How can we calculate man-days needed to complete a job?

[22990atinesh]

Let M=Men, D=Days, H=Hours, η=efficiency, W=Work

$MDH \propto W$

$MDH = kW$

$\frac {MDH}{W} = k$

$\frac {M_1D_1H_1}{W_1} = \frac {M_2D_2H_2}{W_2} = \frac {M_3D_3H_3}{W_3}$

Which means total total Men hour per unit of work is constant and we say

η=total work done in one day (or 1 hour)

My doubts
1. Does η=W/MDH ?
2. Does here η refers to Rate of work.

 

[Mark44]

Let M=Men, D=Days, H=Hours, η=efficiency, W=Work

$MDH \propto W$

$MDH = kW$

$\frac {MDH}{W} = k$

$\frac {M_1D_1H_1}{W_1} = \frac {M_2D_2H_2}{W_2} = \frac {M_3D_3H_3}{W_3}$

Which means total total Men hour per unit of work is constant and we say

η=total work done in one day (or 1 hour)

My doubts
1. Does η=W/MDH ?
2. Does here η refers to Rate of work.

I don't think that efficiency figures in here. Your MDH = kW formula says that if you multiply the number of men by the number of days and the number of hours in a day (wouldn't the number of hours per day be constant?), you get W (or a multiple of W).

If you want to incorporate efficiency, you are going to have to have a more sophisticated formula.

For another thing, how do the units match up? On the left you have (men) * (days) * (hours/day), assuming that H represents hours/day and not hours. So the units on the left are man-hours. Are the units of W also in man-hours?

 

[22990atinesh]

I don't think that efficiency figures in here. Your MDH = kW formula says that if you multiply the number of men by the number of days and the number of hours in a day (wouldn't the number of hours per day be constant?), you get W (or a multiple of W).

If you want to incorporate efficiency, you are going to have to have a more sophisticated formula.

For another thing, how do the units match up? On the left you have (men) * (days) * (hours/day), assuming that H represents hours/day and not hours. So the units on the left are man-hours. Are the units of W also in man-hours?

Considering H=Hours/day still $\frac {MDH}{W}$ refers to total Men hours required to do 1 unit of work. And here Efficiency (η) refers to 1 day work or 1 hour work, then it refers to rate at which work is done. Actually I've seen some people using these words interchangeably but still I'm not sure.

 

[Mark44]

Considering H=Hours/day still $\frac {MDH}{W}$ refers to total Men hours per day.

No, it doesn't.
The units of MDH are (men) * (days) * (hours/day), so the units of MDH are men-hours (usually written as man-hours).

You haven't said what the units of W are in. Possibly it's the number of hours it would take one man to do a job. If that's what W is here, then the units of MDH/W would be (man-hours)/(hours), which would tell you how many men it would take to do a certain job.

For example, if a certain job would take one man 16 hours, and you have two men, then the job could be done in one day (of 8 hours).

And here Efficiency (η) refers to 1 day work or 1 hour work, then it refers to rate at which work is done.

This makes no sense to me. Your formula with MDH assumes that each worker is working at the same efficiency. In other words, if it would take worker A 8 hours to paint something, then worker B could paint the same thing in the same time.

As I said before, if you want to incorporate efficiency of each worker into your formula, it's going to have to be quite a bit more complicated. You'll have different expressions for each worker for the amount of work produced in a unit of time.

 

[22990atinesh]

@Mark44 Please read the above comment I've edited it.

 

[Mark44]

Let M=Men, D=Days, H=Hours, η=efficiency, W=Work

$MDH \propto W$

$MDH = kW$

$\frac {MDH}{W} = k$

$\frac {M_1D_1H_1}{W_1} = \frac {M_2D_2H_2}{W_2} = \frac {M_3D_3H_3}{W_3}$

Which means total total Men hour per unit of work is constant and we say

η=total work done in one day (or 1 hour)

My doubts
1. Does η=W/MDH ?
2. Does here η refers to Rate of work.

This is confusing. At the top, you say η = efficiency. Later on you say η = total work done in one day (or 1 hour), which is really the work rate. If η is efficiency, it would be a number somewhere between 0 and 1.0 (or 0% to 100%). You also haven't defined what "work" means in your formulas. I asked earlier if work represents man-hours, but you haven't answered me.

Also, your formula would be much simpler if you got rid of "days" and just used "hours".

A simple example might help clarify things. A general contractor has two painters, Al and Ben. Al can paint a "standard" house in 16 hours. Ben can paint the same sized house in 20 hours. How long would it take both of them working together to paint a standard sized house?

Al's work rate: 1/16 of a standard house per hour
Ben's work rate: 1/20 of a standard house per hour

Let t = number of hours for both working together to paint a standard house.

