Mixing
Optimizing number of production runs?
$begingroup$
I am having trouble with the following problem:
A manufacturer of hospital supplies has a uniform annual demand for $180, 000$ boxes of bandages. It costs $20$ dollars to store one box of bandages for one year and $320$ dollars to set up production. How many times a year should the manager decide to produce boxes of bandages in order to minimize the total cost of storage and setup?
I think this problem is a little bit unclear. If it costs $20$ dollars to store one box for a year, I'm not sure that this implies a one time flat fee of $20$ dollars, or if it means that storing it for half a year will only cost $10$ . I can see the tradeoff between making too many or too few bandages in one run, but I can't quite put it into equations.
calculus algebra-precalculus optimization economics
asked Nov 6 '15 at 19:24
OviOvi
12.5k1040116
$endgroup$
add a comment |
$begingroup$
I am having trouble with the following problem:
A manufacturer of hospital supplies has a uniform annual demand for $180, 000$ boxes of bandages. It costs $20$ dollars to store one box of bandages for one year and $320$ dollars to set up production. How many times a year should the manager decide to produce boxes of bandages in order to minimize the total cost of storage and setup?
I think this problem is a little bit unclear. If it costs $20$ dollars to store one box for a year, I'm not sure that this implies a one time flat fee of $20$ dollars, or if it means that storing it for half a year will only cost $10$ . I can see the tradeoff between making too many or too few bandages in one run, but I can't quite put it into equations.
calculus algebra-precalculus optimization economics
asked Nov 6 '15 at 19:24
OviOvi
12.5k1040116
$endgroup$
$begingroup$
Do the bandagest cost nothing to produce?
$endgroup$
– 5xum
Nov 6 '15 at 19:27
$begingroup$
@5xu't The problem doesn mention anything about production cost
$endgroup$
– Ovi
Nov 6 '15 at 19:31
$begingroup$
Ah, yes, it does not matter, since you have to produce 180,000 of them anyway. Well, I think that the 20 dollars per year means that it's 10 dollars for half a year, yes. Otherwise I don't see the point of the exercise.
$endgroup$
– 5xum
Nov 6 '15 at 19:33
$begingroup$
@5xum So I can think of the problem intuitively, but could you help me get down the actual formulas? I know I am supposed to end up with a function of the cost and find critical points to find the relative min
$endgroup$
– Ovi
Nov 6 '15 at 19:46
$begingroup$
Say $x$ "times" a year, where $x$ need not be an integer. So each time we produce $K/x$, where $K=180000$. Now storage cost is tricky, I would lean to $20cdot frac{K}{x}cdot frac{1}{2x}cdot x$ since on average a box is stored for $1/2$ of $1/x$ of a year. That gives cost function $320x+frac{10K}{x}$.
$endgroup$
– André Nicolas
Nov 6 '15 at 20:22
add a comment |
$begingroup$
I am having trouble with the following problem:
A manufacturer of hospital supplies has a uniform annual demand for $180, 000$ boxes of bandages. It costs $20$ dollars to store one box of bandages for one year and $320$ dollars to set up production. How many times a year should the manager decide to produce boxes of bandages in order to minimize the total cost of storage and setup?
I think this problem is a little bit unclear. If it costs $20$ dollars to store one box for a year, I'm not sure that this implies a one time flat fee of $20$ dollars, or if it means that storing it for half a year will only cost $10$ . I can see the tradeoff between making too many or too few bandages in one run, but I can't quite put it into equations.
calculus algebra-precalculus optimization economics
asked Nov 6 '15 at 19:24
OviOvi
12.5k1040116
$endgroup$
I am having trouble with the following problem:
A manufacturer of hospital supplies has a uniform annual demand for $180, 000$ boxes of bandages. It costs $20$ dollars to store one box of bandages for one year and $320$ dollars to set up production. How many times a year should the manager decide to produce boxes of bandages in order to minimize the total cost of storage and setup?
