Q#1:

(a) Profit is set to be at least $120,000. Only M3 and M4 satisfies this
condition. Also maximum loss is set to be $40,000. From M3 and M4, only M4
satisifies this condition. So only M4 satisfies both of the aspiration
level constraints. Our decision will be M4.

(b) According to the expert the most probable market need for the product
is 6 years that corresponds to S3, with the probability of 0.4. With this
approach our decision will be M3 since it earns most for the market
condition of S3.

(c) The expected values of each machine can be determined as follows:

	M1 = (100*0.2+90*0.3+80*0.4+70*0.1) * 1000 = $ 86,000
	M2 = (60*0.2+110*0.3+110*0.4+90*0.1) * 1000 = $ 98,000
	M3 = ((-50)*0.2+90*0.3+150*0.4+140*0.1) * 1000 = $ 91,000
	M4 = ((-30)*0.2+80*0.3+100*0.4+200*0.1) * 1000 = $ 78,000

So our decision will be M2.

(d) Since we are using the laplace criterion we will assume the nature to be
indifferent, and assign the probabiltiy of 1/n,(which corresponds to 1/4
for this problem) to each future. So we will look at the average profit for
each machine such as:

	M1 = (100+90+80+70) * 1000*0.25 = $85,000
	M2 = (60+110+110+90) * 1000*0.25 = $ 92,500
	M3 = ((-50)+90+150+140) * 1000 *0.25= $ 57,500
	M4 = ((-30)*+80+100+200) * 1000 *0.25= $ 87,500

So our decision would be M2.

(e) Maximum profits and minimum profits for each machine is as follows:

        Alternative     Min     Max
        M1              70      100
        M2              60      110
        M3              -50     150
        M4              -30     200
	
So our Maximin decision would be M1, and Maximax decision would be M4.

(f) For the Hurwicz rule, alpha is given to be 0.4. Hurwicz rule requires
the computation of the following equation: 
	
Alternatives	 
M1	(0.4*100+0.6*70)*1000= 82,000
M2	(0.4*110+0.6*60)*1000= 80,000
M3	(0.4*150+0.6*(-50))*1000= 20,000
M4	(0.4*200+0.6*(-30))*1000= 62,000

So our decision would be M1.