The amount of work done is (work rate) * (time), or (houses/hour) * (hours), so
Al's work done + Ben's work done = 1 house painted

In symbols,
1/16 * t + 1/20 * t = 1, or
(1/16 + 1/20)t = 1
(5/80 + 4/80)t = 1
(9/80)t = 1
t = 80/9 ≈ 8.9
Working together, Al and Ben can paint a "standard" house in about 8.9 hours.

 

[22990atinesh]

This is confusing. At the top, you say η = efficiency. Later on you say η = total work done in one day (or 1 hour), which is really the work rate. If η is efficiency, it would be a number somewhere between 0 and 1.0 (or 0% to 100%). You also haven't defined what "work" means in your formulas.

Don't get confused with η = efficiency that we read in Science and Engineering. Here it is different it means Rate. Follow this link

http://www.bankexamstoday.com/2013/04/time-and-work-shortcuts-and-tricks.html

Here they have represented η = efficiency as percent
η(A)=20%=1/5 => 1 day work => which represents rate right

I asked earlier if work represents man-hours, but you haven't answered me.

Yes Work represents man hours

Also, your formula would be much simpler if you got rid of "days" and just used "hours".

The reason why I mentioned both, because sometime in question only days is mentioned i.e work done in some days and sometime hours/day is mentioned.

 

[Mark44]

Like you said, efficiency is well understood in science and engineering. When you use the same term to mean something different, you (or the guy in the Indian Bank Prep web site), you run the risk of confusing people.

When you set up an equation that represents a physical system, it's important that the units are consistent on both sides of the equation. If you throw around terms such as efficiency and work, and aren't clear on what the underlying units are, then any equation you write will be meaningless. In all of my replies here, I've been trying to get you to think in terms of the units attached to the terms in your formulas.

Some of the things you wrote are pretty much meaningless, such as this:

η(A)=20%=1/5 => 1 day work => which represents rate right

You (and Ramandeep Singh) are clouding the water by using efficiency and the amount of work per unit time synonymously. I agree that 20% = 1/5, but how does that imply "1 day work" and how is "1 day work" a rate?

A rate is the quotient of two quantities, like distance/time (velocity rate) or mass/volume (density). If you say something is a rate, but don't say what quantities are being divided, you aren't being clear.

I'm not impressed by the quality of the website you posted. In the 3rd question, about filling a tank, he writes in his solution (Method 1),

=> Efficiency of filling pipe = 20 minutes = 1/3 hour = 300%
=> Efficiency of leakage = 60 minutes = 100%

By what mathematics is 1/3 hour the same as 300%?
By what reasoning is 60 minutes or 1 hours the same as 100%?
If "efficiency" here is a rate, what rate (i.e., ratio) are we talking about?

What he's selling on his website is a way of solving problems on a banking exam, with absolutely no understanding required.

 

[22990atinesh]

I agree that 20% = 1/5, but how does that imply "1 day work" and how is "1 day work" a rate?

I think you did't read the question in the picture carefully. In question it is mentioned

'A' can complete a Job(unit of Work) in 5 days
5 days -> 1 (1 represents unit of work)
1 day -> 1/5th of Work this implies in 1 day 1/5th of work is done i.e 1 day work

 

[Mark44]

No, I did carefully read the question in the image you posted, and I looked at the web page in the link, and read what Mr. Singh had to say for the first three examples. I understand his reasoning for the percentages, but when he writes mathematical equations that are nonsense, I question his ability to explain things.

=> Efficiency of filling pipe = 20 minutes = 1/3 hour = 300%
=> Efficiency of leakage = 60 minutes = 100%

The above came from the third example on Singh's page. It's nonsense.

 

[micromass]

Follow this link

http://www.bankexamstoday.com/2013/04/time-and-work-shortcuts-and-tricks.html

This is a pretty horrible site. It might help you to pass the exam, but it doesn't help people understand the solution. In fact, it encourages misunderstanding and memorization. Not my favorite way to do math and science.

Sure, it will help you reach the right answer. But the right answer isn't the most important thing in science and math (although it might be the most important thing if you're doing the exam).

 

[22990atinesh]

This is a pretty horrible site. It might help you to pass the exam, but it doesn't help people understand the solution. In fact, it encourages misunderstanding and memorization. Not my favorite way to do math and science.

Sure, it will help you reach the right answer. But the right answer isn't the most important thing in science and math (although it might be the most important thing if you're doing the exam).

I agree with you.

 

[22990atinesh]

Ok let's forget about the term Efficiency (η) for a while. Now does

$\frac {MDH}{W} = k$

represents constant rate.
Example: Suppose Rates for sets of people doing the same work separately

Rate for 1st set of people $\frac {M_1D_1H_1}{W_1}$
Rate for 2nd set of people $\frac {M_2D_2H_2}{W_2}$
Rate for 3rd set of people $\frac {M_3D_3H_3}{W_3}$

Now if we consider Efficiency (η) then some people write formula like this

$\frac {η MDH}{W} = k$

What does this represent, if η is considered.

