PRODUCTION MANAGEMENT II (PRODUCTION LOT SIZE MODEL)

ILLUSTRATION

Suppose a firm has an annual demand for goods produce at 6,500 units.  The set up cost for each production run is assumed to be N200.  Inventory carrying cost per unit per year is N3.20k, the production rate is 12,500 units per year.

Assume 250 operating days per year:

i.             Calculate the optimal production lot size

ii.            The length in days between the start of each production

iii.           The optimal numbers of production run 

SOLUTION:

 

First of all, let’s take a look at the formula:

(1)        Q^opt  = √2COD

              Cc(1-d)

                       r

(2)        N^opt= D

         Q

(3)        T^opt = No of days

               N^opt

Above are the formulae:  Please, cram them.

Now before we solve, let’s bring out the variables:

D=6,500 Set up cost/Co = 200, Cc = 3.20, Production Rate (r) = 12,500, No of days = 250.

Once you have brought out the variables like this, you are done. All you need do is to substitute.

Solution to i:

Q^opt  = √2COD

 Cc(1-d)

          r

Q^opt  = √(2) (200) (6,500)

 (3.20) (1-6,500)

                12,500

Above, we just substitute figures. That (2) is constant. Then 200 we picked it from the variables. Likewise other figures.  It is only (2) and the 1 below that is constant.

Q^opt  = √2,600,000

         (3.20) (0.48)

Q^opt  = √2,600,000

               1.536

                

Q^opt  = √1,692,708.33

Q^opt  = 130/units

Solution to ii.

N^opt= D

           Q

N^opt= 6,5000

             130

N^opt= 5 times

This one is easy.  D is demand for goods while Q is the quantity. All you need do is to substitute the figure of demand and quantity, then divide.

Lastly, Solution to iii:

T^opt = No of days

N^opt

T^opt = 250

              5

T^opt = 50 days

Again, this last one is easy.  No of days in the question is 250, while N^opt, is 5 times which we have solved for before this very last question.

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