Problem 1

C = 1.5 $/pound

D = 25000 pounds

C_o = $80

C_c = 0.20 $/pounds

L = 20 days

W = 365 days

Discount rules:

Q < 10000 pounds – 0 % discount – C₁ = C = $1.5

Q ≥ 10000 pounds – 3 % discount – C₂ = 1.5 * (1 + 0.03) = $1.455

Q ≥ 20000 pounds – 5 % discount – C₃ = 1.5 * (1 + 0.05) = $1.425

Q_opt?

ROP?

K = (2 * C_o)/ C_c = 2*80/0.2 = 800 pounds

K < D, –> Q^* = = 4472 pounds

For z₁ apply Q = Q^* because 0 < Q^* < 10000

z₁ = C₁ *D + (C_c * Q)/2 + (C_o * D)/ Q = 1.5*25000 + (0.2*4472)/2 + (80*25000)/4472 = 38394.43

For z₂ apply Q = 10000 because 10000 is minimal border of order

z₂ = C₂ *D + (C_c * Q)/2 + (C_o * D)/ Q = 1.455*25000 + (0.2*10000)/2 + (80*25000)/10000 = 37575

For z₂ apply Q = 20000 because 20000 is minimal border of order

z₃ = C₃ *D + (C_c * Q)/2 + (C_o * D)/ Q = 1.425*25000 + (0.2*20000)/2 + (80*25000)/20000 = 37725

We achieve minimal costs in 3 % discount case.

Q_opt = 10000 pounds

ROP = Q_opt *(L/T) = Q_opt *L*D/(Q_opt *365) = (L*D)/365 = 1370 pounds

Problem 2

D = 15000 kits of tools

C = 55 $/ kit of tools

P = 60000 kits of tools

C_s = $150

C_c = $20

d = D/365 = 41.1 k/day

p = P/365 = 164.38 k/day

K = (2 * C_o)/ C_c = 2*150/20 = 15 kits of tools

Q_opt = 547.72

Z_opt = 8616

N =27

T_opt = 13.33

Month

Q = 15000/12=1250

M = Q*(P – D)/P = 1250*(60000 – 15000)/60000 = 938

Z = (C_c*M)/2 + (C_s*D)/Q = (20*938)/2 + (150*15000)/1250 = 11180

Quarter

Q = 15000/4=3750

M = Q*(P – D)/P = 3750*(60000 – 15000)/60000 = 2813

Z = (C_c*M)/2 + (C_s*D)/Q = (20*2813)/2 + (150*15000)/ 3750= 28730

Half a year

Q = 15000/2=7500

M = Q*(P – D)/P = 7500*(60000 – 15000)/60000 = 5625

Z = (C_c*M)/2 + (C_s*D)/Q = (20*5625)/2 + (150*15000)/ 7500= 56550

Loses month = 2959

Z_mon = CM/2 + C_sD/Q + price*D/12 = 79925

Problem 3

C₁ = $175

C₂ = $300

C₃ = $ 43.75

Q → Q + 1 D ≤ Q P(Q) ∆Z₊ = –(C₁ – C₃)

D > Q (1 – P(Q)) ∆Z₊ = C₂ – C₁

= P(Q)*(C₁ – C₃) + (C₂ – C₁)*(1 – P(Q)) ≥ 0

Q → Q – 1 D ≤ Q – 1 P(Q – 1) ∆Z_– = C₁ – C₃

D > Q – 1 (1 – P(Q – 1)) ∆Z_– = C₁ – C₂

= P(Q –1)*(C₁ – C₃) + (C₁ – C₂)*(1 – P(Q – 1)) ≥ 0

P(Q^* –1) ≤ (C₂ – C₁)/ (C₂ – C₃) ≤ P(Q^*)

P(Q^* –1) ≤ 0.48 ≤ P(Q^*)

x	10 – 40	50 – 74	75 – 90	91 – 100
p(x)	0.3	0.4	0.2	0.1
P(x)	0.3	0.7	0.9	1

0.48

Следует заказать от 50 до 74 кресел.

= 80

= 20

P {(Q* – )/} = 0.48

так как 0.48 < 0.5, берем

(Q* – )/ = –0.05

Q* = + (–0.05) = 79