 

[Mark44]

Ok let's forget about the term Efficiency (η) for a while. Now does

$\frac {MDH}{W} = k$

represents constant rate.

Rate of what? I keep asking this question in an effort to get you to state things with more precision. A rate represents a quotient, such as miles/hrs (AKA mph) or meters/seconds, and so on.

Before going further, let's simplify things a bit by getting rid of D in the formula, and just working with hours. So M represents the number of men (all with equal ability) and H represents the number of hours they are all working. With all of them working for H hours, they will accomplish MH man-hours of work.

If a job requires 60 man-hours of work, then it will take one man 60 hours, or two men 30 hours, or four men 15 hours, and so on. In this formulation MH is proportional to W, with k = 1. More to the point, MH = W.

Now, let's change things up, and redefine what W means. Suppose the job is to paint a building that has 4000 sq. ft. of surface to be painted.

So W = 4000 sq. ft. The units of W are area, in sq. ft.

Suppose Al and Ben are painters, where Al's work rate, R_A, is 200 sq. ft./hour, and Ben's work rate, R_B, is 250 sq. ft./hour. Notice that the units for the work rates are quotients - sq. ft./hours.

The amount of work done by either painter is (work rate) * time.

We can find Al's time to complete the job from this equation: R_At_A = 4000, or t_A = 20 hours.
For Ben, the equation is R_Bt_B = 4000, or t_B = 16 hours.

Working together, their rates would add, so to find the time for both of them painting, we have
(R_A + R_B)t = 4000, which is pretty easy to solve after putting in their respective work rates.

Example: Suppose Rates for sets of people doing the same work separately

Rate for 1st set of people $\frac {M_1D_1H_1}{W_1}$
Rate for 2nd set of people $\frac {M_2D_2H_2}{W_2}$
Rate for 3rd set of people $\frac {M_3D_3H_3}{W_3}$

Now if we consider Efficiency (η) then some people write formula like this

$\frac {η MDH}{W} = k$

What does this represent, if η is considered.

 

[Mark44]

Example: Suppose Rates for sets of people doing the same work separately

Rate for 1st set of people $\frac {M_1D_1H_1}{W_1}$
Rate for 2nd set of people $\frac {M_2D_2H_2}{W_2}$
Rate for 3rd set of people $\frac {M_3D_3H_3}{W_3}$

Now if we consider Efficiency (η) then some people write formula like this

$\frac {η MDH}{W} = k$

What does this represent, if η is considered.

In my previous post I purposely did not say anything about the equations above. My sense is that you are throwing around equations with little or no understanding of what they are supposed to represent. If you look at my responses in this thread, in each one I have put in a lot of effort at explaining what each variable means, and the units involved. When I used the word "rate," I explained in detail what quantity was changing with respect to some other quantity. It doesn't seem to have gotten across, though, when you write about a "rate" with no further description.

Unless and until you can ask a coherent question, I'm done here.

 

[22990atinesh]

I'm sorry friends. I got confused because of my teacher, and some bad websites. I checked my notes and my teacher also used the term Efficiency for 'η', I think which is wrong. The correct meaning of η here is rate, As 1 day work/ 1 hour work is a rate.

 

[Mark44]

I'm sorry friends. I got confused because of my teacher, and some bad websites. I checked my notes and my teacher also used the term Efficiency for 'η', I think which is wrong. The correct meaning of η here is rate, As 1 day work/ 1 hour work is a rate.

"1 day work/1 hour work" isn't a rate. As near as I can make out, this represents the number of hours in a day, which has nothing to do with efficiency or work rates.

 

[22990atinesh]

A general contractor has two painters, Al and Ben. Al can paint a "standard" house in 16 hours. Ben can paint the same sized house in 20 hours. How long would it take both of them working together to paint a standard sized house?

Al's work rate: 1/16 of a standard house per hour
Ben's work rate: 1/20 of a standard house per hour

@Mark44 See your above comment, You also said the same thing. I'm not saying the individual Work rate is total work done/ day or hour. I'm saying Work Rate = Work done by individual / day or an hour.

 

[Mark44]

@Mark44 See your above comment, You also said the same thing. I'm not saying the individual Work rate is total work done/ day or hour. I'm saying Work Rate = Work done by individual / day or an hour.

OK, this is a little closer to being clear. To be more clear, the individual work rate = (work done by the individual)/hour. (Let's get rid of days, and just keep track of time by hours alone. Having days in there as well just confuses the issue.)

What you wrote before, and what I was commenting on, was this:

Rate for 1st set of people $\frac {M_1D_1H_1}{W_1}$

This is not a rate.

