Monday

March 30, 2015

March 30, 2015

Posted by **Dom** on Tuesday, July 9, 2013 at 8:49pm.

Sub Rent()

Dim Rent As Variant

Dim increase As Integer

Dim units As Object

Dim Profit As Double

Rent = InputBox("The rent to occupy all the units")

increase = InputBox("The increase in rent that results in a vacant unit")

Rent = Rent + increasea

units = 50

Profit = units * (600 + increase)

Outputbox " " & units & " units and " & Profit & " when profit is maximized."

End Sub

- Computer Programming -
**MathMate**, Wednesday, July 10, 2013 at 5:39amYour code seems to be in reasonable shape, except that it does not yet implement the algorithm to find the maximum revenue.

So the current revenue is 600*50=30,000$ a month.

Your program will increase the rent by $40 steps and recalculate the revenue.

When the revenue starts to decrease, then you have found the maximum revenue.

The programming part will include a for loop for implementing the increase, a method to calculate the new revenue, a variable to keep the old revenue, and an if statement to jump out of the loop when the revenue decreases.

The pseudocode would look something like this:

Define all variables

1. Input current rent (Rent=600$), number of units (Units=50), and critical increase (Increase=40$)

Current revenue (Revenue)=Units*Rent

2. Loop over increases

For inc=40 to Units*40 step40

NewRent=Rent+inc

NewUnits=Units-(inc/40)

NewRevenue=NewRent*NewUnits

If NewRevenue<revenue then exit loop

Revenue=NewRevenue

next inc

Rem: Found max. increase

Output revenue

Output NewUnits-1

If you implement the above pseudocode, you should get the correct result of $1300 as the most profitable rent with a revenue of $42250.

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