The term applied to the amount of cost to transfer to expense resulting from a decline in the utility of intangible assets is:__
1 answer:
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Romans Food Market, located in Saratoga, New York, carries a variety of specialty foods from around the world. Two of the stores leading products use the Romans Food Market name: Romans Regular Coffee and Romans DeCaf Coffee. These coffees are blends of Brazilian Natural and Columbian mild coffee beans, which are purchased from a distributor from New York City. Because Romans purchases large quantities the coffee beans may be purchased om an as need basis for the price of 10% higher than the market price the distributor pays for the beans. The current market price is $0.47 per pound for Brazilian Natural and $0.62 per pound for Columbian Mild The composition of each coffee blend are as follows:
Bean Regular DeCaf Blend
Brazilian Natural 75% 40%
Columbian Mild 25% 60%
Romans sells the Regular blend for $3.60 per pound and the DeCaf blend for $4.40 per pound. Romans would like to place an order for the Brazilian and Colombian coffee beans that will enable the production of 1000 pounds of Romans Regular coffee and 500 pounds of Romans DeCaf coffee. The production cost is $0.80 per pound for the Regular blend. Because of the extra steps required to produce DeCaf, the production cost for the DeCaf blend is $1.05 per pound. Packaging costs for both products are $0.25 per pound. Formulate a linear programming model that can be used to determine the pounds of Brazilian Natural and Colombian Mild that will maximize the total contribution to profit.
Answer:
Explanation:
From the given information:
The total revenue can be illustrated as :
Total revenue = 3.6 BR + 4.4 BD + 3.6 CR + 4.4 CD
On the other hand; the total cost of the beans is:
= 1.1 (0.47 BR + 0.47 BD + 0.62 CR + 0.62 CD)
= 0.517 BR + 0.517 BD + 0.682 CR + 0.682 CD
Also; The total production cost is :
= 0.8 BR + 1.05 BD + 0.8 CR + 1.05 CD
The total profit = Total revenue - Total Cost of Beans - Total Production Cost
The total profit =
The total profit = 2.033 BR + 2.583 BD + 1.868 CR + 2.418 CD
Therefore the linear programming model represents the Objective function of the total profit as:
Answer:
Check the explanation
Explanation:
Yt = Kt + Nt
Taking output per worker, we divide by Nt
Yt/Nt = Kt/Nt + 1
yt = kt + 1
where yt is output per worker and kt is capital per worker.
a) With population being constant, savings rate s and depreciation rate δ.
ΔKt = It - δKt
dividing by Nt, we get
ΔKt/Nt = It/Nt - δKt/Nt ..... [1]
for kt = Kt/Nt, taking derivative
d(kt)/dt = d(Kt/Nt)/dt ... since Nt is a constant, we have
d(kt)/dt = d(Kt/Nt)/dt = (dKt/dt)/Nt = ΔKt/Nt = It/Nt - δKt/Nt = it - δkt
thus, Capital accumulation Δkt = i – δkt
In steady state, Δkt = 0
That is I – δkt = 0
S = I means that I = s.yt
Thus, s.yt – δkt = 0
Then kt* = s/δ(yt) = s(kt+1)/(δ )
kt*= skt/(δ) + s/(δ)
kt* - skt*/(δ) = s/(δ)
kt*(1- s/(δ) = s/(δ)
kt*((δ - s)/(δ) = s/(δ)
kt*(δ-s)) = s
kt* = s/(δ -s)
capital per worker is given by kt*
b) with population growth rate of n,
d(kt)/dt = d(Kt/Nt)/dt =
=
=
= ΔKt/Nt - n.kt
because (dNt/dt)/Nt = growth rate of population = n and Kt/Nt = kt (capital per worker)
so, d(kt)/dt = ΔKt/Nt - n.kt
Δkt = ΔKt/Nt - n.kt = It/Nt - δKt/Nt - n.kt ......(from [1])
Δkt = it - δkt - n.kt
at steady state Δkt = it - δkt - n.kt = 0
s.yt - (δ + n)kt = 0........... since it = s.yt
kt* = s.yt/(δ + n) =s(kt+1)/(δ + n)
kt*= skt/(δ + n) + s/(δ + n)
kt* - skt*/(δ + n) = s/(δ + n)
kt*(1- s/(δ + n)) = s/(δ + n)
kt*((δ + n - s)/(δ + n)) = s/(δ + n)
kt*(δ + n -s)) = s
kt* = s/(δ + n -s)
.... is the steady state level of capital per worker with population growth rate of n.
3. a) capital per worker. in steady state Δkt = 0 therefore, growth rate of kt is zero
b) output per worker, yt = kt + 1
g(yt) = g(kt) = 0
since capital per worker is not growing, output per worker also does not grow.
c)capital.
kt* = s/(δ + n -s)
Kt*/Nt = s/(δ + n -s)
Kt* = sNt/(δ + n -s)
taking derivative with respect to t.
d(Kt*)/dt = s/(δ + n -s). dNt/dt
(dNt/dt)/N =n (population growth rate)
so dNt/dt = n.Nt
d(Kt*)/dt = s/(δ + n -s).n.Nt
dividing by Kt*
(d(Kt*)/dt)/Kt* = s/(δ + n -s).n.Nt/Kt* = sn/(δ + n -s). (Nt/Kt)
using K/N = k
plugging the value of kt*
n
thus, Capital K grows at rate n
d) Yt = Kt + Nt
dYt/dt = dKt/dt + dNt/dt = s/(δ + n -s).n.Nt + n.Nt
using d(Kt*)/dt = s/(δ + n -s).n.Nt from previous part and that (dNt/dt)/N =n
dYt/dt = n.Nt(s/(δ + n -s) + 1) = n.Nt(s+ δ + n -s)/(δ + n -s) = n.Nt((δ + n)/(δ + n -s)
dYt/dt = n.Nt((δ + n)/(δ + n -s)
dividing by Yt
g(Yt) = n.(δ + n)/(δ + n -s).Nt/Yt
since Yt/Nt = yt
g(Yt) = n.(δ + n)/(δ + n -s) (1/yt)
at kt* = s/(δ + n -s), yt* = kt* + 1
so yt* = s/(δ + n -s) + 1 = (s + δ + n -s)/(δ + n -s) = (δ + n)/(δ + n -s)
thus, g(Yt) = n.(δ + n)/(δ + n -s) (1/yt) = n.(δ + n)/(δ + n -s) ((δ + n -s)/(δ + n)) = n
therefore, in steady state Yt grows at rate n.