Question

A jewelry firm buys semiprecious stones to make bracelets and rings. The supplier quotes a price of $8.00 per stone for quantities of 600 stones or more, $8.30 per stone for orders of 400 to 599 stones, and $10 per stone for lesser quantities. The jewelry firm operates 175 days per year. Usage rate is 25 stones per day, and ordering costs are $48. a. If carrying costs are $2 per year for each stone, find the order quantity that will minimize total annual cost. (Round your intermediate calculations and final answer to the nearest whole number.) Order quantity stones b. If annual carrying costs are 30 percent of unit cost, what is the optimal order size? (Round your intermediate calculations and final answer to the nearest whole number.) Optimal order size stones c. If lead time is 10 working days, at what point should the company reorder? Reorder quantity stones

Answer 1

This is a problem where the economic order quantity (Q) needs to be calculated. Let's start from the basics:

Let P be the unit cost, and D be the annual demand quantity.

The purchase cost will be P*D

Suppose we order Q items. Then, the number of orders we would place would be D/Q per year. If the fixed cost per order is K, then the total ordering cost could be (D*K)/Q

The average inventory can be assumed to be Q/2, as the inventory varies between Q and 0. Suppose the annual holding cost per unit is h. Then, the holding, or inventory cost would be Q*h/2.

Summing up, we get

Total Cost (TC) = P*D + (D*K)/Q + (Q*h)/2. (Formula 1)

If the function TC were differentiable, then one could take the derivative of the function, set it to zero, and then compute the optimal Q (Q*).

Let's do that -

Total Cost (TC) = P*D + (D*K)/Q + (Q*h)/2.

Differentiation w.r.t. Q, we get

0 = (-D*K)/(Q^2) +h/2=0

or, (D*K)/(Q^2) = h/2

Q^2 = (2*D*K)/h

Q=sqrt((2*D*K)/h) (Formula 2)

This is a neat little formula. Unfortunately, in our case, the function is not smooth, because the value of P changes with the value of Q. So how do we work on this problem?

Let us go back one step, and examine the derivative, which describes that rate of change of the total cost function.

The derivative is (-D*K)/(Q^2) +h/2.

In our problem, D (annual demand) = 175 * 25 = 4,375

K = 48 (given)

h=2 (given)

Thus, our expression becomes (-4375 * 28) / (Q^2) +2/2

= -210000/Q^2 + 1

This function is a continually increasing function, when Q is positive. As Q increases, the negative expression falls, and the expression increases.

At Q=0, the derivative is -?.

At Q=400, the derivative is -210000/160000 + 1 = -0.3125

Thus, between 0 and 400, the derivative has stayed negative, and the total cost function (TC) has fallen. Thus, the minimum in the interval [0,400) will be at Q=399. (Lemma 1)

At Q=600, the derivative is -210000/360000 + 1 = 0.4167.

Thus, between 400 and 600, the derivative has changed from negative to positive. Thus, there is a local minimum within this interval. The minimum in this interval is obtained by setting the derivative (-D*K)/(Q^2) +h/2 to zero , so we are back to the old formula 2 of Q=sqrt((2*D*K)/h) .

For the interval where Q is greater than 600, i.e. Q ? [600,   ?), the minimum will be at Q=600, as for Q> 600, the value will be higher due to to the positive derivative. (Lemma 2)

By plugging in the values, we get the minimum value = sqrt(2 * 4375 * 48 / 2) = sqrt (210000) = 458.26.

As the order quantity cannot be fractional, the value must be either 458 or 459.

Let us compute the total cost, using formula (1), for Q = 458:

P*D + (D*K)/Q + (Q*h)/2.

Remember that P = 8.30 for this range.

= 8.30 * 4375 + (4375 * 48 ) / 458 + 458 * 2 / 2

= 37229.015

For Q = 459, we get

8.30 * 4375 + (4375 * 48 ) / 459 + 459 * 2 / 2

TC = 37229.016

The value for Q = 458 is slightly lower.

But what about the other intervals?

For the interval where Q is between 0 to 399, the minimum cost will be

P*D + (D*K)/Q + (Q*h)/2

Remember that P=10, when Q is 10. And we already showed in Lemma 1, the minimum will occur when Q = 399.

So, TC = 10 * 4375 + (4375 * 48 ) / 399 + 399 * 2 / 2

= 44675.32.

This is higher than the value for Q = 458 (Actually, it was possible to eliminate this computation by arguing that the function at Q=399 is much greater than at Q=400, because the value of P is much larger, for Q<400.

What about Q=600? (Remember that we get the minimum value for total cost at Q=600, for Q>=600).

