health care reform is expected to significantly reduce the
number of uninsured in the Roanoke region. Expected patients visits in 3 years changes to
6,745 at cherry hill, 5,750 at Eastgate, 5,825 at Kenwood.
To accommodate the patient demands cherry hill will bring addition staff and increasing the
total cost to $685,000
Now the charity amount is increased. During 2
nd
year: 1000000 + 1000000 * 15/100 = 1150000
3
rd
year: 1150000 + 1150000 * 15/100 = 172500 = 1322500
Now analysis based on 3-year operation:
Location
Cherry Hill
Eastgate
Kenwood
Fixed cost (FC)
340,000
410,000
378,000
Annual cost as salary
275,000+increased
amount
275,000
275,000
Total fixed cost = (FC + Annual cost of salary)
685,000 (Increased
given in the
question)
685,000
653,000
Variable cost
20*6745=134,900
27*5750= 155,250
24*5825= 139,800
Total cost= FC+VC
819,900
840,250
792,800
Income from visits
10*6745=67,450
10*5750=57,500
10*5825=58250
Income from donation
1322500
1322500
1,322,500
Total income (income from visits + income from donation)
1,389,950
1,380,000
1,380,750
Profit = (total income-
total cost)
1,389,000-
819,900=570,050
1,380,000-
840,250=539,750
1,380,750-
792,800=587,950
c.
Which location should be chosen based on total costs in year 3 of operations?
Answer: Kenwood is the lowest in all three with total cost of $792,800 in 3 years operation.
d.
Which location should be chosen based on total profits in year 3 of operations?
Answer: Kenwood has the highest profit in 3 year operation with $587,950
e.
Determine the sensitivity of the decision in part (d) to a ± 10 percent fluctuation in visits.
Now let us assume 10% increase
Increase
in No.of visit = 5825 + (5825*10/100=583) = 6408
Total variable cost = 24 * 6408 = 153792
Total cost = 653,000 + 153792 = 806,792
Income from patients: 10*6408= 64080 Income from donation = 1,322,500
Total income = 1,386,580
Profit = 1,386,580 – 806,792 = $579,788
After 10% increase in population Kenwood is the least in the total cost and highest profit with
$806,792 and $579,788 respectively
Now let us consider with the 10% decrease in population.
Decrease in No. of visit = 5825 - (5825*10/100=583) = 5242
Total variable cost = 24 * 5242 = 125,808
Total cost = 653,000 + 125808 = 778,808
Income from patients: 10*5242= 52420 Income from donation = 1,322,500
Total income = 1,374,920
Profit = 1,374,920 – 778,808 = $596,112
After 10% decrease in population Kenwood is the least in the total cost and highest profit with
$778,808 and $596,112 respectively.
Even if there is increase or decrease in number of visits by 10%, the decision is not changed as in
both the conditions, decision d (Kenwood Location) is more profitable compared to
Decision a (Cherry Hill Location)
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6
th
solution:
Given that:
Community hospital is planning to expand its services to three new services in the medical diagnostics categories (MCD-2, MDC-19, & MDC-21)
Five common resources must be allocated among these three new services lines is considered as the constraints.
The 5 resources are beds, nursing staff, radiology, laboratory, and operating room.
Resource Category
MDC
-2
MDC-
19
MDC-
21
Available
Resources
Average revenue
8885
10143
12711
LOS
3.3
6.1
4.4
19710
Nursing Hours
3
5
4.5
16200
Radiology Procedures
0.5
1
0
3000
Laboratory
Procedures
1
1.5
3
6000
Operating Room
2
0
4
1040
a.
Develop an optimization model to formulate this resource planning problem. Define decision variables, objective function, and constraints.
Answer: let us assume decision variables are MCD-2 as x, MCD-19 as y, & MCD-21 as z
objective function: average revenue is the A= 8885x + 10143y + 12711z is the objective function for the above problem
Constraints are: 3.3x + 6.1y + 4.4z <= 19710
3x + 5y + 4.5z <= 16200
0.5x + 1y + <= 3000
1x + 1.5y + 3z + <= 6000
2x + 4z + <= 1040
# Set up data analysis and preparation functions import pandas as pd
import numpy as np
# Max 8885x + 10143y + 12711z
s.t.
