ME 403N

PRODUCTION, PLANNING and CONTROL

 

 

Lecture notes for:

EOQ model with shortages permitted

 

Date : 27^th Sept 2002

 

 

Considering the Economic Order Policy (EOQ) model, when the shortages are permitted.

 

Assumptions:

Ø      Demand is known with certainty and is fixed at λ units / time.

Ø      Lead time for delivery is instantaneous.

Ø      No time discounting of money is there.

Ø      Costs involved are,       k = Ordering cost (fixed cost / order)

                                                h = Inventory holding cost / unit held / time

                                                p = Back ordering cost / unit / time

                                                C = purchase cost / unit

 

(Note that,

Ø      Advantages of allowing shortages :

·        Cycle time increases and hence, per unit ordering cost decreases.

·        Inventory reduces.

Ø      Disadvantages of allowing shortages :

·        One looses on customer satisfaction and hence

·        Future sales decrease )

 

 

Plot of inventory at any time Vs time:

 

    I(t)

 

 

           

  Q       S

 

                                                                                                                                        time

 

 

                        Q / λ

 

Where,             X axis: t = Time

Y axis: I (t) = Inventory at time, t

 

Here,                Q = Quantity ordered

                        S = Amount of inventory left after back logging

                        λ = Deterministic constant demand

 

In such kind of model, one needs to decide:

1. How much to order (Q)? and
2. When to order? (Earlier this decision was fixed, because the demand was fixed and no backlogging/shortages were allowed.)

This decision will in turn decide the amount of “S”, left after back logging.

 

S and Q are independent of each other and S can take any value, independent of value of Q.

 

 

Cost / Cycle     = (Purchase cost / Cycle) + (Fixed cost / order) + (Inventory holding cost)

   + (Back ordering cost)

Cost / Cycle     = {C * Q} + k + {(1/2 * S * S / λ) * h} + {p * (Q – S)² / (2 * λ)}      ….. (i)

Time / Cycle     = Q / λ                                                                                                 ….. (ii)

Cost / time        = (Cost / Cycle) / (Time / Cycle)

	Cost / time        = λ * C + k * λ / Q + 0.5 * h * S2 / Q + 0.5 * p * (Q – S)2 / Q

 

 

 

The Optimal point: Minimum cost / time

Differentiating the above equation, w.r.t S & Q and solving, we get the Optimal point as:

 

	Optimal S:                    S* = (2 * k * λ / h)1/2 * (p / (p + h) )1/2   Optimal Q:                   Q* = (2 * k * λ / h)1/2 * ((p + h) / p )1/2   Optimal time:                t* = Q* / λ = (2 * k / (λ * h))1/2 * ((p + h) / p )1/2

 

 

 

 

 

 

 

As we allow back ordering,

• S decreases w.r.t Q and
• T (cycle time) increases.

 

 

Example: The company that has to store 3 items, has a warehouse of, with total space of warehouse = L. How much quantity of each item should be stored, so as to minimize the cost associated?

 

Solution: Here, one additional constraint has been added                       Total space = L

 

Assuming that no shortages are permitted,

Cost / time = Σ (( h_i * Q_i / 2) + (k_i * λ_i / Q_i ))                                       ….. (i)

{Summing over i = 1 to3}

 

One can not order more than L, so        Q1 + Q2 + Q3 ≤ L                                         ….. (ii)

 

Method 1:

• Differentiate equation (i) and find out the corresponding Q1^*, Q2^*, Q3^*
• See if Q1^*, Q2^* and Q3^* satisfy equation (ii) or not
• If YES              Q1^*, Q2^* and Q3^* are the answers

If NO               Do additional mathematics to get Q1^*, Q2^* and Q3^* that satisfy

equation (ii).

 

Method 2:

• Introduce a new variable “Lagrangine variable (λ)”, such that:

 

Σ (2 * k_i * λ_i / (h_i + 2 * λ))^1/2 = L

 

ü      If λ = 0, violation of constraints

ü      So increase the value of λ, such that the constraint is exactly followed.

	Qi* = (2 * ki * λi / (hi + 2 * λ))1/2

 

                                                                                                            Answer.

 

This is a non-linear programming problem, as the 2^nd term in the equation (i) is non linear. But as this is a convex-bowl shaped problem, one is able to solve it.

	cost

 

 

 

 

 

 

                                                            Optimal point

 

 

 

	Quantity

 

 

                           Q <= L

	One has to order in this region

 

Wagnen – Whitin model:       Model with deterministic time varying demand.

 

The demand is no longer constant, it changes with time, but it is still deterministic.

 

Here

ü      (r1, r2, r3, r4 ………. rn) be the known requirements for a single product over “n” periods.

ü      (y1, y2, y3, y4 ………. yn) be the amounts produced in the respective years.

ü      Lead times are deterministic

Hence the difference between total production and consumption gives the Inventory level at that time.

 

Assumptions:

Ø      Shortages are not permitted

Ø      Starting Inventory = 0

Ø      Linear cost of holding are present

Holding costs                  h_t (x_t) 

Where, x_t is inventory at any time t.

Ø      Fixed costs are present

Fixed cost / unit              c_t (y_t)

            Where, y_t is amount produced at any time t.

 

 

So the objective is to minimize cost

Cost = Σ (c_t (y_t) + h_t (x_t))          {Summing over t = 1 to n}

 

Such that, inventory level:                     

·        x_t = Σ (y_j – r_j)                     {Summing over j = 1 to t}

·        x_t => 0                               t = 1,2,3……..n

·        x₀ = 0                                 {assumed}

 

Taking h_t (x_t) to be linear (concave)

                        h_t (x_t) = h_t * x_t

 

Taking c_t (y_t) to be concave

                        c_t (y_t)    = k + c_t * y_t                              if y_t  > 0

                                    = 0                                           if y_t = 0

 

Result: An optimal ordering policy is developed that has a property

 

	yt  * xt-1 = 0

 

                                                                                    For t = 1,2,3,……….n

 

Note that, if inventory   0 < x_t  < r_t+1

Then one can have a better solution by having

·        x_t  = 0

i.e. producing more during early stage, so as to get inventory at end of t = 0

·        x_t  = r_t+1 

i.e. producing less during early stage, so as to leave inventory equal to demand in next period.

 

	yt  = 0, rt , rt + rt+1, rt  + rt+1 + rt+2 , ……………………., rt  + rt+1 +…………….. + rt+1

 

 

 

Or in other words we have to find the shortest distance between different points. This shortest distance between two points will be the feasible solution.

 

(to be continued in next lecture……..)