Economic Order Calculation and Interpretation

a)SITUATIONS WHERE THE ECONOMIC ORDER QUANTITY MAY NOT APPLY

        I.            If holding costs are unknown and keep changing.

     II.            If ordering costs are unknown and keep changing.

   III.            If the supply lead time is unknown and keep changing.

  IV.            If price and cost per unit keep changing.

     V.            If the annual demand is uncertain, keeps changing and not continuous over time.

b)

i)                     L_q = λ²/µ x µ-λ

Average time = 3² /((5 x (5 – 2))

                     = 0.6 of a min or 36seconds

ii)                   L = L_{q  +}  λ/µ

=0.6 + 2/5

= 1 client

                                                   iii)

                                                          Where     x = No. of successes

                                                           ⋋ = mean no. of the successes in the sample (⋋ = np)

                                     e = 2.718

                                P(3) =  2.718^-33³/3!

                                        =0.075

c)

PRODUCT			Maximum available
(PER MONTH)	A                       B	C
Machine hours	2                       3	1	400
Special component	1                      x	1	150
Special alloy	2                y	4	200
Contribution	800                    500	1000

P = 800A + 500B + 1000C

Subject to ;

Machine hours 2A + 3B + 4C ≤ 400hrs

B < 50

A + C < 150,      B + C + Bx ≤ 150

2A + 4C < 200,   2A + 4C + By ≤ 200

A,B,C > 0

ii)Interpretation

	x	y	z	S1	S2	S3	S4	b
S1	-0.5	0	0	1	-1	-0.75	0	100
z	1	0	1	1	2	0	0	150
y	0.5	1	0	0	0	0.25	0	50
x	-0.5	0	0	0	0	-0.25	1	0
	450	0	0	0	1000	125	0	175000

ü  The solution is optimal as no negative value exist in the bottom row i.e. they are greater than or equal to zero.

ü  The optimal value, basic and non-basic variables are as follows;

·         X = 0 ( non-basic variable)

·         Y = 50 (basic variable )

·         Z = 150 (basic variable)

·         S1 = 0 (non-basic variable)

·         S2 = 0 (non-basic variable)

·         S3 = 0 (non-basic variable)

·         S4 = 0 ( basic variable)

ü   The maximum optimal  value is 0 and found at ( 0,50,150) of the objective function.