# Help with Max Demand Sample PE Exam Question ### Help Support Engineer Boards: #### wfg42438

##### Active member
In the question below we are given the max demand for 4 different 15 min time periods.

We are also told that the max demand at t=0 is 35kW

With the given information for the average consumption at time interval 1 from that i understand refers to from t=0 to t=15 min which from the graph is 25
from t=15 to 30 i see it rises to 50 kW so the avg 25 +50/2 which matches the solution
Time interval 3 goes from 50 to 75 and then 50 so the avg is 175/3=58.3 however they are saying the average is 62.25 kW.

Can anyone clarify how intervals 3 to 5 are calculated?

Additionally for the max demand i cant tell how the values for time intervals from 2 to 5 are obtained.  #### wfg42438

##### Active member
@main197 you happen to have an idea of how to solve this question ?

#### Rafis

##### New member
This is my understanding of the problem:

Only max average demand is registered:
t=15min: Average: 25kw | Max remains at 35 kW
t=30min: Average: (50-25)/2+25=37.5 | New Max 37.5 kW
t=45min: Average: {[(75-50)/2+(75-50)/2]/2}+50=62.5kW | New Max at 62.5 kW
t=60min: Average: 50KW | Max remains at 62.5 kW