Which of the following is not included in the interest calculation using adjusted balance method?
2 answers:
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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.
Answer:
(Q, R) = (1555, 1400)
shortage imputed = $0.388
Explanation:
Lot size-reorder point system is one of the multi period models. This system is denoted by decision variables (Q, R). This multi period model is implemented when there is uncertain demand in inventory control.
nevertheless, in the simple EOQ model, demand is known and fixed. But when the demand is random, these lot size-reorder point (Q, R) systems allow random demand.
There are two decision variables in a (Q, R) system:
Order quantity, Q and
Reorder point, R
Additional steps are attached as files
