Light bulb problem

I am going to buy some bulbs for our cleanroom, 48 inch tubes, one lasts 20,000 hours, consumes 32 watts, costs $2.00 each.
The other lasts 24,000 hours, called the super miser, consumes 30 watts but costs three times as much, $6.00 each.
Power cost is $0.08 /Kwhr. If you have two cleanrooms with say, one hundred of these bulbs in each one, will there ever come a time when the more expensive bulb will start saving money? If so, how much time is that?

32W

31.25 bulbs per hour cost $0.08
200/31.25 = 6.4

to light both rooms for 1 hour costs $0.512 at 32W

30W

33.333 bulbs per hour cost $0.08
light 200 lights for one hour costs $0.48


total cost for 32W = 0.512 * 20000 = 10240 + 400 = 10640

total cost for 30W = 0.48 * 24000 = 11520 + 1200 = 12700

for 100000 hours operation

10640 * 5 = 53200

12700 * 4 = 50880

32W cost on average $0.532 per hour to run for both rooms

30W cost on average $ 0.5088 per hour for both rooms

difference in running cost is 2.32 cents per hour therefore to make up the initial $800 difference is 800/0.0232 is 34483 hours ( 4 years if you round up).

Therefore if you are running these bulbs until they die once, then it is cheaper to get the 32W bulbs initially. However if you stick with the more expensive bulbs long term i.e. 100000 hours approx 11 years, the need for the less set of light bulbs makes the 30W bulbs cheaper. Add to that the time and effort saved in replacing the all the bulb 4 times rather than 5.

Originally posted by sonhouse
I am going to buy some bulbs for our cleanroom, ... one lasts 20,000 hours, ... The other lasts 24,000 hours, called the super miser, consumes 30 watts ... If so, how much time is that?

Why look at the cost of 100 bulbs when the math will be the same for a single bulb (except two magnitudes different)?

Matlab Code:
%Lightbulb Costs
%Variables (in dollars for costs, watts for power, hours for lifetimes and
%kWhrs for powercost)
CostA = 2;
CostB = 6;
PowerA = 32;
PowerB = 30;
LifetimeA = 20000;
LifetimeB = 24000;
PowerCost = 0.08;
Length = 120000;
A=zeros(1,Length);
B=zeros(1,Length);
for i=1:1:Length
A(1,i)= ceil(i/LifetimeA)*CostA + i*PowerCost*PowerA/1000;
B(1,i)= ceil(i/LifetimeB)*CostB + i*PowerCost*PowerB/1000;
end
hold on
plot(A, 'r'😉
plot(B)
xlabel('Time (Hours)'😉;
ylabel('Cost ($)'😉;
title('Cost of Lightbulb after Time'😉;

If you run this you'll see that the above calculations don't give the full picture. You'll get this:
http://img100.imageshack.us/img100/3884/lightbulbsfh9.gif
where the $2 bulb is red and the $6 bulb is blue.

Notice that even as early as ~20000 hours the more expensive bulb can be cheaper (because we haven't had to replace it yet). And notice that this cycle repeats as we continue. In the longer term the blue bulb will be cheaper as the long term behaviour (gradient) comes into force. Note that I chose 120,000 hours as the simulation length as that is the first point at which the replacement times will intersect (i.e. The period length). At 120,000 hours blue is cheaper showing clearly the difference, our period will repeat with the blue line shifted compared to the red line. At (not after!) 120,000 hours buying the more expensive will mean a saving of $1.20. So we can see that after 4 cycles we will have saved $4.80, more than the difference between replacement costs of the bulbs showing that after this point the more expensive bulbs will always be cheaper. And examination of the graph shows that the last point at which the red bulbs are cheaper is at 460,000 hours.

Originally posted by FabianFnas
What does that mean when a light bulb lasts 20,000 hours? Does every one of these last 20 k hour and then die? Is this guaranteed to do that?

In real life I think that 200 k hour mean that some will die quicker and some will last longer. Perhaps some will last only 1 hour and some other will last 100 k hours.