 

[22990atinesh]

OK, this is a little closer to being clear. To be more clear, the individual work rate = (work done by the individual)/hour. (Let's get rid of days, and just keep track of time by hours alone. Having days in there as well just confuses the issue.)

What you wrote before, and what I was commenting on, was this:

This is not a rate.

I agree $\frac {M_1D_1H_1}{W_1}$ doesn't represent rate.

 

[22990atinesh]

As far as formula goes I understand that

$\frac {M_1D_1H_1}{W_1}$ = $\frac {M_2D_2H_2}{W_2}$ = $\frac {M_3D_3H_3}{W_3}$ = $k$
is constant. But I do not understand it intuitively. Please explain the intuitive meaning of $\frac {MDH}{W}$ to be constant. Explain it with following example

A contractor undertook to finish a certain work in 124 days and employed 120 men. After 64 days he found that he had already done $\frac {2}{3}$ ^rd of the work. How many men can be discharged now so that work may be finished in time ?

 

[Mark44]

As far as formula goes I understand that

$\frac {M_1D_1H_1}{W_1}$ = $\frac {M_2D_2H_2}{W_2}$ = $\frac {M_3D_3H_3}{W_3}$ = $k$
is constant. But I do not understand it intuitively.

Which tells me that you don't understand the formula.

Please explain the intuitive meaning of $\frac {MDH}{W}$ to be constant. Explain it with following example

A contractor undertook to finish a certain work in 124 days and employed 120 men. After 64 days he found that he had already done $\frac {2}{3}$ ^rd of the work. How many men can be discharged now so that work may be finished in time ?

For the third time, please stop using MDH! In problems of this type, the time will normally be given either in days or in hours, but not both.

From the information in the problem here, 120 men have been working 64 days, and have completed 2/3 of the job. The work performed so far represents (120 men) * (64 days) = 120*64 man-days. From this information, how many man-days are required for the entire job?
How many man-days have already been used in the first 64 days?
How much time is left to complete the job?
How many man-days are needed to complete the job?
If you know how many man-days are needed to complete the job, and how many days are left, you should be able to figure out how many men are needed, and therefor, how many can be laid off.

The reason this is so difficult for you, I believe, is that you are trying to pick the "right" formula to use, rather than trying to reason things out. Thinking is always harder than plugging numbers into a formula by rote, which part of the reason that we are 22 posts into this thread.

The only "formula" I'm using here is that "work done" is in units of man-days (in this problem), and is calculated by (work done) = (number of men) * (number of days). If the units of time in the problem had been given in terms of hours, then (work done) would be in units of man-hours. Don't use both hours and days, as in MDH. You'll just confuse yourself.

In an example I gave earlier, "work done" was in units of "square feet that are painted". In this case MDH is meaningless.

 

[22990atinesh]

Which tells me that you don't understand the formula.


For the third time, please stop using MDH! In problems of this type, the time will normally be given either in days or in hours, but not both.

From the information in the problem here, 120 men have been working 64 days, and have completed 2/3 of the job. The work performed so far represents (120 men) * (64 days) = 120*64 man-days. From this information, how many man-days are required for the entire job?
How many man-days have already been used in the first 64 days?
How much time is left to complete the job?
How many man-days are needed to complete the job?
If you know how many man-days are needed to complete the job, and how many days are left, you should be able to figure out how many men are needed, and therefor, how many can be laid off.

The reason this is so difficult for you, I believe, is that you are trying to pick the "right" formula to use, rather than trying to reason things out. Thinking is always harder than plugging numbers into a formula by rote, which part of the reason that we are 22 posts into this thread.

The only "formula" I'm using here is that "work done" is in units of man-days (in this problem), and is calculated by (work done) = (number of men) * (number of days). If the units of time in the problem had been given in terms of hours, then (work done) would be in units of man-hours. Don't use both hours and days, as in MDH. You'll just confuse yourself.

In an example I gave earlier, "work done" was in units of "square feet that are painted". In this case MDH is meaningless.

Thanx Mark44, I think I get it. You are trying to say that total Man hour/unit work is constant i.e. MH/W=k. for example we have given that 2 Men working 3 hours/day works for 4 days to complete a work (unit of work). Calculate how many days required by 1 man working 2 hours/day to complete the 1/2 of that work.

Sol: As we know Man hour/unit work is constant. Hence $\frac{M_1H_1}{W_1} = \frac {M_2H_2}{W_2}$
Now we can easily plug data in LHS of the above equation. But for RHS as we know, we have to calculate days required by 1 man working 2 hours/day to complete the 1/2 of that work. so we have to double the total Man hour in RHS i.e.

$\frac{2*(4*3)}{1} = \frac{2*(1*(X*2))}{1}$

$\frac{2*(4*3)}{1} = \frac{1*(X*2)}{1/2}$

$X=6$

Correct If I did something wrong.