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# PS 2

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SVP
Joined: 16 Oct 2003
Posts: 1810
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01 Jun 2004, 17:42
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Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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SVP
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 98 [0], given: 0

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01 Jun 2004, 18:50
assume normally N items are produced at cost C
if additiona n items are produced then actual cost = (N+n) * C

But over production causes savings by x dollars per unit
So savings = (N+n) * x these savings should will be depleted if n itemas are stored for d number of days at y per day

so n*y*d = (N+n)*x
d = x(N+n)/yn = x/y[ 1+N/n ]

none of the answers match this
Director
Joined: 05 May 2004
Posts: 577
Location: San Jose, CA
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Kudos [?]: 59 [0], given: 0

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01 Jun 2004, 21:39
anandnk wrote:
assume normally N items are produced at cost C
if additiona n items are produced then actual cost = (N+n) * C

But over production causes savings by x dollars per unit
So savings = (N+n) * x these savings should will be depleted if n itemas are stored for d number of days at y per day

so n*y*d = (N+n)*x
d = x(N+n)/yn = x/y[ 1+N/n ]

none of the answers match this

I guess we have to assume N=0 for this problem in which case the ans is x/y

What's the OA?
SVP
Joined: 16 Oct 2003
Posts: 1810
Followers: 4

Kudos [?]: 136 [0], given: 0

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02 Jun 2004, 06:27
Total savings xn
Total cost of storage yn

OA xn/yn = x/y
Director
Joined: 25 Jan 2004
Posts: 728
Location: Milwaukee
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05 Jun 2004, 20:17
total amount saved for n units XN
total amount spent on storing N units = YND where D is # of days

there fore when YND = XN

D = X/N
_________________

Praveen

05 Jun 2004, 20:17
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