Eoq Solution

Q.1a) The following graph is EOQ model with planned shortages.

Let the parameters from the basic EOQ model.

d = constant demand rate
K = setup cost for placing one order
Q = order quantity h = inventory holding cost per unit of product per unit of time p = shortage cost per unit of product per unit of time
S = inventory level just after an order of size Q arrives

So, Q– S = Shortage in inventory just before an order of Q units is added Production or ordering cost per cycle = K+cQ , where cQ is the unit cost During each cycle, the inventory level is positive for a time: = Sd

The average inventory level during this time is: = ( S + 0)2= S2 units

The corresponding cost is = hS2per unit time

Therefore, the Holding cost per cycle= hS2 × Sd=hS22d

During each cycle, the shortage occurs for a time: = (Q – S)d

The average amount of Shortages during this time is: = ( 0 + Q - S)2= (Q - S)2 units

The corresponding cost is = p (Q-S)2per unit time

Therefore, the Shortage cost per cycle = p (Q-S)2 × (Q-S)d=p(Q-S)22d
Q.1a) Hence, Total cost per cycle = K+cQ+ hS22d+ p(Q-S)22d Total Cost per unit of time:

T= K+cQ+ hS22d+ pQ-S22dQd=dKQ+dQ+ hS22Q+ p(Q-S)22Q

Q.1b) By setting the partial derivatives∂T∂S=0and ∂T∂Q=0

So that, ∂T∂S= 0+ 2hS2Q- p [2 Q – S (-1)]2Q =0

∂T∂S= hSQ- p (Q - S)Q =0

hS= p (Q - S)

S= pQh+p ------- (1)

Also, ∂T∂Q= - 2dK2Q2 – 2hS4Q2+ ( 2p Q – S2Q- 2p Q – S24Q2) =0

∂T∂Q= - dKQ2 – hS2Q2+ p Q – SQ- p Q – S22Q2 =0

dKQ2 + hS2Q2 =p Q – SQ- p Q – S22Q2

2dkQ2 +hS =2Qp Q – S- p Q – S2

Q2=2dKp+ h+ppS2 ------- (2)

Putting (1) into (2): Q*2=2dKp+ h+pp(pQh+p)2

Q*2=2dKh ∙ p+hp Q*= 2dKh∙ p + hp ------- (3)
Q.1b) Putting (3) into (1): S*= ph+p ( 2dKh∙ p + hp)

S*= 2dKh∙ pp+h

Q.1c) With refer to Part (Q.1a):

The Holding cost per unit of time = hS22d/ ( Qd)

The Shortage cost per unit of time