3.3x + 6.1y + 4.4z + <= 19710
3x + 5y + 4.5z <= 16200
0.5x + 1y + <= 3000
1x + 1.5y + 3z + <= 6000
2x + 4z + <= 1040
# include and enable Gurobi
import gurobipy as gp
from gurobipy import GRB
# Freeing default Gurobi environment
gp.disposeDefaultEnv()
import gurobipy as gp
from gurobipy import GRB
# Create a new model
model = gp.Model()
# Define decision variables
x = model.addVar(vtype=GRB.CONTINUOUS, name="x")
y = model.addVar(vtype=GRB.CONTINUOUS, name="y")
z = model.addVar(vtype=GRB.CONTINUOUS, name="z")
# Set objective function: Maximize revenue
model.setObjective(8885*x + 10143*y + 12711*z, sense=GRB.MAXIMIZE)
# Add constraints
model.addConstr(3.3*x + 6.1*y + 4.4*z <= 19710, "c1")
model.addConstr(3*x + 5*y + 4.5*z <= 16200, "c2")
model.addConstr(0.5*x + 1*y <= 3000, "c3")
model.addConstr(1*x + 1.5*y + 3*z <= 6000, "c4")
model.addConstr(2*x + 4*z <= 1040, "c5")
# Optimize the model
model.optimize()
# Print optimal solution
if model.status == GRB.OPTIMAL:
print('Optimal Solution:')
print('x =', x.x)
print('y =', y.x)
print('z =', z.x)
print('Objective Value =', model.objVal)
else:
print('No solution found')
# Save the model for inspection
model.write('resource_planning.lp')
Gurobi Optimizer version 11.0.0 build v11.0.0rc2 (mac64[arm] - Darwin 23.1.0 23B81)
CPU model: Apple M1
Thread count: 8 physical cores, 8 logical processors, using up to 8 threads
Optimize a model with 5 rows, 3 columns and 13 nonzeros
Model fingerprint: 0xfef18efe
Coefficient statistics:
Matrix range [5e-01, 6e+00]
Objective range [9e+03, 1e+04]
Bounds range [0e+00, 0e+00]
RHS range [1e+03, 2e+04]
Presolve removed 1 rows and 0 columns
Presolve time: 0.00s
Presolved: 4 rows, 3 columns, 10 nonzeros
Iteration Objective Primal Inf. Dual Inf. Time
0 5.6939502e+07 4.466787e+03 0.000000e+00 0s
4 3.3733860e+07 0.000000e+00 0.000000e+00 0s
Solved in 4 iterations and 0.01 seconds (0.00 work units)
Optimal objective 3.373386000e+07
Optimal Solution:
x = 0.0
y = 3000.0
z = 260.0
Objective Value = 33733860.0
# Define the number of cases served for each DRG
num_cases_served = {
'MDC-2': 3.3 * x.x + 6.1 * y.x + 4.4 * z.x,
'MDC-19': 3 * x.x + 5 * y.x + 4.5 * z.x,
'MDC-21': 0.5 * x.x + 1 * y.x
}
# Print the number of cases served for each DRG
print("Number of Cases Served:")
for mdc, cases_served in num_cases_served.items():
print(f"{mdc}: {cases_served}")
Number of Cases Served:
MDC-2: 19444.0
MDC-19: 16170.0
MDC-21: 3000.0
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# Optimize the model
model.optimize()
# Print optimal solution
if model.status == GRB.OPTIMAL:
print('Optimal Solution:')
print('x =', x.x)
print('y =', y.x)
print('z =', z.x)
print('Objective Value =', model.objVal)
print()
# Get shadow prices (dual values)
shadow_prices = {c.constrName: c.pi for c in model.getConstrs()}
# Print shadow prices
print("Shadow Prices (Dual Values):")
for constraint, shadow_price in shadow_prices.items():
print(f"{constraint}: {shadow_price}")
else:
print('No solution found')
Gurobi Optimizer version 11.0.0 build v11.0.0rc2 (mac64[arm] - Darwin 23.1.0 23B81)
CPU model: Apple M1
Thread count: 8 physical cores, 8 logical processors, using up to 8 threads
Optimize a model with 5 rows, 3 columns and 13 nonzeros
Coefficient statistics:
Matrix range [5e-01, 6e+00]
Objective range [9e+03, 1e+04]
Bounds range [0e+00, 0e+00]
RHS range [1e+03, 2e+04]
Solved in 0 iterations and 0.00 seconds (0.00 work units)
Optimal objective 3.373386000e+07
Optimal Solution:
x = 0.0
y = 3000.0
z = 260.0
Objective Value = 33733860.0
Shadow Prices (Dual Values):
c1: 0.0
c2: 0.0
c3: 10143.0
c4: 0.0
c5: 3177.75
1
st
solution:
Factors
23233
23259
23236
Importance
Ranking
Min.