TC = P*D + (D*K)/Q + (Q*h)/2

= 8.00 * 4375 + (4375 * 48 ) /600 + (600* 2 )/2

=   35950.0

Aha! The value for Q=600 is lower than the value for Q=458. This happened because the P dropped sharply from 8.30 to 8.00.

Thus, the solution for (a) part would be Q ( number of stones) =600

(Notice that directly using the formula (2) would have produced an incorrect result. )

Part (b)

Part (b)

Here, the annual carrying cost (h) is no longer fixed, but is a function of P, that is, 30% of P, or 0.3 P.

The TC function becomes

P*D + (D*K)/Q + (Q* P * 0.30)/2

= P*D + (D*K)/Q + 0.15 * Q* P

The derivative becomes

(-D*K)/(Q^2) +.15 * P

We know that D= 4,375, and K is 48, so the derivative is

-210000 / (Q ^ 2) + 0.15 * p.

=

Again, for the intervals (0,400), [400,600), and [600, ?), we calculate the behaviour of derivatives, piece-wise.

At Q=1, P=10, so the derivative is -210000 / (1 ^ 2) + 0.15 * 10, which is very negative.

At Q=399, the derivative is -210000 / (399 ^ 2) + 0.15 * 10 = 2.8191, which is positive.

Hence, a local minima is present in the interval.

At Q=400, the derivative is -0.0675

At Q=599, the derivative is 0.6597

Thus, there is a local minima present here too.

At Q=600, the derivative is   -210000 / (600 ^ 2) + 0.15 * 8.0

= 0.61667

Since the expression for the derivative is an increasing function, the derivative will stay positive, and the TC will increase. Thus, the minimum for this interval is at Q=600.

At Q=600, the TC is P*D + (D*K)/Q + 0.15 * Q* P, where P = 8.0

We get TC=36,070.

Compute The minimum for the interval (0,400).

The derivative is -210000 / (Q ^ 2) + 0.15 * p.

Setting this to zero gives

Q= sqrt( 210000/(.15 * p) )

= sqrt( 210000/(.15 * 10 ))

Q= 374.17

We compute the TC for Q=374, and 375.

TC for Q = 374 is

P*D + (D*K)/Q + 0.15 * Q* P

= 10.0 * 4375 + (4375 * 48 ) / 374 + 0.15 * 374 * 10.0

= 44872.497

For Q = 375, the value is

10.0 * 4375 + (4375 * 48 ) / 375 + 0.15 * 375 * 10.0

44872.500

This is higher than for Q=600.

What about the interval [400,599]?

We have a minimum somewhere in this interval. The derivative is

(-D*K)/(Q^2) +.15 * P, where P=8.3

We set it to zero.

(-D*K)/(Q^2) +.15 * 8.3=0,

(D*K)/(Q^2)=1.2450

Q^2 = D*K/1.2450 = 210000/1.2450

Q = 410.70

We check the TC for Q=410, and Q=411.

For both values, we get

TC = P*D + (D*K)/Q + 0.15 * Q* P, with P=8.3

TC = 37335 (approx)

But for Q=600, the total cost is 36,070, which is the lowest of all. So that is the answer.

Part (c)

This is the simplest part of the question.

The reorder point is consumption during lead time + safety stock.

As no safety stock has been mentioned, we can assume it to be zero.

Consumption during lead time = 25 * 10

= 250

The company should order when it is down to 250 stones.

Here's some octave/matlab code which solves the question, as well as helps to visualize the behaviour of the function

function getp=getp(Q)

getp=zeros(1,size(Q)(2));

for i = 1:size(Q)(2)

if (Q(i) <=0)

error ("Q must be positive!");

elseif( Q(i) < 400)

getp(i)=10.0;

elseif Q(i) < 600

getp(i)=8.3;

else

getp(i)=8.0;

endif

end

endfunction

function tc1=tc1(D, K, Q, h)

P=getp(Q);   

tc1= (P .* D ) + (D.*K)./Q + (h.*Q ) / 2;

endfunction

function tc2=tc2(D, K, Q)

P=getp(Q);   

tc2= (P .* D ) + (D.*K)./Q + (0.3*P.*Q ) / 2;

endfunction

D=175* 25;

K = 48;

h = 2;

x=[1]

func1=@(Q)tc1(D,K,Q,h);

func2=@(Q)tc2(D,K,Q);

q1=fminbnd(func1, 1,10000)

tc=func1(q1)

q2=fminbnd(func2, 1,10000)

tc=func2(q2)

x=1:1000;

plot(x,func1(x),";tc;.b");

xlabel("Order Quantity");

ylabel("Total Cost");

axis([0 1000 35000 50000]);

pause();

x=1:1000;

plot(x,func2(x),";tc;.r");

xlabel("Order Quantity");

ylabel("Total Cost");

axis([0 1000 35000 50000]);

pause();

The graphs are:

Part(a):

Part(b)