I think this problem is a little bit unclear. If it costs $20$ dollars to store one box for a year, I'm not sure that this implies a one time flat fee of $20$ dollars, or if it means that storing it for half a year will only cost $10$ . I can see the tradeoff between making too many or too few bandages in one run, but I can't quite put it into equations.
calculus algebra-precalculus optimization economics
calculus algebra-precalculus optimization economics
asked Nov 6 '15 at 19:24
OviOvi
12.5k1040116
asked Nov 6 '15 at 19:24
OviOvi
12.5k1040116
asked Nov 6 '15 at 19:24
OviOvi
12.5k1040116
asked Nov 6 '15 at 19:24
OviOvi
12.5k1040116
asked Nov 6 '15 at 19:24
OviOvi
12.5k1040116
12.5k1040116
$begingroup$
Do the bandagest cost nothing to produce?
$endgroup$
– 5xum
Nov 6 '15 at 19:27
$begingroup$
@5xu't The problem doesn mention anything about production cost
$endgroup$
– Ovi
Nov 6 '15 at 19:31
$begingroup$
Ah, yes, it does not matter, since you have to produce 180,000 of them anyway. Well, I think that the 20 dollars per year means that it's 10 dollars for half a year, yes. Otherwise I don't see the point of the exercise.
$endgroup$
– 5xum
Nov 6 '15 at 19:33
$begingroup$
@5xum So I can think of the problem intuitively, but could you help me get down the actual formulas? I know I am supposed to end up with a function of the cost and find critical points to find the relative min
$endgroup$
– Ovi
Nov 6 '15 at 19:46
$begingroup$
Say $x$ "times" a year, where $x$ need not be an integer. So each time we produce $K/x$, where $K=180000$. Now storage cost is tricky, I would lean to $20cdot frac{K}{x}cdot frac{1}{2x}cdot x$ since on average a box is stored for $1/2$ of $1/x$ of a year. That gives cost function $320x+frac{10K}{x}$.
$endgroup$
– André Nicolas
Nov 6 '15 at 20:22
add a comment |
$begingroup$
Do the bandagest cost nothing to produce?
$endgroup$
– 5xum
Nov 6 '15 at 19:27
$begingroup$
@5xu't The problem doesn mention anything about production cost
$endgroup$
– Ovi
Nov 6 '15 at 19:31
$begingroup$
Ah, yes, it does not matter, since you have to produce 180,000 of them anyway. Well, I think that the 20 dollars per year means that it's 10 dollars for half a year, yes. Otherwise I don't see the point of the exercise.
$endgroup$
– 5xum
Nov 6 '15 at 19:33
$begingroup$
@5xum So I can think of the problem intuitively, but could you help me get down the actual formulas? I know I am supposed to end up with a function of the cost and find critical points to find the relative min
$endgroup$
– Ovi
Nov 6 '15 at 19:46
$begingroup$
Say $x$ "times" a year, where $x$ need not be an integer. So each time we produce $K/x$, where $K=180000$. Now storage cost is tricky, I would lean to $20cdot frac{K}{x}cdot frac{1}{2x}cdot x$ since on average a box is stored for $1/2$ of $1/x$ of a year. That gives cost function $320x+frac{10K}{x}$.
$endgroup$
– André Nicolas
Nov 6 '15 at 20:22
$begingroup$
Do the bandagest cost nothing to produce?
$endgroup$
– 5xum
Nov 6 '15 at 19:27
$begingroup$
Do the bandagest cost nothing to produce?
$endgroup$
– 5xum
Nov 6 '15 at 19:27
$begingroup$
@5xu't The problem doesn mention anything about production cost
$endgroup$
– Ovi
Nov 6 '15 at 19:31
$begingroup$
@5xu't The problem doesn mention anything about production cost
$endgroup$
– Ovi
Nov 6 '15 at 19:31
$begingroup$
Ah, yes, it does not matter, since you have to produce 180,000 of them anyway. Well, I think that the 20 dollars per year means that it's 10 dollars for half a year, yes. Otherwise I don't see the point of the exercise.
