(Goods 1,2 and 3 are non-durable, goods 4 and 5 are durable; in association with non-durable goods µ = 1 and in association with durable goods µ = 0.04 in the fifth economy and µ = 0.03 in the sixth economy. The net national product of the fifth economy is the unit of measurement of prices, wage, capital, saving and investment in all the six economies. For saving and investment functions and calculation of the ratio between net ouput per unit of labour of any two of the economies prices are expressed at r = R₅).

Rate of profit is determined by wage-profit relation and investment function subject to the constraint of saving:

r = R_i ^{n i}√1-ω (i=1,2,3,4,6)

r = R₅ (1- ω) I = a₀ + a₁ KPr + a₂ KP/L r/ ⁿ√1- ω I ≤ S = β_{0 +} β₁s₁ KPr + β₂s₂ L ω

Where (µ+R) KP = γ P, y′ = γ P-µ KP, and ω = ω L y′₅

We assume a₀ = 0, a₁=1, a₂=1, β₀=0, β₁=1, β₂=1, s₁=0 and s₂ = 0.01 for convenience of calculations in absence of empirically given values of these.

(µ + r) (2600 p₁ + 120p₂) + 48 ω = 5520 p₁

(µ + r) (2200 p₁ + 480p₂) + 32 ω = 690 p₂

NNP (1) = 720 p₁ + 90 p₂

NNP (1) = 0.29 SNP (5) = 0.29

R₁ = 720 p₁ + 90 p₂ = 0.15

4800 p₁ + 600 p₂

When I₁ = s₁, we have 0.8 ω = 0.00237 or ω = 0.00296

. ω = 0.00296 = 0.59 . . 0.005

We have r = 0.15 ⁿ√1-0.59 r = 0.30 (1-0.59)

Taking the right hand sides of the equations and solving for n we have

n = 4.49 and 1 = 0.22 n

With the value of n obtained we have r = 0.15 (1- ω)^0.22 r = 0.30 (1- ω) .

. . (1- ω) ^0.78 - 0.5 = 0

By binomial expansion and solution of the resulting equation we have only one admissible value:

ω = 0.6

r = 0.12

We, thus, find that the two curves have a single switch-point at

The equilibrium rate of profit is

(µ+r) (2600p₁+120p₂) + 60 ω = 5760p₁

(µ+r) (2200p₁+480p₂) + 40 ω = 720p₂

NNP (2) = 960p₁+120p₂

NNP (2) = 0.38 SNP (5) = 0.38

R₂ = 960p₁ + 120p₂ = 0.20

4800p₁+600p₂

y₂′ = NNP (2) = 0.38 = 0.0038

L (2) 100 ^ ^

KP (2) =KP (3) =1.9 ^ ^

KP (2) = KP (3) =1.9 = 0.0158 ^ L(3) 120

L(2)

I₂ =1.9r + 0.0158 x 0.20 = 1.9r + 0.00316

S₂ =1.9r +0.01 x 100w = 1.9r+1 ω

When I₂ = S_2, we haveω = 0.00316 ω = 0.00316 = 0.65 0.005

Putting the value of ω into the wage profit equations

we have

r = 0.20 ⁿ√1-0.65

r = 0.30 (1-0.65)

Solving we have

With the value of n obtained we have

r = 0.20 (1 – ω)^0.61

.

. . (1 – ω)^0.39 – 0.67 =0

By binomial expansion and solution of the resulting equation we have only one admissible value:

r = 0.078 or 0.08

We, thus, find that the curves of the economy II and economy V meet at

r = 0.08 and ω = 0.74

The equilibrium rate of profit is

r = 0.20 ^1.63√1-0.65 = 0.11

(µ + r) (2600p₁120p₂) + 72 ω = 6000p₁

(µ + r) (2200p₁ + 480p₂) + 48 ω =750p₂

NNP (3) = 1200p₁ + 150p₂

NNP (3) = 0.48SNP (5) = 0.48

R_{3 =} 1200p₁ + 150p₂ = 0.25

4800p₁ + 600p₂

y′₃ = NNP (3) = 0.48 = 0.004

L(3) 120

KP(3) = NNP(3) = 0.48 = 1.92

R₃ 0.25

KP (3) = 1.92 = 0.016

L(3) 120

I₃ = 1.92r + 0.016 x 0.25 = 1.92r + .004

S₃ = 1.92r + 0.01 x 120 ω = 1.92r + 1.2 ω

When I_{3 =} S_3, we have

1.2 ω = 0.004

or ω = 0.0033

.