The question here is - does this remark ...[text shortened]... cal question with no bearing in real life and the answer is well defined until the last decimal?

OR
get the cheaper ones, because in 20000 hours time (2+ years) the energy saving ones will be cheaper as they are produced more because of environmental concerns, and then switch to energy saving ones.

OR
get the cheaper ones, and put the rest of the money into a savings account
(what is the minimum interest rate required to offset the extra energy cost?)

Originally posted by aging blitzer
OR
get the cheaper ones, because in 20000 hours time (2+ years) the energy saving ones will be cheaper as they are produced more because of environmental concerns, and then switch to energy saving ones.

OR
get the cheaper ones, and put the rest of the money into a savings account
(what is the minimum interest rate required to offset the extra energy cost?)

OR
get the expensive ones if energy prices are going to increase

OR
get some of each and check how long their average life really is

Note:
Are all the lights always on 24 hours a day, because if not, it takes longer for the energy savers to catch up.

Originally posted by sonhouse
I am going to buy some bulbs for our cleanroom, 48 inch tubes, one lasts 20,000 hours, consumes 32 watts, costs $2.00 each.
The other lasts 24,000 hours, called the super miser, consumes 30 watts but costs three times as much, $6.00 each.
Power cost is $0.08 /Kwhr. If you have two cleanrooms with say, one hundred of these bulbs in each one, will there eve ...[text shortened]... come a time when the more expensive bulb will start saving money? If so, how much time is that?

Originally posted by aging blitzer
OR
get the expensive ones if energy prices are going to increase

OR
get some of each and check how long their average life really is

Note:
Are all the lights always on 24 hours a day, because if not, it takes longer for the energy savers to catch up.

Originally posted by David113
Let's assume you need only one bulb (it makes no difference).

If you use the cheap ones, then you initially spend 2 dollars, and then every hour you spend 0.00256 dollars for the electricity + 0.0001 dollars for new bulbs (since on average every hour you buy 1/20000 of a bulb). So after x hours you spend 2 + 0.00266x dollars.

If you use the expensive ...[text shortened]... money when 6 + 0.00265x < 2 + 0.00266x, that is, when x > 400000 (which is about 46 years
🙄)

My mistake it should have been 120000 hours, was obviously half asleep doing this. As for switching the lights on and off to save energy, maybe be worth looking into whether this will shorten the life of the bulbs and if so by how much. may actually being more harm than good although i doubt it.

You could also introduce an interest rate to take into account the time value of money. Assuming electricity is paid every hour, the cost for each bulb can be broken down as follows:

C(hardware) = B + B/(1+i)^p + B/(1+i)^2p + ...

= B/(1-1/(1+i)^p)

C(electricity) = 1/(1+i)*(E + E/(1+i) + E/(1+i)^2 + ...)

= 1/(1+i)*(E/(1-1/(1+i))

C(total) = C(hardware) + C(electricity)

where:

B = bulb cost
p = bulb replacement period
i = interest (hourly)
E = electricity cost per hour

C(electricity) is shifted back one hour, because I assumed you don't pay for the hour until you use it so an interest adjustment must be made. However, you can remove this without much difficulty or sensitivity to the answer.

Comparing the two cost using Excel, I found that the break even point for the interest rate is about 4.3% annually, or 0.00049% hourly. Below that, where the effect of interest is much less, the expensive bulb performed better, while above that the cheaper bulb was a better bet.

Originally posted by PBE6
You could also introduce an interest rate to take into account the time value of money. Assuming electricity is paid every hour, the cost for each bulb can be broken down as follows:

C(hardware) = B + B/(1+i)^p + B/(1+i)^2p + ...

= B/(1-1/(1+i)^p)

C(electricity) = 1/(1+i)*(E + E/(1+i) + E/(1+i)^2 + ...)

= 1/(1+i)*(E/(1-1/(1+i))

C(total) = C(hard ...[text shortened]... less, the expensive bulb performed better, while above that the cheaper bulb was a better bet.