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One feasible solution is to assign the sea vessel and one aircraft to the highest valued target (3). This results in an expected survival value of <math> 20(0.6)(0.5)= 6 </math>. One could then assign the remaining aircraft and 2 tanks to target #2, resulting in expected survival value of <math> 10 (0.4)(0.8)^2 = 2.56 </math>. Finally, the remaining 3 tanks are assigned to target #1 which has an expected survival value of <math> 5 (0.7)^3 = 1.715 </math>. Thus, we have a total expected survival value of <math> 6 + 2.56 + 1.715 = 10.275 </math>. Note that a better solution can be achieved by assigning 3 tanks to target #1, 2 tanks and sea vessel to target #2 and 2 aircraft to target #3, giving an expected survival value of <math> 5(0.7)^3 +10(0.5)(0.8)^2 + 20(0.5)^2 = 9.915 </math>.
Note that better solution can be achieved by assigning 5 tanks to target #1, and 2 aircraft and 1 sea vessel to target #2 <math> 5(0.7)^5 +10(0.4)^2(0.5) =
==See also==
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