Written Assignment on Economics

Date:  2021-03-05 15:26:35
4 pages  (1157 words)
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Question 3

Ed notices that your plan does not call for any Hereshoffs to be produced. He then says that he has promised 2 of these to special customers. What can you tell Ed about the cost of his promise?

Use the solver. The restrictions may be added that the required number of Hereshoffs is 2. As you know, Hereshoffs cost $ 200.

But because of this Ed will have to produce a smaller number of Large Plows: 90-83=7 not manufactured Large Plows

And Ed will produce more Danforths: 579-570=9 additional pieces

Respectively, Ed will get less profit, as there will not be optimum use of resources: $73 200-$73140 = $60 loss

The strong demand for Danforths has led Ed to think of raising the price, which would result in a $6 increase per anchor in contribution. Ed has evidence that such a small increase in price will not affect the demand for Danforths. What will be the effect on the optimal solution (i.e., the production plan) of such a change?

Name Maximum demand Start Demand with 2 Hereshoffs Danforth 800 570 579

Small Plow 300 300 300

Large Plow 200 90 83

Herreshoff 100 0 2

We use the Solver. We have changed the price of Danforths 50 to 56 dollars.

The optimal plan changes. As we see revenue increase, while the number of anchors will be made the same: 76614-73140=$3474 income

As a result of price increases, the demand doesnt change, but this increase Eds revenues.

Ed has the chance to hire some part-time welding support from another business, but it is expensive. The "standard" welding hour charge which he used to compute the contribution margins was $25/hr, but he will have to pay $75/hr to buy additional welding support. How many hours should Ed buy at this price? (Note that the units of measurement in the solution are minutes that must be converted to hours to complete this part of the question.)

Weld/ Assy: 4D+6SP+8LP+10H<=4800

Name Weld time (min) Demand Need time (min)

Danforth 4 579 4*579=2316

Small Plow 6 300 6*300=1800

Large Plow 8 83 8*83=664

Herreshoff10 2 10*2=20

4800/60 =80 hours = Hours available

Cost of the "standard" welding 80 hr * $25/hr= $2000

Cost of additional welding support 80 hr * $75/hr= $6000

Profit: 76 614-6 000= $70 614

d) Ed's son has agreed to help out by grinding for 30 hours. How will Ed's plan change by having more grinding hours available?

We use the Solver. More available grinding hours mean that we can change restrictions.

Grinding: 8D+20SP+20LP+20H<= (240+30)*60

Ed's plan will not change by having more grinding hours available

e) Recent competition in the small plow end of the market is causing Ed to consider cutting back on the price, and hence the profit contribution of this anchor type. How far can the contribution margin be cut before the production plan (optimal solution) will change?

With an initial analysis of the stability of the system, analyze.

The allowable increase / decrease (in table Adjustable cells) shows the changes in the boundaries of the coefficients of the objective function, which is stored in the set of variables in the optimal solution.

The profit of the anchor can be changed within the:

Name Allowable Increase Allowable Decrease

Danforth 6.67 6.67

Small Plow almost unlimited 20

Large Plow 20 8.57

Herreshoff30 almost unlimited

Substitute adjusted coefficients in the objective function.

(50-6.6)*570+(110-20)*300+(130-8.5)*90+0=

= 43.34*570+90*300+121.43*90+0= $ 62 632.5 - This borderline system.

f) Suppose the metal cutter breaks down and 8 hours of cutting time are lost. What can you say about the effect on profit contribution?

We have changed the limit. We use the Solver.

Hours available for cutting has decreased: 120-8= 112 (112*60=6720 min)

But the cutting time increased by 8 hours due to a breakdown: 5D+10SP+15LP+30H+8*60

Our profit will be $63180. This is 70614-63180=$7434 less than Ed could get.

Question 4 Linear programming models are used by many Wall Street and Bay Street firms to select a desirable bond portfolio for their clients. The following is a simplified version of such a model. TAL Private Investments has $1 million to invest for a client and is considering an investment in four bonds. The expected annual return, the worst-case annual return and the duration of each bond are given below. (The duration of a bond is a measure of its sensitivity to interest rate changes.)

Expected return Worst-case return Duration

Bond 1 13% 6% 3

Bond 2 8% 8% 4

Bond 3 12% 10% 7

Bond 4 14% 9% 9

TAL wants to maximize the expected return from the investment subject to three constraints that have been placed on it by the client.

The worst-case annual return of the portfolio should be at least 8%

The average duration of the portfolio must be no more than 6. (For example, a portfolio that was made up of $600,000 in Bond 1 and $400,000 in Bond 4 has an average duration of It may also help you to recognize that can be rewritten as or .)

At most, 40% of the portfolio can be in invested in a single bond.

Formulate a linear model that will help TAL make the portfolio composition decision. Be sure to define your decision variables clearly including the units of each.

B1 - the amount Tal invested in the first bond

B2 - the amount Tal invested in the second bond

B3 - the amount Tal invested in the third bond

B4 - the amount Tal invested in the fourth bond

Plan objectives: B=(B1, B2, B3, B4)

The objective function is showing how the maximum profit from the investment can be obtained at the set restrictions and the coefficients of the variables.

The objective function is:

13%*B1+8%*B2+12%*B3+14%*B4 = 0.13*B1+0.08*B2+0.12*B3+0.14*B4 max

Formulate a system of restrictions:

6%*B1+8%*B2+10%*B3+9%*B4B1+B2+B3+B48

(The worst-case annual return of the portfolio should be at least 8%

3*B1+4*B2+7*B3+9*B4B1+B2+B3+B46(The average duration of the portfolio must be no more than 6.)

B1+B2+B3+B4=1000000

(TAL Private Investments has $1 million to invest for a client and is considering an investment in four bonds.)

B1, B2, B3, B4 400000(At most, 40% of the portfolio can be in invested in a single bond)

B1, B2, B3, B4 >0

(Be sure to define your decision variables clearly including the units of each)

Linear programming is as follows:

0.13*B1+0.08*B2+0.12*B3+0.14*B4 max

6%*B1+8%*B2+10%*B3+9%*B4B1+B2+B3+B48

3*B1+4*B2+7*B3+9*B4B1+B2+B3+B46B1+B2+B3+B4=1000000

B1, B2, B3, B4 400000B1, B2, B3, B4 >0

Solve the problem using Excel and Solver and submit a printout of your model as is done in Question 1 above.

We use solver. We make all the necessary restrictions.

Unfortunately, limitations in the system we can not indicate that B1 is strictly greater than 0.

And as we see, it is not profitable to invest in bond 4.

But the limitation is said that we must invest in each bond.

That's why we have an additional constraint:

B1, B2, B3, B4 1

As we can see, adding additional constraints objective function value has changed. All constraints are satisfied. The search for solutions has found the optimal plan using the "search for solving linear simplex method"

The optimal plan is as follows:

0.13*400000+0.08*1+0.12*199999+0.14*400000=132000

The optimal portfolio is:

Investments in the Bond1 = $ 400 000

Investments in the Bond2 = $ 1

Investments in the Bond3 = $ 199 999

Investments in the Bond4 = $ 400 000

The income that we get from investments $ 132 000.

 

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