$endgroup$
– 5xum
Nov 6 '15 at 19:33
$begingroup$
Ah, yes, it does not matter, since you have to produce 180,000 of them anyway. Well, I think that the 20 dollars per year means that it's 10 dollars for half a year, yes. Otherwise I don't see the point of the exercise.
$endgroup$
– 5xum
Nov 6 '15 at 19:33
$begingroup$
@5xum So I can think of the problem intuitively, but could you help me get down the actual formulas? I know I am supposed to end up with a function of the cost and find critical points to find the relative min
$endgroup$
– Ovi
Nov 6 '15 at 19:46
$begingroup$
@5xum So I can think of the problem intuitively, but could you help me get down the actual formulas? I know I am supposed to end up with a function of the cost and find critical points to find the relative min
$endgroup$
– Ovi
Nov 6 '15 at 19:46
$begingroup$
Say $x$ "times" a year, where $x$ need not be an integer. So each time we produce $K/x$, where $K=180000$. Now storage cost is tricky, I would lean to $20cdot frac{K}{x}cdot frac{1}{2x}cdot x$ since on average a box is stored for $1/2$ of $1/x$ of a year. That gives cost function $320x+frac{10K}{x}$.
$endgroup$
– André Nicolas
Nov 6 '15 at 20:22
$begingroup$
Say $x$ "times" a year, where $x$ need not be an integer. So each time we produce $K/x$, where $K=180000$. Now storage cost is tricky, I would lean to $20cdot frac{K}{x}cdot frac{1}{2x}cdot x$ since on average a box is stored for $1/2$ of $1/x$ of a year. That gives cost function $320x+frac{10K}{x}$.
$endgroup$
– André Nicolas
Nov 6 '15 at 20:22
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
This exercise can be managed by using the EOQ-formula.
In your case the total costs are
$TC=frac{K cdot D}{Q}+frac{Qcdot h}{2}=frac{320 cdot 180,000}{Q}+frac{Qcdot 20}{2}$
$P$ = store costs per year
$Q$ = produce quantity
$D$ = annual demand quantity
$K$ = Set up production cost
$h$=storage cost per unit
The derivative w.r.t Q is
$frac{partial TC}{partial Q}=-frac{K cdot D}{Q^2}+frac{ h}{2}=0$
Solving for Q
$frac{K cdot Dcdot 2}{h}=Q^2$
$Q^*=sqrt{frac{K cdot Dcdot 2}{h}}$
After you have calculated the optimal produce quantity the number of cost minimizing set ups is $frac{D}{Q^*}$
answered Nov 7 '15 at 2:37
callculuscallculus
18.7k31428
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
This exercise can be managed by using the EOQ-formula.
In your case the total costs are
$TC=frac{K cdot D}{Q}+frac{Qcdot h}{2}=frac{320 cdot 180,000}{Q}+frac{Qcdot 20}{2}$
$P$ = store costs per year
$Q$ = produce quantity
$D$ = annual demand quantity
$K$ = Set up production cost
$h$=storage cost per unit
The derivative w.r.t Q is
$frac{partial TC}{partial Q}=-frac{K cdot D}{Q^2}+frac{ h}{2}=0$
Solving for Q
$frac{K cdot Dcdot 2}{h}=Q^2$
$Q^*=sqrt{frac{K cdot Dcdot 2}{h}}$
After you have calculated the optimal produce quantity the number of cost minimizing set ups is $frac{D}{Q^*}$
answered Nov 7 '15 at 2:37
callculuscallculus
18.7k31428
$endgroup$
add a comment |
$begingroup$
This exercise can be managed by using the EOQ-formula.
