Selling Books
200,000 is the quantity of a particular book which a publisher plans to sell in a year. It costs $3750 to set up the printing. The actual printing of each book costs $3. And to store 1000 books for a year cost $600.
What size printing runs will minimize the publisher's cost?
What size printing runs will minimize the publisher's cost?
posted by Zaux at
6:04 AM
17 Comments:
Standard optimization problem - solve with plotting an area representing cost and then calculate the minimum area.
Although I should point out that it you don't know how quickly the books will sell...
Print no books...? That is the most minimal cost, although, probably the least profitable business opportunity...
Hi Karl...
the optimum solution... print no books... heh heh... I like it
Zaux. Are the books taken out of storage continuously, or daily? Is the storage charge a continuous function or is it e.g. such that if a book is stored for one second, tht you pay for a whole week?
Are they taken out in multiples of 1000, or individually? Are they constrained to be published in multiples of 1000 or could e.g. 17 be printed as a run? Is the storage cost in daily/weekly/monthly increments? i.e. if a book is stored for 1 second, do you pay for a week?
How many days are there in a year?
In response to Chis, on behalf of Zaux....
How many days are there in a year?
Lets take D as the number of days, where Y represents 1 year...
Actually, there are too many unknowns...
Why not print books by the 1000, or on request? Thereby decreasing your storage costs.
The implication of storage costs is that you will produce all 200,000 books - $6,003,750 and will need to initially store them at $0.00164383 per day, each...
So, how fast do they sell? (Killing Flies Walking Loudly from the other day would probably sell out in days...)
Hi Karl, just watch this space ;)
Hmmmmmm....thought I understood this one ... maybe not. I will attempt to present the published solution. As I was reading through it, I hit a snag ... but here goes:
This is basically an inventory problem. Let x represent the size of each printing in units of 1000 books. The storage (holding) cost is:
H = (annual storage cost per 1000)*(x/2)
That puzzles me... they took the storage cost per 1000 times half the x value (units of 1000). My question is why did they multiply by half of x.
But I continue...
H(x)= 600(x/2) = 300x
The ordering cost is:
O(x) = (cost per printing)(q/x)
Note: q is in thousands
O(x)= 3750(200/x)= 750,000x^-1
The total printing cost, $3 x 200,000 = $600,000 can be IGNORED because it is a fixed cost per book.
C(x)=H(x)+O(x)
=300x + 750,000x^-1
C!(x)=300 - 750,000x^-2
0=300 - 750,000/x^2
750,000/x^2 = 300
300x^2 = 750,000
x^2 = 2500
x = 50
X represents printing lots of 1000
Thus to minimize cost, each printing should be 50,000 copies.
In the first step above:
H(x)=(annual storage cost per 1000)(x/2), I am hoping one of you guys will understand and explain why they used x/2 rather than x.
This post has been removed by the author.
I shall assume that the books are sold and that the storage
charges can be treated in a continuous manner.
Let the number of books sold/year = R = 200000
Consider a batch of size B being stored. It will be reduced at a
uniform rate of R over a time T, at which point the stock = 0.
So B = RT. The average number of books be B/2.
The cost of storing one book for a year 600/1000 = 0.6
NB all times are in years. The cost of storing books over a year
is 0.6B/2 = 0.3B
Let P be the number of batches produced over the year.
Then PB = 200000 = R (conveniently), so P = R/B
The cost of making a batch of size B is 3750 +3B. P batches will
cost P times that. Using P = R/B => cost of producing the books is
R(3750+3B)/B = 3750R/B + 3R.
The total cost is therefore C = 3750R/B + 3R + 0.3B
dC/dB = -3750R/B² + 0.3 = 0 at optimum =>
B² = 3750R/0.3 => B = 50000
exactly ...
Hi Zaux. The factor of a half is due the the stock level being a sawtooth shape. If the batch size is B, the average storage size is B/2. If you like, it's because the area of a triangle is 1/2 base * perpendicular height.
Thanks Chris ... that makes sense ... wish I had your math background
I don't think I would have gone through the pain of solving if the problem required a discrete, rather than a continuous analysis. Altough the answer would have een similar.
Printing no books isn't consistent with producing 200000 books. The rate of sale is a forecast, and is the 200000 figure.
I believe the intent, of the "print no books" comment, was humor.
I know, but it didn't really work, sorry.
Didn't say it worked...simply commented on the intent ... heh heh
.. but I commented on the content ;)
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