Given data:

The normal interest rate (i) is 8 %.

Formula used:

Formula to calculate the present worth for the given cash flow diagram is,

PW=[−$ 7000(PA, i, n)+$ 2000(PA, i, n)(PF, i, n)+$ 5000(PA, i, n)(PF, i, n)−$ 7000(PF, i, n)] (1)

Here,

P is the present cost,

A is the uniform series payment,

F is the future cost,

i is the normal interest rate,

n is the number of years.

Refer the Problem 20.18 Figure in the textbook, the equation (1) is rewritten with respect to the given number of years is,

PW=[−$ 7000(PA, i, 5)+$ 2000(PA, i, 5)(PF, i, 2)+$ 5000(PA, i, 2)(PF, i, 7)−$ 7000(PF, i, 10)] (2)

Formula to calculate the present cost for the given uniform series payment is,

(PA, i, n)=[(1+i)n−1i(1+i)n] (3)

Formula to calculate the present cost for the given future value is,

(PF, i, n)=1(1+i)n                                                                                                     (4)

Formula to calculate the annual worth for the given cash flow diagram is,

AW=PW(AP, i, n) (5)

Here,

PW is the present worth.

Refer the Problem 20.18 figure, the equation (5) is rewritten with respect to the given number of years is,

AW=PW(AP, i, 10) (6)

Formula to calculate the uniform series payment for the given present cost is,

(AP, i, n)=[i(1+i)n(1+i)n−1] (7)

Formula to calculate the future worth for the given cash flow diagram is,

FW=PW(FP, i, n) (8)

Refer the Problem 20.18 figure, the equation (8) is rewritten with respect to the given number of years is,

FW=PW(FP, i, 10) (9)

Formula to calculate the future cost for the given present cost is,

(FP, i, n)=(1+i)n                                                                                                    (10)

Calculation:

Case (i): Equivalent present worth.

Substitute 8 % for i, and 5 for n in equation (3) to find (PA, i, 5).

(PA, i, 5)=[(1+(8 %))5−1(8 %)(1+(8 %))5]

(PA, i, 5)=[(1+0.08)5−10.08(1+0.08)5]=3.993

Substitute 8 % for i, and 2 for n in equation (4) to find (PF, i, 2).

(PF, i, 2)=1(1+(8 %))2=1(1+0.08)2=0.857

Substitute 8 % for i, and 2 for n in equation (3) to find (PA, i, 2).

(PA, i, 2)=[(1+(8 %))2−1(8 %)(1+(8 %))2]=[(1+0.08)2−10.08(1+0.08)2]=1.783

Substitute 8 % for i, and 7 for n in equation (4) to find (PF, i, 7).

(PF, i, 7)=1(1+(8 %))7=1(1+0.08)7=0.583

Substitute 8 % for i, and 10 for n in equation (4) to find (PF, i, 10).

(PF, i, 10)=1(1+(8 %))10=1(1+0.08)10=0.463

Substitute 3.993 for (PA, i, 5), 0.857 for (PF, i, 2), 0.583 for (PF, i, 7), 1.783 for (PA, i, 2), and 0.463 for (PF, i, 10) in equation (2) to find PW.

PW=−$ 7000(3.993)+$ 2000(3.993)(0.857)+$ 5000(1.783)(0.583)−$ 7000(0.463)=−$ 19150.55

Case (ii): Equivalent annual worth.

Substitute 8 % for i, and 10 for n in equation (7) to find (AP, i, 10).

(AP, i, 10)=[(8 %)(1+(8 %))10(1+(8 %))10−1]=[0.08(1+0.08)10(1+0.08)10−1]=0.149

Substitute −$ 19150.55 for PW, and 0.149 for (AP, i, 10) in equation (6) to find AW.

AW=(−$ 19150.55)(0.149)=−$ 2853.43

Case (iii): Equivalent future worth.

Substitute 8 % for i, and 10 for n in equation (10) to find (FP, i, 10).

(FP, i, 10)=(1+(8 %))10=(1+0.08)10=2.159

Substitute −$ 19150.55 for PW, and 2.159 for (FP, i, 10) in equation (9) to find FW.

FW=(−$ 19150.55)(2.159)=−$ 41346

Therefore, the equivalent present worth, annual worth, and future worth of the cash flow diagram is −$ 19150.55, −$ 2853.43, and −$ 41346.

Conclusion:

Thus, the equivalent present worth, annual worth, and future worth of the cash flow diagram is −$ 19150.55, −$ 2853.43, and −$ 41346.