In your case the total costs are
$TC=frac{K cdot D}{Q}+frac{Qcdot h}{2}=frac{320 cdot 180,000}{Q}+frac{Qcdot 20}{2}$
$P$ = store costs per year
$Q$ = produce quantity
$D$ = annual demand quantity
$K$ = Set up production cost
$h$=storage cost per unit
The derivative w.r.t Q is
$frac{partial TC}{partial Q}=-frac{K cdot D}{Q^2}+frac{ h}{2}=0$
Solving for Q
$frac{K cdot Dcdot 2}{h}=Q^2$
$Q^*=sqrt{frac{K cdot Dcdot 2}{h}}$
After you have calculated the optimal produce quantity the number of cost minimizing set ups is $frac{D}{Q^*}$
answered Nov 7 '15 at 2:37
callculuscallculus
18.7k31428
$endgroup$
add a comment |
$begingroup$
This exercise can be managed by using the EOQ-formula.
In your case the total costs are
$TC=frac{K cdot D}{Q}+frac{Qcdot h}{2}=frac{320 cdot 180,000}{Q}+frac{Qcdot 20}{2}$
$P$ = store costs per year
$Q$ = produce quantity
$D$ = annual demand quantity
$K$ = Set up production cost
$h$=storage cost per unit
The derivative w.r.t Q is
$frac{partial TC}{partial Q}=-frac{K cdot D}{Q^2}+frac{ h}{2}=0$
Solving for Q
$frac{K cdot Dcdot 2}{h}=Q^2$
$Q^*=sqrt{frac{K cdot Dcdot 2}{h}}$
After you have calculated the optimal produce quantity the number of cost minimizing set ups is $frac{D}{Q^*}$
answered Nov 7 '15 at 2:37
callculuscallculus
18.7k31428
$endgroup$
This exercise can be managed by using the EOQ-formula.
In your case the total costs are
$TC=frac{K cdot D}{Q}+frac{Qcdot h}{2}=frac{320 cdot 180,000}{Q}+frac{Qcdot 20}{2}$
$P$ = store costs per year
$Q$ = produce quantity
$D$ = annual demand quantity
$K$ = Set up production cost
$h$=storage cost per unit
The derivative w.r.t Q is
$frac{partial TC}{partial Q}=-frac{K cdot D}{Q^2}+frac{ h}{2}=0$
Solving for Q
$frac{K cdot Dcdot 2}{h}=Q^2$
$Q^*=sqrt{frac{K cdot Dcdot 2}{h}}$
After you have calculated the optimal produce quantity the number of cost minimizing set ups is $frac{D}{Q^*}$
answered Nov 7 '15 at 2:37
callculuscallculus
18.7k31428
edited Nov 7 '15 at 2:43
answered Nov 7 '15 at 2:37
callculuscallculus
18.7k31428
answered Nov 7 '15 at 2:37
callculuscallculus
18.7k31428
answered Nov 7 '15 at 2:37
callculuscallculus
18.7k31428
18.7k31428
add a comment |
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$begingroup$
Do the bandagest cost nothing to produce?
$endgroup$
– 5xum
Nov 6 '15 at 19:27
$begingroup$
@5xu't The problem doesn mention anything about production cost
$endgroup$
– Ovi
Nov 6 '15 at 19:31
$begingroup$
Ah, yes, it does not matter, since you have to produce 180,000 of them anyway. Well, I think that the 20 dollars per year means that it's 10 dollars for half a year, yes. Otherwise I don't see the point of the exercise.
$endgroup$
– 5xum
Nov 6 '15 at 19:33
$begingroup$
@5xum So I can think of the problem intuitively, but could you help me get down the actual formulas? I know I am supposed to end up with a function of the cost and find critical points to find the relative min
$endgroup$
– Ovi
Nov 6 '15 at 19:46
$begingroup$
Say $x$ "times" a year, where $x$ need not be an integer. So each time we produce $K/x$, where $K=180000$. Now storage cost is tricky, I would lean to $20cdot frac{K}{x}cdot frac{1}{2x}cdot x$ since on average a box is stored for $1/2$ of $1/x$ of a year. That gives cost function $320x+frac{10K}{x}$.
$endgroup$
– André Nicolas
Nov 6 '15 at 20:22