ChE 356/384 Exam #1 – Part A (in-class)

(due 2/26/96)

1. Develop a derivation that shows the optimal value of mA in Example 1.6 is in fact

(20) the average of the data for mB minus the average of data for mC. Assume three data points: (mB1, mB2, mB3) and (mC1, mC2, mC3), and use a general objective function similar to (b) in Example 1.6.

The example states that the model is given by

We develop the objective function as the sum of squared errors between the model value, , and the variable to be optimized, . This gives

The only constraint is .

To find the optimum, we take the derivative of the objective function with respect to , and set it equal to zero

Since we know that the averages of the data are given by

We have proven that

2. In the inventory problem described in Example 1.7, the carrying cost assumes

(20) that the inventory has a maximum value of D, then declines to zero over a period of time, when the next run is carried out (sawtooth pattern). So in effect the average inventory determines the carrying cost. Suppose that you require a minimum inventory level,, so that you have a “cushion” of product when the next run is made. In this case, the inventory cost would be How does this modified assumption change the value of Dopt (cf. eqn (f) on p. 23)?

The modified total cost function is taken from the above change to the inventory cost and the cost per run given in eqn (a) on p.21

Taking the first derivative with respect to D and setting it equal to zero gives

We can see that there is no change to the optimal number of units per run, as compared with the original optimal point in eqn (f).

3. A trucking company has borrowed $600,000 for new equipment and is

(20) contemplating three kinds of trucks. Truck A costs $10,000, truck B $20,000, and truck C $23,000. How many trucks of each kind should be ordered to obtain the greatest capacity in ton-miles per day based on the following data?

Truck A requires one driver per day and produces 2100 ton-miles per day.

Truck B requires two drivers per day and produces 3600 ton-miles per day.

Truck C requires two drivers per day and produces 3780 ton-miles per day.

There is a limit of 30 trucks and 145 drivers.

Formulate a complete mathematical statement of the problem, and label each individual part, identifying the objective function and constraints (including non-negativity) with the correct units ($, days, etc.). Make a list of the variables by names and symbol plus units, and identify which ones are integer vs. continuous variables. Do not solve.

Let

number of trucks of type A no units, integer variable

number of trucks of type B no units, integer variable

number of trucks of type C no units, integer variable

Objective function

(ton-miles/day)

Constraints

1. ($)

2. (drivers)

3. (trucks)

4. (physical requirement)


ChE 356/384 Exam #1 – Part B (take-home)

(due 2/28/06)

4. Figure 2.4 shows that the Nusselt number (y) can be correlated with Reynolds

(20) number (x) on a log-log plot such that y = axb.

(a) fit a and b to at least 10 points using nonlinear regression (Excel).

(b) take the log of both sides of y = axb and fit a and b using linear regression. Compare the fitted values for the two approaches. Are the results different? Why or why not?

Part (a)

The data points I used in my regression are given below

Re / Nu
0.2 / 0.55
1.0 / 0.90
2.0 / 1.10
20 / 2.90
100 / 5
400 / 10
2000 / 23
6000 / 40
20000 / 80
200000 / 400

Using nonlinear regression to minimize the sum of squared error between the model prediction and the data I get the following results:

The model fit looks like this on a regular plot

And like this on a log/log plot

Part (b)

Transforming the data and using linear regression gives the following results

The model fit looks like this on a log-log plot

And like this on a regular plot

You can see that the nonlinear regression fits the points at higher Re and Nu, while the linear regression fits the points at low Re and Nu better. This is due to the fact that the logarithm makes large errors seem smaller (when you take the log of a large number, it is less than the original number). Linear regression on the log transformed model weights points at low Re and Nu more than points at high Re and Nu. While it appears to be a better fit on the log-log plot, the large error at high Re and Nu is apparent on the regular plot. The error apparent on the log-log plot for the nonlinearly regressed model is actually not very large when you view it on the regular plot.

5. You are the manufacturer of PCl3, which you sell in barrels at a rate of P barrels

(20) per day. The cost per barrel produced is

C = 50 + 0.1P + 9000/P in dollars/barrel

For example, for P = 100 barrels/day, C = $150/barrel. The selling price per barrel is $300. Determine for a basis of one day.

(a) The production level giving the minimum cost per barrel.

(b) The production level which maximizes the profit per day.

(c) The production level at zero profit.

(d) Why are the answers in (a) and (b) different?

(e) Give a possible explanation for the shape of the C vs. P curve.

Part (a)

The cost per barrel is given by . Taking the first derivative with respect to P and setting it equal to zero gives

Part (b)

The profit per barrel is given by . To convert this into profit per day, we multiply by the production rate giving

We take the first derivative with respect to P and set it equal to zero

Part (c)

The production level at zero profit (per day) is given by

We can see here that there are two production levels that will result in zero profit per day.

Part (d)

The answers in parts (a) and (b) are different because when you minimize cost, you fail to take into account the fact that additional barrels sold can offset the cost of producing them. When we account for this fact mathematically, we see that the optimum production level is higher.

Part (e)

The C vs. P curve is shown below

The behavior is partly explained by economies of scale. We can see that at very low production rates, per barrel costs are relatively high. Capital costs such as cost of equipment dominate here. You have some initial costs (like equipment and labor) that remain the same regardless of your production level, so if you produce little, the cost per barrel is high. We reach a minimum at 300 bbl/day, after which costs begin to increase again. This is likely due to costs associated with additional labor or equipment necessary to deal with a higher throughput. It is also likely that the efficiency of the process drops when production is increased, resulting in a higher per barrel cost.