. . ω = 0.0033 = 0.66

0.005

we have r = 0.25 ⁿ√1-0.66 r = 0.30 (1-0.66)

Solving for n we have n = 1.2 and 1 = 0.83 n

With the value of n obtained we have

r = 0.25 ^1.2√1- ω

. . (1- ω)^0.17 – 0.83 = 0

By binomial expansion and solution of the resulting equation we have only one admissible value:

we have

r = 0.07

The two curves, thus, meet at

The equilibrium rate of profit is

r = 0.25 (1-0.66)^0.83 = 0.10

(µ + r) (3200 p_{1 +} 200 p₂) + 90 ω = 7140 p₁

(µ + r) (2400 p₁ + 600 p₂) + 70 ω =1020 p₂

NNP (4) = 1540 p₁ + 220 p₂

NNP (4) = 0.67 SNP (5) = 0.67

R₄ = 1540 p₁ + 220 p₂ = 0.275

5600 p₁ + 800 p₂

y′₄ = 0.67 = 0.00418

160

KP(4) = NNP(4) = 0.67 = 2.4

R₄ 0.275

KP (4) = 2.4 = 0.015

L(4) 160

I₄ = 2.4 r + 0.015 x 0.275 = 2.4 r + 0.004125

S₄ = 2.4 r + 0.01 x 160 ω = 2.4 r + 1.6 ω

When I_{4 =} S_4, we have

1.6 ω = 0.004125

or ω = 0.002578

.

. . ω = 0.002578 = 0.52

0.005

r = 0.275 ⁿ√1-0.52

r = 0.30 (1- 0.52)

Solving we have

n = 1.11 and 1 = 0.9

n

With the value of n obtained we have

r = 0.275 ^1.11√1- ω

r = 0.30 (1- ω).

. . (1- ω)^0.1 – 0.92 = 0

By binomial expansion and solution of the resulting equation we have only one admissible value:

r = 0.28

The result shows that there is no meeting point between the two curves within the range of the maximum value of r, that is, 0.275.

The equilibrium rate of profit is

r = 0.275 (1-0.52)^0.9 = 0.15

(µ + r) (140p₁+ 4p₂ + 6p₃ + 4p₄ + 3p₅) + 60 ω = 715p₁

(µ + r) (90p₁+ 12p₂ + 12p₃ +8p₄ + 9p₅) + 40 ω = 104p₂

(µ + r) (120p₁+ 6p₂ + 12p₃ +12p₄ + 8p₅) + 50 ω = 117p₃

(µ + r) (105p₁+ 36p₂ + 12p₄ + 18p₅) + 30 ω = 17p₄

(µ + r) (95p₁+ 22p₂ + 24p₃ +14p₄ + 12p₅) + 20 ω = 17p₅

GNP = 715 p₁ + 104 p₂ + 117p₃ + 65p₄ + 65p₅

(The machines are 50 each in use, their depreciation at 0.04 amounts to 2 machines so that the 50 depreciated machines are equivalent to 48 new ones and these added to 17 newly produced machines make a total of 65 in the GNP.)

NNP (5) = SNP(5) = 165 p₁ + 24 p₂ + 27p₃ + 15p₄ + 15p₅ = 1

R_{5 =} 165p₁ + 24 p₂ + 27p₃ + 15p₄ + 15p₅ = 0.30

550p₁ + 80p₂ + 90p₃ + 50p₄ + 50p₅

y′₅ = NNP (5) = 1 = 0.005

L(5) 200

KP₅ = NNP(5) = 1 = 3.3

R₅ 0.30

KP (5) = 3.3 = 0.0165

L(5) 200

I₅ = 3.3r + 0.0165 x 0.3 = 3.3r + 0.00495

S₅ = 3.3r + 0.01 x 200 ω = 3.3 r + 2 ω

When I_{5 =} S_5, we have

2 ω = 0.00495

or ω = 0.002475

and ω = 0.002475 = 0.495

0.005

n = 1

Therefore r_{5 =} 0.3 (1-0.495) = 0.1515 or 0.15

(µ + r) (140p₁+ 2p₂ + 4p₃) + 60 ω = 714p₁

(µ + r) (50p₂+ 40p₃) + 50 ω = 112p₂

(µ + r) (30p₂ + 50p₃) + 40 ω = 126p₃

(µ + r) (10p₂+ 8p₃ + 25p₄) + 30 ω = 16p₄

(µ + r) (8p₂ + 8p₃ +25p₅) + 20 ω = 16p₅

NNP of the basic goods sector = SNP (6) = 32 p₂ + 36p₃

NNP (6) = 0.99 SNP (5) = 0.99

R_{6 =} 32 p₂ + 36p₃ = 0.40

80p₂ + 90p₃

y′₆ = NNP (6) = 0.99 = 0.00495

L(6) 200

KP(6) = 2.11 SNP(5) = 2.11

KP (6) = 2.11 = 0.01055

L(6) 200

I₆ = 2.11r + 0.01055 x 0.4 = 2.11r + 0.00422

S₆ = 2.11r + 0.01 x 200 ω = 2.11r + 2 ω

When I_{6 =} S_6, we have

2 ω = 0.00422

or ω = 0.00211

and ω = 0.00211 = 0.42

0.00495

we have

r = 0.30 (1-0.42)

Solving we have

n = 0.65 and 1 = 1.54 n

With the value of n obtained we have

r = 0.40 ^0.65√1- ω

. . (1- ω)^0.54 – 0.75 = 0

By binomial expansion and solution of the resulting equation we have only one admissible value:

r = 0.18

The equilibrium rate of profit is

r = 0.40^0.65√1-0.42 = 0.17

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