A company is considering two insurance plans with coverage and premiums shown in

Question

A company is considering two insurance plans with coverage and premiums shown in the table. (For example, this means that $50 Fire/Theft buys one unit of plan A, consisting of $10,000 fire and theft insurance and $180,000 of liability insurance.) Answer parts a and b. Policy APolicy B $15,000 180,000 120,000 $10,000 isbiy $50 $40 a. The company needs at least $450,000 fire and theft insurance and at least $6,600,000 liability from these plans. How many units should be purchased from each plan to minimize the cost of the premiums? What is the minimum premium? The company should purchase units of Policy A and units of Policy B, for a premium of s b. Suppose the premium for policy A is reduced to $25. Now how many units should be purchased from each plan to minimize the cost of the premiums? What is the minimum premium? The company should purchase units of Policy A andunits of Policy B, for a premium of S

Explanation / Answer

a. let us assume the company should purchase x1 units of plan A and x2 units of plan B. Then our problem will be solving the linear constraint optimization problem as stated below...........

Minimize 50*x1 + 40* x2

with subject to...

10000*x1 + 15000*x2 >=450000

180000*x1 + 120000*x2 >=6600000

We can solve this in solver in excel/libreoffice or can use any programming language.

Below is the code in python.......

###Python Code

from scipy.optimize import minimize

def cost(x):
return (50*x[0]+ 40*x[1])

cons = ({'type':'ineq','fun' : lambda x: np.array([10000*x[0]+15000*x[1]-450000])},
{'type':'ineq','fun' : lambda x: np.array([180000*x[0]+120000*x[1]-6600000])},
{'type':'ineq','fun':lambda x:x[0]},
{'type':'ineq','fun':lambda x:x[1]},
)

x0=np.array([0,0])

res=minimize(cost,x0,constraints=cons,method='SLSQP',options={'disp':True})

res.x

###

which will give output as [30,10] that means 30 units of policy A and 10 units of policy B

that optimized output is 1900

So the answer of first question is 30, 10 and 1900

b. Here the solving procedure will be same. only the function will change.

Minimize 25*x1 + 40* x2

with subject to...

10000*x1 + 15000*x2 >=450000

180000*x1 + 120000*x2 >=6600000

In the python code, we will change the cost function...........

def cost(x):
return (25*x[0]+ 40*x[1])

The remaining portion will be exactly same. After optimization the output will be 45 and 0

and the optimized value will be 1125.

So the answer of second question will be 45, 0 and 1125

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