Acceptable
Level
Rent Estimate
1380
1325
1415
4
≤$1,500
Population Size
31097
23466
30487
3
≥21,000
Annual Expected Visits
7073
5839
6849
1
≥1,545
% of Population with ≥ High School
0.92
0.93
0.94
7
≥90%
Average Household Income
73518
87275
90170
5
≥$61,950
% of Population Insured
0.88
0.94
0.94
2
≥85%
Expected Annual Population Growth
0.019
0.034
0.02
6
≥1%
Factors
Relative scores
Weights
Rent Estimate
25
0.15
Population Size
30
0.18
Annual Expected Visits
50
0.30
% of Population with ≥
High School
1
0.006
Average Household
Income
15
0.09
% of Population Insured
40
0.24
Expected Annual
Population Growth
3
0.01
Sum of relative scores
164
1.0
Relative score = evaluated outcome/most desirable outcome
Factors
23233
23259
23236
Rent Estimate
97
93
100
Population Size
100
75
98
Annual Expected Visits
100
82
96
% of Population with ≥ High School
97
99
100
Average Household Income
81
96
100
% of Population Insured
93
100
100
Expected Annual Population Growth
55
100
58
Sum of relative scores
623
645
652
Factors
Weights
23233
23259
23236
Rent Estimate
0.15
97*0.15=14.55
93*0.15=14.88
100*0.15=15
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Population Size
0.18
100*0.18=18
75*0.18=13.5
98*0.18=17.64
Annual Expected Visits
0.30
100*0.3=30
82*0.3=24.6
96*0.3=28.8
% of Population with ≥
High School
0.006
97*0.006=0.58
99*0.006=0.59
100*0.006=0.6
Average Household
Income
0.09
81*0.09=7.29
96*0.09=8.64
100*0.09=9
% of Population Insured
0.24
93*0.24=22.32
100*0.24=24
100*0.24=24
Expected Annual
Population Growth
0.01
55*0.01=5.5
100*0.01=10
58*0.01=5.8
Sum of relative scores
98
96
100
a.
Which zip code should be selected using the minimum attribute satisfaction procedure?
Answer:
The minimum attribute satisfaction procedure is a method used in multi-criteria decision
making to select the best alternative based on the minimum satisfaction level of the
attributes.
Rent estimation:
Now let’s talk about the problem here it is given that rent estimation in all the zip codes
should be <=$1,500,
1380<=1500 for 23233
1325<=1500 for 23259
1415<=1500 for 23236
As you see all the three zip codes satisfies this condition
Population size: here the minimum acceptable level is >= 21,000
31097>=21000 for 23233
23466>=21000 for 23259
30487>=21000 for 23236
This factor also satisfies the minimum acceptable level mentioned in the problem
Annual expected visits: minimum acceptable here is >=1,545
7073>=1545 for 23233
5839>=1545 for 23259
6849>=1545 for 23236
% of population with >= high school: minimum acceptable here is >=90%
92%>=90% for 23233
93%>=90% for 23259
94%>=90% for 23236
Satisfied all the conditions.
Average household income: minimum acceptable here is >=$61,950
73518>=61950 for 23233
87275>=61950 for 23259
90170>=61950 for 23236
Satisfies all the conditions.
% of population insured: minimum acceptable here is >=85%
88%>=85% for 23233
94%>=85% for 23259
94%>=85% for 23236
Satisfies all conditions.
Expected annual population growth: minimum acceptable here is >=1%
1.9%% >= 1% for 23233
3.4% >= 1% for 23259
2% >= 1% for 23236
Satisfies all the conditions.
As you can see all the factors are within the minimum acceptable conditions. So, therefore all
the zip codes 23233, 23259, & 23236
are selected using minimum attribute satisfaction
procedure.
b.
Which zip code should be selected using the dominance procedure?
c.
Which zip code should be selected based on the most important attribute procedure?
To calculate this, we need to use the information importance ranking for each criterion.
From the table given above:
Rent estimator has a ranking of 4.
Population size has a ranking of 3.
Annual expected visits have a ranking of 1
% of population with >= high school has a ranking of 7
Average household income has a ranking of 5.
% of population insured has a ranking of 2 Expected annual population growth has a ranking of 6
From all the mentioned above annual expected visits has the highest importance ranking
of 1, indicating that it is the most critical factor in decision making. So now let’s evaluate
zip codes on this ranking. 23233 has 7073 visits, 23259 has 5839 visits, & 23236 has
6849 visits. Therefore, zip code 23233
has the highest score for annual expected visits is
the best option for the most attribute procedure.
d.
The practice manager has assigned the following relative scores for each of the factors (in
the Excel file). Calculate the relative weight of each factor, and then determine the
location of the new women’s health center based on each location’s composite factor
scores.
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